Digitized by tine Internet Arciiiye in 2010 witii funding from Boston Library Consortium IVIember Libraries littp://www.archive.org/details/elementarytrigonOOhall ELEMENTARY TRIGONOMETRY. •9- ■n '!: r .>-"-:;V- V- ii. ELEMENTARY TRIGONOMETRY BY H. S. HALL, M.A., FOKMEELY SCHOLAR OF CHRIST'S COLLEGE, CAMBRIDGE; MASTER OF THE I\riLITARY SIDE, CLIFTON COLLEGE. AND S. R. KNIGHT, B.A., M.B., Ch.B., FORMERLY SCHOLAR OF TRINITY COLLEGE, CAMBRIDGE; LATE ASSISTANT MASTER AT MARLBOROUGH COLLEGE. BOSTOI? COLLEGE LIBRARl CHit8TlSUT HILL iMASS, Hon^on: MaI^. ^*i MACMILLAN AND CO. AND NEW YOEK. 1893 {All Rights reserved.'] CDamhrtUsE : PRINTED BY C. J. CLAY, M.A, AND SONS, AT THE UNIVERSITY PRESS. 150028 PREFACE. THE following pages will be found to comprise all the parts of Elementary Trigonometry which can con- veniently be treated without the use of infinite series and imaginary quantities. The chapters have been subdivided into short sections, and the examples to illustrate each section have been very carefully selected and arranged, the earlier ones being easy enough for any reader to whom the subject is new, while the later ones, and the Miscellaneous Examples scattered throughout the book, will furnish sufficient practice for those who intend to pursue the subject further as part of a mathematical education. No substantial progress in Trigonometry can be made until the fundamental properties of the Trigonometrical Ratios have been thoroughly mastered. To attain this object very considerable practice in easy Identities and Equations is necessary. We have therefore given special prominence to examples of this kind in the early pages; with the same end in view we have postponed the sub- ject of Radian or Circular Measure to a later stage than is usual, believing that it is in every way more satis- VI PREFACE. factory to dwell on the properties of the trigonometrical ratios, and to exemplify their use in easy problems, than to bewilder a beginner with an angular system the use of which he cannot appreciate, and which at this stage fur- nishes nothing but practice in easy Arithmetic. The subject of Logarithms and their application has been treated very fully, and illustrated by a selection of carefully graduated Examples. It is hoped that the ex- amples worked out in this section may serve as useful models for the student, and may do something to cure that inaccuracy in logarithmic work which is so often due to clumsy arrangement. In the experience of most teachers it is found extremely difficult to get boys to handle problems in Heights and Distances with any degree of confidence and skill. Ac- cordingly we have devoted much thought to the exposition of this part of the subject, and by careful classification of the Examples we have endeavoured to make Chapters vi. and XVII. as easy and attractive as possible. Yery little advance can be expected in Trigonometry until the principal formulae can be quoted readily, but whether it is advisable for learners to have lists of formulae compiled for them, so as to be easily accessible at all times, is a matter upon which teachers hold different views. In our own opinion it is distinctly mischievous to furnish such lists ; it encourages indolent habits, and fosters a spurious confidence which leads to disaster when the student has to rely solely upon his own knowledge. In the general arrangement and succession of the different parts of the subject we have been mainly guided by our own long experience in the class room ; but as the manuscript and proof-sheets have been read by several skilled teachers, and have been frequently tested by pupils PREFACE. Vll in all stages of proficiency, the hope is entertained that our treatment is such as to enable beginners to take an intelligent interest in the subject from the first, and to acquire a sound elementary knowledge of practical Trigo- nometry before they encounter the more theoretical diffi- culties. At the same time, as each chapter is, as far as possible, complete in itself, it will be easy for teachers to adopt a different order of treatment if they prefer it ; the full Table of Contents will facilitate the selection of a suitable course of reading, besides furnishing a useful aid to students who are rapidly revising the subject. "We are indebted to several friends for valuable criticism and advice; in particular, we have to thank Mr T. D. Davies of Clifton College for many useful hints, and for some in- genious examples and solutions in Chapters xxiv. and xxv. H. S. HALL. S. R. KNIGHT. November, 1893. CONTENTS. Chapter I. ^^ieasurement of angles. Page Definition of Angle , . . . 1 Sexagesimal and Centesimal Measures . . . , . 2 Formula — = — . . . .. . . , , . S Chapter II. trigonometrical ratios. Definitions of Ratio and Commensurable Quantities ... 5 Definitions of the Trigonometrical Ratios 6 Sine and cosine are less than unity, secant and cosecant are greater than unity, tangent and cotangent are unrestricted . 7 The trigonometrical ratios are independent of the lengths of the lines which include the angle . . . . . 9 Definition of function 10 Chapter III. relations between the ratios. The reciprocal relations 12 Tangent and cotangent in terms of sine and cosine ... 13 Sine-cosine, tangent-secant, cotangent-cosecant formula . . 14 Easy Identities .16 Each ratio can be expressed in terms of any of the others . 21 Chapter IV. trigonometrical ratios of certain angles. Trigonometrical Ratios of 45°, 60°, 30° 24 Definition of complementary angles 27 Each function of an angle is equal to the corresponding co- function of its complement 27 Easy Trigonometrical Equations . . . . . . 30 Miscellaneous Examples. A. . . . . . ■. . 32 CONTENTS. Chapter V. solution of right-angled triangles. PAGE Case I. When two sides are given 35 Case II. When one side and one acute angle are given . . 36 Case of triangle considered as sum or difference of two right- angled triangles 38 Chapter VI. easy problems. Angles of Elevation and Depression The Mariner's Compass Chapter VII. radian or circular measure. Definition of Eadian .... Circumference of circle = 27r (radius) . All radians are equal .... IT radians =: 2 right angles = 180 degrees Eadian contains 57 '2958 degrees D e Formula 180 Values of the functions of TT TT TT 4' 3 ' 6 Ratios of the complementary angle ^-6 Eadian measure of angles of a regular polygon _, ,. , , subtending arc Eadian measure of an angle = ^^-^ radius Eadian and Circular Measures are equivalent Miscellaneous Examples. B. . Chapter VIII. ratios of angles of any magnitude. Convention of Signs (1) for line, (2) for plane surface, (3) for angles Definitions of the trigonometrical ratios of any angle . Signs of the trigonometrical ratios in the four quadrants . Definition of Coterminal Angles ....... The fundamental formulae of Chap. iii. are true for all values of the angle The ambiguity of sign in cos A= ^ aJi - sin"-^ A can be removed when A is known 41 45 49 50 51 52 52 53 55 56 56 58 58 61 64 66 68 68 70 71 CONTENTS. XI Chapter IX. variations of the functions. Definition of liinit ..... Functions of 0° and 90° ... . Changes in the sign and magnitude of sin A Changes in the sign and magnitude of tan A Definition of Circular Functions Note on the old definitions .... Miscellaneous Examples. C. . PAGE 74 74 76 78 79 80 81 Chapter X. circular functions of allied angles. Circular functions of 180° -A Definition of supplementary angles Circular functions of 180° + ^ Circular functions of 90° + -dt Circular functions of - ^ . Definition of even and odd functions ..... Circular functions of 90° - A for any value of ^ Circular functions of n . 360° ^A The functions of any angle can be expressed as the same func tions of some acute angle ...... The number of angles which have the same trigonometrical ratio is unlimited ....... 82 83 84 85 86 87 88 89 90 91 Chapter XI. functions of compound angles. Expansions of the sine and cosine oi A + B and A-B sin (^ + J5) sin (^ - B) = sin2yf - sin^J? . Expansions of tan {A-\-B) and cot (^ +5) . Expansions of sin {A + B+G) and tan {A+B + C) Converse use of the Addition FormulsB Functions of 2^ Functions of 3^ Value of sin 18° Chapter XII. transformation of products and sums Transformation of products into sums or differences . Transformation of sums or differences into products . Belations when ^+£ + (7=180° 94 96 98 99 100 102 105 106 110 112 118 Xll CONTENTS. Chapter XIII. relations between the sides and angles OF A TRIANGLE. PAGE ^ ' 123 sin A sin B sin G a? = If- + c=^ - 2hc cos yi , and cos ^ = " jT ~ ^ . . . .124 a = 6 cos C + c cos jB . 124 The above sets of formulffi are not independent .... 125 Solution of Triangles without logarithms 126 Case I. When the three sides are given 126 Case II. When two sides and the included angle are given . 127 Case III. When two angles and a side are given . . . 128 Case IV. When two sides and an angle opposite to one of them are given. Sometimes this case is ambiguous . . 130 The Ambiguous Case discussed geometrically .... 131 The Ambiguous Case discussed by first finding the third side . 135 Miscellaneous Examples. D. 138 Chapter XIV. logarithms. a* = A'' and a; = loga^ are equivalent 139 ^logai^^^ is identically true 139 Logarithm of a product, quotient, power, root .... 140 The characteristic of a common logarithm may be written down by inspection 142 The logarithms of all numbers which have the same significant digits have the same mantissa 143 logj-N'=, xloga2^, and log6axlog„6 = l .... 146 •■■Oga " Exponential Equations 148 Miscellaneous Examples. E. . 150 Chapter XV. the use of logarithmic tables. Eule of Proportional Parts 152 Use of Tables of Common Logarithms ..... 152 Use of Tables of Natural and Logarithmic Functions . , 156 CONTENTS. Xlll Chapter XVI. solution op triangles with logarithms. PAGE Functions of the half-angles in terms of the sides Sin A in terms of the sides Solution when the three sides are given Solution when two sides and the included angle are given Solution when two angles and a side are given . Solution when two sides and the angle opposite to one of are given Adaptation of a- + &- to logarithmic work . Adaptation of a^ + &^ - 2ab cos C to logarithmic work . Chapter XVII. heights and distances. Measurements in one plane Problems dependent on Geometry .... Measurements in more than one plane . . them 164 166 167 170 174 175 177 178 184 189 193 Chapter XVIII. properties of triangles and polygons. Area of a triangle Radius of the circum-circle of a triangle Radius of the in-circle of a triangle .... Eadii of the ex-circles of a triangle .... Some important relations established by Geometry . Inscribed and circumscribed polygons Area of a circle and sector of a circle .... The Ex-central Triangle The Pedal Triangle Distances of in-centre and ex-centres from circum-centre Distance of orthocentre from circum-centre Area of any quadrilateral Diagonals and circum-radius of a cyclic quadrilateral Miscellaneous Examples. F 198 200 201 202 204 208 210 212 214 216 218 220 222 228 Chapter XIX. general values and inverse functions. Formula for all angles which have a given sine .... 232 Formula for all angles which have a given cosine . . . 233 Formula for all angles which have a given tangent . . . 234 Formula for angles both equi-sinal and equi-cosinal . . . 234 XIV CONTENTS. PAGE General solution of equations .236 Inverse Circular Functions 238 Solution of equations expressed in inverse notation . . . 244 Miscellaneous Examples. G 246 Chapter XX. functions of submultiple angles. Trigonometrical Batios of - 247 8 A A Given cos A to find sin — and cos —...... 248 A A To express sin — and cos — in terms of sin A . . . . 250 a it Variation in sign and magnitude of cos ^ - sin ^ . . . . 254 Sine and cosine of 9° 254 A To find tan — when tan A is given 256 Given a function of A to find the functions of — . . . 258 A Given cos A to find cos — 259 o Chapter XXI. limits and approximations. TV If d< ~ , sin d, d, tan 6 are in ascending order of magnitude . 261 . 262 0^ ^-, . - sin(9 ^ , tan<9 , When ^=0, —-— = 1, and —~~ = 1 cos ^ > 1 - — , and sin ^ > ^ - . Value of sin 10" .... sin^ 265 266 266 267 269 cos - cos -r cos - cos —p. — ^ .... 2 4 8 16 6 sin 6 ^ . ,,2^, „ ^,T —z — decreases from 1 to — as ^ mcreases from to — d TT 2 Distance and Dip of the Visible Horizon . Chapter XXII. geometrical proofs. Expansion of tan (^+5) .273 Formulae for transformation of sums into products . . . 274 CONTENTS. XV PAGE Proof of the 2A formulge 276 Value of sin 18° 277 Proofs by Projection 278 General analytical proof of the Addition Formula . . . 282 Miscellaneous Examples. H 283 Graphs of sin 6, tan ^, sec ^ 285 Chapter XXIII. summation of finite sekies. If Uj,=^v^.^^-Vj., then *S' = v,j^.j-V]^ 288 Sum of the sines and cosines of a series of n angles in a. p. . 289 2k7r When the common difference is , the sum is zero . . 290 n Sum of the squares and cubes of the sines and cosines of a series of angles in a. p 293 Chapter XXIV. miscellaneous tkansformations and identities. Symmetrical Expressions. S and 11 notation .... 296 Alternating Expressions 303 Allied formulae in Algebra and Trigonometry .... 306 Identities derived by substitution 308 Chapter XXV. miscellaneous theorems and examples. InequaHties. Maxima and Minima .... Elimination Application of Trigonometry to Theory of Equations Application of Theory of Equations to Trigonometry Miscellaneous Examples. I 313 319 326 328 336 ELEMENTARY TRIGONOMETRY. CHAPTER I. MEASUREMENT OF ANGLES. 1. The word Trigonometry in its primary sense signifies the measurement of triangles. From an early date the science also included the establishment of the relations which subsist between the sides, angles, and area of a triangle ; but now it has a much wider scope and embraces all manner of geometrical and algebraical investigations carried on through the medium of certain quantities called trigonometrical ratios, which will be defined in Chap. II. In every branch of Higher Mathematics, whether Pure or Applied, a knowledge of Trigonometry is of the greatest value. 2. Definition of Angle. Suppose that the straight line OP in the figure is capable of revolving about the point 0, and suppose that in this way it has passed successively from the posi- tion OA to the positions occupied by OB, 00, OB, ..., then the angle between OA and any position such as OC is measured by the amount of revolution which the line OP has undergone in passing from its initial position OA into its final position 00. Moreover the line OP may make any number of complete re- volutions through the original posi- tion OA before taking up its final position. H. K. E. T. 05 2 ELEMENTARY TRIGONOMETRY. [CHAP. It will thus be seen that in Trigonometry angles are not re- stricted as in Euclid, but may be of any magnitude. The point is called the origin^ and OA the initial line; the revolving line OP is known as the generating line or the radius vector, 3. Measurement of Angles. We must fii'st select some fixed unit. The natural unit would be a right angle, but as in practice this is inconveniently large, two systems of measure- ment have been established, in each of which the unit is a certain fraction of a right angle. 4. Sexagesimal Measure. A right angle is divided into 90 equal parts called degrees, a degree into 60 equal parts called minutes, a minute into 60 equal parts called seconds. An angle is measured by stating the number of degrees, minutes, and seconds which it contains. For shortness, each of these three divisions, degrees, minutes, seconds, is denoted by a symbol ; thus the angle which contains 53 degrees 37 minutes 2*53 seconds is expressed symbolically in the form 53° 37' 2-53". 5. Centesimal Measure. A right angle is divided into 100 equal parts called grades, a grade into 100 equal parts called minutes, a minute into 100 equal parts called seconds. In this system the angle which contains 53 grades 37 minutes 2*53 seconds is expressed symbolically in the form 53^ 37' 2*53" . It wiU be noticed that different accents are used to denote sexagesimal and centesimal minutes and seconds ; for though they have the same names, a centesimal minute and second are not the same as a sexagesimal minute and second. Thus a right angle contains 90 x 60 sexagesimal minutes, whereas it contains 100 X 100 centesimal minutes. Sexagesimal Measure is sometimes called the English System, and Centesimal Measure the French System. 6. In numerical calculations the sexagesimal measure is always used. The centesimal method was proposed at the time of the French Eevolution as part of a general system of decimal measurement, but has never been adopted even in France, as it would have made necessary the alteration of Geographical, Nautical, Astronomical, and other tables prepared according to the sexagesimal method. Beyond giving a few examples in transformation from one system to the other which afford exercise in easy Arithmetic, we shall after this rarely allude to centesimal measure. I.] MEASUREMENT OF ANGLES. 3 In theoretical work it is convenient to use another method of measurement, where the unit is the angle subtended at the centre of a circle by an arc whose length is equal to the radius. This system is known as Circular or Radian Measure, and will be fully explained in Chapter VII. An angle is usually represented by a single letter, different letters ^, ^, (7, ..., a, j3, y, ..., ^, 0, a/',..., being used to distin- guish different angles. For angles estimated in sexagesimal or centesimal measure these letters are used indifferently, but we shall always denote angles in circular measure by letters taken from the Greek alphabet. 7. If the numher of degrees and grades contained in an angle T) C he D and G respectively, to prove that -k = jk' In sexagesimal measure, the given angle when expressed as the fraction of a right angle is denoted by — . In centesimal measure, the same fraction is denoted by -^-—: ; D _ G D _G •'• 90-100' ^^^*'^' 9-l0' 8. To pass from one system to the other it is advisable first to express the given angle in terms of a right angle. In centesimal measure any number of grades, minutes, and seconds may be immediately expressed as the decimal of a right angle. Thus 23 grades = ^^ of a right angle = "23 of a right angle ; 15 minutes =^^Q^g^ of a grade = '15 of a grade = '0015 of a right angle ; .-. 23^ 15' = -2315 of a right angle. Similarly, 15^ 7' 53-4'' = -1507534 of a right angle. Conversely, any decimal of a right angle can be at once ex- pressed in grades, minutes, and seconds. Thus •2173025 of a right angle = 21 -73025^ = 21« 73-025* = 2P73' 2-5*\ In practice the intermediate steps are omitted. 1—2 4 ELEMENTARY TRIGONOMETRY. [CHAP. I. Example 1. Eeduce 2^13' 4'5'* to sexagesimal measure. __ This angle = -0213045 of a right angle -0213045 of a right angle = 1° 55' 2-658". 1-917405 degrees 60 55-0443 minutes 60 2-658 seconds. Obs. In the Answers we shall express the angles to the nearest tenth of a second, so that the above result would be written 1°55'2'7". Example 2. Eeduce 12° 13' 14-3" to centesimal measure. Thisangle^ -13578487., .of aright angle 601M1 seconds _ 60 ) 13-238333- -minutes — 13^57 84-9 . 90 ) 12-2206 388... degrees 13578487.. .ofarightangle. EXAMPLES. I. Express as the decimal of a right angle 1. 67° 30'. 2. 11° 15'. 4. 2° 10' 12". 5. 8° 0' 36". Reduce to centesimal measure 7. 69° 13' 30". 8. 19° 0' 45". 10. 43° 52' 38-1". 11. 11° 0' 38-4". 13. 12' 9". 14. 3' 26-3". Reduce to sexagesimal measure 15. 56^ 8r 50'\ 16. 39^ 6' 25". 17. 40= r 25-4". 18. P 2^ 3". 19. 3^ 2' 5^\ 20. 8^ 10' 6-5'\ 21. & 25'\ 22. 37' 5'\ 23. The sum of two angles is 80^ and their diflference is 18° ; find the angles in degrees. 24. The number of degrees in a certain angle added to the number of grades in the angle is 152 : what is the angle ? 25. If the same a-ngle contains in English measure .r minutes, and in French measure y minutes, prove that 50^=27?/. 26. If s and t respectively denote the numbers of sexa- gesimal and centesimal seconds in any angle, prove that 250s = 81 t 3. 37^ 50\ 6. 2^ 4' 4-5". 9. 50° 37' 5-7". 12. 142° 15' 45". CHAPTER II. TRIGONOMETRICAL RATIOS. 9. Definition. Ratio is the relation "which one quantity bears to another of the same kind, the comparison being made by considering what multiple, part or parts, one quantity is of the other. To find what multiple or part A in oi B we divide A by B ; hence the ratio of A to B may be measured by the fraction ^ . In order to compare two quantities they must be expressed in terms of the same unit. Thus the ratio of 2 yards to 2 X 3 X 12 8 27 inches is measured by the fraction — — or - . Obs. Since a ratio expresses the nuiiiber of times that one quantity contains another, every ratio is a numerical quantity. 10. Definition. If the ratio of any two quantities can be expressed exactly by the ratio of two integers the quantities are said to be commensurable ; otherwise, they are said to be incommensurable. For instance, the quantities 8i and 5^- are commensurable, while the quantities y/2 and 3 are incom- mensurable. But by finding the numerical value of x/2 we may express the value of the ratio ^2 : 3 by the ratio of two com- mensurable quantities to any required degree of approximation. Thus to 5 decimal places ^2 = 1 '41421, and therefore to the same degree of approximation V2 : 3 = 1-41421 : 3 = 141421 : 300000. Similarly, for the ratio of any two incommensurable quantities. ELEMENTARY TRIGONOMETRY. [chap. Trigonometrical Ratios. 11. Let PAQ be any acute angle ; in AP one of the bound- ary lines take a point B and draw BC perpendicular to AQ. Thus a right-angled triangle BAG is formed. With reference to the angle A the following definitions are em- ployed. The ratio -r^^ or -^ — is called the sine of A. AB hypotenuse „, ,. AC adjace7it side . ^^ ■, ,-, . - . The ratio -rr=, or -~ — is called the cosine of A. AB hypotenuse The ratio -— ^ or -^ : — r^- is called the tangent of A. AC adjacent side ^, ,. AC adjacent side . ^^ ^ ,^ . i. r a The ratio -^^ or — - — -. t-j- is called the cotangent of A. BC opposite side ^, ,. AB hypotenuse . ■,^ j .■, j. r a The ratio ~rj^ or —f^^ — - — rj- is called the secant of A. AC adjacent side The ratio ^^ or '^^ . ^^ is called the cosecant of A. BC opposite side These six ratios are known as the trigonometrical ratios. It will be shewn later that as long as the angle remains the same the trigonometrical ratios remain the same. [Art. 19.] 12. Instead of writing in full the words sine, cosine, tangent, cotangent, secant, cosecant, abbreviations are adopted. Thus the above definitions may be more conveniently expressed and arranged as follows : . , BC sm A = -r^ . AB cos -4 AC AB' tan^=j^,, cosec A sec J.= cot J.= AB BC AB AC AC BC II.] TRIGONOMETRICAL RATIOS. 7 In addition to these six ratios, two others, the versed sine and coversed sine are sometimes used ; they are written vers A and covers A and are thus defined : vers J. = 1 - cos J[, covers J^ = 1 — sin ^. 13. In Chapter VIII. the definitions of the trigonometrical ratios will be extended to the case of angles of any magnitude, but for the present we confine our attention to the consideration of acute angles. 14. Although the verbal form of the definitions of the trigonometrical ratios given in Art. 11 may be helpful to the student at first, he will gain no freedom in their use until he is able to write down from the figure any ratio at sight. In the adjoining figure, PQR is a right-angled triangle in which P^ = 13, PR = b, QR = \± Since PQ is the greatest side, R is the right angle. The trigonometrical ratios of the angles P and Q may be written down at once ; for example, . ^ PR b ^ OR 12 ^^^^=i^=l3' ^"^^ = 7^ = 13' tanP-^-1? cosecP-^-^ 15. It is important to observe that the trigonometrical ratios of a7i angle are numerical quantities. Each one of them re- presents the ratio of one length to another^ and they must them- selves never be regarded as lengths. ^o" 16. In every right-angled triangle the hypotenuse is the greatest side; hence from the definitions of Art. 11 it will be seen that those ratios which have the hypotenuse in the denominator can never be greater than unity, while those which have the hypotenuse in the numerator can never be less than unity. Those ratios which do not involve the hypotenuse are not thus restricted in value, for either of the two sides which subtend the acute angles may be the greater. Hence the sine and cosine of an angle can never he greater than 1 ; the cosecant and secant of an angle can never he less than 1 ; the tangent and cotangent may have any numerical 'value. 8 ELEMENTARY TRIGONOMETRY. [chap. 17. Let ABC be a right-angled triangle having the right angle at A ; then by Euc. i. 47, the sq. on BC = sum of sqq. on AC and AB, or, more briefly, BC^ = AC^ + AB\ When we use this latter mode of ex- pression it is understood that the sides AB, AC, BC are expressed in terms of some common unit, and the above statement may be regarded as a iiumerical relation connecting the numbers of units of length in the three sides of a right-angled triangle. It is usual to denote the numbers of units of length in the sides opposite the angles A, B, (7 by the letters a, 6, c respectively. Thus in the above figure we have a^ = h^-\-c^, so that if the lengths of two sides of a right-angled triangle are known, this equation will give the length of the third side. Example 1. ABC is a right-angled triangle of which C is the right angle; if a=3, & = 4, find c, and also sin A and cot B. Here c2 = a2 4- &2 = (3)2 +(4)2^ 9 ^16 = 25; .'. c = 5. Also . . BC ^ 3 5' 3 4' Example 2. A ladder 17 ft. long is placed with its foot at a distance of 8 ft. from the wall of a house and just reaches a window- sill. Find the height of the window-sill, and the sine and tangent of the angle which the ladder makes with the wall. Let AG he the ladder, and BC the wall. Let X be the number of feet in BC; then .T- = (17)2- (8)2 =(17-1- 8) (17 -8) = 25x9; .-. x = 5 x3 = 15. Also . ^ AB 8 ^^"^=Jc = r7 ^ ^ AB 8 ^^^^=50 = 15 II.] TRIGONOMETRICAL RATIOB. 9 18. The following important proposition depends upon the property of similar triangles proved in Euc. vi. 4. The student who has not read the sixth Book of Euclid shoidd not fail to notice the result arrived at, even if he is unable at this stage to understand the proof. 19. To prove that the trigonomet-rical ratios remain unaltered so long as the angle remains the same. Let A OP be any acute angle. In OP take any points B and D, and draw BG and DE perpendicular to OA. Also take any point F in OP and draw FG at right angles to OP. BG From the triangle BOG, sin PDA = -^-„ ; DE from the triangle DOE, sin POA = from the triangle FOG, sin POA = OD' FG OG' But the triangles BOG, DOE, FOG are equiangular ; BG^^DE^FG ''' 0B~ 0D~ OG' [Euc. VI. 4.] Thus the sine of the angle POA is the same whether it is obtained from the triangle BOG, or from the triangle DOE, or from the triangle FOG. A similar proof holds for each of the other trigonometrical ratios. These ratios are therefore independent of the length of the revolving line and depend only on the magnitude of the angle. 10 ELEMENTARY TRIGONOMETRY. [chap. 20. If A denote any acute angle, we have proved that all the trigonometrical ratios of A depend only on the magnitude of the angle A and not upon the lengths of the lines which bound the angle. It may easily be seen that a change made in the value of A will produce a consequent change in the values of all the trigonometrical ratios of A. This point will be discussed more fully in Chap. IX. Definition. Any expression which involves a variable quantity x, and whose value is dependent on that of x is called a function of x. Hence the trigonometrical ratios may also be defined as trigonometrical functions ; for the present we shall chiefly em- ploy the term ratio ^ but in a later part of the subject the idea of ratio is gradually lost and the term function becomes more appropriate. 21. The use of the principle proved in Art. 19 is well shewn in the following example, where the trigonometrical ratios are employed as a connecting link between the lines and angles. Example. ABC is a right-angled triangle of which A is the right angle. BD is drawn perpendicular to BC and meets CA produced in D: a AB = 12, AC =16, BC =20, &nd BD and CD. From the right-angled tri- angle CBD, BD BC = tan C ; from the right-angled triangle ABC, AB AC = tan C ; BD BC BD 20 AB AG 12 16* -;rT- = 5-7i ; whence BD = 15. Again, CD ^ BC — =secO = — ; CD 2Q , ^^' — — = -— ; whence CD = 25. 20 lb The same results can be obtained by the help of Euc. vi. 8. II.] TRIGONOMETRICAL RATIOS. 11 EXAMPLES. II. 1. The sides AB, BO, CA of a right-angled triangle are 17, 15, 8 respectively ; write down the values of sin A, sec A, tan B, sec B. 2. The sides PQ, QR, RP of a right-angled triangle are 13, 5, 12 respectively : write down the values of cot P, cosec $, cos Q, cos P. 3. ABC is a triangle in which A is a right angle; if 6=15, c = 20, find a, sin C, cos B, cot C, sec C. 4. ABC is a triangle in which ^ is a right angle; if a = 24, 5 = 25, find c, sin (7, tan^, cosec A. 5. The sides ED, EF, DF of a right-angled triangle are 35, 37, 12 respectively: write down the values of sec E, secF, coiE, sin jP. 6. The hypotenuse of a right-angled triangle is 15 inches, and one of the sides is 9 inches : find the third side and the sine, cosine and tangent of the angle opposite to it. 7. FiiMi the hypotenuse AB of a right-angled triangle in which J (7= 7, BC=2^. Write down the sine and cosine of J., and shew that the sum of their squares is equal to 1. 8. A ladder 41 ft. long is placed with its foot at a distance of 9 ft. from the wall of a house and just reaches a window-sill. Find the height of the window-sill, and the sine and cotangent of the angle which the ladder makes with the ground. 9. A ladder is 29 ft. long ; how far must its foot be placed from a wall so that the ladder may just reach the top of the wall which is 21 ft. from the ground 1 Write down all the trigono- metrical ratios of the angle between the ladder and the wall. 10. ABCD is a square ; C is joined to E, the middle point of AD : find all the trigonometrical ratios of the angle ECD. 11. ABCD is a quadrilateral in which the diagonal ^C is at right angles to each of the sides AB, CD : if AB = 15, AC =36, AD = 85, find sin ABC, sec ACB, cos CD A, cosec D AC. 12. PQRS is a quadrilateral in which the angle PSR is a right angle. If the diagonal PR is at right angles to RQ, and RP=20, BQ = 2l, RJS=ie, find sinPRjS, tanRPS, cosRPQ, cosec PQR. CHAPTER III. RELATIONS BETWEEN THE TRIGONOMETRICAL RATIOS. 22. Reciprocal relations between certain ratios. (1) Let ABO be a triangle, right-aogled at C; then and . , BC a cosec A = AB_ c BG~a' sm A X cosec A = - x - == 1. c a Thus sin A and cosec A are reciprocals ; I cosec A ' sin A = and cosec J. =-^ — -r. SlTlA (2) Again, , AO b , , AB c .-. cos J. xsec^i=- X y=l ; c 6 . • . cos ^1 = 7 , and sec A = y . sec A cos J. (3) Also tan.l=-j^=^, and cotJ=-^^=-, .-. tan^ X cot^l = Y X - = 1 ; a .'. tan^l= — — T) ^^^^ cotyl=7 ---. cot A tan A EELATIONS BETWEEN THE TRIGONOMETRICAL RATIOS. 13 23. To express tan A mid cot A in terms of sin A and cos A. From the adjoining figure we have , _ BC a ah AC b c ' c = sin J.-^cos^ ; sin^ cos A ' AC_h_h^a, BC a c ' c :cos^-^sin>4 ; Again, tan A = cot^ = , , cos A .•. cotJ.= . — J-: sm J. which is also evident from the reciprocal relation cot A Example. Prove that cosec A tan A = sec A . tan J. cosec A tan A = — sin 4 sin A cos A cos A = secA. 24. We frequently meet with expressions which involve the square and other powers of the trigonometrical ratios, such as (sinJ^)2, (tanJ.)^, ... It is usual to write these in the shorter forms sin^ A^ tan^ A, ... Thus tan2.4 = (tan^)2^f ^ ^cos J^ (sin^)2_ sin^A (cos -4)2 cos^^* Example. Shew that sin^^ sec A cot^^ =cos A 1 =8m^A X / cos^ Y cos^ \sin^y cos^^ cos^ sin^^ = 008^, by cancelling factors common to numerator and denominator. 14 ELEMENTARY TRIGONOMETRY. [chap. 25. To prove that sin^ ^ + cos^ A = \. Let BAG be any acute angle ; draw BC perpendicular to AO, and denote the sides of the right-angled triangle ABC by «, 6, c. By definition, . , BC a and cos J^ AC_b ''AB~c 2 62 ^2+52 •. sin2^+cos2^==— + -^ = = 1. Cor. sin2 ^ = l - cos2 A , sin A = \/1-gos^A ; cos2^ = l-sin2^, cosA = \/l-Bm^A. Example 1. Prove that cos^ A - sin^ A = eos^ A - sin^ A . cos^ A - sin^ A = (cos^ A + sin2 ^) (cos2 ^ - sin2 ^) = cos2^-sin2^, since the first factor is equal to 1. Example 2. Prove that cot a Jl-cos^a = cos a. cot a J\- cos^a = cot a x sin a cos a sm a X sm a = cos a , III.] RELATIONS BETWEEN THE TRIGONOMETRICAL RATIOS. 15 26. To prove that sec^ A = l+ tan^ A. With the figure of the previous article, we have secA = -j-^=p = H-tan2J. CoR. sec2 A - tan2 ^1 = 1, sec A = VT+tanM , tan^ A = sec2 il - 1, tan A = Vsec^ il - 1. Example. Prove that cos ^ ^ysec^ A-l = sin ^ . cos A sj^.ec'^ ^ - 1 = cos A x tan A , sin A = cos ^ X r COB^ = sin A. 27. To prove that cosec^ A = \-\- cot^ A . With the figure of Art. 25, we have , AB c cosec A = ^^,, = - ; BO a „ . ^(P- a^-\-h'^ . • . cosec^ A = — „ = 5— = l+cot2J. CoR. cosec^ A - cot^ J^ = 1 , cosec ^ = V 1 + cot^ J., cot^ J = cosec^ A — 1, cot J^ = Vcosec^ A — I. Example. Prove that cot"* a-l = cosec^ a - 2 cosec^ a. cot* a - 1 = (cot2 a + 1) (cot2 a - 1) = cosec^ a (cosec^ a - 1 - 1) = cosec^ a (cosec- a - 2) = cosec* a - 2 cosec^ a. 16 ELEMENTARY TRIGONOMETRY. [CHAP. 28. The formulse proved in the last three articles are not independent, for they are merely different ways of expressing in trigonometrical symbols the property of a right-angled triangle proved in Euc. i. 47. 29. It will be useful here to collect the formulae proved in this chapter. 1 . 1 I. cosec J. xsin J = l, cosecJ.=-^ — r, sin^= -; sm A cosec A sec ^ X cos J, = 1, secJ.= r-, cosJ. = j ; cos A sec A cot A X tan ^ == 1, cot A = ■ j- , tan A tan A ' '' cot A ° TT 1 . sin J. . . cos A II. tan^ = 7^, cot^= ^ — J. cos A sm A III. sin2^+cos2^ = l, sec2 J. = l+tan2 J, cosec^ ^4 = 1 + cot^ ^4 , Easy Identities. 30. We shall now exemplify the use of these fundamental formulse in proving identities. An identity asserts that two expressions are always equal, and the proof of this equality is called "proving the identity." Some easy illustrations have already been given in this chapter. The general method of procedure is to choose one of the expressions given (usually the more complicated of the two) and to shew by successive transformations that it can be made to assume the form of the other. Example 1. Prove that sin^ A cot^ A + cos- A tan" A = l. Here it will be found convenient to express all the trigonometrical ratios in terms of the sine and cosine. The first side=sin2^ . . „ ^ +008^^ . — — -- sm- A cos- 4 =cos^^+sin2^ III.] EASY IDENTITIES. 17 Example 2. Prove that sec^ 6 - sec^ d = tan^ d + tan^ 6. The form of this identity at once suggests that we should use the secant-tangent formula of Art. 26 ; hence the first side = sec^ d (sec^ ^ - 1) = (l + tan2^)tan2d.+cosec^. 26. cos e (tan ^ + 2) (2 tan ^ + 1) - 2 sec (9 + 5 sin 6. 27. (tan^+sec^)2=i-±^. ^ 1— sm 6 -,_ 2 sin 6 cos ^ — cos 6 , , 28. ^j • ^ , ■ 0^ 2^ = cot (9. 1 - sm 6 + sin^ 6 — cos^ ^ 29. cot^^.gg^+sec^a. 7^-^ = 0. l+sin(9 1 + sec^ [T/te folloiving examples contain functions of two angles; in each case the two angles are quite independent of each other. 1 30. tan2a + sec2j3 = sec2a + tan2/3. -- tan a + cot /3 _ tan a _ tan a — cot /3 _ cot /3 cota+tan/3 tan/3' ' cot a — tan /3 cota' 33. cot a tan /3 (tan a + cot /3) = cot a + tan /3. 34 . sin^ a cos2 /3 - cos^ a sin^ /3 = sin^ a — sin^ /3. 35. sec2 a tan2 /3 - tan2 a sec^ ^ = tan2 18 - tan2 a. 36. (sin a cos ^ + cos a sin /S)^ + (cos a cos ^ — si n a sin ^)2 = I . III.] RELATIONS BETWEEN THE TRIGONOMETRICAL RATIOS. 21 32. By means of the relations collected together in Art. 29, all the trigonometrical ratios can be expressed in terms of any- one. Example 1. Express all the trigonometrical ratios of A in terms of tan^. We have cot A = tan 4 ' sec A=zjl + tan2 A ; cos^ = = . ; sec^ Jl + io.u'^A . , mrxA , , , , tan^ sm A = -cos A = tan A cos A = cos^ 7l + tan2^ 1 Jl + tan^Z cosec A = -: — - = ^ — . sm A tan A Obs. In writing down the ratios we choose the simplest and most natm-al order. For instance, cot A is obtained at once by the recipi-ocal relation connecting the tangent and cotangent : sec A comes imme- diately from the tangent-secant formula ; the remaining three ratios now readily follow. 5 Example 2. Given cos A=y5, find cosec 4 and cot^. 1 1 1 V169 1 "12" 13 1 1 13 MS)' y- 25 169 "12 cot A = — — ^ = cos A X cosec A smA _5_ 1B_5_ ~13^12~12* 33. It is always possible to describe a right-angled triangle when two sides are given: for the third side can be found by Euc. 1. 47, and the construction can then be effected by Euc. i. 22. We can thus readily obtain all the trigonometrical ratios when one is given, or express all in terms of any one. 22 ELEMENTARY TRIGONOMETRY. [chap. Example 1. Given GoaA = :r-^, find cosec A and cot^l. Take a right-angled triangle PQR, of which Q is the right angle, having the hypotenuse PR = 13 units, and PQ = 5 units. Let QR = X units ; then a;2= (13)2 - (5)2= (13 + 5) (13 - 5) = 18x8 = 9x2x8; .-. a; = 3x4 = 12. Now so that Hence and cos ijpa=||= A, lRPQ = A. . PR 13 cosec^ = P = j2. ^°*^ = P = 12- [Compare Art. 32, Ex. 2.] Example 2. Find tan^ and cos A in terms of cosec A. Take a triangle PQR right-angled at Q, and having z RPQ = A. For shortness, denote cosec -4 by c. Then cosec y4=c=-; but cosec -4 = pr^ ; '• Q^ ~ 1 ' Let QE be taken as the unit of measurement ; thenQE = l, and therefore PJJ = c Let PQ contain x units ; then a;2=c2 - 1, so that x=Jc^-l. 1 1 Hence tan^=^-- = and eos^ PQ Jc^'-i 7cosec2^-l _PQ _ Jc'-l _ ^cosec2^-l PR cosec A III.] RELATIONS BETWEEN THE TRIGONOMETRICAL RATIOS. 23 EXAMPLES. III. c. 1. Given sin ^4 = -, find sec J and cot^-1. 4 2. Given tan^l=-, find sin yl and cos^. o 3. Find cot^ and sin ^ when sec ^=4. 4. If tana=-, find sec a and coseca. 5. Find the sine and cotangent of an ande whose secant is 7. 6. If 25 sin ^1 = 7, find tan J^ and sec^. 7. Express sinyl and tan^ in terms of cosy4. 8. Express coseca and cos a in terms of cot a. 9. Find sin 6 and cot 6 in terms of sec Q. 10. Express all the trigonometrical ratios of A in terms of sin A. 11. Given sin yl — cos xl = 0, find cosec^. 12. If sin A — — ^ prove that sjn^ - m^ . tan A = m. 13. If jo cot 6=^'\l(f' -p^, find sin 6. ^oj, -4- 1 14. When secvl= — — ■. find tanxl and sin^. 2m 15. Given tan^ = -^^„, find cos^ and cosecJ.. ■lo Tif 13 „ , ,. , „ 2 sin a— 3 COS a lb. it seca=-p-, find the value of - — -. . o 4 sin a — 9 cos a 17. If cot«=«, find the value „{P^^^«fl^ CHAPTER IV. TRIGONOMETRICAL RATIOS OF CERTAIN ANGLES. 34. Trigonometrical Ratios of 45". Let BA C be a right-angled isosceles triangle, with the right angle at C; so that B = A=45°. Let each of the equal sides contain I units, then Also AC^BC=l. AB^ = P + P = 2r2; .'. AB = ly/2. • ^^o BC I 1 sm45 =^-j^^ = ^', o AC cos 45 = -7-^ = tan 45= AB BC I W2 2 J2 I AC I ^' The other three ratios are the reciprocals of these ; thus cosec 45° = V2, sec 45° = ^2, cot 45° = 1 ; or they may be read off from the figure. TRIGONOMETRICAL RATIOS OF CERTAIN ANGLES. 25 35. Trigonometrical Ratios of 60° and 30°. Let ABC be an equilateral triangle ; thus each of its angles is 60°. Bisect iBAOhy AD meeting BC at D ; then l BAD='S0\ By Euc. L 4, the triangles ABB, ACD are equal in all re- spects ; therefore BD = (7Z>, and the angles at D are right angles. In the right-angled triangle ADBj let BD = l; then AB = BC==^l', .-. AD = l^/2. AD IJZ V3 Again, sin 60° = cos 60' AB BD AB ■ n 2^ 1 2' . «P,o BD I 1 sm30=^=^ = -; AD^V3_x/3 ' AB 2? ~ 2 cos 30' , _-„ BD I 1 tan 30=-^ = ^ = -. The other ratios may be read off from the figure. 26 ELEMENTARY TRIGONOMETRY. [chap. 36. The trigonometrical ratios of 45°, 60°, 30° occur very frequently ; it is therefore important that the student should be able to quote readily their numerical values. The exercise which follows will furnish useful practice. At first it will probably be found safer to make use of the accompanying diagrams than to trust to the memory. Fig. 1. Fm. 2. The trigonometrical ratios of 45° can be read off from Fig. 1 ; those of 60° and 30° from Fig. 2. Examiile 1. Find the values of sec^ 45° and sin 60° cot 30° tan 45°. sec3 45° = (sec 45°)3= (^2)3=^^2x^2x^2 = 2^2. sin 60°oot 30° tan 45°=^ x ^3 x 1 = | . Example 2. Find the value of 2 cot 45° + cos3 60° - 2 sin^ 60° + 1 tan^ 30°. Thevalue = (2 x 1) + Q' - 2 (^^y + ?( J^)' o 1 9 1 -, ^8 8^4 ^ EXAMPLES. IV. a. Find the numerical value of 1. tan2 60°+2tan2 45°. 3. 2cosec2 45°-3sec2 30°. 2. tan3 45°+4cos3 60°, 4. cot 60° tan 30° + sec2 45' IV.] COMPLEMENTARY ANGLES. 27 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15. 2 sin 30° cos 30° cot 60°. tan2 45° sin 60° tan 30° tan2 60°. tan2 60° + 4 cos2 45° -!- 3 sec^ 30°. J cosec2 60° + sec2 45° - 2 cot2 60°. tan2 30° + 2 sin 60° + tan 45° - tan 60° + cos^ 30°. cot2 45° + cos 60° - sin2 60° - 1 cot2 60°. 3 tan2 30° + 1 cos2 30° - 1 sec2 45° - 1 sin2 60°. cos 60° - tan2 45° + 1 tan2 30° + cos2 30° - sin 30°. I sin2 60° - ^ sec 60° tan2 30° + ^ sin2 45° tan^ 60°. If tan2 45° - cos2 60° = x sin 45° cos 45° tan 60°, find x Find X from the equation cot2 30° sec 60° tan 45° X sin 30° cos2 45° = cosec2 45° cosec 30° 37. Definition. The complement of an angle is its defect from a right angle. Two angles are said to be complementary when their sum is a right angle. Thus in every right-angled triangle, each acute angle is the complement of the other. For in the figure of the next article, if B is the right angle, the sum of A and G is 90°. .-. (7=90°-^, and ^ = 90°-(7. Trigonometrical Ratios of Complementary Angles. 38. Let ABC be a right-angled triangle, of which B is the right angle ; then the angles at A and C are com- plementary, so that C=90°-^. AB .'. sin(90°-^)=sin(7=-j-^=cos J.; BC and cos (90° - J. ) = cos C= -rri= sin A. AO Similarly, it may be proved that tan(90°-^) = cot^, | and sec (90°-^) = cosec ^, cot(90°-^) = tan^; j cosec (90°-^) = sec J. 28 ELEMENTARY TRIGONOMETRY. [CHAP. 39. If we define the co-sine, co-tangent, co-secant, as the co-functions of the angle, the foregoing results may be embodied in a single statement : each function of an angle is equal to the corresponding co-function of its complement. As an illustration of this we may refer to Art, 35, from which it will be seen that sin 60° = cos 30°=^; sin 30° = cos 60° =i; tan 60° = cot 30° = ^3. Example 1. Find a value of A when cos 2A = sin SA. Since cos 2 A = sin (90° -2 A), the equation becomes sin (90° -2A) = sin SA ; /. 90° -2.1 =3^; whence ^ = 18°. Thus one value of A which satisfies the equation ia A = 18°. In a later chapter we shall be able to solve the equation more completely, and shew that there are other values of A which satisfy it. Example 2. Prove that sec A sec (90° -A)=tanA+ tan (90° -A). Here it will be found easier to begin with the expression on the right side of the identity. The second side = tan A + cot A _ sin A cos A _ sin^ A + cos^ A ~ cos A sin A ~ cos A sin A cos A sin A = sec -4 cosec J^ = sec .1 sec (90° -A). EXAMPLES. IV. b. Find the complements of the following angles : 1. 67° 30'. 2. 25° 30". 3. 10°!' 3". 4. 45°-^. 5. 45° +5. 6. 30° -.B. IV.] COMPLEMENTARY ANGLES. 29 7. In a triangle C is 50° and A is the complement of 10° ; find^. 8. In a triangle A is the complement of 40° ; and B is the complement of 20° : find C Find a value of A in each of the following equations : 9. sin ^= cos 4^. 10. cos 3^4 = sin 7^. 11. tan ^ = cot 3^. 12. cotJ^=tan^. 13. cot J. = tan 2 J. 14. sec5^ = cosec^. Prove the following identities : 15. sin (90° -^) cot (90°-^) = sin ^. 16. sin A tan (90° - .4) sec (90° - ^)=cot A. 17. cos A tan A tan (90° -A) q.q^^q. (90° - ^) = 1. 18. sin A cos (90° - .4) + cos A sin (90° -A) = \. 19. cos(90°-^)cosec(90°-^) = tan^. 20. cosec2 (90° - .4 ) = 1 + sin2 A cosec2 (90° - .4 ) . 21. sin A cot ^ cot (90° -^) sec (90° - ^) = 1. 22. sec (90° -A)-q,o\.A cos (90° - .4) tan (90° - A)^ sin A. 23. tan2 A sec2 (90° -A)- sin2 A cosec2 (90° - ^ ) = 1. 24. tan (90° - ^ ) + cot (90° - yl ) = cosec A cosec (90° - ^ ). sec (90 -A) cos>4 cosec2 A tan2 A cot A „ ,^ . „ , . , 26. TTHTTs — JV-- — ^2-7=sec2(90°-^)-l. cot (90 -A) sec^J. ^ ^„ cot (90°-^) sec^cot3.4 ,- — ^ , , 28. '52!!fi2!^) = i+3i„(90°-4). vers A ^ ' „^ cot2^sin2(90°-^) ^ ,^^„ ,, , 29. i— i — ^ -, — ^ = tan(90°-^)-cosyl. cot A + cos A 30. If X sin (90° - ^) cot (90° - .4) - cos (90° - .4), find x. 31. Find the value of x which will satisfy sec A cosec (90° - ^) - .r cot (90° - ^) = 1. 30 ELEMENTARY TRIGONOMETRY. [cHAP. Easy Trigonometrical Eauations. 40. As a further exercise in using the formulsB of Art. 29 and the numerical values of the functions of 45°, 60°, 30°, we shall now give some examples in trigonometrical equations. Example 1. Solve 4 cos ^ = 3 sec ^. By expressing the secant in terms of the cosine, we have 4 cos A = i , cos 4 4cos2^ = 3, ■■• cobA=^ (1). or cos^= -^ (2). Since cos 30°=^ , we see from (1) that ^ = 30°. The student will be able to understand the meaning of the nega- tive result in (2) after he has read Chap. VIII. Example 2. Solve 3 sec^ ^ = 8 tan ^ - 2. Since sec^ ^ = 1 + tan^ 6 , we have 3 (l + tan2^) = 8tan^-2, or 3tan26'-8tan^ + 5 = 0. This is a quadratic equation in which tan 6 is the unknown quantity, and it may be solved by any of the rules for solving quad- ratic equations. Thus (tan 6' - 1) (3 tan ^ - 5) = 0, therefore eif/ier tan^-l = (1), or 3tan^-5=:0 (2). From (1), tan 5 = 1, so that 5=45°. From (2), tan 5= - , a result which we cannot interpret at present. 41. When an equation involves more than two functions, it will usually be best to express each function in terms of the sine and cosine. IV.] EASY TRIGONOMETRICAL EQUATIONS. 31 Example. Solve 3 tan ^ + cot ^ = 5 cosec 6. ^^ - 3 sin ^ cos 6 5 We nave ~ + -. — ^ = -;: — - , cos sm 6 sm 6 3sin2^ + cos2^ = 5cos^, 3 (1 - cos^ 0) + cos2 ^ = 5 cos 0, 2cos2^ + 5cos^-3=0, (2cos^-l) (cos^ + 3)=:0; therefore either or 2cos6'-l = cos^ + 3 = 0, .(2). From (1), cos 6' = -, so that 5 = 60°. From (2), cos ^=: - 3, a result which must be rejected as impossible, because the numerical value of the cosine of an angle can never be greater than unity. [Art. 16.] EXAMPLES. IV. c. Find a solution of each of the following equations : tan ^ = 3 cot ^. sec B — cosec ^ = 0. cosec^ 9 — A. tan ^ = 2 sin 6. cosec2(9 = 4cot^^. sec2^ + tan2<9 = 7. 2(cos2^-sin2(9)-l. 6 cos'^ 6 = 1 + cos 6. 2sin2<9 = 3cos^. 1 . 2 sin 6 = cosec 6. 2. 3. sec ^ = 4 cos ^. 4. 5. 4 sin ^=3 cosec 6. 6. 7. V2cos(9 = cot^. 8. 9. sec2^ = 2tan2^. 10. 11. sec2^ = 3tan2^-l. 12. 13. cot2^+cosec2^ = 3. 14. 15. 2cos2^ + 4sin2^ = 3. 16. 17. 4sin^-12sin2(9-l. 18. 19. tan (9 = 4-3 cot ^. 20. cos2 6 - sin2 ^ = 2 - 5 cos B. 21. cot (9 + tan ^ = 2 sec B. 22. 23. tan ^ - cot ^ = cosec 6. 24. 25. 2sin(9tan(9 + l=tan<9 + 2sin^. 26. 6tan^-5V3sec^+12cot(9 = 0. 27. If tan (9 + 3 cot ^ = 4, prove that tan (9 = 1 or 3 28. Find cot B from the equation cosec2 ^ +cot2 ^ = 3 cot <9. 4 cosec ^ + 2 sin ^ = 9. 2 cos ^ + 2^2 = 3 sec ^. 32 ELEMENTARY TRIGONOMETRY. [cHAP. MISCELLANEOUS EXAMPLES. A. 1. Express as the decimal of a right augle (1) 25^ 3r 6-4" ; (2) 63° 21' 36". 2. Shew that sin A cos A tan A + cos A sin A cot A = l. 3. A ladder 29 ft. long just reaches a window at a height of 21 ft. from the ground : find the cosine and cosecant of the angle made by the ladder with the ground. 17 4. If cosec A = —, find tan A and sec A. 5. Shew that cosec^ A - cot A cos A cosec ^ — 1=0. 6. Beduce to sexagesimal measure (1) 17« 18' 75" ; (2) -0003 of a right angle. 7. ABC is a triangle in which ^ is a right angle ; if c = 9, °. In the right-angled triangle ^i5D, AB -— ^ = cosec ABD ; AD :. AB = AD cos.BC ABD — o cosec PiQ'- 2 _ 10 _ 10 v/3 " N^3~^/3" 3 • Also ^ = coi ABD \ AD BD^ AD cot ABD = 5cot60°r=-^^ ./3 In the right-angled triangle ADC, lDAC = 45° = .: DC = DA^ I DC A 5. v.] SOLUTION OF RIGHT-ANGLED TRIANGLES, 39 Thus BC=C'Z)+5D = 5 + ^ = ^^#^^. o o AC And — ^ = cosec A CD ; .: AC = AD cosec ACD = 5 cosec 45° = 5 fJ2. Finally, z .54 C= 180° -60° -45°:= 75°. Thus a = 'l±l^, . = 5^2, e = i5^^ .4 = 75°. EXAMPLES. V. b. 1. ABC is a triangle, and BD is perpendicular to J.C pro- duced : find 52), given ^ = 30°, (7=120°, i<7=ga 2. If BB is perpendicular to the base AC of a, triangle ABCj find a and c, given ^ = 30°, 6'= 45°, BD = 10. 3. In the triangle ABC, AD is drawn perpendicular to BC making BD equal to 15 ft. : find the lengths of AB, AC, and AD, given that B and C are equal to 30° and 60" respectively. 4. In a right-angled triangle PQR, find the segments of the hypotenuse PR made by the perpendicular from Q ; given QR = ^, L QRP=m% L qPR=^ZO\ 5. If PQ, is drawn perpendicular to the straight line QRS^ find RS, given p^=36, lRPQ=zo% Lspq=m\ 6. If PQ is drawn perpendicular to the straight line QRS, find RS, given PQ = 20, zPi?>9= 135°, z P>S'P=30°. 7. In the triangle ABC, the angles B and 6' are equal to 45° and 120° respectively; if «. = 40 find the length of the perpen- dicular from A on BC produced. 8. If CD is drawn i^erpendicular to the straight line DBA, find DC and BD, given ^5=59, LCBD=Ab% iGAB=^'dr bO', cot 32° 50' = 1 -59. CHAPTER VI. EASY PROBLEMS. 48. The principles explained in the previous chapters may now be applied to the solution of problems in heights and distances. It will be assumed that by the use of suitable instruments the necessary lines and angles can be measured with sufficient accuracy for the purposes required. After the practice afforded by the examples in the last chapter, the student should be able to write down at once any side of a right-angled triangle in terms of another through the medium of the functions of either acute angle. In the present and subsequent chapters it is of great importance to acquire readiness in this respect. For instance, from the adjoining figure, we have a=csin^, a=ccos B, a = h cot B, a=b t&ii A, c = a sec -5, 6 = atan^. These relations are not to be committed to memory but in each case should be read off from the figure. There are several other similar relations connecting the parts of the above triangle, and the student should practise himself in obtaining them quickly. EASY PROBLEMS. 41 Example. Q, R, T are three points in a straight line, and TP is drawn perpendicular to QT. If PT=a, iPQT = ^, lPRT=2^, express the lengths of all the lines of the figure in terms of a and ^. Q R T By Euc. I. 32, IQPR^ IPRT- iPQR', :. iQPR = 2^-p=^= IPQR; .-. QR=PR. In the right-angled triangle PRT, PR = a cosec 2j3 ; .*. QR = a cosec 2/3. Also TE = acot2j3. Lastly, in the right-angled triangle PQT, QT = acot^, PQ = a Gosec^. 49. Angles of elevation and depression. Let OP be a horizontal line in the same vertical plane as an object Q, and let OQ he joined. O P angle of Repression Fig.2 In Fig. 1, where the object Q is ahove the horizontal line OP, the angle F0<^ is called the angle of elevation of the object Q as seen from the point 0. In Fig. 2, where the object Q is helow the horizontal line OP, the angle POQ, is called the angle of depression of the object Q as seen from the point 0. 42 ELEMENTARY TRIGONOMETRY. [chap. Example I. A flagstaff stands on a horizontal plane, and from a point on the ground at a distance of 30 ft. its angle of elevation is 60° : find its height. Let AB be the flagstaff, C the point of observation ; then AB = JSC tan 60° = 30 ^^3 = 30x1-732 = 51-96. Thus the height is 51-96 ft. EXAMPLES. VI. a. [The results should be expressed in a form free from surds by using the approximations /y/2 = 1-414, ^3 = 1-732.] 1. The angle of elevation of the top of a chimney at a distance of 300 feet is 30° : find its height. 2. From a ship's masthead 160 feet high the angle of depression of a boat is observed to be 30° : find its distance from the ship. 3. Find the angle of elevation of the sun when the shadow of a pole 6 feet high is 2 ^f3 feet long. 4. At a distance 86-6 feet from the foot of a tower the angle of elevation of the top is 30°. Find the height of the tower and the observer's distance from the top. 5. A ladder 45 feet long just reaches the top of a wall. If the ladder makes an angle of 60° with the wall, find the height of the wall, and the distance of the foot of the ladder from the wall. 6. Two masts are 60 feet and 40 feet high, and the line joining their tops makes an angle of 33° 41' with the horizon : find their distance apart, given cot 33° 41' = 1 "5. 7. Find the distance of the observer from the top of a cliff which is 132 yards high, given that the angle of elevation is 41° 18', and that sin 41° 18'= -66. 8. One chimney is 30 yards higher than another. A person standing at a distance of 100 yards from the lower observes their tops to be in a line inclined at an angle of 27° 2' to the horizon: find their heights, given tan 27° 2' = '51. YI.] EASY PROBLEMS. 43 Example II. From the foot of a tower the angle of elevation of the top of a column is 60°, and from the top of the tower, which is 50 ft. high, the angle of elevation is 30° : find the height of the column. Let AB denote the column and CD the tower ; draw GE parallel to DB. Let^-B = a;; then AE = AB~BE^x-50. Let DB=CE = y. From the right-angled triangle ADB, cc y = x cot 60° = -^. From the right-angled triangle ACE, y = {.X - 50) cot 30° = ^/3 (.r - 50). /. j-^ = s/S{x-50), x = S{x-50); whence x = 75. Thus the column is 75 ft. high. 9. The angle of elevation of the top of a tower is 30° ; on walking 100 yards nearer the elevation is found to be 60" : find the height of the tower. 10. A flagstaff stands upon the top of a building; at a distance of 40 feet the angles of elevation of the tops of the flagstaff and building are 60° and 30° : find the length of the flagstaff. 11. The angles of elevation of a spire at two places due east of it and 200 feet apart are 45° and 30° : find the height of the spire. 12. From the foot of a post the elevation of the top of a steeple is 45°, and from the top of the post, which is 30 feet high, the elevation is 30° ; find the height and distance of the steeple. 13. The height of a hill is 3300 feet above the level of a horizontal plane. From a point A on this plane the angular elevation of the top of the hill is 60°. A balloon rises from A and ascends vertically upwards at a uniform rate ; after 5 min- utes the angular elevation of the top of the hill to an observer in the balloon is 30° : find the rate of the balloon's ascent in miles per hour. 44 ELEMENTARY TRIGONOMETRY. [chap. Example III. From the top of a cliff 150 ft. high the angles of depression of two boats which are due South of the observer are 15° and 75° : find their distance apart, having given cot 15°= 2 + ^3 and cot 75° =2-^3. Let OA represent the cliff, B and C the boats. Let OP be a horizontal line through ; then ZP0C' = 15° and zPOJ5 = 75°; .-. z OCA = 15° and z 05^ = 75°. Let CB = x, AB = y; then GA = x + y. From the right-angled triangle OB A , y = 150 cot 75° = 150 (2 - ;^3) = 300 - 150 ^'S. From the right-angled triangle OCA, :c + 2/ = 150 cot 15° = 150 (2 + JS) = 300 + 150 J3. By subtraction, x — 300 ,^3 = 519-6. Thus the distance between the boats is 519'6 ft. 14. From the top of a monument 100 feet high, the angles of depression of two objects on the ground due west of the monument are 45° and 30° : find the distance between them. 15. The angles of depression of the top and foot of a tower seen from a monument 96 feet high are 30° and 60°: find the height of the tower. 16. From the top of a cliff 150 feet high the angles of depression of two boats at sea, each due north of the observer, are 30° and 15° : how far are the boats apart ? 17. From the top of a hill the angles of depression of two consecutive milestones on a level road running due south from the observer are 45° and 22° respectively. If cot 22° = 2 '475 find the height of the hill in yards. 18. From the top of a lighthouse 80 yards above the horizon the angles of depression of two rocks due west of the observer are 75° and 15°: find their distance apart, given cot 75° = -268 and cot 15° = 3-732. VL] THE MARINER'S COMPASS. 45 50. Trigonometrical Problems sometimes require a know- ledge of the Points of the Mariner's Compass, which we shall now explain. In the above figure, it will be seen that 32 points are taken at equal distances on the circumference of a circle, so that the arc between any two consecutive points subtends at the centre of the circle an angle equal to %¥'-°, that is to llj°. The points North, South, East, West are called the Cardinal Points, and with reference to them the other poiiits receive their names. The student will have no difficulty in learning these if he will carefully notice the arrangement in any one of the principal quadrants. 51. Sometimes a slightly dififerent notation is used; thus N. 11 j° E. means a direction 11^° east of north, and is therefore the same as N. by E. Again SrW. by S. is 3 points from south and may be expressed by S. 33|° W., or since it is 5 points from west it can also be expressed by W. 56^-° S. In each of these cases it wiU be seen that the angular measurement is made from the direction which is first mentioned. 46 ELEMENTARY TRIGONOMETRY. [chap. 52. The angle between the directions of any two points is obtained by multiplying 11J° by the number of intervals between the points. Thus between S. by W. and W.S.W. there are 5 intervals and the angle is 56^°; between N.E. by E. and S.E. there are 7 intervals and the angle is 78|°. 53. If B lies in a certain direction with respect to A, it is said to bear in that direction from A ; thus Birmingham bears N.W. of London, and from Birmingham the bearing of London is S.E. Example 1. From a lighthouse L two ships A and B are observed in directions S.W. and 15° East of South respectively. At the same time B is observed from ^ in a S.E. direction. If LA is 4 miles find the distance between the ships. Draw L8' due South; then from the bearings of the two ships, lALS'=4:5°, A BLS' = 15°, so that lALB = QO°. Through A draw a line NS pointing North and South ; then ANAL= /.ALS' = 4:5°, and lBAS = 4:5°, since B bears S.E. from A ; hence z5^L = 180°-45°-45°=90°. In the right-angled triangle ABL, AB=AL tan ALB = ^ tan 60° = 4^3 = 6-928. Thus the distance between the ships is 6-928 miles. Example 2. At 9 a.m. a ship which is sailing in a direction E. 40° S. at the rate of 8 miles an hour observes a fort in a direction 50° North of East. At 11 a.m. the fort is observed to bear N. 20° W.: find the distance of the fort from the ship at each observation. fort. Let A and C be the first and second positions of the ship ; B the Through A draw lines towards the cardinal points of the compass. From the observations made lEAG= 40°, z EAB = 50°, so that iBAG= 90°. VI.] THE mariner's COMPASS. 47 Through C draw C2V' towards the North; then lBGN' = '2.QP, for the bearing of the fort from C is N. 20° W. Also Z A GN' = A CAS =90°- 40" = 50° ; .-. iAGB= lACN'- LBGN' = bO° - 20° = ^[)-. In the right-angled triangle AGB, AB = AG iQ.n A GB -= 16 tan 30° = -y^ =: — ^ = 9-237 nearly ; and BG = AG&eGAGB=\& sec 30°=- 16 x ~ = ^?^ = 18-475 nearly. Thus the distances are 9*237 and 18-475 miles nearly. EXAMPLES. VI. b. 1. A person walking due E. observes two objects both in the N.E. direction. After walking 800 yards one of the objects is due N. of him, and the other lies N.W. : how far was he from the objects at first ? 2. Sailing due E. I observe two ships lying at anchor due S. ; after sailing 3 miles the ships bear 60° and 30° S. of W.; how far are they now distant from me ? 3. Two vessels leave harbour at noon in directions W. 28° S. and E. 62° S. at the rates 10 and 10^ miles per hour respectively. Find their distance apart at 2 p.m. 48 ELEMENTARY TRIGONOMETRY. [CHAP. VI. 4. A lighthouse facing N. sends out a fan-shaped beam extending from N.E. to N.W. A steamer sailing due W. first sees the light when 5 miles away from the hghthouse and continues to see it for 30 J 2 minutes. What is the speed of the steamer ? 5. A ship sailing due S. observes two lighthouses in a line exactly W. After sailing 10 miles they are respectively N.W. and W.N.W. ; find their distances from the position of the ship at the first observation. 6. Two vessels sail from port in directions N. 35° W. and S. 55° W. at the rates of 8 and 8^/3 miles per hour respectively. Find their distance apart at the end of an hour, and the bearing of the second vessel as observed from the first. 7. A vessel sailing S.S.W. is observed at noon to be E.S.E. from a lighthouse 4 miles away. At 1 p.m. the vessel is due S. of the lighthouse: find the rate at which the vessel is sailing. Given tan 67^° = 2-41 4. 8. A, B, G are three places such that from A the bearing of G is N. 10° W., and the bearing of B is N. 50° E. ; from B the bearing of G is N. 40° W. If the distance between B and G is 10 miles, find the distances of B and G from A. 9. A ship steaming due E. sights at noon a lighthouse bearing N.E., 15 miles distant; at 1.30 p.m. the lighthouse bears N.W. How many knots per day is the ship making? Given 60 knots = 69 miles. 10. At 10 o'clock forenoon a coaster is observed from a lighthouse to bear 9 miles away to N.E. and to be holding a south-easterly coiu-se; at 1 p.m. the bearing of the coaster is 15° S. of E. Find the rate of the coaster's sailing and its distance from the lighthouse at the time of the second observation. 11. The distance between two lighthouses, A and B, is 12 miles and the hne joining them bears E. 15° N. At midnight a vessel which is sailing S. 15° E. at the rate of 10 miles per hour is N.E. of A and N.W. of B : find to the nearest minute when the vessel crosses the line joining the lighthouses. 12. From A to B, two stations of a railway, the line runs W.S.W. At ^ a person observes that two spires, whose distance apart is 1-5 miles, are in the same line which bears N.N.W. At B their bearings are N. 7^° E. and N. 37^° E. Find the rate of a train which runs from J. to ^ in 2 minutes. CHAPTER YII. RADIAN OR CIRCULAR MEASURE. 54. We shall now return to the system of measuring angles which was briefly referred to in Art. 6. In this system angles are not measured in terms of a submulfciple of the right angle, as in the sexagesimal and centesimal methods, but a certain angle known as a radian is taken as the standard unit, in terms of which all other angles are measured. 55, Definition. A radian is the angle subtended at the centre of any circle by an arc equal in length to the radius of the circle. In the above figure, ABC is a circle, and its centre. If on the circumference we measure an arc AB equal to the radius and join OA, OBj the angle AOB is a radian. 56. In any system of measurement it is essential that the unit should be always the same. In order to shew that a radian, constructed according to the above definition, is of constant magnitude, we must first establish an important property of the circle. H. K. E. T. 4 50 ELEMENTARY TRIGONOMETRY. . [CHAP. 57. The circumferences of circles are to one another as their radii. Take any two circles whose radii are r^ and r^^ and in each circle let a regular polygon of n sides be described. Let A^B-^ be a side of the first, J. 2^3 a side of the second polygon, and let their lengths be denoted by a^, a^. Join their extremities to 0^ and 0^ the centres of the circles. We thus obtain two isosceles triangles whose vertical angles are equal, each being - of four right angles. n Hence the triangles are equiangular, and therefore w^e have by Euc. VI. 4, that is, oia. na radius Let AOG be any angle at the centre of a circle, and A OB a radian ; then radian measure of /.AOG ^lAOC ~ lAOB arc AC B^rcAB arc AC radius ' since arc J.jS= radius ; . 1 , • . 1 -1 • r i y^.y subtendinfif arc that IS, the radian measure or lAOC= :n — . radius 71. If a be the length of the arc which subtends an angle of B radians at the centre of a circle of radius r, we have seen in the preceding article that = - , and therefore a = rd. r The fraction — ^. — is usually called the circidar measure of radius the angle at the centre of the circle subtended by the arc. The circidar measure of an angle is therefore equal to its radian Tneasure^ each denoting the number of radians contained in the angle. We have preferred to use the term radian measure exclusively, in order to keep prominently in view the unit of measurement, namely the radian. Note. The term circular measure is a survival from the times when Mathematicians spoke of the trigonometrical functions of the arc. [See page 80.] Example 1. Find the angle subtended by an arc of 7*5 feet at the centre of a circle whose radius is 5 yards. Let the angle contain 9 radians ; then ' arc _ 7-5 _ 1 ~ radius" 15 ~ 2" Thus the angle is half a radian. VII.] RADIAN MEASURE. 59 Example 2. In running a race at a uniform speed on a circular course, a man in each minute traverses an arc of a circle winch sub- tends 2f radians at the centre of the course. If each lap is 792 yards, r 22~| how long does he take to run a mile ? tt = — . Let r yards be the radius of the circle ; then 27rr = circumference = 792 ; r= 792 _ 792 X 7 2^~ 2x22 :=126. Let a yards be the length of the arc traversed in each minute j then from the formula a=rd, a = 126x2f = i^?^ = 360; that is, the man runs 360 yds. in each minute. ^, ^. 1760 44 . ^ .'. the time = -ttt^t or -- minutes. obO 9 Thus the time is 4 min. 534 sec. Example 3. Find the radius of a globe such that the distance measured along its surface between two places on the same meridian 22 whose latitudes differ by 1° 10' may be 1 inch, reckoning that ir=-^ . Let the adjoining figure represent a sec- tion of the globe through the meridian on which the two places P and Q lie. Let be the centre, and denote the radius by r inches. Now — ^T—^ = number of radians in iPOQ; radius but arc P(3 = l inch, and /.P0Q = 1°1Q'; :. - = number of radians in li° whence = 1 _7r__7 22 1 _ 11 ^^i80~6^ 7 ^180 "540 r= 540 IT' :49t\ Thus the radius is 49xt inches. 60 ELEMENTARY TRIGONOMETRY. [CHAP. EXAMPLES. VII. c. 1. Find the radian measure of the angle subtended by an arc of 1 "6 yards at the centre of a circle whose radius is 24 feet. 2. An angle whose circular measure is '73 subtends at the centre of a circle an arc of 219 feet ; find the radius of the circle. 3. An angle at the centre of a circle whose radius is 2 "5 yards is subtended by an arc of 7*5 feet; what is the angle? 4. What is the length of the arc which subtends an angle of 1*625 radians at the centre of a circle whose radius is 3-6 yards ? 5. An arc of 17 yds. 1 ft. 3 in. subtends at the centre of a circle an angle of 1 "9 radians ; find the radius of the circle in inches. 6. The flywheel of an engine makes 35 revolutions in a second ; r 22"! how long will it take to turn through 5 radians 1 tt = — . 7. The large hand of a clock is 2 ft. 4 in. long ; how many r 22~ inches does its extremity move in 20 minutes ? f] 8. A horse is tethered to a stake ; how long must the rope be in order that, when the horse has moved through 52-36 yards at the extremity of the rope, the angle traced out by the rope may be 75 degrees ? 9. Find the length of an arc which subtends 1 minute at the centre of the earth, supposed to be a sphere of diameter 7920 miles. 10. Find the number of seconds in the angle subtended at the centre of a circle of radius 1 mile by an arc 5^ inches long. 11. Two places on the same meridian are 145*2 miles 22 apart; find their difierence in latitude, taking 7r=i=-, and the earth's diameter as 7920 miles. 12. Find the radius of a globe such that the distance measured along its surface between two places on the same meridian whose 22 latitudes differ by 1^° may be 1 foot, taking '7r=Tr-. VII.] MISCELLANEOUS EXAMPLES. B. 61 MISCELLANEOUS EXAMPLES. B. 1. Express in degrees the angle whose circular measui-e is •15708. 2. If C= 90°, A = 30°, c = 1 10, find h to two decimal places. 3. rind the number of degrees in the unit angle when the IStt angle -- is represented by If. 4. What is the radius of the circle in which an arc of 1 inch subtends an angle of 1' at the centre ? 5. Prove that (1) (sin a + cos a) (tan a + cot a) = sec a + cosec a ; (2) ( V3 + 1 ) (3 - cot 30°) = tan3 60° - 2 sin 60°. 6. Find the angle of elevation of the sun when a chimney 60 feet high throws a shadow 20 ^/3 yards long. 7. Prove the identities : (1) (tan^ + 2)(2tan^ + l) = 5tan^ + 2sec2^; cot^a (2) 1+- =coseca. 1 + cosec a 8. One angle of a triangle is 45° and another is -3- radians ; 8 express the third angle both in sexagesimal and radian measure. 9. The number of degrees in an angle exceeds 14 times the 22 number of radians in it by 51. Taking 7r = -=- , find the sexa- gesimal measure of the angle. 10. If 5=30°, C=90°, 6 = 6, find a, c, and the perpendicular from G on the hypotenuse. 11. Shew that (1) cot^ + cot f --^) = cosec^cosecl l^-^j; (2) cosec^ Q 4- cosec2 / - - ^ J = cosec^ 6 cosec^ \- — d\. 62 ELEMENTARY TRIGONOMETRY. [CHAP. 12. The angle of elevation of the top of a pillar is 30°, and on approaching 20 feet nearer it is 60° : find the height of the pillar. 13. Shew that tan^A - sin^^ =sin*^ secM. 14. In a triangle the angle A is S.v degrees, the angle B is X grades, and the angle C is 7-— radians : find the number of oKjyJ degrees in each of the angles. 15. Find the numerical value of sinS 60° cot 30° - 2 sec^ 45° + 3 cos 60° tan 45° - tan2 60°. 16. Prove the identities : (1) (1 4-tan ^)2 + (1 +cot ^)2 = (sec .4 + cosec yl )2 ; (2) (sec a - 1)2 - (tan a - sin a)2 = (1 - cos a)"-. 17. Which of the following statements is possible and which impossible ? a^+b^ 1 (1) cosec^= ^ ; (2) 2sin<9 = « + -. 18. A balloon leaves the earth at the point A and rises at a uniform pace. At the end of 1 '5 minutes an observer stationed at a distance of 660 feet from A finds the angular elevation of the balloon to be 60° ; at what rate in miles per hour is the balloon rising ? 19. Find the number of radians in the angles of a triangle which are in arithmetical progression, the least angle being 36°. 20. Shew that sin^a sec2/3 + tan^/S cos^a = sin^a + tan^/S. 21. In the triangle ABC if ^=42°, 5=116° 33', find the perpendicular from Cupon AB produced; given c = 55, tan 42° = -9, tan 63° 27' = 2. VII.] MISCELLANEOUS EXAMPLESo B. 68 22. Prove the identities : sin a (1) cota+^ = coseca: ^ ^ 1 + cosa (2) cosec a (sec a — 1 ) - cot a ( 1 — cos a) = tan a - sin a. «o oi ,^ , /l+cot60°\2 l+co»30° 23. Shew that ' ^ 1 - cot 607 1 - cos 30° ' 24. A man walking N. W. sees a windmill which bears N. 15° W. In half-an-hour he reaches a place which he knows to be W. 15° S. of the windmill and a mile away from it. Find his rate of walk- ing and his distance from the windmill at the first observation. 25. Find the number of radians in the complement of -^ . 8 26. Solve the equations : (1) 3sin<9 + 4cos2^ = 4|; (2) tan ^ + sec 30° = cot (9. 27. If 5 tan a = 4, find the value of 5 sin a — 3 cos a sina + 2cosa 28. Prove that 1 — sin A cos A sinM — cos^^ X . „ . ;^7=sinJ^. cos A (sec A — cosec A) sin^A + cos^A 29. Find the distance of an observer from the top of a cliff which is 195*2 yards high, given that the angle of elevation is 77° 26', and that sin 77° 26' = -976. 30. A horse is tethered to a stake by a rope 27 feet long. If the horse moves along che circumferencb of a circle always keeping the rope tight, find how far it will have gone when the [22~1 CHAPTER YIIL TRIGONOMETRICAL RATIOS OF ANGLES OF ANY MAGNITUDE. 72. In the present chapter we shall find it necessary to take account not only of the magnitude of straight lines, but also of the direction in which they are measured. Let be a fixed point in a horizontal line XX\ then the position of any other point P in the line, whose distance from is a given length a, will not be determined unless we know on which side of the point P lies. X' O X But there will be no ambiguity if it is agreed that distances measured in one direction are positive and distances measured in the opposite direction are negative. Hence the following Convention of Signs is adopted : lines Tneasured from to the right are positive, lines measured from to the left are negative. X' Q O P X Thus in the above figure, if P and Q are two points on the line XX' at a distance a from 0, their positions are indicated by the statements OP=+a, OQ=-a. 73. A similar convention of signs is used in the case of a plane surface. Let be any point in the plane ; through draw two straight lines XX' and YY' in the horizontal and vertical direction re- spectively, thus dividing the plane into four quadrants. CONVENTION OF SIGNS. 65 Then it is universally agreed to consider that (1) horizontal lines to the right of YY' are positive^ horizontal lines to the left of YY' are negative ; (2) vertical lines above XX' are positive, vertical lines heloio XX' are negative. Y M3 M2 O P3 IVI4 M^ X Y' Thus OJ/j, Oi/4 are positive, OM.^, OM^ are negative; MiP^, M^P^ are positive, M.^P^, ^^iP^ are negative. 74. Convention of Signs for Angles. In Art. 2 an angle has been defined as the amount of revolution which the radius vector makes in passing from its initial to its final position. In the adjoining figure the straight line OP may be supposed to have arrived at its present position from the position occupied by OA by revolution about the point in either of the two directions indicated by the arrows. The angle AOP may thus be regarded in two senses according as we suppose the revolution to have been in the same direction as the hands of a clock or in the opposite direction. To distinguish between these cases we adopt the following convention : when the revolution of the radius vector is connter-clocTcivise the angle is positive, when the revolution is clochwise the angle is negative. H. K. E. T. 5 O 66 ELEMENTARY TRlGONOMETilY. [chap. Trigonometrical Ratios of any Angle. 75. Let XX' and YY' be two straight lines intersecting at right angles in 0, and let a radius vector starting from OX revolve in either direction till it has traced out an angle J., taking up the position OP. V Y X' o M X X' M O From P draw PM perj)endicular to XX'; then in the right- angled triangle 0PM, due regard being paid to the signs of the lines, sin A = cos A = MP OP' OM OP' A OP cosec A = -,"7^^- MP' ^^"""^-QM' sec A = cot J.: OP OM' OM 'MP' The radius vector OP wUicJi only fixes the boundary of the angle is considered to he always positive. VIII.] TRIGONOMETRICAL RATIOS OF ANY ANGLE. 67 From these definitions it will be seen that any trigono- metrical function will be positive or negative according as the fraction which expresses its value has the numerator and de- nominator of the same sign or of opposite sign. 76. The four diagrams of the last article may be con- veniently included in one. With centre and fixed radius let a circle be described ; then the diameters XX' and YY' divide the circle into four quadrants XOY^ YOX, X'OP, Y'OX, nsuned Jii^stj second, third, fourth respectively. Let the positions of the radius vector in the four quadrants be denoted by OP^, OP. 2, OP^, OP^, and let perpendiculars PiJ/i, P2M2, P33/3, P4J/4 be drawn to XX' ; then it will be seen that in the first quadrant all the lines are positive and there- fore all the functions of A are positive. In the second quadrant, OP2 and M^P^ are positive, OM^ is negative ; hence sin A is positive, cos A and tan A are negative. In the third quadrant, OP^ is positive, OJ/3 and J/3P3 are negative ; hence tan A is positive, sin A and cos A are negative. In the fourth quadrant, OP4 and OM^ are positive, M^P^ is negative ; hence cos A is positive, sin A and tan A are negative, 5—2 68 ELEMENTAKY TRIGONOMETRY. [chap. 77. The following diagrams shew the signs of the trigono- metrical functions in the four quadrants. It will be sufficient to consider the three principal functions only. sine cosine tangent 4- The diagram below exhibits the same results in another use- ful form. sine positive cosine negative tangent negative tangent positive sine negative cosine negative all the ratios positive cosine positive sine negative tangent negative 78. When an angle is increased or diminished by any multiple of four right angles, the radius vector is brought back again into the same position after one or more revolutions. There are thus an infinite number of angles which have the same boundary line. Such a,ngles are called coterminal angles. If n is any integer, all the angles coterminal with A may be represented by n. 360° + ^. Similarly, in radian measure all the angles coterminal with B may be represented by 27i7r + 6. From the definitions of Art. 75, we see that the position of the boundary line is alone sufficient to determine the trigono- metrical ratios of the angle; hence all coterminal angles have the same trigonometrical ratios. For instance. sin {n . 360° + 45°)= sin 45° = --^ ; and cos f 2w7r 4- — j viil] TRIGONOMETRICAL RATIOS OF ANY ANGLE. 69 Example. Draw the boundary lines of the angles 780°, - 130°, - 400°, and in each case state which of the trigonometrical functions are negative. (1) Since 780 = (2 x 360) + 60, the radius vector has to make two complete revolutions and then turn through 60°. Thus the bound- ary line is in the first quadrant, so that all the functions are positive. (2) Here the radius vector has to revolve through 130° in the negative direction. The boundary line is thus in the third quadrant, and since OM and MP are negative, the sine, cosine, cosecant, and secant are negative. (3) Since -400= -(360 + 40), the radius vector has to make one complete revolution in the negative dhection and then turn through 40°. The boundary line is thus in the fourth quadrant, and since MP is negative, the sine, tangent, cosecant, and cotangent are negative. M / / / 1/ P EXAMPLES. VIIL a. State the quadrant in which the radius vector lies after describing the following angles : 1. 5. 135°. 2. 265°. 3. -315°. 4. -120°. 27r Stt _ IOtt o IItt 3-. b. --. 7.-3-. 8. --^. For each of the following angles state which of the three principal trigonometrical functions are positive. 9. 470°. 10. 330°. 11. 12. -230°. 13. -620°. 14. 47r -_ IStt 15. 17. 575°. - 1200°. IStt 6 • 70 ELEMENTARY TRIGONOMETRY. [chap. In each of the following cases write down the smallest positive coterminal angle, and the value of the expression. 18. sin 420°. 21. sec 405°. 24. cot^^. 19. cos 390°. 20. tan (-315°) 22. cosec ( - 330°). 23. cosec 4380°. 257r „„ / 57r 25. sec ' 26. tan 79. Since the definitions of the functions given in Art. 75 are applicable to angles of any magnitude, positive or negative, it follows that all relations derived from these definitions must be true universally. Thus we shall find that the fundamental formulce given in Art. 29 hold in all cases ; that is, sin ^ X cosec J^ = 1, cos J x sec J. = 1, tan^4 xcot^ = l ; tan A sin J. cos^ ' cot A = cos J. sin A ' sin^ A + cos^ A^l, l+tan^.4-secM, 1 + cot^ A = cosec^ A . It will be useful practice for the student to test the truth of these formulae for different positions of the boundary line of the angle A. We shall give one illustration. 80. Let the radius vector revolve from its initial position OX till it has traced out an angle A and come into the position OP indicated in the figure. Draw F3I perpendicular to XX'. In the right-angled triangle OMP, MP'- + OIP = OP^ (1). Divide each term by OP^ ; thus /i/P\2 /Oj¥^ X' M \0P + OP o Y' that is, sin^ A+cos^ A = l. viil] trigonometrical ratios of any angle. 71 Divide each term of (1) by 02P; thus \om) '^^~\om) ' that is, tan2^4 + l = sec2 J.. Divide each term of (1) by 3/P2; thus (OMy _ fopy '^[mfJ ~\MPJ ' that is, 1 + cot2 A = cosec^ A, It thus appears that the truth of these relations depends only on the statement OF^=MF^+OJI^ in the right-angled triangle OMP, and this will be the case in whatever quadrant OP lies. Note. OM" is positive, although the line 031 in the figure is negative. 81. In the statement cos J =\/l — siu'^^I, either the positive or the negative sign may be placed before the radical. The sign of the radical hitherto has always been taken positively, because we have restricted ourselves to the consideration of acute angles. It will sometimes be necessary to examine which sign must be taken before the radical in any particular case. Example 1. Given cos 126° 53' = - 1 , find sin 126° 53' and cot 126° 53'. Since sin-^ + cos-^ = 1 for angles of any magnitude, we have sin ^ = ± a/1 - cos^^ . Denote 126° 53' by A ; then the boundary line of A lies in the second quadrant, and therefore siu A is positive. Hence the sign + must be placed before the radical ; .-. sinl26°53'=+^l4. + ^| = ^; fioAo^o, COS 126° 53' / 3\ . /4\ 3 cot 126° 53 = ^.^ ^^^, .3. =. (^ - -gj -^ (5 J - - i . ^ The same results may also be obtained by the method used in the following example. The appropriate signs of the lines are she^m in the fisrure. 72 ELEMENTARY TRIGONOMETRY. [CHAP. VIII. 15 Example 2. If tan A= --^, find sin A and cos A. o The boundary line of A will lie either in the second or in the fourth quadrant, as OP or OF', In either position, the radius vector = ^{15f+{8f = ^289 = 17. Hence sin XOP=i|, cos ZOP=-^ 17 15 17 8 and sin XOP' = - t^ , cos XOP' - . Thus corresponding to tan^, there are two values of sin A and two values of cos J^. If however it is known in which quadrant the boundary line of A lies, sin A and cos A have each a single value. EXAMPLES. VIII. b. 1. Given sin 120° ='^ , find tan 120°. 2. Given tan 135° = - 1, find sin 135°. 3. Find cos 240°, given that tan 240° = ^3- 4. If A = 202° 37' and sin ^ = - -^ , find cos .1 and cot A, 5. If -4 = 143° 8' and cosec ^4 = 1| , find sec A and tan A. 6. If ^ = 216° 52' and cos J = - ^ , find cot A and sin A. 7. Given sec — = - 2, find sin — and cot — . 8. Given sin - t- = — pr , find tan — - and sec — - . 4 V^ 4 4 12 9. If cos -4 = r^ , find sin A and tan A. lo CHAPTER IX. VARIATIONS OF THE TRIGONOMETRICAL FUNCTIONS. 82. A CAREFUL perusal of the following remarks will render the explanations which follow more easily intelligible. Consider the fraction - in which the numerator a has a certain X fixed value and the denominator x is a quantity subject to change; then it is clear that the smaller x becomes the larger does the value of the fraction - become. For instance X ^=10a, -1-=1000^^ 1 =10000000«. 10 1000 10000000 By making the denominator x sufficiently small the value of the fraction - can be made as large as we please; that is, as x approaches to the value 0, the fraction - becomes infinitely great. The symbol oo is used to express a quantity infinitely great, or more shortly infinity, and the above statement is concisely written" ivhen x=0. the limit of - = co . •' X Again, if x is a quantity which gradually increases and finally becomes infinitely large, the fraction - becomes infinitely small ; that is, when ^= 00 , the limit of ~ — 0. ' '' X 74 ELEMENTARY TRIGONOMETRY. [CHAP. 83. Definitiox. If y is a function of x, and if when a; approaches nearer and nearer to the fixed quantity a, the value of y approaches nearer and nearer to the fixed quantity b and can be made to difier from it by as little as we please, then h is called the limiting value or the limit of y when x = a. 84. Trigonometrical Functions of 0°. Let XOF be an angle traced out by a radius vector OP of fixed len^h. M X Draw PJ/ perpendicular to J"; then sinPOJ/=^^. If we suppose the angle FOM to be gradually decreasing, MP will also gradually decrease, and if OP ultimately come into coincidence with OM the angle POM vanishes and MP—0. Hence sin 0° = , ^ = 0. Again, cos POM— -^ ; but when the angle POM vanishes OP becomes coincident with OM. Hence cosO =7=r7> = l. OM Also when the angle POM vanishes, tan 0° = YTT? = 0. And cosec 0° = -. — -^ = :^ = oo : sm secO°=:--~=?^ = l; cos 1 cot =- :==- = cC. tanO IX.] VARIATIONS OF THE FUNCTIONS. 75 85. Trigonometrical Functions of 90° or « • Let XOP be an angle traced out by a radius vector of fixed leiitrth. Y P O M M X Draw PM perpendicular to OX, and OY perpendicular to OX. By definition, MP OM sin POM=-^ , cos POJi[= ^^, , tan POM=- -^ OM If we suppose the angle POM to be gradually increasing, MP will gradually increase and OM decrease. When OP conies into coincidence with Y the angle POM becomes equal to 90", and OJ/ vanishes, while MP becomes equal to OP. Hence And OP sm90° = ^=l; c<«90°=-^ = 0; , Q.. MP OP cot90° = -^— = i-==0: tan 90 cc sec 90° = cosec 90° — cos 90° 1 = - = QC sin 90° 1. 76 ELEMENTARY TRIGONOMETRY. [chap. 86. To trace the changes in sign and magnitude of sin A as A increases from 0° to 360°. Let XX' and TY' be two straight lines intersecting at right angles in 0. With centre and any radius OP describe a circle, and suppose the angle A to be traced out by the revohition of OP through the four quadrants starting from OX. Draw Pi¥ perpendicular to OX and let OP = r; then sin A = MP and since r does not alter in sign or magnitude, we have only to consider the changes of MP as P moves round the circle. When .1=0°, i/P=0, and sinO° = -=0. r In the first quadrant, MP is positive and increasing; .•. sin A is positive and increasing. When ^ = 90°, MP=r, and sin 90° = - = 1. In the second quadrant, 21 P is positive and decreasing ; .*. sin A is positive and decreasing. When .4 = 180°, MP = 0, and sinl80° = ^=a IX.] VARIATIONS OF THE FUNCTIONS. In the third quadrant^ MP is negative and increasing ; .*. sin A is negative and increasing. When A — 270°, MP is equal to r, but is negative ; hence sin 270° 77 -"=-1. r In the fourth quadrant^ MP is negative and decreasing ; .'. sin^ is negative and decreasing. When J =360°, MP = 0, and sin 360° = - = 0. 87. The results of the previous article are concisely shewn in the following diagram : sin go—1 sin A positive and decreasing sin lSo=o sin A negative and iyicreasing sin A positive and increasing — stn o=-o sin A negative and decreasing stn 2JO '=-1 88. We leave as an exercise to the student the investi- gation of the changes in sign and magnitude of cos ^ as ^ increases from 0° to 360°. The following diagram exhibits these changes. cos go — o cos A negative and increasing cosj8o—~1 cos A negative and decreasing cos A positive and decreasifirg cos o- i cos A positive and increasing cos 2jo=o 78 ELEMENTARY TRIGONOMETRY. [CHAP. 89. To trace the changes in sign and magnitude of tan A as A increases from 0° to 360°. MP With the figure of Art. 86, tan A = ^^ , and its changes will therefore depend on those of JfP and OM. When A = 0° 3fP= 0, 03/= r; . • . tan 0° = - = 0. r In the first quadrant, MP is positive and increasing, OM is positive and decreasing ; .'. tan J. is positive and increasing. When .1 - 90°, MP = r, OM^ ; . '. tan 90° = |^ = oo . In tJie second quadrant^ MP is positive and decreasing, OM is negative and increasing ; .*. tan^ is negative and decreasing. When A = 180°, MP=0; .-. tan 180° = 0. In the third quadrant, MP is negative and increasing, OM is negative and decreasing ; .*. tan J. is positive and increasing. When A = 270°, 0M= ; . • . tan 270° - oo . In the fourth quadrant, MP is negative and decreasing, OM is positive and increasing ; .'. tan J. is negative and decreasing. When A = 360°, MP=0; . • . tan 360° = 0. Note. When the numerator of a fraction changes continually from a small positive to a small negative quantity the fraction changes sign by j)assing through the value 0. When the denomi- nator changes continually from a small positive to a small negative quantity the fraction changes sign by passing through the value oo . For instance, as A passes through the value 90°, OM changes from a small positive to a small negative quantity, hence -^r^, that is cos^, changes sign by passing through the value 0, while yyjTj-, that is tan^, changes sign by passing through the value oo . IX.] VARIATIONS OF THE FUNCTIONS. 79 90. The results of Art. 89 are shewn in the following diagram : tan gd—Qo ^£ini8o=o tan A negative and decreasitvg tan A positive and increasing tan A positive andJAicreasing tan o^=-o tan A negative ayid decreasing tan 2jo°:=.<:;o The student will now "have no difficulty in tracing the variations in sign and magnitude of the other functions. 91. In Arts. 86 and 89 we have seen that the variations of the .trigonometrical functions of the angle XOP depend on the position of P as P moves roimd the circumference of the circle. On this account the trigonometrical functions of an angle are called circular functions. This name is one that we shall use frequently. EXAMPLES. IX. Trace the changes in sign and magnitude of 1. cot A J between 0° and 360°. 2. cosec 6^ between and tt. 3. cos 6, between it and Stt. 4. tan J, between —90° and -270°. 5. sec 6, between - and '-z- . 2 2 Find the value of 6. cos 0° sin2 270° - 2 cos 180° tan 45°. 7. 3 sin 0° sec 1 80° + 2 cosec 90° - cos 360°. 8. 2 sec2 TT cos + 3 sin^ —- - cosec - . 2 2 9. tan IT cos — - + sec 27r — cosec -r- . 2 2 80 ELEMENTARY TRIGONOMETRY. [chap. Note on the old definitions of the Trigonometrical Functions, Formerly, Mathematicians considered the trigonometrical functions with reference to the arc of a given circle, and did not regard them as ratios but as the lengths of certain straight lines drawn in relation to this arc. Let OA and OB be two radii of a circle at right angles, and let P be any point on the circumference. Draw PM and PN perpendicular to OA and OB respectively, and let the tangents at A and B meet OP produced in T and t respectively. The lines PM, AT, OT, AM were named respectively the sine, tangent, secant, versed-sine of the arc AP, and PN, Bt, Ot, BN, which are the sine, tangent, secant, versed-sine of the complementary arc BP, were named respectively the cosine, cotangent, cosecant, coversed-sine of the arc AP. As thus defined each trigonometrical function of the arc is equal to the corresponding function of the angle, which it subtends at the centre of the circle, multiplied by the radius. Thus AT —-= tan PO^; that is, ^r=0.4 x tanPO^; OA and —- = seG BOP = coseGPOA; that i&, Ot= OB xcoseo POA. Oh The values of the functions of the arc therefore depended on the length of the radius of the circle as well as on the angle subtended by the arc at the centre of the circle, so that in Tables of the functions it was necessary to state the magnitude of the radius. The names of the trigonometrical functions and the abbreviations for them now in use were introduced by different Mathematicians chiefly towards the end of the sixteenth and during the seventeenth century, but were not generally employed until their re-introduction by Euler. The development of the science of Trigonometry may be considered to date from the publication in 1748 of Euler's Introductio in analysin Infinitorum. The reader will find some interesting information regarding the progress of Trigonometry in Ball's Short History of Mathematics. IX.] MISCELLANEOUS EXAMPLES. C. 81 MISCELLANEOUS EXAMPLES. C. 1. Draw the boundary lines of the angles whose tangent is 3 equal to - -j , and find the cosine of these angles. 2. Shew that cos A (2 sec A + tan A) (sec A- 2 tan A) = 2 cos ^ - 3 tan A. 3. Given C=90°, 6 = 10-5, c = 2l, solve the triangle. 25 4. If secJ.= -— , and A lies between 180° and 270°, find cot A. 5. The latitude of Bombay is 19° N. : find its distance from the equator, taking the diameter of the earth to be 7920 miles. 6. From the top of a cliff 200 ft. high, the angles of de- pression of two boats due east of the observer are 34° 30' and 18° 40': find their distance apart, given cot 34° 30' = 1 -455, cot 18° 40' = 2-96. 7. If A Hes between 180° and 270°, and 3 tan ^=4, find the value of 2 cot A- 5 cos J. -fsin A. 8. Find, correct to three decimal places, the radius of a circle in which an arc 15 inches long subtends at the centre an angle of 71° 36' 3-6". 9. Shew that tan^^ cot^^ _ 1 - 2 sin^ 3 cos^ $ l+tan2^ 1 + cot^ ~ sin ^ cos ^ * 10. The angle of elevation of the top of a tower is 68° 11', and a flagstaflF 24 ft. high on the summit of the tower subtends an angle of 2° 10' at the observer's eye. Find the height of the tower, given tan 70° 21' = 2-8, cot 68° 11' = '4. H. K. E. T. CHAPTER X. CIRCULAR FUNCTIONS OF CERTAIN ALLIED ANGLES. 92. Circular Functions of 180° - A. Take any straight line , XOX\ and let a radius vec- tor starting from OX revolve until it has traced the angle A, taking up the position x' M' O MX OP. Again, let the radius vector starting from OX revolve through 180° into the position OX' and then hack again through an angle A taking up the final position OF. Thus XOP' is the angle 180° -A. From P and P' draw PM and P'M' perpendicular to XX' ; then by Euc. i. 26 the triangles 0PM and OP'M' are geometrically equal. By definition, M'F sin (180°-^) = -^^; but M'P' is equal to MP in magnitude and is of the same sign ; .'. sm(180 -^) = yTp = sm J. Again, cos (180°- J.) = -yp;r; and OM' is equal to OM in magnitude, but is of opposite sign ; ,,n^o AS -OM OM .', cos(180 -^)=-^^p-= --yp=-cos^. M'F MP MP Also tan (180° " ^) = -^^ = raF= " Qjr - *^^ ^- CIRCULAR FUNCTIONS OF CERTAIN ALLIED ANGLES. 83 93. In the last article, for the sake of simplicity we have supposed the angle A to be less than a right angle, but all the formulse of this chapter may be shewn to be true for angles of any magnitude. A general proof of one case is given in Art. 102, and the same method may be applied to all the other cases. 94. If the angles are expressed in radian measure, the formulse of Art. 92 become sin {u-6) = sin 6, cos (tt — ^) = — cos 6j tan (tt — ^) = — tan 6. ExaTivple 1. Find the sine and cosine of 120°. sin 120° = sin ( 180° - 60°) = sin 60° = ^ . COS 120°= COS (180° - 60°) = - cos 60° = - ^ . Example 2. Find the cosine and cotangent of -s- D cos-g=cosU-^j^-cos-=-^. ,57r cot-g = cot f TT- |J= - cot^= -^3. 95. Definition. When the sum of two angles is equal to two right angles each is said to be the supplement of the other and the angles are said to be supplementary. Thus if A is any angle its supplement is 180° — ^. 96. The results of Art. 92 are so important in a later part of the subject that it is desirable to emphasize them. We therefore repeat them in a verbal form : the sines of supplementary angles are equal in magnitude and are of the same sign; the cosines of supplementary angles are equal in magnitiide hut are of opposite sign; the tangents of supplementary angles are equal in magnitude hut are of opposite sign. 6—2 84 ELEMENTARY TRIGONOMETRY. [chap. 97. Circular Functions of 180°+ A. Take any straight line A'' OX' and let a radius vector starting from OX revolve until it has traced the angle A, taking up the position OP. X' Again, let the radius vec- tor starting from OX revolve through 180° into the position OX', and then further through an angle A, taking up the final position OP'. Thus XOF is the angle 180° + ^. From P and P' draw Pifand P'M' perpendicular to XX'; then OP and OP' are in the same straight line, and by Euc. i. 26 the triangles OPIf and OP'M' are geometrically equal. By definition. M'P' sin (180°+^)=-^^ and M'P' is equal to MP in magnitude but is of oj)posite sign ; Again, -MP .'. sin (180°+^) = -^ cos (180° + ^): MP OP = —sin J. OM' OP' and OM' is equal to OM in magnitude but is of opposite sign ; .-. cos(180°+^) = - 03f_ 0M_ , Also tan (180°+^): M'P' MP MP = -?r^^ = \sxi A. OM' - OM OM Expressed in radian measure, the above formulae are written sin(7r + ^)= -sin ^, cos(7r+^)= -cos^, tan (7r + ^)=tan ^^ In these results we may draw especial attention to the fact that an angle may be increased or diminished by two right angles as often as we please without altering the value of the tangent. Example. Find the value of cot 210°. cot 210° = cot (180° + 30°) = cot 30° = ^3. A M X X.] CIRCULAR FUNCTIONS OF CERTAIN ALLIED ANGLES. 85 98. Circular Functions of 90° + A. Take any straight line XOX\ , Y and let a radius vector starting from OX revolve until it has traced the angle A^ taking up the position OP, Again, let the radius vec- \\-P°° tor starting from OX revolve \ ^\-7 through 90° into the position I M^^-^^v 07, and then further through X' MO an angle A^ taking up the final position OF. Thus XOF' is the angle 90° + A. From P and P' draw PM and FM' perpendicular to XX' ; then L M'FO= l F0Y=A = z P03L By Euc. I. 26, the triangles 0PM and OFM' are geometrically equal; hence MF' is equal to OM in magnitude and is of the same sign, and OM' is equal to MP in magnitude but is of opposite sign. By definition, sm(90 +.4) = -^^= _=cos.l; cos (90° + ^) = tan (90° + -.!) = OM' -MP OF M'F OP OM MP OP — — sin A ; OM'~-MP~ MP~ ^^''^^• sin - Expressed in radian measure the above formulas become /^|+(9^=cos^, cos(^|+^^= -sin^, tan ^| + ^j= -cot(9. Example 1. Find the value of sin 120°. sin 120° = sin (90° + 30°) = cos 30° = ^ . 86 ELEMENTARY TRIGONOMETRY. [CHAP. Example 2, Find the values of tan (270° + A) and cos ( -— + ^ j . tan (270° + J) = tan (180° + 90° + ^) = tan (90° + .4)= -cot^; cos ( -|^ + ^ ) = cos ( TT + ~ + ^ j = - cos d + ^ j = sin (9. 99. Circular Functions of -A. Take any straight line OX and let a radius vector starting from OX revolve until it has traced the angle A J taking up the position OP, Again, let the radius vector start- ing from OX revolve in the opposite direction until it has traced the angle J, taking up the position OP'. Join PP' ; then MP' is equal to MP in magnitude, and the angles at M are right angles. [Euc. i. 4.] By definition, MP -MP sm(-J.) = ^p- = ^^^^=-sm^; , ,, OM OM cos{-A) = -^, = ^ = cosA; ^ . ,, MP' -MP ^ , It is especially worthy of notice that ice may change the sign of an angle ivithout altering the value of its cosine. Example. Find the values of cosec ( - 210°) and cos {A - 270°). cosee ( - 210°) = - cosec 210° = - cosec ( 180° + 30°) = cosec 30° = 2. cos (4 - 270°) = cos (270° ~A) = cos (180° + 90° - A) = - cos (90° -4)= -sin 4. X.] CIRCULAR FUNCTIONS OF CERTAIN ALLIED ANGLES. 87 100. If /(^) denotes a function of A which is unaltered in magnitude and sign when —A is written for A, then /(J.) is said to be an even function of ^. In this case/( - A)=f{A). If when —A is written for A, the sign oi f{A) is changed while the magnitude remains unaltered, f{A) is said to be an odd function of ^, and in this case/( - J.)= -f{A). From the last article it will be seen that cos J. a7id sec J. are even functions of A^ sin ^, cosecJ., tan^, q,o\j A are odd functions of A . EXAMPLES. X.a. Find the numerical value of 1. cos 135°. 2. sin 150°. 3. tan 240°. 4. cosec225°. 5. sin (-120°). 6. cot (-135°). 7. cot 315°. 8. cos (-240°). 9. sec (-300°). 10. tan — - . 11. sin — - . 12. sec — . 4 3 3 13. cosec ( - ^ ) • 14. cos ( - ~T ) • 15. cot ( - ^ Express as functions of A : 16. cos (270° + .!). 17. cot (270°-.!). 18. sin (^-90°). 19. sec (^-180°). 20. sin (270°-^). 21. cot (^-90°). Express as functions of 6 : 22. sin('^-|V 23. tan(^-7r). 24. iieof~-S\. Express in the simplest form : 25. tan (180° + ^ ) sin (90° + ^) sec (90° -A). 26. cos (90° + ^) + sin (180°-^) -sin (180° + ^)- sin (-^). 27. sec (180°+^) sec (180° -^) + cot (90° + ^) tan (180° + ^). 88, ELEMENTARY TRIGONOMETRY. [chap. 101. In Art. 38 we have established the relations which subsist between the trigonometrical ratios of 90° — A and those of A, when A is an acute angle. We shall now give a general proof which is applicable whatever be the magnitude of J.. 102. Circular Functions of 90° - A for any value of A. X'M X X' Let a radius vector starting from OX revolve until it has traced the angle A, taking up the position OP in each of the two figures. Again, let the radius vector starting from OX revolve through 90° into the position 03^ and then back again through an angled, taking up the final position OF' in each of the two figures. Draw PM and P'3I' perpendicular to XX'; then whatever be the value of A, it will be found that iOFM'= i POM, so that the triangles OMP and OM'P' are geometrically equal, having MP equal to OM', and OM equal to M'P', in magnitude. When P is above XX', F is to the right of YT, and when P is below XX', P' is to the left of YT. When P' is above XX', P is to the right of YY', and when P' is below XX', P is to the left of YT. Hence MP is equal to OM' in magnitude and is always of the same sign as OM' ; and M'P' is equal to OM in magnitude and is always of the same sign as OM. X.] CIRCULAR FUNCTIONS OF CERTAIN ALLIED ANGLES. 89 By definition, sin (90 -A)=-^=-^=cosA\ ,_ ,. OM' MP . , cos (90 - A) = -^ = -^ = ^m A', M'P' OM tan(90°-^)=^ = ^=cot.4. A general method similar to the above may be applied to all the other cases of this chapter. 103. Circular Functions of n . 360° + A. If n is any integer, n . 360° represents 71 complete revolutions of the radiiLS vector, and therefore the boundary line of the angle n . 360" + 4 is coincident with that of A. The value of each func- tion of the angle n . 360° + A is thus the same as the value of the corresponding function of A both in magnitude and in sign. 104. Since the functions of all coterminal angles are equal, there is a recurrence of the values of the functions each time the boundary line completes its revolution and comes round into its original position. This is otherwise expressed by saying that the circular functions are periodic^ and 360° is said to be the amplitude of the period. In radian measure, the amplitude of the period is Stt. Note. In the case of the tangent and cotangent the amplitude of the period is half that of the other circular functions, being 180° or TT radians. [Art. 97.] 105. Circular Functions of n . 360° - A. If n is any integer, the boundary line of n . 360° — ^ is co- incident with that of -A. The value of each function of n . 360° — ^ is thus the same as the value of the corresponding function of - A both in magnitude and in sign ; hence sin (;i . 360° - ^) = sin ( - ^) = - sin ^ ; cos {n . 360° - A) = cos {-A) = cos A; tan {n . 360° - ^) =tan ( - ^) = - tan A. 90 ELEMENTARY TRIGONOMETRY. [CHAP. 106. We can always express the functions of any angle in terms of the functions of some positive acute angle. In the arrangement of the work it is advisable to follow a uniform plan. (1) If the angle is negative, use the relations connecting the functions of — ^ and Jl. [Art. 99.] Thus sin(-30°)=-sin30°=-^; cos (-845°) = cos 845°. (2) If the angle is greater than 360°, by taking off multiples of 360° the angle may be replaced by a coterminal angle less than 360°. [Art. 103.] Thus tan 735° = tan (2x360° + 15°) = tan 15°. (3) If the angle is still greater than 180°, use the relations connecting the functions of 180° + J. and A. [Art. 97.] Thus cot 585° = cot (360° + 225°) = cot 225° = cot(180° + 45°)=cot45° = l. (4) If the angle is still greater than 90°, use the relations connecting the functions of 180° — udL and A. [Art. 92.] Thus cos675° = cos(360° + 315°) = cos315° = cos (180° + 135°) = - cos 135° = -cos (180° -45°)= cos 45° = 45- Example. Express sin ( - 1190°), tan 1000°, cos ( - 3860°) as func- tions of positive acute angles. sin (-1190°)= -sin 1190°= -sin (3x360°+ 110°)= -sin 110° = - sin (180° - 70°) = - sin 70°. tan 1000°= tan (2 x 360° + 280°) = tan 280° = tan (180° + 100°) = tan 100° = tan (180° - 80°) = - tan 80°. cos ( - 3860°) = cos 3860° = cos (10 x 360° + 260°) = cos 260° = cos (180° + 80°) = - cos 80°. X.] CIRCULAR FUNCTIONS OF CERTAIN ALLIED ANGLES. 91 107. From the investigations of this chapter we see that the number of angles which have the same circular function is unhmited. Thus if tan^ = l, 6 may be any one of the angles coterminal with 45° or 225°. /3 Example. Draw the boundary lines of A when sin^ = ^ , and write down all the angles numerically less than 360° which satisfy the equation. V3 Since sin 60° : , if we draw OP making z TOP = 60°, then OP is one position of the boundary line. Again, sin 60°= sin (180° - 60°) = sin 120°, so that another position of the boundary line will be found by making ZOP' = 120°. There will be no position of the boundary line in the third or fourth" quadrant, since in these quadrants the sine is negative. Thus in one complete revolution OP and OP' are the only two positions of the bound- ary line of the angle A. Hence the positive angles are 60° and 120°; and the negative angles are - (360° - 120°) and is, - 240° and - 300°. (360° -60°); that EXAMPLES. X. b. Find the numerical value of 1. 4. 7. 10. 13. 16. 19. cos 480°. sin (-870°). cosec ( — 660°). cos 4005°. sec 2745°. IStt 4 • IGtt 3 ' sm cot 2. 5. 8. 11. 14. 17. 20. sin 960°. sec 900°. cot 840°. cot 990°. cos 2400°. 2^ 4 * cot sec 3. cos ( - 780°). 6. tan ( - 855°). 9. cosec ( - 765°). 12. sin 3015°. 15. sec ( - 5895°). 18. sec -r- . +1 92 ELEMENTARY TRIGONOMETRY, Find all the angles numerically less than 360° which satisfy the equations : 21. cos^=^. 22. sin^=-|. 2 2 23. tan(9=-v/3. 24. cot^=-l. If A is less than 90°, prove geometrically 25. sec (^ - 1 80°) = - sec ^. 26. tan (270°+^)=- cot ^. 27. cos (.4 -90°) = sin .1. 28. Prove that tan A + tan (180° - A) + cot (90°+.l) = tan (360° - A). 29. Shew that sin (180°-^) cot (90° -J) cos (3 60° -^) _ . . tan (180° + ^) • tan (90° +.4) * "sin ( - A) ~^^" " Express in the simplest form sin(- ^) tan (90°+^) cos^ sin (180° + ^) cotyl sin(90° + yl)' cosec(180°-.l) cos (-.4) sec (180°+^) • cos(90° + JL)' cos (90°+^) sec (-.4) tan (180° -vl) sec (360° + .4) sin (1 80° + xt ) cot (90° -A)' 33. Prove that sin C| + (^^ cos (tt - ^) cot ('^ + ^^ = sin(|-^)sin(^-^)cotg + , IItt 34. AVheu a=— r— , find the numerical value of 4 sin^ a — cos^ a + 2 tan a — sec^ a. CHAPTER XI. FUNCTIONS OF COMPOUND ANGLES. 108. When an angle is made up bj the algebraical sum of two or more angles it is called a compound angle ; thus A + B, A — B, and A + B — C are compound angles. 109. Hitherto we have only discussed the properties of the functions of single angles, such as A, B, a, 6. In the present chapter we shall prove some fundamental properties relating to the functions of compound angles. We shall begin by finding expressions for the sine, cosine, and tangent of ^1+^ and A — B in terms of the functions of A and B. It may be useful to caution the student against the prevalent mistake of supposing that a function of ^4 +^ is eqvial to the sum of the corresponding functions of A and B, and a function of ^ - ^ to the difference of the corresponding functions. Thus sin (A -\-B) is not equal to sin A +sin B^ and cos {A - B) is not equal to cos A — cos B. A numerical instance will illustrate this. Thus if ^ = 60°, B = 30°, then A+B= 90°, so that cos {A+B) = cos 90° = ; but cos^+cosi?=cos60° + cos30° = i-f ^. Hence cos {A + B) is not equal to cos A + cos B. In like manner, sin (J ■\-A) is not equal to sin ^ +sin A ; that is, sin 2^ is not equal to 2 sin A. Similarly tan ZA is not equal to 3 tan ^. 94 ELEMENTARY TRiaONOMETRY. [chap. 110. To prove theformulce sin {A+B)= sin A cos B + cos A sin B^ cos {A-^B) = cos A cos ^ - sin A sin 5. Let zZOif= J, and lMON=B; then lLON=A-\-B. In OiV, ^Ae boundary line of the compound angle A+B, take any point P, and draw FQ and Pi? perpendicular to OL and OJf respectively; also draw P>S' and i? 7^ perpendicular to OL and PQ respectively. By definition, RS+PT ^RS PT OP ~OP'^OP _BS OR PT PR ~ OR' OP'^ PR' OP = sin A . cos B + cos jTPP . sin B. But z TPR=90°- L TRP= l TR0= l R0S=A ; .'. sin (^4 +5)= sin ^ cos ^ + cos ^ sin ^. OS TR Also cos {A+B) = y-yp = OP OP OS OR TR PR OP OR 'OP PR' OP = cos A . cos B — sin TPR . sin B = cos A cos B — sin A sin B. XI.] FUNCTIONS OF COMPOUND ANGLES. 111. To prove the form/uXcB sin {A — B)= sin A cos B — cos A sin B, cos {A—B) = cos A cos B + sin J. sin B. Let z Z(9i/= ^ , and z if OiY = ^ ; then z ZOi\^= ^ - 5. 95. In OJV, the boundary line of the compound angle A—B, take any j)oint F, and draw P§ and PR perpendicular to OL and OM respectively; also draw RS and ET perpendicular to OL and QP respectively. By definition, ... J.. PQ RS-PT RS OP OP PT OP _RS^ OR_PT PR "OR 'OP PR 'OP = sin A . cos B - cos TPR . sin B. But z TPR = 90°- L TRP= z MRT= z MOL=A ; .'. sin(^ -.5)=sin J. cosJ5— cos J. sin5. , , ^, OQ OS+RT OS RT Also cos(^-^)=^= — oF^^OP^OP _qS OR RT RP ~ OR' OP^ RP' OP = cos A . cos B + sin TPR . sin B = cos A cos .5 -f sin J^ sin 5. 96 ELEMENTARY TRIGONOMETRY. [CHAP. 112. The expaiisions of sin (^+5) and cos (^+5) are fre- quently called the "Addition Formulje." We shall sometimes refer to them as the "J. + ^" and ".4 -^" formulae. 113. In the foregoing geometrical proofs we have supposed that the angles A, B, A+B are all less than a right angle, and that J. — ^ is positive. If the angles are not so restricted some modification of the figures will be required. It is however unnecessary to consider these cases in detail, as in Chap. XXII. we shall shew by the Method of Projections that the Addition Formulae hold universally. In the meantime the student may assume that they are always true. Example 1. Find the value of cos 75°. cos 75° = cos (45° + 30°) = cos 45° cos 30°- -sin 45° sin 30° 1 V3 1 V2' 2 V2" 1 V3-1 2~ 2^2 ■ Exa uiple 2. 4 5 If sin A =- and sin J5 = -^ , find sin {A - -B). sin (A-B) — sin A co s J5~cos^ s in 5. 3 ~5' V-i- But cos A = /y/l - sin^^ : and cosB=^l-sm^B = ^ 1-—= — ; . ,, „, 4 12 3 5 33 3 12 Note. Strictly speaking cos^=±^ and cosJ5 = ±— , so that o lo sin {A - B) has four values. We shall however suppose that in similar cases only the positive value of the square root is taken. 114. To prove that sin {A +B) sin {A - B) = sin^ A - sin^ B. The first side = (sin A cos B + cos A sin B) (sin A cos B - cos A sin B) = sin^ A cos^ B — cos- A sin^ B = sinM (l-sin2^)-(l-sin2^)sin2^ =sdn2^ - sin^i?. XI.] FUNCTIONS OF COMPOUND ANGLES. 97 EXAMPLES. XL a. [The examples printed in more prominent type are important^ and should be regarded as standard forrmdce.j Prove that L sin(^+45°) = -^(sin^l+cos^). 2. cos(^ + 45°)=-T-(cos^-sm^4). 3. 2siu(30°-^) = cos.l-V3sin.4. 4 3 4. If cos ^ =- , cos ^ = - , find sin {A + B) and cos {A - B). o o 3 12 5. If sin -4 =-, cos 5= — ) find cos {A +B) and sin (A - B). O xO 17 5 6. If sec^4=-^, cosec^ = -r» find sec (.4+5). 8 4' • ' Prove that 7. sin 75° = cos 15°=^^. 8. sin 15° = cos 75° ='^^. sin(a+^) . , . a 9. ^ ^,=tana+tanR cos a cos /S ,. sin(a — S) , o i. 10. . ^ . ^X = cotB-cota. smasinjS 11. 55iO:^]=cot^+taria. COS a sm p 12. cos (A + B) cos ( A - B) = cos^ A - sin^ B. 13. sin (A + B) sin ( A - B) = cos^ B - cos2 A. 14. COS (45°-^)- sin (45°+^) =0. 15. cos(45° + ^) + sin(^-45°) = 0. 16. cos {A-B)- sin {A+B) = (cos ^ - sin ^ ) (cos B - sin B). 17. cos {A+B)-\- sin {A-B)=^ (cos ^ + sin ^ ) (cos B - sin B). H. K. E. T. 7 98 ELEMENTARY TRIGONOMETRY. [chap. Prove the following identities : _ 18. 2sin(^+45°)sin(^-45°) = sinM-cosM. 19. 2 cos (7 + ") <^^^ (j — «)=cos2a — sin^ a. 20. 2sin(^|+ajcosr^ + ^j = cos(a+/3) + sin(a-i3). 21. s^QO-y) I sin(y-a ) ^ sin(a-3) ^^ cos/3 cos-y cos y COS a cos a cos/3 115. 2^0 expand tan (^ + ^) m ^erm^ of tan ^ anc? tan B. , , . „, _sin(J.+^)_sin^ cos 5 + cos J. sin 5 cos {A +B) cos A cos B - sin J. sin B ' To express this fraction in terms of tangents, divide each term of numerator and denominator by cos A cos B ; sin -4 . sin 5 tan(^+J5) = . cos A cos B 1- sin A sin B ' tan(^+^) = cos A * cos B tan ^ + tan B that is, .,«,x.,-- 1 --/ — ^ , J. „. ' ^ ' 1— tan^tan^ A geometrical proof of this result is given in Chap. XXII. Similarly, we may prove that , , . „. tan A — tan B tan (A—B) = - — ^— ^ . ^ ^ l+tan^tan5 Example. Find the value of tan 75°. tan 45° + tan 30° tan 75°= tan (45° + 30°) = 1- tan 45° tan 30° 1 + -73 v/3^ ^3 + 1 v/3-1 (V3 + 1)(V3 + 1) _ 4 + 2V3 3-1 2 = 2 + ^3. XI.] FUNCTIONS OF COMPOUND ANGLES: 99 116. To expand cot {A -\-B) in terms of cot A and cot B. ,,,,„, cos(^+^) cos J. cos j5 — sin J. sin 5 cot (A-{-B)= -. — )-. ~ = — — -. fz .— . — 5^ . sin(J.+^) sin J.cos^ + cos^ sm5 To express this fraction in terms of cotangents, divide each term of numerator and denominator by sin A sin B ; cos A cos B , , , „, sin A sin B cot A cot ^ — 1 .-. cot(J.+^)= „ ■ = -r^ 1— r . ^ cos^ cos J. cot ^ + cot ^ sin B sin A Similarly, we may prove that i(i-B') _ ^Q^-^'^Q^-^ + l ' cot ^ — cot ^ 117. To find the expansion of sin {A +^+ C)- ii\n{A-\-B+C) = iim{{A-\-B)-\-C] =sin {A +B) cos C'+cos {A +B) sin C = (sin A cos B -\- cos A sin B) cos C + (cos A cos B — sin A sin B) sin O = sin A cos B cos 6'+ cos A sin B cos (7 + cos A cos ^ sin C — sin ^1 sin B sin (7. 118. :7b find the expansion of tan {A+B-\- C). tan J. + tan ^ , ^ :+tan C 1 — tan A tan .^ tan J. + tan ^ ^ 1 — tan J. tan B ' _ tan A + tan 5 + tan O— tan ^ tan B tan (7 1 - tan A tan ^ — tan B tan C'— tan (^' tan A ' Cor. If^+^+(7=180°, thentan(^-f5+(7) = 0; hence the numerator of the above expression must be zero. .'. tan A + tan 54-tan (7= tan A tan B tan C. 7—2 100 ELEMENTARY TRIGONOMETRY. [CHAP. EXAMPLES. XI. b. [The examples printed in more prominent type are important^ and should he regarded as standard formulceJl 1. Find tan {A + B) when tan -4 = ^ , tan B=-. 2. If tan ^ = I , and 5 = 45°, find tan {A - B). 3. If cot ^=^, cot 5 = ^, find cot (^+^) and tan (.4 -5). 7 o 4. If cot J[ =^ , tan B=^ , find cot {A - B) and tan {A +B). 5. tan(45"+A) = Ltt^. J"^ ^ cot^ + I -J 4- /tt , .\ cot^-1 7. «ot(4-^) = c-ot^^- ^- ^^Hi+^j^cotT+I- 9. tanl5°=2-V3. 10. cotl5°=2+x/3. 11. Find the expansions of cos (^ +^+C^ and sin (^ -5+ 0). 12. Express tan (J. - ^ — (7) in terms of tan A^ tan B^ tan C. 13. Express cot (^ +5+ C) in terms of cot J, cot ^, cot G. 119. Beginners not unfrequently find a difficulty in tlie con- verse use of the A-\-B and A — B formulae ; that is, they fail to recognise when an expression is merely an expansion belonging to one of the standard forms. Example 1. Simplify cos (a - /3) cos (a + /3) - sin (a - /3) sin (o + /S). This expression is the expansion of the cosine of the compound angle (a + /3) + (a-/3), and is therefore equal to cos {(a + jS) + (a-/3}} ; that is, to cos 2a. -n, 7 n CI- XT. i tan^ + tan24 Example 2. Shew that :j — ^t j-r- — jr^=tan3u4. J. — tan ^ ran ^^ By Art. 115, the first side is the expansion of tan (^ +2.4), and is therefore equal to tan 3x1. XI.] CONVERSE USE OF THE ADDITION FORMULAE. 101 Example 3. Prove that cot 2 A + tan A =GOsec 2A. _,, ^ ^ . -, cos 2A sin A cos 2A cos A + sin 2A sin A The first side = . ^ . + = r-^. -. sm 2^ cos^ sin2J.cos-4 __cos (2.4 -^) _ cos^ ~ sin 2^ cos -4 ~ sin 2^ cos ^ 1 sin 2^ = cosec2^. Example 4. Prove that cos 4:9 cos 9 + sin 49 sin 6 = cos 2^ cos 9 - sin 2^ sin 9. The first side = cos (4^ -9)= cos 30 = cos {29 + 9) = cos 29 cos - sin 29 sin 0. EXAMPLES. XL c. Prove the following identities : L cos (^ + 5) cos ^+ sin (J. + 5) sin 5=008:4. 2. sin 3 A cos A — cos 3 A sin 4 = sin 2 A. 3. cos 2a cos a + sin 2a sin a = cos a. 4. cos(30° + .4)GOs(30°-4)-sin(30° + 4)sin(30°-4)=^. 5. sin(60°-yl)cos(30°+.4) + cos(60^-4)sin(30° + 4) = l. cos 2a sin 2a 6. =cos3a. sec a cosec a tan(a-^) + tan^ I • '^ ; ;; T^n 7^ = ZBjU a. 1 — tan (a — /3) tan /3 8. «ot(a+fflcota+l ^^ _ cot a — C0t(a + /3) Q tan 4 J. — tan 3 J. _ , 1+ tan 44 tan 3 J. 10. cot ^- cot 2^= cosec 2^. 11. 1 + tan 2^ tan ^ = sec 2^. 12. 1 4- cot 2^ cot ^= cosec 2^ cot ^. 13. sin 26 cos B + cos 2B sin S = sin 4^ cos 6 - cos 4^ sin 6. 14. cos 4a cos a — sin 4a sin a = cos 3a cos 2a - sin 3a sin 2a. 102 ELEMENTARY TRIGONOMETRY. [CHAP. Functions of Multiple Angles. 120. To express sin 2^ in terms of sin A and cos A . sin 2A = sin {A+A) = sin A cos A + cos A sin A ; that is, sin 2 A =2 sin vl cos J.. Since A may have any value, this is a perfectly general formula for the sine of an angle in terms of the sine and cosine of the half angle. Thus if 2A be replaced by 6, we have sm 6 = 2 sm - cos - . 2 2 Similarly, sin 4A = 2 sin 2 A cos 2 A = 4 sin ^ cos A cos 2^. 121. To express cos 2 J. z?i terms of cos ^ awo? sin ^. cos 2^ =cos {A + J.) = cos A cos ^1 —sin A sin A : that is, cos2^=cos2 J. — sin^ J. (1). There are two other useful forms in which cos 2^1 may be expressed, one involving cos A only, the other sin A only. Thus from (1), cos 2A = cos2 A-{1 - cos^ A) ; that is, cos 2 J. = 2 cos- ^ - 1 (2). Again, from (1), cos 2^ = (1 - sin^ ^4) - sin^ J ; that is, cos 2 J. = 1 - 2 sin2 J (3). From formulae (2) and (3), we obtain by transposition l + cos2^ = 2cos2^ (4)^ and l-cos2J=2sin2.4 (5). By division, =— --.=tan2^ (6). •^ ' l + cos2^ Example. Express cos 4a in terms of sin a. From (3), cos 4a= 1 - 2 sin22a= 1 - 2 (4 sin^a cos^a) = 1-8 sin^a (1 - sin^a) = l-8sin2a + 8sin4a. XI.] PUNCTIONS OF 2A. 103 122. The six formulae of the last article deserve special attention. They are universally true so long as one of the angles involved is double of the other. For instance, cos a = cos-^ - — sm^ - , cosa=2 cos^-- 1, cosa=l -2sin2-, A A l+cos^ = 2 cos^-, l-cos^ = 2sin2-. A 2i 6 Example. If cos 6 = "28, find the value of tan — . ^g_l-cosg_ l--28 _ -72 _ 72 _ 9 ^ ^^"2 ~ 1 + coT^ ~ 1 + •28 ~ r28 ~ 128 ~ 16 ' .-. tan-=±-. 123. To express tan 2 J. in terms 0/ tan A. r. A 1 , A 4N tan^+tan^ tan2^ = tan(^+^) = ^_^^^^^^^^ ; . , . . . o . 2 tan il that IS, tan '±A = ~ — 7 — g^j • 124. To express sin 2^ and cos 2 A in terms of tan A. sin 2^ = 2 sin ^ cos J. = 2 7 cos^ ^ = 2 tan ^ cos^ A ; cos J. . g i _ 2 tan ^ _ 2 tan A "sec^J. l-ftan^J.* Again, cos 2A = cos2 A - sin2 A = cos^ A {I- tan^ A) ; 1 - tanM _ 1 - tanM . * . cos JtJx — n — '. — v~; 7 ty A ' sec^^ 1+tan^J. . c^^. .^. A. 1 - tan2{45° - ^) . ^ . Example. Shew that q — - — 5^7^^ — -{ = sm2A. ^ l + tan^(45°-^) The first side = cos 2 (45° - ^) =cos (90° - 24) = sin 2A. 104 . ELEMENTARY TRIGONOMETRY. [CHAP. EXAMPLES. XL d. \The examples printed in more prominent type are important, and should be regarded as standard formulce.l 1. If cos J. = ^, find cos 2^. o 2 2. Find cos 2 A when sin J.=-. 5 3. If sin ^=? find sin 24. o 4. If tan ^=^, find tan 2^. o 5. If tan ^ = - , find sin 2$ and cos 26. 4 a 6. If cos a=- , find tan - . o ^ 7. Find tan A when cos 2A = "96. Prove the following identities : 1+C0S2A 1-C0S2A ,^ 1-cosA . A 11 1 + cosA ^ .A 10. — -. — T— = tan7^. IL .^ A =cot7c-. sin A 2 smA 2 12. 2 cosec 2a = sec a cosec a. 13. tan a + cot a = 2 cosec 2a. 14. cos^a — sin^a = cos2a. 15. cot a - tan a = 2 cot 2a. ir. a.nA cot-A-1 T„ cotJ[-tan4 16. cot2A=-^s — TTH—' 17. — . . , , j = cos2X 2 cot A cot J. + tan A _ 1 + C0t2.1 „ . .Q C0t2yl+1 18. -t; — — — =cosec24. 19. .o . — - = sec24. 2 cot 4 cot^ A- I -. 1+sec^ „ „^ „- sec^-1 „ . ,^ 20. ^ =2cos2-. 21. ^- = 2sm2-. sec^ 2* ' sec(9 2 -sec2^ _. ^„ cosec2^- 5-r- = cos 2(9. 23. ox sec2 6 cosec^ 6 „ 2-sec2^ ^. __ cosec2^-2 .. 22. :r-r- = cos2(9. 23. 5-^—= cos 2^. XI.] FUNCTIONS OF 3A. 105 Prove the following identities : 24. (sin-rt +cos-^ j =l + sinA. 25. f sin rt^ - cos -rt ) =l-siiiA. cos 2a , .._o V 26. ., . . _ =tan (45 -a). l+sin2a __ cos 2a . / 4,^0 \ 27. q r-^r-= cot (45° - a). 1 — sm 2a 28. sin 8 J. = 8 sin J. cos A cos 24 cos 44. 29. cos44 = 8cosM-8cos2 4 + l. 30. sin4 = l-2sin2/'45°-^' 31. cos^ f - — a j - sin^ f - — a j=sin2a. 32. tan (45° + 4) - tan (45° -A) = 2 tan 2 A. 33. tan (45° + 4) +tan (45° -4) = 2 sec 24. 125. Functions of 3A. sin 34 = sin (24 + 4) = sin 24 cos A + cos 24 sin 4 — 2 sin 4 cos^ 4 + (1 - 2 sin^ 4) sin 4 = 2 sin 4 (1 - sin2 4) + (1 - 2 sin^ 4) sin 4 ; =3 sin 4-4 sin^ 4. Similarly it may be proved that cos 34 = 4 cos^ 4 — 3 cos 4. Again, tan 3A = tan (2A + A) = ^^Tu^^AuA , .,. 1 r. A 2 tan 4 by putting tan2^ = j-^^^, we obtain on reduction 3 tan 4 — tan^ 4 tan 34 ■■ l-3tan2 4 106 ELEMENTARY TRIGONOMETRY. [CHAP. These formulse are perfectly general and may be applied to cases of any two angles, one of which is three times the other ; thus cos 6a = 4 cos^ 2a — 3 cos 2a ; sin 9^ =3 sin 3^1 - 4 sin^ 3 A. 126. To find the value of sin 18°. Let ^ = 18°, then bA = 90°, so that 2A = 90° - 3.4. .-. sin 2^= sin (90° -3^)= cos 3^; .*. 2sin J. cos J. = 4cos3 J. — 3 cos J.. Divide by cos A (which is not equal to zero) ; .-. 2sin^ = 4cosM-3=4(l-sinM)-3; .-. 4sinM+2sin.4-l=0; . , -2 + V4 + 16 -l±x/5 ... s,n.4 = --- = — ^— Since 18° is an acute angle, we take the positive sign ; .-. sin 18° = ^^^. 4 Example. Find cos 18° and sin 54°. cos ^80 = Vl^sSn8"=^l _6_^.Vi»+V^ Since 54° and 36° are complementary, sin 64°= cos 36°. Now cos36° = l-2sinn8°=l-^^^^|^^=^^; lb 4 sin 54' >_n/5 + 1 4 EXAMPLES. XLe. 1. If cos J. =-, find cos 3^. o 3 2. Find sin 3^ when sin J. = - . 5 3. Given tan ^ = 3, find tan ZA. XI.] FUNCTIONS OF 2 A AND 3 A. 107 Prove the following identities : . sin 3 J. cos 3^ ^ _ ^ „ , cot^ ^4-3 cot A 4. — — -. 7-= 2. 5. cot 3x1=—^ — To-j — i — • sm A cos A 3 cot-^ -d - 1 ^ 3cosa + cos3a ,„ _ sinSa + sin^a , 6. ^— . ^-^ = cot3a. 7. — o ^ = cota. 3sma — sin3a cos^ a — cos 3a Q cos^ a — cos 3a sin^ a + sin 3a cos a sm a 1 9. sinl8°+sin30°=sin54°. 10. cos 36° -sin 18° = -. 11. cos2 36° + sin2l8° = ?. 12. 4 sin 18° cos 36° = 1. 4 127. The following examples further illustrate the formulae proved in this chapter. 3 Example 1 . Shew that cos^ a + sin^ a = 1 - ^ sin^ 2a. The first side = (cos^ a + sin^ a) (cos^ a + sin"* a - cos^ a sin^ a) = (cos2a + sin-a)2-3 cos^asin^a 3 = 1 - 2 (4 cos- a sin^ a) = 1--; sin22a. 4 Example 2. Prove that ; : — -= sec 2^ -tan 2 J. cos A + sm A -,, . 1 . . 1 1 sin 2^ 1 - sin 24 The right side = ^^ ^. = — - , cos 2A cos 2 A cos 24 and since cos24 = cos-4 -sin24 = (cos4 + sin4)(cos4-sin4), this suggests that we should multiply the numerator and denominator of the left side by cos A - sin A ; thus the first side=^"^^^-^^"^^^"^^^-^^^^) (cos A + sm A) (cos A - sin 4) _eos24 + sin24 -2cos4 sin4 ~ cos^^-sin^J. 1 - sin 24 n A ^ c^A = — -— =see24 -tan 24. cos 24 108 ELEMENTARY TRIGONOMETRY. [CHAP. Example 3. Shew that - — jr-^ — r -, -— — --: =cot 2 A. tan SA - tan A cot SA - cot A The first side=« sin BA sin A cos 3^ cos A cos 84 cos 4 sin 34 sin 4 cos SA cos A sin 6 A sin A sin 3A cos A - cos SA sin 4 cos 34 sin A - sin 34 cos 4 _ cos 34 cos 4 + sin 34 sin 4 ~ sin 34 cos 4 - cos 34 sin 4 cos (34 - 4) cos 24 , « , = . ,Q. 7v^ ■ ^. =cot24. sm (34 - 4) sm 24 Note. This example has been given to emphasize the fact that in identities involving the functions of 24 and 34 it is sometimes best not to substitute their equivalents in terms of functions of 4. EXAMPLES. XI.f. Prove the following identities : 1. tan 24 — sec 4 sin A = tan A sec 24. 2. tan 24 + cos A cosec A = cot A sec 24. „ l-cos2^+sin2^ ^ . 3. :; ?r;i -. — ^TT, = tan 6. l+cos2^+sm2^ 1+COS^ + COS- ': 4. =cot- . ij 9 - sin 6 + sin 5 , 5. cos^ a - sin^ a= cos 2a ( 1 - ^ sin^ 2a j . 6. 4(cos6^ + sin6^) = l+3cos2 2^. _ cos3a+sin3a , , „ . „ 7. -— = 1 + 2 sm 2a. cos a — sm a - cos 3a — sin 3a , ^ . ^ 8. : = l-2sm2a. cos a + sm a - cosa + sina , ^ « " 9, -. — = tan 2a + sec 2a. cos a — sm a XI.] FUNCTIONS OF 2A AND 3^. 109 10. Prove the following identities : cot a - 1 _ 1 — sin 2a cot a + 1 cos 2a A A 1+tan- - cot- + l 1 + sin^ 2 cos<9 2 cos^ , ^ ^' * l-sin(9 ,6 ^ 1 - tan - cot -x—y 13. sec A - tan A = tan ( 45° — - 14. tan^+sec^=cotf45°--y l-sm^ \4 2/ 16. (2 cos ^ + 1) (2 cos .4-1) = 2 cos 2^1 -Hi. sin 2^ cos^ , A 1 Y — tan l + cos2^'l+cos^ 2* sin 2^ 1 - cos ^ . J. 18. = ^-7. ^— = tan- . 1 — cos 2 J. cos J. 2 19. 4 sin^ a cos 3a + 4 cos' a sin 3a = 3 sin 4a. \Put 4 sin^ a = 3 sin a - sin 3a and 4 eos^ a = 3 cos a + cos 3a.] 20. cos' a cos 3a + sin' a sin 3a = cos' 2a. 21. 4 (cos' 20° + cos' 40°) = 3 (cos 20° + cos 40°). 22. 4 (cos' 10°+ sin' 20°) = 3 (cos 10° + sin 20°). 23. tan 3.1 - tan 2 A - tan A = tan ZA tan 2 A tan A . [ Use tan 3 A = tan {2A +A).] rt. cot 6 tan ^ ii4. \- = 1 . cot ^ - cot 3^ tan ^- tan 3^ * tan 3^+ tan 6 ~ cot 3^ + cot 0~^^ CHAPTER XII. TRANSFORMATION OF PRODUCTS AND SUMS. Transformation of products into sums or differences. 128. In the last chapter we have proved that sin A cos B + cos A sin B = sin {A + B), and sin A cos B — cos A sin B = sin {A — B). . By addition, 2sinAcosB = &in{A + B) + sin{A-B) (1); by subtraction 2cos^sin^=sin(^+i?)-sin(^-^) (2). These formulae enable us to express the product of a sine and cosine as the sum or difi'erence of two sines. Again, cos A cos B — sin A sin B = cos {A + B), and cos J. cos ^4- sin J. sin ^ = cos (^4 — ^). By addition, 2 cos A cos B=cos {A+B) + cos {A - B) (3) ; by subtraction, 2 sin ^ sin 5= cos (^-^)- cos (.4+^) (4). These formulae enable us to express (i) the product of two cosines as the sum of two cosines ; (ii) the product of two sines as the difference of two cosines. 129. In each of the four formulae of the previous article it should be noticed that on the left side we have any two angles A and B, and on the right side the sum and difference of these angles. TRANSFORMATION OF PRODUCTS INTO SUMS. Ill For practical purposes the foUowing verbal statements of the results are more useful. 2 sin A cos B = sin (sum) + sin {difference) ; 2 cos A sin J5=sin {sum) — sin {difference) ; 2 cos A cos B = cos {sum) + cos {difference) ; 2 sin A sin ^=cos {diference) — co^ {sum). N.B. In the last of these formulae, the difference precedes the sum. Example 1. 2 sin lA cos 44 = sin [sum) + sin {difference) = sin 114 + sin 34. Example 2. 2 cos 30 sin 6d = sin (30 + 66) - sin (30 - 60) = sin 90 -sin (-30) = sin90 + sin30. 34 54 1 ( /34 54\ fSA 5A\} Example 3. cos — cos — = - jcos f _ + — j + cos ( — - -j \\ — - {cos4^ + cos( -4)} — ^ (cos 44 + cos .4). Example 4. 2 sin 75° sin 15° = cos (75° - 15°) - cos (75° + 15°) = cos 60° -cos 90° =^« 1 ~2' 130. After a little practice the student will be able to omit some of the steps and find the equivalent very rapidly. Example 1. 2 cos ( 7 + ^ ) cos f - - j = cos ^ + cos 20 = cos 20. Example 2. sin (a - 2^) cos (a + 2jS) = - {sin 2a + sin ( - 4/3) } = - (sin 2a - sin 4^). a 112 ELEMENTARY TRIGONOMETRY. . [CHAP. i EXAMPLES. XII. a. Express in the form of a sum or difference 1. 2 sin 3d cos 0. 3. 2 cos 7J-Cos5^ 5. 2 cos 56 sin 46. 7. 2 sin 9^ sin 36. 9. 2 cos 2a cos 11a. 11. sin 4a cos 7a. 13. A . 3A cos 2 sin 2 . 15. 2^ 5^ 2 cos — cos — . o o 2. 2 cos 66 sin 3^. 4. 2 sin 3 A sin 2 A. 6. 2 sin 4^ cos 8^. 8. 2 cos 9^ sin 76. 10. 2 sin 5a sin 10a. 12. sin 3a sin a. 14. . 6A 7A sm 2" cos 2 • 16. . 6 . 36 sin -7 sm — - . 4 4 18. 2 sin 3a sin (a 4 17. 2 cos 2/3 cos (a - /S). 19. 2 sin (2^. + <^) cos (^-20). 20. 2cos(3^+<^)sin(^-2(/)). 21. cos(60° + a)sin(60°-a). Transformation of sums or differences into products. 131. Since sin {A+B) = sin yl cos ^ + cos A sin B, and sin {A — B)^ sin ^ cos B - cos ^ sin ^ ; by addition, sin(^+^)+sin(^-5) = 2sin^cos5 (1); by subtraction, sin(^+^)-sin(^-^) = 2cos^sin5 (2). Again, cos {A+B) = cos J. cos ^ - sin A sin B, and cos(^-5) = cos^ cos-S + sin^sin^. By addition, cos(^+^) + cos(^-^) = 2cos^cos5 (3); by subtraction, cos(^+5)-cos(^— ^)= -2sin^sini> = 2sin^ sin(-5) (4). XII.] TRANSFORMATION OF SUMS INTO PRODUCTS. 113 Let A-]-B=C, and A-B=D; then A = — - — , and ij = — By substituting for A and B in the formulse (1), (2), (3), (4), we obtain sm 6 + sin Z* = 2 sm — - — cos -r — , 9 2 am (7 - sm i>= 2 cos — - — sm — - — , 2 2 cos C + COS Z> = 2 cos ^r — COS — - — , COS (7— cos /> = 2 sm ^r — sm — ^r — . 2 2 132. In practice, it is more convenient to quote the formu- lae we have just obtained verbally as follows : sum of two sines = 2 sin {half -sum) cos {half-difference) ; difference of two sines = 2 cos {half -sum) sin {half-difference) ; sum of two cosines = 2 cos {half -sum) cos {half-difference) ; difference of two cosines = 2 sin {half-sum) sin {half-difference reversed) ^ Example 1. sm 14^ + sm 6^ = 2 sm — cos — =2 sin 10^ cos 4^. . . „. « 94 + 7^ . ^A-IA Example 2. sm 9^ - sm 7^ = 2 cos ^ — sm — ^ — = 2 cos 8^ sin^. ^A f lA • Example 3. cos A + cos 8^ = 2 cos -^ cos ( - -9- 94 74 = 2 cos -^ cos -jr- . Example 4. cos 70° - cos 10° = 2 sin 40° sin ( - 30°) = - 2 sin 40° sin 30° = - sin 40°. H. K. E. T. 8 114 ELEMENTARY TRIGONOMETRY. [CHAP. EXAMPLES. X II. b. Express in the form of a product 1. sin 8^+ sin 4^. 2. sin 6B - sin 6. 3. cos 7^ + cos 3^. 4. cos 9^ — cos 11^. 5. sin 7a — sin 5a. 6. cos 3a + cos 8a. 7. sin 3a + sin 13a. 8. cos 5a— cos a. 9. cos 2x4+ cos 9^. 10. sin 3 J. — sin 11 A. 11. cos 10° - cos 50°. 12. sin 70° + sin 50°. Prove that cos a -cos 3a ^ ^ _. sin 2a + sin 3a , 13. —. — i^ ■■ — = tan 2a. 14. ^ — = cot a sin 3a -sin a "'^^ ""' cos 2a — cos 3a ~""2' cos 4^ -cos (9 ,5(9 -_ cos 2^ -cos 12(9 , _, 15. -^-^ r-— . = tan— . 16. • .ozi . • oa = ^^^^^- sm 6 - sm 4:6 2 sm 12^ + sm 26 17. sin (60° + . 4) - sin (60°-^) = sin ^. 18. cos (30° - xl ) + cos (30° + ^ ) = ^/3 cos A . 19. cos f ^+aj-cos ( --aj= -s/2sina. cos (2a - 3^) + cos 3/3 _^^^^ sin (2a- 3^)+ sin 3/3 cos(^-30)-cos(3^ + 0)^ sin(3^ + (^) + sm(^-3(^) ^ ^ ^^ sin( a+^)-sin4/3 ^^^^ a-3/ 3 ' cos(a + /3) + cos4/3 2 ' 133. The eight formulae proved in this chapter are of the utmost importance and very httle further progress can be made until they have been thoroughly learnt. In the first group, the transformation is from products to sums and differences ; in the second group, there is the converse transformation from sums and differences to products. Many examples admit of solution by applying either of these transformations, but it is absolutely necessary that the student should master all the formulae and apply them with equal readiness. XII.] TRANSFORMATION OF PRODUCTS AND SUMS. 115 134. The following examples should be studied with gi'eat care. Example 1. Prove that sin 5A + sin 2^ - sin A — sin 2A (2 cos SA + 1). The first side = (sin 5 A - sin A) + sin 2 A = 2 cos 3 A sin 2A + sin 2A = sin 2A (2 cos SA + 1). Example 2. Prove that cos 2d cos 6 - sin 4:6 sin 6 = cos 3^ cos 26. The first side = - (cos 3^ + cos 6)-^ (cos S6 - cos 56) — ^ (cos ^ + cos 5^) = COS 3^ COS 20. Example 3. Find the value of COS 20° + COS 100° + COS 140°. The expression = cos 20° + (cos 100° + cos 140°) = cos 20° + 2 COS 120° cos 20° (-» =cos20°+2( --1 cos 20° = cos 20° -cos 20° =0. Example 4. Express as the product of four sines sin ()S + 7 - a) + sin (7 + a - j3) + sin (a + /3 - 7) - sin (a + /3 + 7). The expression = 2 sin 7 cos (j8 - a) + 2 cos (a + ^) sin ( - 7) = 2 sin 7 {cos (/3 - a) - cos (a + /3) } = 2 sin 7 (2 sin /S sin a) = 4 sin a sin /3 sin 7. Example 5. Express 4 cos a cos /3 cos 7 as the sum of four cosines. The expression = 2 cos a {cos (/S + 7) + cos (/S - 7) } = 2 cos a cos (^ + 7) + 2 cos a cos (j3- 7) =cos (a +/S + 7) + cos (a - /3 - 7) + cos (a + j3 - 7) + cos (a - jS + 7) = cos (a+^-f 7) + C0S (j8 + 7 - a) +COS (7 + a - 18) + cos (a + j3 - 7). 8—2 116 ELEMENTARY TRIGONOMETRY. [CHAP. Example 6. Prove that sin^Sa; - sm2 3a; = sin 8x sin 2x. First solution. amP5x - sin^Sa; = (sin ox + sin 3a;) (sin 5x - sin 3a;) = (2 sin 4a; cos a;) (2 cos 4a; sin a;) = (2 sin 4a; cos 4a;) (2 sin x cos x) = sin 8a; sin 2a;. Second solution. sin 8a; sin 2a; = - (cos 6a; - cos 10a;) = 5 {1 - 2 sin23a; - (1 - 2 sin25a;)} = sin2 5a;-sin2 3.r. Third solution. By using the formula of Art. 114 we have at once sin^ 5a; - sin^ 3a; = sin (5a; + 3a;) sin (5a; - 3a;) = sin 8a; sin 2a;. EXAMPLES. XII. c. Prove the following identities : 1. cos 3 J. -f sin 2 J. - sin 4,A = cos 3^1 (1 - 2 sin A). 2. sin 3^ - sin (9 - sin 5^ = sin 3^ ( 1 - 2 cos 2^). 3. cos ^ + cos 2^ + cos be = cos 2<9 (1 + 2 cos 3^}. . . . _ • « J • o 3a 4. sina-sm 2a + sin 3a=4sm-cos a cos -r-. 2 2 . ^ • w • -./^ M ' ,r *ici 3a 5. sm 3a + sm la + sm 10a = 4 sin 5a cos — cos — . 6. sin J. + 2 sin 3^ + sin 5^ = 4 sin 3^ cos^^. sin 2a + sin 5a — sin a ^ ^ 7. IT = tan 2a. cos 2a + cos 5a + cos a - sina4-sin2a+sin4a4-sin 5a , _ o. z ^ = tan oa. cosa+cos2a+ cos 4a + cos 5a Q cos7^ + cos3^ — cos5^ — cos^_ - sin 7^ - sin 3^ — sin 5^ + sin ^ ~ 10. cos 3 A sin 2yl — cos 4 J. sin A = cos 2 A sin A . XII.] TRANSFORMATION OF PRODUCTS AND SUMS. 117 Prove the following identities ; 11. cos 5 A cos 2 A - cos 4:A cos 3^ = — sin 2 A sin A. 12. sin 46 cos — sin 3d cos 2d = sin $ cos 26. 13. cos 5° -sin 25° = sin 35°. [C/se8in25^=cos65°.] 14. sin 65° + cos 65° = ^2 cos 20°. 15. cos 80° + cos 40° - cos 20° = 0. 16. sin 78° - sin 18° + cos 132° =0. 17. sin^ 5 A — sin^ 2 A = sin *7A sin 3A. 18. cos 2 A cos 6 A =cos2 — sin^ -— . 2 2 19. sin(a+j3+y)+sin(a — /3-y)+sin(a+^-y) 4-sin(a — /3 + y) = 4sin a cos /3 cosy. 20. cos (/S + y- a)-COs('y + a-/3)+cos (a + jS-y) — cos (a+/3 + y)=4 sin a cos )3 sin y. 21. sin 2a + sin 2/3 + sin 2y - sin 2 (a + /3 + y) =4 sin i/^ + y) sin (y +a) sin (a+/3), 22. cosa+cos/34-cosy + cos(a + /3 + y) . /3+y y + a a-f/3 = 4 cos — —- cos ^ — cos ~-~ . AAA 23. 4 sin ^ sin (60° + ^) sin (60° - ^) = sin 3.4. 24. 4 cos <9 cos ( ^ + ^ j cos \^-6\ = cos 3^. 25. cos(9 + cosry-<9j+cosry + ^j = 0. 26. cos2^ + cos2(60° + .4) + cos2(60°-^)=|. [Pwt 2eo8M = l + cos2A] 27. sin2.4+sin2(120° + ^l) + sin2(120'-^) = |. 28. cos 20° cos 40° cos 80° = ^ . o 29. sin 20° sin 40° sin 80° = 3^3- o 118 ELEMENTARY TRIGONOMETRY. [CHAP. 135. Many interesting identities can be established connect- ing the functions of the three angles A, B, C, which satisfy the relation A + B + C=180°. In proving these it will be necessary to keep clearly in view the properties of complementary and supplementary angles. [Arts. 39 and 96.] From the given relation, the sum of any two of the angles is the supplement of the third ; thus sin {B-\-0)= sin A, cos {A + B) = - cos C, tan(C+^)= -tan^, cos B ^ ~ cos (C + A), sin (7= sin {A + B), cotA = - cot {B + C). Again, -^ + -^ + — =90°, so that each half angle is the comple- AAA ment of the sum of the other two ; thus A-^B . C . C+A B ^ B+C M cos— 2 — =sin-, sm— 2— =cos-, tan— ^= cot -^j C . A+B . A B+G ^ B ^C+A cos-=sm-^— , sm — =cos-^— , tan- = cot-^^. Example 1. IiA + B + G= 180°, prove that sin 2A + sin 2B + sin 2C = 4 sin A sin B sin C. The first side = 2 sin {A + B) cos {A-B) + 2 sin C cos G = 2 sin cos (^ - £) + 2 sin (7 cos G = 2 sin C {cos {A-B) + cos C] -2sixLG{Goa{A-B)-GOs{A+B)] = 2 sin C X 2 sin -4 sin B = 4 sin j4 sin B sin C. Example 2. IfA+B+G== 180°, prove that tan ^ + tan 5 + tan (7 = tan ^ tan £ tan (7. Since A + B is the supplement of C, we have tan {A+B)= - tan G ; tan A + tan B . ^ .• = — tan G ; 1 - tan A tan B whence by multiplying up and rearranging, tan^+tanB +tan C = tan^ tan Stan C. XII.] RELATIONS WHEN A+B+ 0=180°. 119 Example 3. liA + B + C=z 180°, prove that A Ti C COS A + cos B + cos C = 1 + 4 sin— sin — sin — . ^ Z d The first side = 2 cos — - — cos — - — |- cos C ^ . G A-B ^ . . „C = 2 sm — cos — X — + 1-2 sm^ — , ^ . C / A-B . C = 1 + 2 sm — ( cos — 2 sm — ^ ^ . G f A-B A+B\ = 1+2 sin— (cos— ^ °°^~2 — / , ^ . G ( ^ . A . B = 1 + 2 sm — ( 2 sin — sin ^ , , . A . B . G = 1 + 4 sm — sin — sm — . EXAMPLES. XII. d. If .4 +5+ (7= 180°, prove that 1. sin ^A - sin 2-S + sin 2(7= 4 cos A sin B cos C. 2. sin 2 J. - sin '2B — sin 2C= — 4 sin ^4 cos B cos C. 3. sinA+smB4-smC=4cos-rt- cos n cos rt- 4. sm A + sm B — sm 6 = 4 sm — sm — cos — . 2 2 2 5. cos A — cos j5+ cos 6'=4 cos — sin — cos ^ - 1. ji Ji ji sin 5 + sin C- sin J. ^ B^ G 6. -. — -. . „ . — -r — ^ = tan — tan - . sm^+sin5+smC/ 2 2 „ ^ B^ C , . C. A , . A. B - 7. tan^tan^ + tan^- tan y + tan ^^tan^ — 1- ^^. ^ A+B ,G ,,, . , A+B ^ G ,, [Use tan — % — = cot -^ , a/ta tnerefore tan — ^r — tan — = 1. J 120 ELEMENTARY TRIGONOMETRY. [CHAP. lfA+B + C= 180°, prove that -^ 1+cos J.-cos^ + cosC , B ^C 8. -^ -, T, ry= ■tan — cot - . 1 + cos A + cos B - cos C 2 2 9. cos 2A + cos 2B + cos 2(7+ 4 cos ^ cos J5 cos C+ 1 = 0. 10. cot BcotC+ cot 6' cot ^ + cot ^ cot J5 = 1 . 11. (cot ^ + cot Cf) (cot C+cot A) (cot J. + cot B) = cosec A cosec B cosec C. 12. cos^ J. + cos2^ + cos2(7+2cos J. cos5cosC=l. [Use 2cos2^ = l + cos24.] 13. siii2 - + sin2 - + sm2 - = 1 - 2 sm - sm - sm - . 14. cos2 2 A + cos2 2B + cos^ 2C= 1 + 2 cos 2A cos 2B cos 2^ cot 5 + cot G cot (7+ cot A cot ^ + cot B _ ianB+tsinC tanC+tan^l tanJ. + tan5 16. tan.4+tan^ + tan(7 ^an-tan-tan- (sin A + sin ^ + sin (7)^ 2 cos A cos ^ cos C " 136. The following examples further illustrate the formulae proved in this and the preceding chapter. 4 cos 24 Example 1. Prove that cot {A + 15°) - tan {A - 15°) = . ^ . , ., . The first side = ""^ j^ + ]^^ - '^^^^=^, sm {A + 15°) cos {A - 15°) _ cos (A + 15°) cos (4 - 15°) - sin (A + 15°) sin {A - 15°) sin {A + 15°) cos {A - 15°) cos {(4 + 15°) + (4 -15°)} ~ sin {A + 15°) cos {A - 15°) _ 2 cos 24 _ 2 cos 2A "■ 2 sin (4 + 15°) cos {A - 15°) ~ sin 2A + sin 30° 4 cos 2 A ~2 sin 24 + 1 '. Note. In dealing with expressions which involve numerical angles it is usually advisable to effect some simplification before substituting the known values of the functions of the angles, especially if these contain surds. xil] miscellaneous identities. 121 Example 2. Prove that A B G , ir-A T-B w-O cos — + COS — + COS -pr = 4: COS — -. — cos 7— COS — -. — . 2 2 2 4 4 4 The second side = 2 cos — j— cos - — —^ + cos — j — TT-A TT + A „ ir-A B-G = 2 cos — , - cos — H 2 cos — z — cos — z — 4 4 4 4 / TT ^\ o B + G B-G = I cos -+ cos TT J + 2 cos — , — cos — i — \ 2 2/ 4 4 A B G = cos — + cos — + cos - , EXAMPLES. XII. e. Prove the following identities : 1. cos (a + /3) sin (a - /3) + cos O + y) sin O - y) + COS (y + S) sin (y - S) + cos {8+ a) sin (S - a) =0. 2 sin(/3-y) ^ sin(y-a) ^ si n(a-^) ^^ sinjSsiny sin y sin a sin a sin /3 _ sin a + sin /3 + sin (a +i3) ^a .3 3. —. ; ^ ; 7 ^= cot -- cot ^ . sina+sm/3-sm (a+/3) 2 2 4. sin a cos (/3 + y ) — sin /3 cos (a + y) = cos y sin (a - /3). 5. cos a cos (/3 + y ) — cos /3 cos (a + y ) = sin y sin (a — /3). 6. (cos ^4 — sin .4) (cos 2 A — sin 2A) = cos J. — sin ZA. 7. If tan d = - y prove that a cos 2^ + 6 sin 2^ = a. [See Art. 124.] a T> +1. ^ • o < , o < (l+tan^)2-2tanM 8. Prove that sm 2 A + cos 2 A = ^ — r—o—r • l + tan^^ rt T. ^1 X • ^ ^ 4tan^ (l-tan^^lj 9. Prove that sin 4^1 = — ; ^ ,,„ — ^ . (l+tan2^)2 10. liA+B= 45°, prove that (1 + tan A) a +tan 5)=2. 122 ELEMENTARY TRIGONOMETRY. [CHAP. XII. Prove the following identities : ^_ A COS 2_4 11. cot(15°-J)+tan(15° + J) = j-2^.^23- 13. tan(4+30»)ta„(4-30") = j-p^^;5j^. 14. (2 COS .4 + 1) (2 cos ^ - 1) (2 cos 2^ - 1) = 2 cos 4^ + 1. 15. tanO-'y)+tan(y — a) + tan(a-j3) = tan (/3 — y) tan (y - a) tan (a - /3). 16. sin(/3-y) + sin(y-a) + sin(a-^) . ^ — y . y — a . a-/3 + 4 sin —77-^ sm ^-7^— sin — — ^ = 0. 2 2 2 17. cos2 - y) + C0s2 (y - a) + cos2 (a - /3) = 1 + 2 cos {j3 — y) cos (y - a) cos (a — /3). 18. cos2 a+cos2 /3 - 2 cos a cos /3 cos (a + /3) =sin2 (a+^). 19. sin2 a + sin2 /3 + 2 sin a sin ^ cos (a + /3) = sin^ (a + j3). 20. cos 12° + cos 60° + cos 84° = cos 24° + cos 48°. liA + B-\-C= 180°, shew that A B ^ C , B+G C+A A+B 21. cos — +COS — +cos- = 4cos cos — 7— cos — —— . A B C , TT + A TT-B TT + C 22. cos — - cos — + COS - =4 cos — - — cos — —- cos — — . . A . B , G , , . TT-A . TT-B . TT-G 23. sm — + sin 77 + sin -= 1 +4 sm -— — sm — - — sm — -— . 2 2 2 4 4 4 If a+/3+y = ^, shew that _. sin 2a + sin 2/3 + sin 2y , ,^ 24. -. — . t: ; — -/-=cotacot/3. sm 2a + sm 2/3 — sm 2y 25. tan y3 tan y + tan y tan a +tan a tan /3 = 1. * CHAPTER XIII. RELATIONS BETWEEN THE SIDES AND ANGLES OF A TRIANGLE. 137. In any triangle the sides are 'proportional to the sines of the opposite angles; that is^ sin A sin B sin C (1) Let the triangle ABC be acute-angled. From A draw AD perpendicular to the opposite side ; then AD=AB&inABD=G sin B, and ^i)=.^(7sin^C'i) = 6sinC'; .-. 6 sin (7= c sin ^, h c that is, sin B sin G ' (2) Let the triangle ABC have an obtuse angle B Draw AD perpendicular to CB produced; then AD=AC &m ACD=h&inC, and AD = ABsinABD =c sin (180° -B) = c sin B ; .•. h sin(7=csin^; h c that is, sin^ sinC In like manner it may be proved that either of these ratios is equal to — sin A ' Thus a sin A sin B sin G ' 124 ELEMENTARY TRIGONOMETRY. [chap. 138. To find an expression for one side of a triangle in terms of the other two sides and the included angle. (1) Let ABG be an acute-angled triangle. Draw BD perpendicular to AC; then by Euc. ii. 13, AB^=BC^+GA^-2AC.CD', .-. c'^=a^ + h'^—2h,aco&G = a^ + 62 _ 2a6 cos C. (2) Let the triangle ABC have an obtuse angle C. Draw BD perpendicular to AC produced ; then by Euc. ii. 12, AB'^=BG^+CA'^ + 2AC. CD; .-. c^ = a^ + b^ + 2b. a cos BCD = a^ + b^+2ab cos (180° - C) = a^ + b'^ — 2ab cos G. Hence in each case, c^=a^ + b^ — 2ab cos G. Similarly it may be shewn that a^=b^-{-c^ — 2bccosA, and b^==G^ + a^ — 2ca cos B. 139. From the formulae of the last article, we obtain 52 + c2-a2 „ c2 + a2_62 «2 + 52_c2 cos^l= -^ — — : cosB = ;: : COS 6 = 26c 2ca 2ab These results enable us to find the cosines of the angles when the numerical values of the sides are given. 140. To express one side of a triangle in t&rms of the adjacent angles and the other two sides, (1) Let ABG be an acute-angled triangle. Draw AD perpendicular to BC\ then BG=BD+GD =AB cos ABD+ AG cos ACD; that is, a = c cos B-\-b cos G. XIII.] RELATIOJ^S BETWEEN THE SIDES AND ANGLES. 125 (2) Let the triangle ABC have an obtuse angle C. Draw AD perpendicular to BC produced; then BC=BD-CD =AB coH ABB -AC cos ACD; ^- .-. a=c cos B-b cos {180° -C) =c cos B-\- b cos C Thus in each case a=b cos (7+ c cos B. Similarly it may be shewn that b=c cos A +a cos C, and c=acosB + bcos A. Note. The formula we have proved in this chapter are quite general and may be regarded as the fundamental relations subsisting between the sides and angles of a triangle. The modified forms which they assume in the case of right-angled triangles have already been considered in Chap. V. ; it will therefore be unnecessary in the present chapter to make any direct reference to right-angled triangles. 141. The sets of formulae in Arts. 137, 138, and 140 have been established independently of one another ; they are how- ever not independent, for from any one set the other two may be derived by the help of the relation A+B + C=18(f. For instance, suppose we have proved as in Art. 137 that a b c sin A sin B sin C ' then since sin A = sm{B+C) = sin B cos C+ sin C cos B ; sin 5 sin (7 „ .. 1 = ——-, cos C+ . , cos B ; smA sin A . • . 1 = - cos C -] — cos B ; a a .'. a=b cos C+c cos B. Similarly, we may prove that b=c cos A +a cos C, Sind c= a cos B + b cos A. Multiplying these last three equations by a, b, —c respec- tively and adding, we have a^ -f 6^ — c^ = 2ab cos C ; .-. c^-=a^ + b^--2ab cos C. Similarly the other relations of Art. 138 may be deduced. 126 ELEMENTARY TRIGONOMETRY. [cHAP. Solution of Triangles. 142. When any three parts of a triangle are given, provided that one at least of these is a side, the relations we have proved enable us to find the numerical values of the unknown parts. For from any equation which connects four quantities three of which are known the fourth may be found. Thus if c, a, B are given, we can find h from the formula Z)2 = c^ + a^ — 2ca cos ^ ; and if B, C, h are given, we find c from the formula c h sin G sin B ' "We may remark that if the three angles alone are given, the formula a h c sin A sin B sin C enables us to find the ratios of the sides but not their actual lengtJis, and thus the triangle cannot be completely solved. In such a case there may be an infinite number of equiangular triangles all satisfying the data of the question. [See Euc. vi. 4.] 143. Case I. To solve a triangle having given the three sides. The angles A and B may be found from the formula? , &2_j_g2_^2 C2_j.^2_52 then the angle Cis known from the equation 6'= 180° — ^ —B. Example 1. If a = 7, & = 5, c=8, find the angles A and B, having given that cos 38° ll' = r-r • 14 cos A-. 62 + c2- a^ 52 + 82- .72 40 1 2bc 2x5x 8 2 X 5x 8 ~2' • '. ^ = 60 B: C^ + 0?- 2ca .62_ 82+72- 2x8x •52 7 88 11 2 x8x 7 14' /. JB = 38° 11'. XIII.] SOLUTION OF TRIANGLES. 127 Example 2. Find the greatest angle of the triangle whose side are 6, 13, 11, having given that cos 84° 47' = r-T • Let a = 6, 6 = 13, c = ll. Since the greatest angle is opposite to the greatest side, the required angle is B. , , ^ c2 + a2-&2 112 + 62-132 -12 And cos B = 2ca ~ 2x11x6 ~2xllx6' .'. cosB=-i=-cos84°4r; .-. £ = 180° -84° 47' = 95° 13'. Thus the required angle is 95° 13'. 144. Case II. To solve a tnaiigle having given two sides and the included angle. Let Z>, c, A be given ; then a can be found from the formula a^ = 62 -j. c2 _ 25c cos A. We may now obtain B from either of the formulae „ c2+a2_52 5sin.l cosi;= ;; , or smi> = ; 2c(X a then G is knov^n from the equation C= 180° — A — B. Example. If a = 3, & = 7, C = 98°13', solve the triangle, having given cos 81° 47' = = . c2 = a2 + 62 _2a& cos C = 9 + 49-2x3x7cos98°13'. But 98° 13' is the supplement of 81° 47'; .'. c2 = 58 + (2 X 3 X 7 cos 81° 47') = 58 + f2x3x7xi^=58 + 6 = 64J .-. c = 8. c2 + a2-62 64 + 9-49 24 1 COSJB: 2ca ~ ~ 2x8x3 ~2x8x3~2' .-. 5 = 60°. C = 180° - 60° - 98° 13' = 21° 47'. 128 ELEMENTARY TRIGONOMETRY. [CHAP. 145. Case III. To solve a triangle having given two angles and a side. Let J5, r", a be given. The angle A is found from A — 180° — B— C; and the sides b and c from , a sin B " a sin C = ^. — J- and c=—. — j- . sm A sin A Example. If ^ = 105°, C=60°, &=4, solve the triangle. 5 = 180° -105° -60° =15°. _ 6sina _ 4sin60° _4^3 _2_n/2^ _ _4x/6_ ■■ ^~ sin£ ~ sin 15° "T' ' ^B-l~ ^S-1 .-. c = 6^/2 + 2^6. _ 6 sm 4 _ 4 sin 105° _ 4 sin 75° *~ sin 5 ~ sin 15° sin 15° _ V3 + 1 2^2 _ 4(V3 + 1) . -'''' 2^2 \/3-l" n/3-1 ' .-. a=4 (2 + ^/3). EXAMPLES. XIIL a. L If a=15, 6 = 7, c=13, find C 2. Ifa=7, 6-3, c=5, find^. ^ 3. If a = 5, 6 = 5 a/S, c = 5, find the angles. 4. Ifa=25, 6 = 31, c=7V2, find A 5. The sides of a triangle are 2, 2|, 3J, find the greatest angle. 6. Solve the triangle when a=x/3 + I, 6 = 2, c=a/6. 7. Solve the triangle when a=^J% 6=2, c=x/3 — 1. I— 8. If a=8, 6 = 5, c=^/19,findC; given cos 28° 56'=^ . o 9. If the sides are as 4 : 7 : 5, find the greatest angle ; given cos 78° 27'=^. i XIII.] SOLUTIONS OF TRIANGLES. 129 10. If a = 2, 5=^3 + 1, (7=60°, find c. 11. Given a = 3, c=5, ^ = 120°, find h. ' 12. Given 6 = 7, c = 6, J. = 75° 31', find a; given cos 75° 31' = -25. 13. If 6 = 8, c = 1 1, ^ = 93° 35', find a ; given cos 86° 25' = -0625. 23 14. If a = 7, c = 3, i? = 123° 12', find 6; given cos 56° 48' = —. 15. Solve the triangle when a = 2 JQ, c = 6 — 2 sj^^ B = 75°. 16. Solve the triangle when A = 72°, 6 = 2, c = ^5 + 1. 17. Given^ = 75°, 5=30°, 6 = V8, solve the triangle. 18. If 5 = 60°, C=15°, 6=^/6, solve the triangle. 19. If ^ = 45°, B = 105°, c = V2, solve the triangle. 20. Given A = 45°, B = 60°, shew that c : a = ^3 + 1 : 2. 21. If (7=120", c = 2^3, a =2, find &. 22. If5=60°,a = 3, 6 = 3^3, find c. 23. Given (a+6+c)(6+c-a) = 36c, find A. 24. Find the angles of the triangle whose sides are 3 + v/3, 2V3, V6. 25. Find the angles of the triangle whose sides are V3 + 1 x/3-1 V3 2^2 ' 2^2 ' 2 *. 26. Two sides of a triangle are — rr jz^ and .^ ._ , and the y6— v/2 v6+v2 included angle is 60° : solve the triangle. 146. When an angle of a triangle is obtained through the medium of the sine there may be ambiguity, for the sines of supplementary angles are equal in magnitude and are of the same sign, so that there are two angles less than 180° which have the same sine. When an angle is obtained through the medium of the cosine there is no ambiguity, for there is only one angle . less than 180° whose cosine is equal to a given quantity. ' Thus if sin ^=^, then J. = 30° or 150°; if cos4=^, then.>l = 60°. H. K. E. T. 9 130 ELEMENTARY TRIGONOMETRY. [CHAP. Example. If (7=60°, & = 2^3, c = 3^2, find ^. ^ T. XI • i. . „ 6 sin C From the equation sin jB = , c , . P 2^3 ^3 1 wehave sm5 = 3^.^ = -^; .-. 5 = 45° or 135°. The value J5 = 135° is inadmissible, for in this case the sum of B and G would be greater than 180°. Thus ^ = 180° -60° -45° = 75°. 147. Case IV. To solve a triangle having given two sides and an angle opposite to one of them. Let a, 6, A be given ; then B is to be found from the equation sin B = - sin^. a (i) If a<6sin J., then >1, so that sin^>l, which is impossible. Thus there is no solution. (ii) Ifa=&sin^, has only the value 90°. (iii) If a > 6 sin A , then < 1, and two values for £ may be found from sin B= . These values are supplementary, so that one angle is acute, the other obtuse. (1) If a 6, then A>B', in either case B cannot be obtuse, and therefore only the smaller value of B is admissible. When B is found, C is determined from C=180° — A-B. CI QT yi / Finally, c may be found from the equation c=— ^ — -j-- From the foregoing investigation it appears that the only case in which an ambiguous solution can arise is when the smaller of the two given sides is opposite to the given angle. (ii) If a=&sin^, then =1, so that sinjB = l, and B XIII.] THE AMBIGUOUS CASE. 131 148. To discicss the Amhigiious Case geometrically. Let a, 6, A be the given parts. Take a line AX unlimited towards X ; make / X2 (7 equal to A, and J. C' equal to h. Draw CD perpendicular to AX, then CD=hB\nA. With centre C and radius equal to a describe a circle. (i) If a < 6 sin A^ the circle will not meet AX', thus no triangle can be constructed with the given parts. (ii) If a = 5sin^, the circle will touch AX at D ; thus there is a right-angled triangle with the given parts. (iii) If a > 6 sin J., the circle will cut ^X in two points B^, B^. (1) These points will be both on the same side of A^ when ah. In this case there is only one solution, for the angle CAB^ is the supplement of the given angle^ and thus the triangle AB^C does not satisfy the data. (3) If a = h, the point B2 coincides with A, so that there is only one solution. 9—2 132 ELEMENTARY TRIGONOMETRY. [CHAP. Example. Given 5 = 45°, c=^12, h=^j8, solve the triangle. w -u ■ ^ csinS 2^3 1 ^/3 .-. (7 = 60° or 120°, and since b<:c, both these values are admissible. The two triangles which satisfy the data are shewn in the figure. Denote the sides BG-^, BC^hj a-^^ a^, and the angles BAG^^, BAG^ lay A-^, A^res^Qciviely. A B • (i) IntheA^^Cj, /.4i = 75°; (ii) Inthe A^BC., lA^^=U°\ hence . &sin^, _2V2 ^3-1,, .,,3 ,. Hence a^-- _-^ _-^. ^-^ = ^^(^8-1). v/2 j'C=60°, or 120°; Thus the complete J ^ _ 750 qj. j^^o . solution is I ' ' (a=v/6 + N/2, or ^6-^2. EXAMPLES. XIII. b. 1. Given «=1, 6 = V3, J. = 30°, solve the triangle. 2. Given 6 = 3^^2, 0=2^/3, C=45°, solve the triangle. 3. If 6'= 60°, a=2, c=V6, solve the triangle. xiil] relations between the sides and angles. 133 4. If J. = 30°, a = 2, c= 5, solve the triangle. 5. liB = 30°, 6 = V6, c = 2 ^3, solve the triangle. 6. liB= 60°, 6 = 3 V2, c = 3 + V3, solve the triangle. 7. If a = 3 + V3, c = 3 - V3, C= 15°, solve the triangle. 8. If ^ = 18°, a = 4, & = 4 + ;v/80, solve the triangle. , 9. If 5=135°, a = 3V2, & = 2 J3, solve the triangle. . 149. Many relations connecting the sides and angles of a triangle may be proved "by means of the formulce we have established. A B — C Example 1. Prove that {b - c) cos -e = ci sin — ^ — . Let ^■=- sin A sin B sin C ' then a = A;sin^, h=ksvQ.B, c^fcsinC; A A .: (b - c) cos ^r = ^ (sin B - sin C) cos — _, ^ B+C . B-G A = 2k cos — ^r — sni — ^-— cos — ^, . A A . B-G = 2k sm — cos — sm — ^r— B-G — k Bin. A sin 2 B-G =asm Example 2. If acos^— +ccos2— = , shew that the sides of the triangle are in a. p. G A Since 2acos2 — + 2c cos^— = 3&, 2, a :. a(l + cosC) + c{l + cos^) = 3&, .•. a + c + (a cos C + ccos^) = 3&, .-. a + c + & = 3&, .-. a + c = 2&. Thus the sides rt, i, c are in A. p. 134 ELEMENTARY TRIGONOMETRY. [chap. Example 3. Prove that (62 _ c2) cot ^ + (c2 - a2) cot J5 + (a' - h^) cot C = 0. Let k = sin A sin B sin G ; then the first side cos^ sin^ + ^ } = ;c2 |(sin2B-sin2C) = ft2 |sin(B + C)sin(B-C)^^+ + I. [Art. 114], Bnt sin {B + C) = sin A, and cos ^ = - cos (B + G); the first side = -/c2{sin(S-C7)cos{£ + (7) + + } 1.2 = _ {(sin 2B - sin 2G) + (sin 2C - sin 2A) + (sin 2A - sin 2B)} =0. EXAMPLES. XIII. c. Prove the following identities : 1. a (sin ^ - sin C) + 6 (sin (7- sin J.) + c (sin A - sin B)=0. 2. 2 {be cos A + ca cos B+ab cos C) = a^ + b^ + c^. 3. a{b cosC—c cos B) = b^—c^. 4. (64-c)cos J. + (c + a)cos^ + (a + 6)cos C=a + b+c. ( C A\ 5. 2 ( a sin^ — +csin2 — j=c + a — 6. cos B _c — b cos J. cos (7 6 — c cos A ' 8. (b + c) sm — = a cos — - — . A A 7. tan A ■■ a sin (7 6 — a cos C 9. a-\-b . „(7 cosJ. + cos5 sm^ - = • c 2 2 10. asin(5-(7) + 6sin(C-^)+csin(^-J5)=0. sin(^-^) a2_j2 ^^ csin(^-5)_a2_62 II sin(^+^) = — „- . 12. 6 sin ((7— J.) c' a-, XIII.] THE AMBIGUOUS CASE. 135 [All articles and examples marked with an asterisk mmj he omitted on the first reading of the subject.] *150. The ambiguous case may also be discussed hj first finding the third side. As before, let a, 6, A be given, then , b^-\-c^ — a^ .*. c^ — 2bcosA.c + h^ — a^ = 0. By solving this quadratic equation in c, we obtain c=bcosA + *Jb'^ cos^ A + a^ — b^ = b cos A + V«^ — ^^ sin2 A. (i) When ab sin Ay there are three cases to consider. (1) Suppose a b, then V«^ - b^ sin2 A > ^b^ - b^ sin^ A ; that is ^/a^ — W sin^ A>b cos A ; hence one value of c is positive and one value is negative, whether A is acute or obtuse, and in each case there is only one solution. (3) Suppose a = 5, then '\/a^ — b^ sin^ A = b cos A ; .'. c=2bcos A or ; hence there is only one solution when A is acute, and when A is obtuse the triangle is impossible. 136 ele;^ientary trigonometry. [chap. L.> Example. If b, <;, B are given, and if 6<:c, shew that (a-i - a^f' + (tti + a^Y t&n^B = 4&2 :^vhere a^^a^ are tlie two values of the third side. From the formula cos .6= , 2ca we have a^ - 2c cob B .a + e^-lr = 0. But the roots of this equation are a^ and ac^\ hence by the theory of quadratic equations . a-^ + a.2=i2c COB B and a^aj = c^ - 6^- .'. (ai-a2)- = (ai + ao)^-4aja2 =4c2cos25-4(c2-&2). .;. K-«2)^+(«i + «2)^tan2J5=4c2cos2£-,4(c2-&2)+4c2cos2£tan2£ =4c2(cos2B + sin2£)-4c2 + 4&2 = 4c2-4c2 + 4&2 / ■ =462. ^EXAMPLES. Xra. d. 1. In a triangle in which each base angle is double of the third angle the base is 2 : solve the triangle. 2. If 5=45°, (7=75°, and the perpendicular from A on BO is. 3, solve the triangle. 3. If a = 2, 6 = 4 - 2 v'3, c = 3 ^2 - V^, solve the triangle. 4. If ul = 1 8°, 6 - a = 2, a& =4, find the other angles. ;.; ■- 5. Given 5=30°, c=150, &=50;v/3, shew that of the two triangles which satisfy the data one will be isosceles and the other right-angled. Find the third side in the greater of these triangles. "Would the solution be ambiguous if the data had been 5=30°, c=150, Q. If ^=36°, a =4, and the perpendicular from (7 upon AB is a/5 — 1, find the other angles. 7. If the angles adjacent to the base of a triangle are 22|° and 112^°, shew that the altitude is half the base. '■ 8. If a=2b and A = 3B, find the angles and express c in terms of a. XIII.] RELATIONS BETWEEN THE SIDES AND ANGLES. 137 9. The sides of a triangle are 2^+3, x^ + 3x+'S, :v^ + 2x: shew that the greatest angle is 120°. Shew that in any triangle -i« /7 \ ^ / Ti i\ • A — B A-\-B 10. (6 - a) cos C + G (cos n — cos A) — c sm — - — cosec — - — . 11. asin ( — +^ )=^ (& + c) sin 12. sin(^5 + -jcos- = ^— cos-cos-^. 1 + CQS (A-B) cos G _ a^ + b^ I -\-cos (A -C) cos B~a^ + c^' 14. If c* - 2 (a2 + 62) c2 + a4 + ^2^,2 + £4 _ q, prove that C is 60° or 120°. 15. If a, b, A are given, and if Cj, c^ are the values of the third side in the ambiguous case, prove that if c-^>c.^, (1) Ci — C2=2acosB^. (2) C03.-i^ = ^. (3) Ci2 -j- C2^ - 2c^c,2 cos 2 A = 4a2 cos^ A. (4) sm — i— — ^ sm -^^-^ — ^ = cos A cos i?i. 16. If J. = 45°, and c^, C2 be the two values of the ambiguous side, shew that cos B^CB2==^Y^- 17. If cos J. + 2 cos C : cos J. +2 cos ^= sin J5 : sin C, prove that the triangle is either isosceles or right-angled. 18. If a, 6, c are in A. P., shew that A .B 2' ^"*2 ^^ ^^ .^ 1 • cot — , cot - , cot — are also m A. p. 19. Shew that a^siJi{B-C) b'^sm{C-A) c^sin{A-B) ^^ sin ^+ sin C sin (7+ sin ^ sin J. + sin ^ 138 ELEMENTARY TRIGONOMETRY. [CHAP. XIIL MISCELLANEOUS EXAMPLES. D. 1. Prove that (1 ) tan 2^ cot ^ - 1 = sec 26 ; (2) sin a — cot 6 cos a = — cosec 6 cos (a + 6). 2. If a = 48, 6 = 35, C=60°, find c. 8 15 3. If cosa = Y7 ^^^ ^°^i^~T7» ^^^ tan (a +^) and cosec (a +/3). 4. If a = ST 5 fiiid the value of -; — - — ; — -. — zrr • 21' sin2a + sinl4a 5. Prove that sin 6 (cos 26 + cos A6 + cos 6^) = sin 3^ cos 4^. 6. If6 = x/2, c=V3 + l, ^ = 45°, solve the triangle. 7. Prove that (1) 2sin2 36° = V5sinl8°; (2) 4sin 36° cos 18°=V5. « -r. J.1 J. sin 3a cos 3a , 8. Prove that -. 1 =4 cos 2a. sm a cos a 9. If 6=c=2, a=V6 — \/2, solve the triangle. 10. Shew that (1 ) cos 2a — cot 3a sin 2a = tan a (sin 2a + cot 3a cos 2a). o (2) COSa + cos2a4-cos3a = 4cosaCOS- COS — — 1. 11. In any triangle, prove that (1) 62 sin 2(7+ c2 sin 25= 26c sin ^; a^m.n{B~C) 62 sin ((7-^) c^&m{A-B) ^ ' sin A sin B sin G 12. If Ay B, C, B are the angles of a quadrilateral, prove that tan ^ + tan 5 + tan (7+ tan i) , i, n. ^. t^ ■ — r— i 7S r-?^ r^r = tan A tan i? tan C tan Z>. cot A + cot ij + cot 6 + cot x> [ ?7se tan (^ + £) = tan (360° - C - D).] CHAPTER XIV. LOGAKITHMS. 151. Definition. The logarithm of any number to a given base is the index of the power to which the base must be raised in order to equal the given number. Thus if a^=N, x is called the logarithm of N to the base a. Example 1. Since 3*= 81, the logarithm of 81 to base 3 is 4. Example 2. Since 10i = 10, 102 = 100, 103 = 1000, the natural numbers 1, 2, 3,... are respectively the logarithms of 10, 100, 1000, to base 10. Example 3. Find the logarithm of '008 to base 25. Let X be the required logarithm ; then by definition, that is, (52)^=5-3, or 52»'= 5-3; whence, by equating indices, 2a;= - 3, and x= - 1*5. 152. The logarithm of JV to base a is usually written log^ iV, so that the same meaning is expressed by the two equations From these equations it is evident that a^o&a-^=iV. Example. Find the value of log .^i '00001. Let log .01 -00001 = X ; then ( -Ol)* = -00001 ; (io^) - 1 11 100000 ' °^ 102=« ~ 105 • .'. 2a;=5, and a; = 2-5. 153. When it is understood that a particular system of log- arithms is in use, the suffix denoting the base is omitted. 140 ELEMENTARY TRIGONOMETRY. [CHAP. Thus in arithmetical calculations in which 10 is the base, we usually write log 2, log 3, instead of log^^o^? logio^j Logarithms to the base 10 are known as Common Log- arithms ; this system was first introduced in 1615 by Briggs, a contemporary of Napier the inventor of Logarithms. Before discussing the properties of common logarithms we shall prove some general propositions which are true for all logarithms independently of any particular base. 154. The logarithm of 1 is 0. For a^ — 1 for all values of a ; therefore log 1 = 0, whatever the base may be. 155. Thelogamthm of the hose itself is 1. For «! = a ; therefore log^ a = 1. 156. To find the logarithm of a product. Let MN be the j)roduct ; let a be the base of the system, and suppose so that ■ a^=J/, ay=]Sf. Thus the product J/iV = a=« x a^' = a^ + 2' ; whence, by definition, \ogaMN~x-{-y ^log„if+log„iV. Similarly, log^ J/'i\^P=log« Jf+log,ji\^+log^P; and so on for any number of factors. Example. log42=log(2 x 3 x7) = log 2 + log3 + log7. 157. To find the logarithm of a fraction. M Let -^ be the fraction, and suppose :r=logaJ/, y=logaiV; so that a===i/; ay=N. M a^ Thus the freiction ^ = — = a«-i' ; N a^ whence, by definition, log^ --y-= x-y = log„if-log«iV. XIV,] LOGARITHMS. 141 15 Example. log (2^) = log — = log 15 - log 7 = log (3 X 5) - log 7 = log 3 + log 5 - log 7. 158. To find the logarithm, of a number raised to any power ^ integral or fraxitional. Let loga (J/^) be required, and suppose jr=loga J/, so that a^=M \ then i/^ = (a=*^)^ = aP^; whence, by definition, log^ ( J/^) = 'px ; that is, loga {MP) =p log^ M. \ 1 Similarly, log^ (J/^) = - log^ M. 159. It follows from the results we have proved that (1) the logarithm of a product is equal to the sum of the logarithms of its factors ; (2) the logarithm of a fraction is equal to the logarithm of the numerator diminished by the logarithm of the de- nominator ; (3) the logarithm of the ^th power of a number is p times the logarithm of the number ; (4) the logarithm of the Hh root of a number is - of the logarithm of the mmaber. Thus by the use of logarithms the operations of multipli- cation and division may be replaced by those of addition and subtraction ; the operations of involution and evolution by those of multiplication and division. a^ Jh Example. Express log —~ in terms of log a, log &, log c. The expression = log (a^ ^&) - log 4/c^ 2 = log a^ -f log ;.y & - - log c 1 2 = 51oga-f--log6--logc. 142 ELEMENTARY TRIGONOMETRY. [CHAP. 160. From the equation 10="=^^, it is evident that common logarithms will not in general be integral, and that they will not always be positive. For instance 8154 >103 and <104; . • . log 3154 = 3 + a fraction. Again, '06 > 10-2 and <10-i; . • . log '06 = — 2 + a fraction. 161. Definition. The integral part of a logarithm is called the characteristic, and the decimal part is called the mantissa. The characteristic of the logarithm of any number to the base 10 can be found by inspection, as we shall now shew. 162. To determine the characteristic of the logarithm of any number greater than unity. It is clear that a number with two digits in its integral part lies between 10^ and 10^; a number with three digits in its integral part lies between 10^ and 10^ ; and so on. Hence a number with n digits in its integral part lies between \0^~^ and 10". Let iV be a number whose integral part contains n digits ; then AT" "1 n(n-l)-|-a fraction , . • . log N— {n-\) + a, fraction. Hence the characteristic is ^i — 1 ; that is, the characteristic of the logarithm of a number greater than unity is less by one than the number of digits in its integral part, and is positive. 163. To determine the characteristic of the logarithm of a decimal fraction. A decimal with one cipher immediately after the decimal point, such as "0324, being greater than '01 and less than "1, lies between IQ-^ and 10-^; a number with two ciphers after the decimal point lies between 10 -^ and IQ-^- and so on. Hence a decimal fraction with n ciphers immediately after the decimal point lies between 10-(" + i) and lO"**. Let /) be a decimal beginning with n ciphers ; then T)=z\()~ («+l)-fa fraction . .• . log Z)=-(n + l) + a fraction. XIV.] LOGARITHMS. 143 Hence the characteristic is —(n+1); that is, the characteristic of the logarith^n of a decimal fraction is greater hy unity than the number of ciphers immediately after the decimal point and is negative. 164. The logarithms to base 10 of all integers from 1 to 200000 have been found and tabulated ; in most Tables they are given to seven places of decimals. The base 10 is chosen on account of two great advantages. (1) From the results already proved it is evident that the characteristics can be written down by inspection, so that only the mantissas have to be registered in the Tables. (2) The mantissse are the same for the logarithms of all numbers which have the same signij&cant digits ; so that it is sufl&cient to tabulate the mantissas of the logarithms of integers. This proposition we proceed to prove. 165. Let N be any number, then since multiplying or dividing by a power of 10 merely alters the position of the decimal point without changing the sequence of figures, it follows that ^x 10^, and iY-^lO^, where p and q are any integers, are numbers whose significant digits are the same as those of N. Now log {Nx 10^) =log N+p log 10 =\ogN+p (1). Again, log(iV^-^10«) = logiy-2'loglO =logiV-^ (2). In (1) an integer is added to logiV, and in (2) an integer is subtracted from log N ; that is, the mantissa or decimal portion of the logarithm remains unaltered. In this and the three preceding articles the mantissas have been supposed positive. In order to secure the advantages of Briggs' system, we arrange our work so as always to heep the mantissa positive, so that when the mantissa of any logarithm has been taken from the Tables the characteristic is prefixed with its appropriate sign, according to the rules already given. 166. In the case of a negative logarithm the minus sign is written over the characteristic, and not before it, to indicate that the characteristic alone is negative, and not the whole expression. 144 ELEMENTARY TRIGONOMETRY, [CHAP. Thus 4-30103, the logarithm of '0002, is equivalent to - 4 + -30103, and must be distinguished from — 4-30103, an expression in 'which both the integer and the decimal are negative. In working with negative logarithms an arithmetical artifice will sometimes be necessary in order to make the mantissa positive. For instance, a result such as — 3-69897, in which the whole expression is negative, may be transformed by subtracting 1 from the integral part and adding 1 to the decimal part. Thus -3-69897= - 4 + (1- -69897) = 4-30103. Example 1. Eequired the logarithms of '0002432. In the Tables we find that 3859636 is the mantissa of log 2432 (the decimal point as well as the characteristic being omitted) ; and, by Art. 163, the characteristic of the logarithm of the given number is - 4 ; .-. log -0002432= 4-3859636. Example 2. Find the cube root of "0007, having given log 7 = -8450980, log 887904 = 5-9483660. Let X be the required cube root ; then log a; = I log (-0007) =^(4-8450980)=: J (6 +2-8450980); o o o that is, log X = 2-9483660 ; but log 887904= 5-9483660; /. a; =-0887904. 167. The logarithm of 5 and its powers can easily be obtained from log 2 ; for log 5 =log Y=log 10 - log 2 = 1 - log 2. Example. Find the value of the logarithm of the reciprocal of 324 4/125, having given log 2 = -3010300, log 3 = -4771213. Since log-= -log a, the required value = - log (3244/125) = - log (22 x 3^ x 5* ) = -(2log2 + 41og3 + |log5) = - 2-9299272 = 3-0700728. 2]og2= *(50'20600 4 log 3=1 •9084852 — log 5= -4193820 2-9299272 XIV.] LOGARITHMS. 145 EXAMPLES. XIV. a. 1. Find the logarithms respectively of the numbers 1024, 81, -125, '01, '3, 100, to the bases 2, ^3, 4, -001, 'i, '01. 2. Find the values of logs 16, log8i243, log.oilO, log^g 343^7. 3. Find the numbers whose logarithms respectively to the bases 49, -25, -03, 1, -64, 100, '1, are 2, ^, -2, -1, -^, 1-5, -4. 4. Find the respective characteristics of the logarithms of 325, 1603, 2400, 10000, 19, to the bases 3, 11, 7, 9, 21. 5. Write down the characteristics of the common logarithms of 3-26, 523-1, -03, 1'5, '0002, 3000-1, '1. 6. The mantissa of log 64439 is -8091488, write down the logarithms of -64439, 6443900, -00064439. 7. The logarithm of 32-5 is 1-5118834, write down the numbers whose logarithms are •5118834, 2-5118834, 4-5118834. [ When required the following logarithms may he used log 2 = -3010300, log 3 = -4771213, log 7 = -8450980.] Find the value of 8. log 768. 9. log 2352. 10. log 35-28. 11. log\/6804. 12. log 4^-00162. 13. log -0217. 14. log cos 60°. 15. Iogsin3 60°. 16. log\/sec45°. Find the numerical value of ir, ^1 15 , 25 ^, 4 17. 21og--log^ + 31og-. H. K. E. T. 10 146 ELEMENTARY TRIGONOMETRY. [CHAP. 18. Evaluate 16 log — - 4 log — - 7 log — . 19. Find the seventh root of 7, given log 1-320469 = -1207283. 20. Find the cube root of -00001764, given log 260315 = 5-4154995. 21. Given log 3571 = 3-5527899, find the logarithm of 3-571 X -03571 x \/3571. 22. Given log 1 1 = 1 '041 3927, find the logarithm of (-00011)^ X (1-21)2 X (13-31)*-M2100000. 23. Find the number of digits in the integral parts of '2i\300 /126\^^'^'' /2i\300 /i26\i V20J ^^^ V125J 24. How many positive integers have characteristic 3 when the base is 7 ? 168. Suppose that we have a table of logarithms of numbers to base a and require to find the logarithms to base h. Let N be one of the numbers, then log^ N is required. Let hy=N^ so that y=\o%^N. .-. loga(6^)=log^iV^; that is, - y log„ h = log^ N; or log5i\^=j— ^xlog^iV^ (1). Now since N and h are given, log^iY and log^ft are known from the Tables, and thus log^ N may be found. Hence it appears that to transform logarithms from base a to base 6 we have only to multiply them all by = ^ ; this is a constant quantity and is given by the Tables ; it is known as the modulus. XIV.] LOGARITHMS. 147 If in equation (1) we put a for N, we obtain logft a=. J- X loga a=T T ; log„6 *" loga& .-. log5axlog«6 = l. 169. The following examples further illustrate the great use of logarithms in arithmetical work. Example 1. Given log 2 = -3010300 and log 4844544 = 6-6852530, find the value of (6-4)tVx (^^256)3-5-^80. Let X be the value of the expression ; then 1 , 64 3 , 256 1 . ^. log ^ = Jo log j^ + ^ log ^Q- 2 log SO = i{log26_l) + |(log28-3)-^{log2=^ + l) = (5 + ^)log2-2i^ = 1-5051500 + -0301030 - 2-85. log a; =i'- 6852530. log 4844544 = 6-6852530, .-. a; =-04844544. Thus But Example 2. Find how many ciphers there are between the decimal point and the first significant digit in (-0504)i0; having given log 2 =-301, log 3 = -477, log 7 = '845. Denote the expression by E ; then 504 logE = 101og^-^ = 10 (log 504 -4) = 10 {log (23x32x7) -4} = 10{31og2 + 21og3 + log7-4) = 10 (2 -702 - 4) = 10 (2-702) = 20 + 7 -02 = 13-02, Thus the number of ciphers is 12. [Art. 163.] 3 log 2= -903 2 log 3= -954 log 7 = -845 2-702 10—2 148 ELEMENTARY TRIGONOMETRY. [CHAP. Exponential equations. 170. If in an equation tlie unknown quantity appears as an exponent, the solution may be effected by the help of logarithms. Example 1. Solve the equation 8^~^* = 12'*-"*, having given log 2 = -30103, and log 3 = -47712. From the given equation, by taking logarithms, we have (5 - 3a;) log 8 = (4 - 2x) log 12 ; .-. 3 (5 -3a;) log 2 = (4-2a;)(21og2 + log3); .-. 151og2-81og2-41og3 = x (9 log 2 - 4 log 2 - 2 log 3) ; _ 7 log 2 - 4 log 3 _ 49873 •'• "^ "" 5 log 2 - 2 log 3 ~ -55091 ' Thus a; =-36 nearly. 7 log 2 =2-10721 4 log 3= 1-90848 -19873 5 log 2=1-50515 2 log 3= -95424 •55091 55091 ) 198730 ( 86 165273 334570 Example 2. Given log 2 = -30103, solve the simultaneous equations 2«.52'=1, 5^1. 22/ = 2. Take logarithms of the given equations ; .-. a;log2 + i/log5 = 0, (a; + 1) log 5 + ?/ log 2 = log 2. For shortness, put log 2 = a, log 5 = 6. Thus ax + by = 0, and b{x + l) + ay = a, or hx + ay = a-b. By eliminating y, x [a^ -h^)= -b{a-h), b log 5 log 5 , „ nr.r.r.r, •■• ^= T = -^ — 7^^^ — i= " rA a = - log 5= - -69897. a + b log2 + log5 loglO And 2/=-^ = ^log5 = a = log2=-30103. o XIV.] LOGARITHMS. 149 EXAMPLES. XIV. b. [ Whe7i required the values of log 2, log 3, log 7 giveti on p. 145 may he used.] Find the value of ("147 Y '?'75\f ,' ' ) , given log 9-076226 =-9579053. IzD X lb / 2. \^^ X Vl08 ^ (^^1008 X v^486), given log 301824=5-4797536. 3. (1080)4 X (-24)^x810, given log 2467266 = 6-3922160. Calculate to two decimal places the values of 4. Iog2o800. 5. log3 49. 6. Iogi25 4000. 7. Find how many ciphers there are before the first signifi- cant digits in (-00378)^" and (-0259)^0. 8. To what base is 3 the logarithm of 11000 ? given log 11 = 1-0413927 and log 222398 = 5-3471309. Solve to two decimal places the equations : 9. 2»:-i=5. 10. 3^-^ = 7. 11. 51-^=6^-3, 12. 5=«=2-J' and 52 + ^=22-^. 13. 2='=3y and 2^ + 1 = 3^-1. 14. Given log 28 = a, log 21 = 5, log25 = c, find log 27 and log 224 in terms of a, b, c. 15. Given log 242 = a, log 80 =6, log45 = c, find log 36 and log 66 in terms of a, b, c. 150 ELEMENTARY TRIGONOMETRY. [CHAP. XIV. MISCELLANEOUS EXAMPLES. E. 1. Prove that cos (30° +^) cos (30° -A)- cos (60° + ^) cos (60° -A)=^. 2. 1{A + B + C= 180°, shew that sin 2^ +sin 25 + sin 2(7 ^ . A . B . C ■ — ; — J— — . — ^- — -. — y^~ = 8 sm — sm - sm - . sm^+sm^ + sinC 222 3. If a = 2, c = v/2, 5=15°, solve the triangle. 4. Shew that cos a + tan — sin a = cot — sin a — cos a. 5. If b cos A = a cos B, shew that the triangle is isosceles. 6. Prove that (1) sin 6 (sin 3(9 + sin 5^ + sin 7^ + sin 9^) = sin 66 sin4^ ; ,^. sin a + sin 3a + sin 5a + sin 7a (2) — ;; rr- = tan 4a. ^ ' cos a + cos 3a + cos 5a + cos 7a rr en j^i x ^os 3a . sin 3a ^ ^ ^ 7. Shew that -. 1 = 2 cot 2a. sm a cos a 8. If 6=a (v/3 - 1), (7= 30°,- find A and B. n ci, 4.1. + 4. 4 4tana-4tan3a 9. Shew that tan 4a = - — —, — ;; -, — t— . l-6tan2a+tan*a 10. In a triangle, shew that (1) a^cos2B + b^cos2A = a^ + b^ — 4:ab sin A sin B ; (A B C\ (2) 4 ( be cos^ ^ + c<^ cos^ l^ + ^^ ^0^^ o ) = {tt + b + cf. 11. If ci4 + 54_,_c4==2c2(a2 + &2)^ prove that (7=45° or 135°. [S'oZ'ye as a quadratic in c^.] 12. If in a triangle cos 3^+ cos 35+ cos 3 (7=1, shew that one angle must be 120°. CHAPTER XV. THE USE OF LOGARITHMIC TABLES. 171. We shall now explain the use of logarithmic Tables to which reference has been made in the previous chapter. In a book of Tables there will usually be found the mantissce of the logarithms of all integers from 1 to 100000; the charac- teristics can be written down by inspection and are therefore omitted. [Art. 162.] The logarithm of any number consisting of not more than 5 significant digits can be obtained directly from these Tables. For instance, suppose the logarithm of 3.36 "34 is required. Opposite to 33634 we find the figures 5267785; this, with the decimal point prefixed, is the mantissa for the logarithms of all numbers whose significant digits are the same as 33634. We have therefore only to prefix the characteristic 2, and we obtain log 336-34 = 2 -5267785. Similarly, log 33634 = 4-5267785, and log -0033634 = 3-5267785. 172. Suppose now that we required log 33634-392. Since this number contains more than 5 significant digits it cannot be obtained directly from the tables ; but it lies between the two consecutive numbers 33634 and 33635, and therefore its logarithm lies between the logarithms of these two numbers. If we pass from 33634 to 33635, making an increase of 1 in the number, the corresponding increase in the logarithm as obtained from the tables is '0000129. If now we pass from 33634 to 33634-392, making an increase of '392 in the number, the in- crease in the logarithm will be -392 x -0000129, provided that the increase in the logarithm is proportional to the increase in the number. 162 ELEMENTARY TRIGONOMETRY. 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CI. !=! •<*« t- (D CO A > co' bo . '^ • ^H 02 CM © -TJ lO P c3 T 02 bO o iC ^ ."S 02 fl 0) rti '35 ©•2'-^ © •rH O C3 ■^ ,?H ^ 1— I ^ «5 •Sl^ J © a .^ "^ Ph "5.2 o © © -p I— I -fj © PI -^ 0000129 •392 •0000050 568 258 61 7 XV.] THE USE OF LOGARITHMIC TABLES. 153 Similarly, log 33651 = 4^5269980, and log 33652 = 4-5270109, the transition in the mantissse from 526... to 527... being shewn by the bar drawn over 0109. This bar is repeated over each of the subsequent logarithms as far as the end of the line, and in the next line the mantissas begin with 527. Example. Find log 33634-392. From the Tables, log 33635 = 4 -5267914 log 33634 = 4-5267785 difference for 1 = ^0000129 Now by the Eule of Proportional Parts, log 33634^392 will be greater than log 33634 by "392 times the difference for 1 ; hence to 7 places of decimals, we have log 33634 = 4^5267785 proportional difference for -392= ^0000051 .•. log 33634^392 = 4^5267836 In practice, the difference for 1 is usually quoted without the ciphers; if therefore we treat the difference 129 as a tvliole number, on multiplying by '392 we obtain the product 50^568, and we take the digits given by its integral part (51 approximately) as the propor- tional increase for '392. 175. The method of calculating the proportional difference for '392 which we have explained is that which must be adopted when we have nothing given but the logarithms of two con- secutive numbers between which lies the number whose logarithm we are seeking. But when the Tables are used the calculation is facilitated by means of the proportional differences standing in the column to the right. This gives the differences for tenths of unity. The difference for "392 is obtained as follows. •392 X 129 = ('^ + i5o + 1^) ^ 129 = 39 -f-11-6-1- •26 = 50-86. The difference for 9 quoted in the margin (really 9 tenths) is 116, and therefore the difference for 9 hundredths is 11*6; and similarly the difference for 2 thousandths is -26. 154 ELEMENTARY TRIGONOMETRY. [chap. In practical work, the following arrangement is adopted. log 33634 =4-5267785 add for 3 39 9 116 2 26 .-. log 33634-392 = 4-5267836 176. The following example is solved more concisely as a model for the student. In the column on the left we work from the data of the question ; in the column on the right we obtain the logarithm by the use of the Tables independently of the two given logarithms. Example. Find log 33 "656208, having given log 33656 = 4-5270625 and log 38657 = 4-5270754. log 33-657 = log 33-656 = diff. for -001 = 1-5270754 1-5270625 129 208 1032 2^80^ diff. for -000208= 26 832 log 33-656 =1-5270625 log 33-656208 = 1-527065^ From the Tables, we have log 33-656 =1-5270625 add for 2 26 103 log 33-656208 = 1-5270652 177. The Rule of Proportional Parts also enables us to find the number corresponding to a given logarithm. Example 1. Find the number whose logarithm is 2*5274023, having given log 3-3683 = -5274108 and log 3 -3682 = -5273979. Let X be the required number ; then log X = 2-5274023 log -033683 = 2'5274108 log -033682 =2 -5273979 diff.= 44 log -033682 = 2-5273979 diff. for -000001= l29 hence x lies between -033682 and -033683, 44 and is greater than -033682 by -— x -000001, that is by -00000034. .-. a; = -03368234. 129 ) 440 ( 34 387 530 516 XV.] THE USE OF LOGARITHMIC TABLES. 155 In working from the Tables, we proceed as follows, log a; =.2*5274023 log •033682 = 2*5273979 44 8 89 50 4 52 a; = •03868234. We are saved the trouble of the division, as the multiples of 129 which occur during the work are given in the approximate forms 39 and 52 in the difference column opposite to the numbers 3 and 4. Example 2. Find the fifth root of •0025612, having given log 2-5612 = •4084435, log 8-0317 = -4816862, log 3 -0318 = -4817005. Let a; = (-0025612) 6; then log a; = i log (-0025612) = ^ (3*4084435) (5 +2 •4084435); = 1*4816887. log a; =1*4816887 log -30317 = 1*4816862 diff.= 25 25 log •30318=1*4817005 log -30317 =1*4816862 diff. for -00001= l43 proportional increase =-— x -00001 = -00000175. 14o Thus a; =-30317175. 143 ) 250 ( 175 143^ 1070 1001 690 715 EXAMPLES. XV. a. 1. Find the value of log 4951634, given that log 49516=4-6947456, log 49517 =4-6947543. 2. Find log 3-4713026, having given that log 347*13 = 2-5404921, log 34714 = 4-5405047. 3. Find log 2849614, having given that log 2-8496 = -4547839, log 2-8497 = *4547991. 156 ELEMENTARY TRIGONOMETRY. [CHAP. 4. Find log57'63325, having given that log 576-33 = 2-7606712, log 5763-4 = 3-7606788. 5. Given log 60814 = 4*7840036, diflf. for 1 = 72, find log 6081465. 6. Find the number whose logarithm is 4-7461735, given log 55740 = 4-7461670, log 55741 = 4-7461748. 7. Find the number whose logarithm is 2-8283676, given log 6-7354 = -8283634, log 67355 = 4-8283698. 8. Find the number whose logarithm is 2-0288435, given log 1068-6 = 3-0288152, log 1-0687 = -0288558. 9. Find the number whose logarithm is 3-9184377, given log 8-2877 = -9184340, log 8287-8 = 3-9184392. 10. Given log 253-19 = 2-4034465, diff. for 1 = 172, find the number whose logarithm is 1 -4034508. 11. Given log 2-0313 = -3077741, log 2*0314 = '3077954, and logl-4271 = -1544544, find the seventh root of 142-71. 12. Find the eighth root of 13-89492, given log 13894=4-1428273, log 138-95 = 2-1428586. 13. Find the value of ^242447, given log 2-4244 = -3846043, difi". for 1 = 179. 14. Find the twentieth root of 2069138, given log 20691 = 4-3157815, difi". for 1 = 210. Tables of Natural and Logaritlxmic Functions. 178. Tables have been constructed giving the values of the trigonometrical functions of all angles between 0° and 90° at intervals of 10". These are called the Tables of natural sines, cosines, tangents,... In the smaller Tables, such as Chambers', the interval is 1'. The logarithms of the functions have also been calculated. Since many of the trigonometrical functions are less than unity XV.] THE USE OF LOGARITHMIC TABLES. 157 their logarithms are negative, and as the characteristics are not always evident on inspection they cannot be omitted. To avoid the inconvenience of printing the bars over the character- istics, the logarithms are all increased by 10 and are then registered under the name of tabular logarithmic sines, cosines,... The notation used is L cos A^ L tan Q ; thus L sin A = log sin A + 10. For instance, Z sin 45° = 10+ log sin 45° = 10+ log -^ = 10 - ^ log 2 = 9-8494850. 179. With certain exceptions that need not be here noticed, the rule of proportional parts holds for the natural sines, co- sines,... of all angles, and also for their logarithmic sines, cosines,.... In applying this rule it must be remembered that as the angle increases from 0° to 90° the functions sine, tangent, secant increase, while the co-functions cosine, cotangent, cosecant decrease. Example 1. Find the value of sin 29° 37' 42". From the Tables, sin 29° 38' = -4944476 sin 29° 37'- -4941948 diff. for 60"= 2528 42 .-. propi increase for 42"= ^ x 2528 = 1770 bO sin 29° 37' = '4941948 .-. sin 29° 37' 42"= -4943718. Example 2. Find the angle whose cosine is -7280843. Let A be the required angle ; then from the Tables, cos 43° 16' = -7281716 cos 43° 16'= -7281716 cos 43° 17' = -7279722 cos^ =-7280843 diff. for 60" 1994 prop' part 873 I fin But cos A is less than cos 43° 16' ; hence A must be greater than 43° 16' 87 S by jw^. X 60", that is by 26" nearly. Thus the angle is 43° 16' 26". 1994 ) 52380 ( 26 3988 12500 11964 158 ELEMENTARY TRIGONOMETRY. [chap. 180. In order to illustrate the use of the tabular logarithmic functions we give the following extract from the table of log- arithmic sines, cosines,.., in Chambers' Mathematical Tables. 27 Deg. 56 57 58 59 60 Sine Diff. Cosec. 9-6570468 9-6572946 9-6575423 9-6577898 9-6580371 9-6706576 9-6708958 9-6711338 9-6713716 9-6716093 2478 2477 2475 2473 2382 2380 2378 2377 10-3429532 10-3427054 10-3424577 10-3422102 10-3419629 10-3293424 10-3291042 10-3288662 10-3286284 10-3283907 Cosine Diff. Secant Secant D. 10-0501191 10-0501835 10-0502479 10-0503124 10-0503770 10-0537968 10-0538638 10-0539308 10-0539979 10-0540651 Cosec. 644 644 645 646 670 670 671 672 D. Cosine / 9-9498809 60 9-9498165 59 9-9497521 58 9-9496876 57 9-9496230 56 9-9462032 4 9-9461362 3 9-9460692 2 9-9460021 1 9-9459349 Sine / 62 Deg. 181. We have quoted here the logarithmic sines, cosecants, secants, and cosines of the angles difiering bj 1' between 27° 0' and 27° 4', and also between 27° 56' and 27° 60'. The same extract gives the logarithmic functions of the complements of these angles, namely those between 62° 0' and 62° 4', and those between 62° 56' and 62° 60'. The column of minutes for 27° is given on the left and increases downwards, the column for 62° is on the right and increases upwards. The names of the functions printed at the top refer to the angle 27°, the names printed at the foot refer to the angle 62°. Thus Z cos 27° 3' = 9-9496876, X sin 62° 2' = 9 -9460692, L cosec 27° 58' = 10-3288662, Z cos 62° 59' = 9-6572946. The first difference column gives the differences in the log- arithms of the sines and cosecants, the second difference column gives the differences in the logarithms of the cosines and secants, each difference corresponding to a difference of 1' in the angle. XV.] THE USE OF LOGARITHMIC TABLES. 159 Example 1. Find L cos 62° 57' 12". From the Tables, L cos 62° 57' = 9 -6577898 L cos 62° 58' = 9 -6575423 diff. for 60" 2475 12 .-. proportional decrease for 12" = 777-x 2475 = 495. 60 " X cos 60 57' = 9-6577898 Subtract for 12" 495 .-. L cos 62° 57' 12"= 9-6577403 Example 2. Given L sec 27° 39' = 10-0526648, diff. for 10" = 110, find A when L sec ^ = 10-0527253. Xsec^ =10-0527253 L sec 27° 39' = 10-0526648 diff. 605 .-. proportional increase = ^—r x 10" = 55". Thus ^ = 27° 39' 55". EXAMPLES. XV. b. 1. Find sin 38° 3' 35", having given that sin 38° 4' = -6165780, sin 38° 3' = '6163489. 2. Find tan 38° 24' 37-5", having given that tan 38° 25' = -7930640, tan 38° 24' = -7925902. 3. Find cosec 55° 21' 28", having given that cosec55° 22' = 1-2153535, cosec 55° 21' = 1-2155978. 4. Find the angle whose secant is 2-1809460, given sec 62° 43' = 2-1815435, sec 62° 42' = 2-1803139. 5. Find the angle whose cosine is -8600931, given cos 30° 41' = -8600007, cos 30° 40' = -8601491. 6. Find the angle whose cotangent is -8766003, given cot 48° 46' = -8764620, cot 48° 45' = -8769765. 7. Find L sin 44° 17' 33", given X sin 44° 18' = 9-8441137, Z sin 44° 17' = 9 -8439842. 160 ELEMENTARY TRIGONOMETRY. [CHAP. 8. Find L cot 36° 26' 16", given L cot 36° 27' = 10-1315840, L cot 36° 26' = 101318483. 9. Find L cos 55° 30' 24", given Xcos 55° 31' =9-7529442, L cos 55° 30' = 9-7531280. 10. Find the angle whose tabular logarithmic sine is 9*8440018, using the data of example 7. 11. Find the angle whose tabular logarithmic cosine is 9*7530075, using the data of example 9. 12. Given L tan 24° 50' = 9*6653662, diff. for 1' = 3313, find Xtan24°50'52*5". 13. Given L cosec 40° 5' = 10*1911808, diflf. for 1' = 1502, find Zcosec40°4' 17*5". 182. Considerable practice in the use of logarithmic Tables will be required before the quickness and accuracy necessary in all practical calculations can be attained. Experience shews that mistakes frequently arise from incorrect quotation from the Tables, and from clumsy arrangement. The student is reminded that care in taking out the logarithms from the Tables is of the first importance, and that in the course of the work he should learn to leave out all needless steps, making his solutions as concise as possible consistent with accuracy. Example 1. Divide 6*6425693 by -3873007. 1-5880475 7 From the Tables, log 6*6425 = -8223316 6 40 9 5 3 9 20 log -38730 7 log -3873007 log 6*6425693= -8223362 log -3873007 = 1-5880483 By subtraction, we obtain 1-2342879 From the Tables, log 17*150 = 1-2342641 238 9 229 90 3 76 Thus the quotient is 17*15093. XY.] THE USE OP LOGARITHMIC TABLES. 161 Example 2. The hypotenuse of a right-angled triangle is 3-14102i and one side is 2-593167; find the other side. Let c be the hypotenuse, a the given side, and x the side required ; then x^ = c^ - a^= {c+ a) (c ~ a) ; .\ 2 log a;=log {c + a)+ log (c - a). From the Tables, log 5-7341 = -7584653 9 68 1 log -54785 =17386ea7 7 55 By addition, •4971394 Dividing by 2, we have logx = •2485697 log 1-7724 = •2485617 80 3 74 60 2 49 Thus the required side is 1^772432. c=3-141024 a = 2-593167 c+a = 5-73-il91 c-a= 'SiTSoS EXAMPLES. XV. c. [In this exercise the logarithms are to be taken from the Tables.] 1. Multiply 300-2618 by •0078915194. 2. Find the product of 235-6783 and 357-8438. 3. Find the continued product of 153-2419, 2-8632503, and -07583646. 4. Divide 1-0304051 by 27-093524. 5. Divide 357-8364 by -00318973. 6. Find x from the equation -0178345:^ = 21-85632. 7. Find the value of 3-78956 X -0536872-f -0072916. H. K. E. T. 11 162 ELEMENTARY TRIGONOMETEY. [chap. 8. Find the cube of -83410039. 9. Find the fifth root of 15063-018. 10. Evaluate v'384-731 and a/ 15-7324. 11. Find the product of the square root of 1034-3963 and the cube root of 353246. 12. Subtract the square of -7503269 from the square of 1-035627. X3, Find the value of (34-7326)^ X \/2-53894 \/4-39682 Example 3. Find a third proportional to the cube of -3172564 and the cube root of 23-32873. Let X be the required third proportional ; then (-3172564)3 : (23-32873)^= (23-32873)^ : x; whence a; = (23-32873)3 -=-(-3172564)3; .-. log a; r= I log 23-32873 - 3 log -3172564. From the Tables, log -31725 = 1-5014016 6 82 4 5 1*5014103 3^ 2-5042311 By subtraction, log x log 255-67 log 23-328 =1-3678775 7 130 3 5 1-3678910|6 2 3 I 2-7357821 -9119274 2-5042311 = 2-4076963 = 2-4076798 165 153 120 .119 Thus the third proportional is 255-6797. XV.] THE USE OP LOGARITHMIC TABLES. 163 14. Find a mean proportional between •0037258169 and -56301078. 15. Find a third proportional to the square of '43607528 and the square root of -03751786. 16. Find a fourth proportional to 56712-43, 29-302564, -33025107. 17. Find the geometric mean between (-035689)^ and (2-879432)t. 18. Find a fourth proportional to V^32-7812, ^357-814, 'J/7836-43. 19. Find the value of sin 27° 13' 12" x cos 46° 2' 15". 20. Find the value of cot 97° 14' 16" X sec 112° 13' 5". 21. Evaluate sin 20° 13' 20" x cot 47° 53' 15" x sec 42° 15' 30". 22. Find the value of ah sin C, when a=324-1368, 6 = 417-2431, 0=113° 14' 16". 23. If a -. 6 = sin J. : sin B, find a, given 6 = 378-25, ^ = 35° 15' 33", 5 = 119° 14' 18". 24. Find the smallest values of 6 which satisfy the equations (1) tan3(9 = -5-; (2) 3 sin2 ^+2sin ^=1. 25. Find x from the equation ^x sec 28° 17' 25" = sin 23° 18' 5" x cot 38° 15' 13". 26. Find ^ from the equation sin^ ^=cos2 a^ot j3, where a = 32° 47' and /3 = 41° 19'. 11—2 CHAPTER XVI. SOLUTION OF TRIANGLES WITH LOGARITHMS. 183. The examples on the solution of triangles in Chap. XIII. furnish a useful exercise on the formulae connecting the sides and angle of a triangle ; but in practical work much of the labour of arithmetical calculation is avoided by the use of logarithms. We shall now shew how the formulae of Chap. XIII. may be used or adapted for use in connection with logarithmic Tables. 184. To find the functions of the half-angles in terms of the sides. We have 2 sin^ — = 1 - cos ^ Let then and 2 sin^ 26c 26c-62-c2 + a2 a2- -(62- -2bc+c^) 26c 26c a^-(J)-cf {a + h- c)(a- -6+c) 2bc 26c a + h+c=2s; a-{-b-c = 2s-2c=2{s-c), a_6+c=2s-26 = 2(s-6) .. ,A A{s-c){s-b)_2{s- b){s- -«). 2 26c bo . A /{s-b){s-c) ■• «^^2 = V 6c SOLUTION OF TRIANGLES WITH LOGARITHMS. 165 A b^+c^ — a^ Again, 2 cos^ — = 1 + cos A = l-\ i {b + cf-a^ { b+G+a){b + c-a) ^ ~ 26c ~ 26c ' ^A 4s (s- a) 2s {s- a) .-. 2cos2— = — )^ -'= r ; 2 26c 6c = x/ A /s(s — a) cos — ' 2 V 6c Also tan — = sin — -f- cos — 2 2 2 =y (s-6)(5-c)^ 6c 6c s{s — a)^ tan — = v^ A_ /(s-b){s-c) 2 V s{s-a) 185. Similarly it may be proved that B 2-\ sin - . . /(^ZM^.^ 3i, O /t:a)is-b) ca ' 2 V a6 B cos- = Aj^-6) C_ /s{s-c), V ca ' '''''^2-V ~^^^6~^ B_ / is-c){s-a) C_ / (s-a){s-b) In each of these formulae the positive value of the square root must be taken, for each half angle is less than 90°, so that all its functions are positive. 186. To find sin J. in terinis of the sides. sm J. = 2 sm — cos — =2 /(iz^)if_-if) X ^(^~^) . V 6c 6c ' .*. sin J. = v- Vs (« — a) (s — 6) (5 - c). We may also obtain this formula in another way which is instructive. 166 ELEMENTARY TRIGONOMETRY. [cHAP. We have siu2 A = l— cos2 A = {1 + cos A) (1 - cos ^1) V 26c ;V 26c ^(6+c)2-a2 ct2-(6-c)2 26c ^ 26c _(6 +c + a)(6 + c-a) (a + 6 -c) (a-6-f c) 4Pc2 _ 16g(g-a)(g-6)(g-c) 4Pc2 ' 2 .-. sin^4 = Y-\/«(s-a)(s— 6)(s-c). The positive value of the square root must be taken, since the sine of an angle of any triangle is always positive. EXAMPLES. XVI. a. Prove the following formulae in any triangle : 1. bcos^^+acos^ — =s. 2. stan — tan— =5-1^. vers^ a(a+c-6) , A ^ . ^B 3. 0=7-77-^ r. 4. 6sm^ — +asm^ — = 5-c. vers^ 6(6 + c-a) 2 2 A B C 5. (« - a) tan — = (s - 6) tan — = (5 - c) tan - . AAA 6. Find the value of tan — , when a=lO, 6=17, c = 21. A c 7. Find cot-, when a = 13, 6 = 14, c = 15. 8. Prove that 1 2^ 1 2^1 „(7 s2 - COS^ TT + T COS^ TT + - COS-*— = -^- . a 26 2c 2a6c 9. Prove that 6-c „A , c-a ^B , a-6 ^C ^ COS'' — H J — COS'* — H — COS-^ -::: = 0. a 2 6 2 c 2 XVI.] SOLUTION OF TRIANGLES WITH LOGARITHMS. 167 187. To solve a triangle when the three sides are given. From the formula v^ A_ /(,-«(.-.) 2 V s{s-a) ' A 1 log tan - = - {log (5 - 6) + log {s-c)- log s - log (s - a)] j whence — may be obtained by the help of the Tables. Similarly B can be found from the formula for tan—, and then C from the equation (7= 180° — A-B. In the above solution, we shall require to look out from the tables four logarithms only, namely those of s, s — a, s — b, s — c; whereas if we were to solve from the sine or cosine formulae we should require six logarithms ; for A cos- = /-^a.acosfV^. SO that we should have to look out the logarithms of the siv quantities s, s—a, s-b, a, b, c. If therefore all the angles have to be found by the use of the tables it is best to solve from the tangent formulae ; but if one angle only is required it is immaterial whether the sine, cosine, or tangent formula is used. In cases where a solution has to be obtained from certain given logarithms, the choice of formulae must depend on the data. Note. We shall always find the angles to the nearest second, so that, on account of the multiplication by 2, the half-angles should be found to the nearest tenth of a second. 188. In Art. 178 we have mentioned that 10 is added to each of the logarithmic functions before they are registered as tabular logarithms ; but this device is introduced only as a convenience for the purposes of tabulation, and in practice it will be found that the work is more expeditious if the tabular logarithms are not used. The 10 should be subtracted mentally in copying down the logarithms. Thus we should write log sin 64° 15'= 1 '9546793, log cot 18° 35' = -4733850, and in the arrangement of the work care must be taken to keep the mantissae positive. 168 ELEMENTARY TRIGONOMETRY. [chap. Example 1. The sides of a triangle are 35, 49, 63; find the greatest angle; given log 2 = -3010300, log 3 = -4771213, L cos 47° 53' = 9-8264910, diff. for 60"= 1397. Since the angles of a triangle depend only on the ratios of the sides and not on their actual magnitudes, we may substitute for the sides any lengths proportional to them. Thus in the present case we may take a = 5, 6 = 7, c = 9; then C is the greatest angle, and cos C_ / s {s-c) _ /21 3 J__ /9_ 2~ V ~~^b~^~ V 2 ''2''5x7~V20' Thus logcos^ = J(21og3-log2-l). log cos ^ = 1-8266063 log cos 47° 53' = 1-8264910 diff. ll53 1153 .-. proportional decrease = :r-^^ x 60" = 49*5" ; .-.^ = 47° 52' 10-5". Thus the greatest angle is 95° 44' 21". 2 log 3= -9542426 1 •3010300 2 ) 1-0532126 r'8266063 1153 60_ 1397 ) 69180 ( 49-5 5588 13300 12573 7270 Example 2. If a = 283, 6 = 317, c = 428, find all the angles. A^ / {s-b)(s-c) _ I 2 V s(s-a\ V tan' 197 X 86 514 X 231 ' log tan 4 = 5 (log 197 + log 86 - log 514 - log 231). From the Tables, log 197 = 2-2944662 log 86 = 1-9344985 4^28"9647 5-0745751 2 ) 1-1543896 log tan 4 = 1*5771948 log tan 20° 41' =1-5 769585 diff. 2363 283 317 428 2 ) 10i!8 log 514 = 2-7109631 log 231 = 2-3636120 5-0745751 514 =S 231 =s- a 197=5-6 86=5- c xvl] SOLUTION OF TRIANGLES WITH LOGARITHMS. 169 But diff. for 60" is 3822, , . 2363 .-. prop', increase = ^g22 X 60" = 37-1"; ... ± = 20° 4:1' 37-1" and ^ = 41° 23' 14". , B / {s-c){s-a) I Agam, tan-=^ ^^^_^^ = ^J \ log tan ^ = J (log 86 + log 231 - log 514 - log 197). 2363 60 3822 ) 141780 ( ST'l 11466 27120 26754 3660 86 X 231 514x197 log 86 = 1-9344985 log 231 = 2-3636120 4-2981105 5-0054293 2 ) 1-2926812 log tan 1 = 1-6463406 log tan 23° 53' = 1-6461988 diff. iilS log 514 = 2-7109631 log 197 = 22944662 5-0054293 But diff. for 60" is 3412 ; , . 1418 .*. prop', increase = 3412 X 60" = 24-9". ■ 2 = 23° 53' 24-9" and 5 = 47° 46' 50". Thus ^=41° 23' 14", 5 = 47° 46' 50", C = 90° 49' 56". 1418 60 3412 ) 85080 ( 24-9 6824 16840 13648 31920 EXAMPLES. XVI. b. 1. The sides of a triangle are 5, 8, 11 ; find the greatest angle ; given log 7 = '8450980, L sin 56° 47' = 9'9225205, L sin 56° 48' = 9-9226032. 2. If a=40, 6 = 51, c=43, find A ; given X tan 24° 44' 13" = 9-6634464, log 1 28 = 2-1072100, log 603 = 2-7803173. 3. The sides a, 6, c are as 4 : 5 : 6, find B ; given log 2, L cos 27° 53' = 9-9464040, difi". for 1' = 669. 170 ELEMENTARY TRIGONOMETRY. FCHAP. , f 4. Find the gi'eatest angle of the triangle in which the sides are 5, 6, 7; given log 6 = '7781513, X cos 39° 14' = 9-8890644, difif. for l' = 1032. • 5. If a = 3, 5 = 1-75, c = 2-75, find C ; given log2, Xtan 32° 18' = 9-8008365, diff. for l'=2796. 6. If the sides are 24, 22, 14, find the least angle; given Z tan 17° 33' = 9-500042, difil for l' = 439. 7. Find the greatest angle when the sides are 4, 10, 11; given log 2, log 3, L cos 46° 47' = 9-8355378, difi'. for 1' = 1345. 8. If a : 6 : c=15 : 13 : 14, find the angles ; given log 2, log 3, log 7, L tan 26° 33' = 9-6986847, diff. for 1' = 3159, L tan 29° 44' = 9-7567587, diff. for l' = 2933. 9. If a : 6 : c = 3 : 4 : 2, find the angles ; given log 2, log 3, X tan 14° 28'= 9-4116146, difi". for 10" = 870, L tan 52° 14' = 10-1 108395, diff". for 10" =435. 189. To solve a tHangle having given two sides and the in- cluded angle. Let the given parts be 6, c, J., and let T_smB sin (7 then h G ' sin B — sin G kb — he h — c sin^+sin(7 kb+kc b+c* ^ B+C . B-C 2 cos -^ sin ^- ^_^ • „ . B+C B-C b+c' 2 sin — - — cos ■ tan 2 2 B-C . B+C b+c' tan — — — XVI.] SOLUTION OF TRIANGLES WITH LOGARITHMS. 171 B-C b-G^ B+C b-c ^A .-. tan — - — = ,— — tan — s^ = itt" ^^^ "H" » 2 6 + c 2 b+c 2' since — - — = 90 ~ — . 2 2 B— G A .'. logtan — - — = log(6-c)-log(6 + c)+logcot 2' JBrom which equation we can find B-0 Also — - — = 90° - — , and is therefore known. By addition and subtraction we obtain B and C. _, , . . b sin ^4 rrom the equation a= — . — jj-, ^ sm^ log a =log 6 -I- log sin A - log sin B ; whence a may be found. Example 1. If the sides a and b are in the ratio of 7 to 3, and the included angle C is 60°, find A and B ; given log 2 = -3010300, log 3 = -4771213, L tan 34° 42' = 9-8403776, diff. for 1'= 2699. ^ 4-JB a-b .C 7-3 4 tan -2-=-^ cot - = ^3 cot 30 =^^^S; 10 -. Iogtan^-2^ = 21og2-l+-log3; ,-. log tan ^^ = 1-8406207 log tan 34° 42'= 1-8403776 diff. 2431 2431 prop', increase = A-B And 2 A+B :90= By addition, and by subtraction, x60" = 54"; 2699 = 34° 42' 54". ^ = 94° 42' 54" 5 = 25° 17' 6". 2 log 2= •6020600 I log 3 = -2385607 •8406207 2431 60 2699 ) 145860 ( 54 13495 10910 10796 172 ELEMENTARY TRIGONOMETRY. [chap. Example 2. If a = 681, c = 243, 5 = 50° 42', solve the triangle, by the use of Tables. . A-C a-c ^B 438 ,„^„„,, *^^-^=^c^°*2=924^°*2^^1' A— G :. log tan —^ = log 438 - log 924 + log cot 25° 21' .-. log tan ^^=-0002383 log tan 45° =-0000000 diff. 2383 And difi. for 68" is 2527 ; /. propi. increase = — — x60"=57"; A-c = 45°0'57". Also 4±^=90°-|- = 64°39'. log 438 = 2-6414741 log cot 25° 21'= -3244362 2-9659103 log 924=2 -9656720 •0002383 2383 60 2527 ) 142980 ( 56-6 12635 16630 15162 14680 By addition, and by subtraction, 4 = 109° 39' 57", G= 19° 38' 3". , . - c sin B .-. log 6 = log c + log sin JB - log sin C = log 243 + log sin 50° 42' - log sin 19° 38' 3" .-. log & = 2-7479012 log 559-63 =2-7479010 .-. 5=559-63. Thus ^ = 109° 39' 57", C=19°38'3", & = 559-63. log sin 19° 38' = 1 '5263387 ^x3540= 177 log sm 19° 38' 3" = 1 '5263564 log 243 = 2-3856063 log sin 50° 42'= 1-8886513 2-2742576 log sin 19° 38' 3" = 1-5263564 2-7479012 190. From the formula , B-C b-c A tan — - — = = — cot — , 2 b+c 2' XVI.] SOLUTION OF TRIANGLES WITH LOGARITHMS. 173 it will be seen that if 6, c, and B— C are known A can be found ; that is, the triangle can be solved when the given parts are two sides and the dift'erence of the angles opposite to them. EXAMPLES. XVI. c. 1 . If a = 9, 6 = 6, (7= 60°, find ^ and ^ ; given log 2, log 3, i tan 19° 6' = 9-5394287, X tan 19° 7' = 9-5398371. 2. If a = 1, c = 9, 5 = 65°, find A and C ; given log 2, Z cot 32° 30'= 10-1958127, X tan 51° 28' = 10-0988763, diff. for l' = 2592. 3. If I7a=1b, C= 60°, find ^ and i5 ; given log 2, log 3, L tan 35° 49' = 9-8583357, diff. for 10" = 2662. 4. If 6 = 27, c = 23, .4 = 44° 30', find X and C ; given log 2, Z cot 22° 15' = 10-3881591, Ztanll° 3'= 9-2906713, diff. for 1' = 6711. 5. Ifc = 210, « = 110, Z=34°42'30", find Cand^; given log 2, Zcot 17° 21' 15" = 10-5051500. 6. Two sides of a triangle are as 5 : 3 and include an angle of 60° 30' : find the other angles ; given log 2, Z cot 30° 15' = 10-23420, Z tan 23° 13'= 9-63240, diff. for 1' = 35. 7. If a = 327, c = 256, ^ = 56° 28', find ^ and (7; given log 7-1 = -8512583, log 5-83 = '7656686, Z tan 61° 46'= 10-2700705, Ztanl2° 46'= 9-3552267, diff. for l' = 5859. 8. If 6 = 4c, A = 65°, find B and G ; given log 2, log 3, Z tan 57° 30' = 10-1958127, Z tan 43° 18'= 9-9742133, diff. for 1' = 2531. 9. If a = 23031, & = 7677, (7=30° 10' 5", find A and B ; given log 2, Ztanl5° 5' = 9-4305727, diff. for 10" = 838, ^ Z cot 61° 41' = 9-7314436, diff. for 10" = 504. 174 ELEMENTARY TRIGONOMETRY. [chap. 191. To solve a triangle having given two angles and a side. Let the given parts be denoted by B, C, a ; then the third angle A is found from the equation A — 180° — B — G^ a&mB and 5= sin J. ' .'. log5=loga-f-logsin^-logsin J.; whence h may be found. Similarly, c may be obtained from the equation log c= log a -{-log sin (7— log sin ^. Example. If 6 = 1000, ^=45°, C = 68° 17' 40", find the least side, having given log 2 = -3010300, log 7-6986 = -8864118, di£f. for 1 = 57, L sin 66° 42'= 9-9630538, diff. for 1'= 544. B = 180° - 45° - 68° 17' 40" = 66° 42' 20". & sin A 1000 sin 45° The least side=a= sinB sin 66° 42' 20"' .-. log a= 3 + log -75 - log sin 66° 42' 20' = 3 - ^ log 2 - log sin 66° 42' 20" = 3 --1135869 log sin 66° 42'= 1-9630538 20 60 X 544 = 181 5 log 2= -1505150 •1135869 /. log a =2-8864131 log 769-86 = 2-8864118 diff. 13 13 .-. prop^ increase=^=*22. Thus the least side is 769-8622. EXAMPLES. XVI. d. 1. If 5=60° 15', (7=54° 30', a=100, find c ; given L sin 54° 30' = 9-9106860, log 8-9646162 = '9525317, Z sin 65° 15' = 9-9581 543. XVI.] SOLUTION OF TRIANGLES WITH LOGARITHMS. 1*75 2. If ^ = 55°, ^ = 65°, c = 270, find a ; given log 2, log 3, log 25538 =4-4071869, L sin 55° = 9-9133645, log 25539 = 4-4072039. 3. If ^=45° 41', (7=62° 5', 6 = 100, find c ; given log 9-2788 = -96749, L sin 62° 5' = 9-94627, X sin 72° 14' = 9-97878. 4. If i?=70°30', (7=78° 10', a = 102, find 6 and c ; given log 2= -301, log 1-02 = -009, log 1-85 = -267, log 1-92 = -283, L sin 70° 30' = 9-974, L sin 78° 10' = 9-990, X sin 31° 20' = 9-716. 5. If « = 123, ^ = 29° 17', (7=135°, find c ; given log2, log 123 = 2-0899051, Z sin 15° 43' = 9*4327777, log 3211 =3-5066403, i)=135. 6. If ^ = 44°, (7=70°, 6 = 1006-62, find a and c ; given Z sin 44° = 9-8417713, log 100662 =5-0028656, Z sin 66° = 9-9607302, log 103543 =5-0151212, Z sin 70° = 9-9729858, log 7654321 = 6-8839067. 7. Ifa=1652, 5=26°30', (7=47° 15', find6andc; Z sin 73° 45' = 9-9822938, log 1-652 =-2180100, Z sin 26° 30' = 9-6495274, log 7-6780 = -8852481, D = 57, Zsin 47° 15' = 9-8658868, log 1-2636 = -1016096, i) = 344. 192. To solve a triangle when two sides and the angle opposite to one of them are given. Let a, 6, A be given. Then from sin ^=- sin A^ we have log sin 5 = log 6 — log a -I- log sin ^ ; whence B may be found ; then C is fovmd from the equation C= 180° — A — B. . . a sin (7 Agam, c=—. — -.-, .-. log c = log a + log sin (7 -log sin ^. If a<6, and A is acute the solution is ambiguous and there will be two values of B supplementary to each other, and also two values of C and c. [Art. 147.] 176 ELEMENTARY TRIGONOMETRY. [chap. Examjple. If & = 63, c = 36, C=29° 23' 15", find B ; given log 2 = -3010300, log 7 = -8450980. L sin 29° 23'= 9-6907721, diff. for 1' = 2243, L sin 59° 10'= 9-9338222, difE. for l' = 755. . ^ b . ^ 63 . sin B =- sin (7 =777^ sm (7 c Ob =^ sin 29° 23' 15"; .-. logsinJB=log7-21og2 + log sin 29° 23' 15"; .-. log sin 5 = 1-9338662 log sin 59° 10' = 1-9338222 diff. 440 440 .*. prop^ increase ==^ X 60" = 85"; .-. J5 = 59°10'35". Also since c < 6 there is another value of B supplementary to the above, namely JB = 120° 49' 25". log sin 29° 23' = 1-6907721 15 60 x2243 = 561 log 7= -8450980 •5359262 2 log 2 = -6020600 1-9338662 440 60 755 ) 26400 ( 35 2265 3750 3775 EXAMPLES. XVI. e. 1. If a = 145, 6 = 178, B=4V 10', find ^; given log 178 = 2-2504200, L sin 41° 10' = 9-8183919, log 145 = 2-1613680, L sin 32° 25' 35" = 9-7293399. 2. If ^ = 26° 26', 6 = 127, a = 85, find B ; given log 1-27 = '1038037, Xsin 26° 26' = 9-6485124, log 8-5 =-9294189, Z sin 41° 41' 28" = 9*8228972. 3. If c= 5, 6 = 4, C=45°, find ^ and ^ ; given log 2, Z sin 34° 26' = 9-7525750, Z sin 34° 27' = 9-7525761. 4. If a = 1 405, b = 1 706, A = 40°, find B ; given log 1-405 = -1476763, log 1706 = -2319790, Z sin 40"° = 9-8080675, Z sin 51° 18' =9 "8923342, diff. for 1'= 1012. xvl] solution of triangles with logarithms. 177 5. If ^=112° 4', 6 = 573, c=394, find A and C; given log 573 = 2-7581546, log 394 = 2'5954962, L sin 39° 35' = 9-8042757, diff. for 60" = 1527, Z cos 22° 4' = 9-9669614. 6. If 6 = 8-4, c=12, ^ = 37° 36', find A ; given log 7 = -8450980, L sin 37° 36' = 9-7854332, X sin 60° 39' = 9-9403381, difi". for 1' = 711. 7. Supposing the data for the solution of a triangle to be as in the three following cases, point out whether the solution will be ambiguous or not, and find the third side in the obtuse angled triangle in the ambiguous case : (i) ^ = 30°, a= 125 feet, c = 250feet, (ii) ^ = 30°, a = 200 feet, c = 250 feet, (iii) ^ = 30°, a = 200 feet, c = 125 feet. Given log 2, log 6-0389 = -7809578, L sin 38° 41' = 9-7958800, log 6-0390 = -7809650, Zsin8° 41' = 9 1789001. 193. Some formulae which are not primarily suitable for working with logarithms may be adapted to such work by various artifices. 194. To adapt the formula & — ci^-\- IP- to logarithnic compu- tation. We have (-S)- Since an angle can always be found whose tangent is equal to a given numerical quantity, we may put -=tan^, and thus obtain c2 = (x2 ( 1 + tan^ 6) = cjfi sec^ B ; . - . c = a sec 6 ; .'. log c = log a -f log sec ^. The angle 6 is called a subsidiary angle and is found from the equation log tan 6 = log b - log a. Thus ani/ expression %ohich can be put into the form of the siom of two squares can be faadily adapted to logarithmic ivorJc. H. K. E. T. 12 178 ELEME^TTARY TRIGONOMETRY. [CHAP. 195. To adapt the formula G^=a'^-\-h^ — ^ab con G to log- arith iiiic computation. From the identities C . G G G cos G= cos^ — — sin^ — , and 1 = cos^ — -f siu^ - , 2 2 2 2 we have c2 = (a2 + 52) ^cos2 ^' + sin2 ^\ - 2ab ("cos^ | - sin^ ^\ = (a2 + 62 - 2a 6) cos2 1' + (a2 ^. 52 + 2«6) sin2 ^ (7 (7 = (a — 6)2 cos2 -^ + (« + &)^ sin2 — Take a subsidiary angle $, such that . . a+b, G tan ^ = r tan — , a — b 2 then c2 = {a- bf cos2 ^ (1 + tan2 6) G = {d- by cos2 — sec2 ; G ,-. c = (a— 6) cos — sec^ ; G .'. log c = log (a - 6)4- log cos — + log sec ^, where 6 is determined from the equation G log tan Q = log ( Since 2\/ab is never greater than a+b and cos— is positive Ji and less than unity, cos^ is positive and less than unity, and thus 6 is an acute angle. EXAMPLES. XVI. f. 1. If a = 8, 6 = 7, c = 9, find the angles ; given log 2, log 3, L tan 24° 5' = 9-6502809, diff. for 60" = 3390, X tan 36° 41' = 9-8721 123, diff. for 60" = 2637. 2. The difference between the angles at the base of a tri- angle is 24°, and the sides opposite these angles are 175 and 337 : find all the angles ; given log 2, log 3, X tan 12° =9-3274745, Z cot 56° 6' 27" -9*8272293. xvl] solution op triangles with logarithms. 181 3. One of the sides of a right-angled triangle is two-sevenths of the hypotenuse : find the greater of the two acute angles ; given log 2, log 7, Zsin 14° 11' = 9*455921, Zsin 14° 12' = 9-456031. 4. Find the greatest side when two of the angles are 78° 14' and 71° 24' and the sides joining them is 2183 ; given log 2-183 = -3390537, log 4-2274= -6260733, i)=103, L sin 78° 14' = 9-9907766, L sin 30° 22' = 9-7037486. 5. If 6 = 2 ft. 6 in., c = 2 ft., ^ = 22° 20', find the other angles ; and then shew that the side a is very approximately 1 foot. Given log 2, log 3, L cot 1 1° 10' = 10-70465, L sin 49° 27' 34" = 9-88079, L sin 22° 20' = 9-57977, L tan 29° 22' 26" = 9-75041. 6. If a = 1-56234, & = -43766, C=58° 42' 6", find ^ and 5; given log 56234=4-75, log cot 29° 21' = -250015, log cot 29° 22'= '249715. 7. If a = 9, Z)=12, ^ = 30°, find the values of c, having given log 12 = 1-07918, Zsin 30° =9-69897, log 9= -95424, Zsin 11° 48' 39" = 9-31108, log 171 = 2-23301, Z sin 41° 48' 39" = 9-82391, log 368 = 2-56635, Z sin 108° 11' 21" = 9-97774. 8. The sides of a triangle are 9 and 3, and the difierence of the angles opposite to them is 90° : find the angles ; having given log 2, Z tan 26° 33' = 9-6986847, Z tan 26° 34' = 9-6990006. 9. Two sides of a triangle are 1404 and 960 respectively, and an angle opposite to one of them is 32° 15': find the angle contained by the two sides ; having given log 2, log 3, log 13 = 1-1139434, Zcosec 32° 15' = 10-2727724, Zsin 21° 23' = 9-5621316, Zsin 51° 18'= 9-8923236. 10. If 6 : c=ll : 10 and ^ = 35° 25', use the formula tani(Z-C) = tan2|cot^ to find ^ and (7; given log 1-1= -041393, Z tan 12° 18' 36"= 9-338891, Z cos 24° 37' 12" = 9-958607, Z cot 17° 42' 30" = 10-495800, Z tan 8° 28' 56-5" = 9-173582. 182 ELEMENTARY TRIGONOMETRY. [CHAP. 11. 1{A= 50°, b = 1071, a = 873, find B ; given log 1 -071 = -029789, log 8-73 = '941014, X sin 50° = 9-884254, X sin 70° =9-972986, Z sin 70° 1' = 9-973032. 12. If a = 3, 6 = 1, (7=53° T 48", find c without determining A and B ; given log 2, log 25298 = 4-4030862, L cos 26° 33' 54" = 9-9515452, log 25299=4-4031034, Z tan 26° 33' 54" = 9-6989700. {In the following Examples the necessary Logarithms must he taken from the Tables.) 13. Given a = 1000, b = 840, c = 1258, find B. 14. Solve the triangle in which a = 525, 6=650, c=777. 15. Find the least angle when the sides are proportional to 4, 5, and 6. 16. UB=90°, ^C= 57-321, AB = 28-58, find A and C. 17. Find the hypotenuse of a right-angled triangle in which the smallest angle is 18° 37' 29" and the side opposite to it is 284 feet. 18. The sides of a triangle are 9 and 7 and the angle between them is 60° : find the other angles. 19. How long must a ladder be so that when inclined to the gromid at an angle of 72° 15' it may just reach a window 42-37 feet from the grovmd ? 20. If a = 31 -95, 6=21-96, (7=35°, find A and B. 21. Find B, C, a when 6 = 25-12, c = 13-83, .1 = 47° 15". 22. Find the greatest angle of the triangle whose sides are 1837-2, 2385-6, 2173-84. 23. When a = 21-352, 6 = 45*6843, c=37-2134, find A, B, and C. 24. If 6=647-324, c=850-273, ^ = 103° 12' 54", find the re- maining parts. 25. If 6=23-2783, ^ = 37° 57', ^=43° 13', find the remaining sides. 26. Find a and 6 when 5=72" 43' 25", C=47° 12' 17", c=2484-,3. XVI.] SOLUTION OP TRIANGLES WITH LOGARITHMS. 183 27. If AB=4ol7, .46'= 150, ^ = 31° 30", find the remaining parts. 28. Find A, B, and b when a = 324-68, c = 421'73, (7=35° 17' 12". 29. Given a=321-V, c = 435-6, yl = 36° 18' 27", find C. 30. If 6=1625, c=1665, B=o2' 19', solve the obtuse-angled triangle to which the data belong. 31. If a = 3795, B=T3° 15' 15", C'=42° 18' 30", find the other sides. 32. Find the angles of the two triangles which have 6=17, c=12, and (7=43° 12' 12". 33. Two sides of a triangle are 2*7402 ft. and "7401 ft. respectively, and contain an angle 59° 27' 5" : find the base and altitude of the triangle. 34. The difierence between the angles at the base of a triangle is 17° 48' and the sides subtending these angles are 105-25 ft. and 76*75 ft.: find the angle included by the given sides. 35. From the following data : (1) .4 = 43° 15', .4.B = 36-5, BC=20, (2) .4 = 43° 15', .45 = 36-5, BC=30, (3) .4=43° 15', .45 = 36-5, 5(7=45, point out which solution is impossible and which ambiguous. Find the third side for the triangle the solution of which is neither impossible nor ambiguous. 36. In any triangle prove that c = (a - b) sec 6, where , , 2a/^ . (7 tan 6 = ,- sm — . a — o z If a = 17-32, 6 = 13-47, (7=47° 13', find c without finding A and B. a + b C C 37. If tan = — . tan — , prove that c={a — b) cos — sec ^. If a = 27-3, 6 = 16-8, (7=45° 12', find 6, and thence find c. CHAPTER XVII. HEIGHTS AND DISTANCES. 198. Some easy cases of heights and distances depending only on the solution of right-angled triangles have been already dealt with in Chap. VI. The problems in the present chapter are of a more general character, and require for their solution some geometrical skill as well as a ready use of trigonometrical formulae. Measurements in one plane. 199. To find the height and distance of an inaccessible object on a horizontal plane. Let A be the position of the observer, CPthe object ; from Pdraw PC perpendicular to the horizontal plane; then it is required to find PC and AC. At A observe the angle of eleva- ^^^ f.p tion PAC. Measure a base line AB ^ in a direct line from A towards the object, and at B observe the angle of elevation PBG. Let LPAC=a, lPBC=^, AB = a. From lPBC, PC =PB sin ^. From hPAB, j^_AB sin PAB _ a sin a sin APB ~sin(^-a)' . • . PC= a sin a sin /3 cosec (/3 - a). Also A C= PC cota = a cos a sin /3 cosec (/3 - a). Each of the above expressions is adapted to logarithmic work ; thus if PC=x, we have log x=\oga + log sin a + log sin j3 + log cosec {^-a). MEASUREMENTS IN ONE PLANE. 185 Note. Unless the contrary is stated, it will be supposed that the observer's height is disregarded, and that the angles of elevation are measured from the ground. Example I. A person walking along a straight road observes that at two consecutive milestones the angles of elevation of a hill in front of him are 30° and 75°: find the height of the hill. • In the adjoining figure, lPAC=SO°, APBC = l5°,AB = lmile; lAPB = 75°-S0°=4:5°. Let X be the height in yards ; then a; = P^ sin 75°; ^ ^ „„ AB sin PAB 1760 sin 30° but PB= — , — = ■ .-o ; sm APB sm 45° _ 1760 sin 30° sin 75° *"' ^ ~ sin 45° = 1760xix,/2x-^-?±i = 440(V3 + 1). If we take (,^3 = 1-732 and reduce to feet, we find that the height is 3606-24 ft. EXAMPLES. XVII. a. 1. From the top of a cliflf 200 ft. above the sea-level the angles of depression of two boats in the same vertical plane as the observer are 45° and 30° : find their distance apart. 2. A person observes the elevation of a mountain top to be 15°, and after walking a mile directly towards it on level ground the elevation is 75° : find the height of the mountain in feet. 3. From a ship at sea the angle subtended by two forts A and B is 30°. The ship sails 4 miles towards A and the angle is then 48° : prove that the distance of B at the second observa- tion is 6*472 miles. 4. From the top of a tower A ft. high the angles of depression of two objects on the horizontal plane and in a line passing through the foot of the tower are 45° — J. and 45° +^4. Shew that the distance between them is 2 A tan 2A. 186 • ELEMENTARY TRIGONOMETRY. [chap. 5. An observer finds that the angular elevation of a tower is A. On advancing a feet towards the tower the elevation is 45° and on advancing 6 feet nearer the eleYation is 90° — A : find the height of the tower. 6. A person observes that two objects A and B bear due N. and N. 30° W. respectively. On walking a mile in the direction N.W. he finds that the bearings of A and B are N.E. and due E. respectively : find the distance between A and B. 7. A tower stands at the foot of a hill whose inclination to the horizon is 9° ; from a point 40 ft. up the hill the tower sub- tends an angle of 54° : find its height. 8. At a point on a level plane a tower subtends an angle a and a flagstaff c ft. in length at the toj) of the tower subtends an angle /3 : shew that the height of the tower is c sin a cosec ^3 cos (a + /3). Example II. The upper three-fourths of a ship's mast subtends at a point on the deck an angle whose tangent is -6 ; find the tangent of the angle subtended by the whole mast at the same point. Let C be the point of observation, and let AFB he the mast, AF being the lower fourth of it. Let .45 = 4a, so that AF = a\ also let ^(7 = &, lACB=^Q, lBCP=p, so that tanj3=-6. From A PGA, from aBCA, tan (6-^) = tan0 = tan^ = 4tan(^-/S) = 3 b' 4a T' c 4 (tan e - tan /S) _ tan^ = On reduction, whence 4 (tan d- 1 + - tan ^ 5 tan20- Stan ^ + 4 = 0; tan = 1 or 4. 1 + tan 6 tan j3 4 (5 tan O-S) 5 + 3 tan Note. The student should observe that in examples of this class we make use of right-angled triangles in which the horizontal base line forms one side. XVII.] MEASUREMENTS IN ONE PLANE. 187 Example III. -A- tower BCD surmounted by a spire DE stands on a horizontal plane. From the extremity J. of a horizontal line BA, it is found that BC and DE subtend equal angles. If J5C=9 ft., CD = 72ft., andD£ = 36ft., find^^. Let lBAC= lDAE = d, lDAB = a, AB = xit. Now 1^(7=9 ft., BD = 81 ft., BE = m ft. BE _irj AB ~~x ^ . ^, BE 117 But tan a BD ~ AB 81 ~ X ' tan^ BC ~AB _9 X ' tan ( a+e)= tan a + tan^ ' 1 - tan a tan d ' 117 _ X ~ 81 — + X 9 X _90 ~ X a;2 X 9 * .r ' .r2-81x9 117a;2-81x9xll7 = 90a;2; .-. 27a;2 = 81x9xll7; .-. a;2 = 81x39; .'. a; = 9^39. But ^39 = 6-245 nearly; .-. a- = 56-205 nearly. Thus AB = 56-2 ft. nearly. 9. A flagstaff 20 ft. long standing on a wall 10 ft. high subtends an angle whose tangent is -5 at a point on the ground : find the tangent of the angle subtended by the wall at this point. 10. A statue standing on the top of a pillar 25 feet high subtends an angle whose tangent is '125 at a point 60 feet from the foot of the pillar : find the height of the statue. 11. A tower BCD surmounted by a spire DE stands on a horizontal plane. From the extremity ^ of a horizontal line BA it is found that BC and BI) subtend equal angles. If 5(7=9 ft., (7i)=280ft., and BE=35ft., prove that BA = 180 ft. nearly. 188 ELEMEXTARY TRIGONOMETRY. [chap. 12. On the bank of a river there is a column 192 ft. high supporting a statue 24 ft. high. At a point on the opposite bank directly facing the column the statue subtends the same angle as a man 6 ft. high standing at the base of the column : find the breadth of the river. 13. A monument ABODE stands on level ground. At a point P on the ground the portions AB, AC^ AD subtend angles a, /3, 7 respectively. Supposing that AB^a^ AC=b, AD=c, AP = x, and a+/3+y=180°, shew that {a^-h + c) x'^ = ahc. Example IV. The altitude of a rock is observed to be 47°; after walking 1000 ft. towards it up a slope inclined at 32° to the horizon the altitude is 77°. Find the vertical height of the rock above the first point of observation, given sin 47° = •731. Let P be the top of the rock, A and B the points of observation; then in the figure I PAG =47°, iBAC=S2°, lPDG= I PEE = 11°, ^5 = 1000 ft. Let X ft. be the height ; then x = PA sin P^ C = P^ sin 47°. We have therefore to find PA in terms of AB. In aP^P, zP^P = 47° -32° = 15°; Z^PP = 77° -47° = 30°; .-. Z^PP= 135°: PA = AB sin ABP smAPB 1000 sin 135° sin 30° = 1000^2; .-. x = PA sin 47° = 1000 J2 x -731 = 731^2. If we take /^2 = 1 '414, we find that the height is 1034 ft. nearly. 14. From a point on the horizontal plane, the elevation of the top of a hill is 45°. After walking 500 yards towards its summit up a slope inclined at an angle of 15° to the horizon the elevation is 75* : find the heisrht of the hill in feet. XVII.] PROBLEMS DEPENDENT ON GEOMETRY. 189 15. From a station B at the base of a mountain its summit A is seen at an elevation of 60° ; after walking one mile towards the summit up a plane making an angle of 30° with the horizon to another station 0, the angle BCA is observed to be 135° : find the height of the mountain in feet. 16. The elevation of the summit of a hill from a station A is a. After walking c feet towards the summit up a slope inclined at an angle /3 to the horizon the elevation is y : shew that the height of the hill is c sin a sin (y — /3) cosec (y — a) feet. 17. From a point A an observer finds that the elevation of Ben Nevis is 60° ; he then walks 800 ft. on a level plane towards the summit and then 800 ft. further up a slope of 30° and finds the elevation to be 75' : shew that the height of Ben Nevis above A is 4478 ft. approximately. 200. In many of the problems which follow, the solution depends upon the knowledge of some geometrical proposition. Example I. A tower stands on a horizontal plane. From a mound 14 ft. above the plane and at a horizontal distance of 48 ft. from the tower an observer notices a loophole, and finds that the portions of the tower above and below the loophole subtend equal angles. If the height of the loophole is 30 ft., find the height of the tower. Let ^J5 be the tower, C the point of observation, L the loophole. Draw CD vertical and CE horizontal. Let AB = x. We have Ci) = 14, AD = EG = ^^, BE = x-U. From A ADC, ^(7-'= (14)2 + (48)2 =2500; .-. AC = 50. From A CEB, CB^= {x - 14)2+ (48)2 = a;2- 28a; + 2500. Now lBCL= lACL; , , ^ ^ BC BL hence by Euc. vi. 3, j^ = ji ; Jx'^ - 28.r + 250 J _ a; - 30 ■ ■ 50 30~ • By squaring, 9 (a:^ - 28a; + 2500) = 25 {x'^ - 60a; + 900). On reduction, we obtain 16a;2- 1248.r; =0; whence a; = 78. Thus the tower is 78 ft. high. 190 ELEMENTARY TRIGONOMETRY. [chap. EXAMPLES. XVILb. 1. At one side of a road is a flagstaff 25 ft. high fixed on the top of a wall 15 ft. high. On the other side of the road at a point on the ground directly opposite the flagstaff and wall sub- tend equal angles : find the width of the road. 2. A statue a feet high stands on a column 3a feet high. To an observer on a level with the top of the statue, the column and statue subtend equal angles : find the distance of the observer from the top of the statue. 3. A flagstaff a feet high placed on the top of a tower h feet high subtends the same angle as the tower to an observer h feet high standing on the horizontal plane at a distance d feet from the foot of the tower : shew that (a-6)c^2 = (a + 6)62_262^-(a-6)A2, Example II. A flagstaff is fixed on the top of a wall standing upon a horizontal plane. An observer finds that the angles sub- tended at a point on this plane by the wall and the flagstaff are a and /3. He then walks a distance c directly towards the wall and finds that the flagstaff again subtends an angle /S. Find the heights of the wall and flagstaff. Let ED be the wall, T>C the flagstaff, A and B the points of observation. Then z CAD = ^= L CBB, so that the four points C, A,B, D are concyelic. .-. ^jBD = 8uppt. of I ACE = 90° + {a + p), from a CAE. Hence in a ADB, I ADB = 180° -a- {90° + (a + ^)} = 90°-(2a+^). AB sin ABD _ c cos (a + /3) sin ADB ~ cos (2a + 'j3) ' Hence in aADE, T^ rr AT^ • c sin a cos (a + /3) DE = AD sm a = .^ ^ „, . cos(2a + /3) And in a CAD, ^_ADsinGAD _ ^D sin/3 _ csin/S " sin A CD " cos (a + |S) "" cos (2a + j8) * XVII.] PROBLEMS DEPENDENT ON GEOMETRY. 191 4. A tower standing on a cliff subtends an angle a at each of two stations in the same horizontal line passing through the base of the cliff and at distances of a feet and b feet from the cliff. Prove that the height of the tower is (a +6) tan a feet. 5. A column placed on a pedestal 20 feet high subtends an angle of 45° at a point on the ground, and it also subtends an angle of 45° at a point which is 20 feet nearer the pedestal : find the height of the column. 6. A flagstaff on a tower subtends the same angle at each of two places A and B on the ground. The elevations of the top of the flagstaff as seen from A and B are a and /3 respectively. If AB = a, shew that the length of the flagstaff is a sin (a + /3 - 90°) cosec (a - j3). Example III. A man walking towards a tower AB on which a flagstaff BG is fixed observes that when he is at a point E, distant c ft. from the tower, the fiagstaff subtends its greatest angle. If jLBEG=a, prove that the heights of the tower and flagstaff are c tan (i-i) and 2c tan a ft. respectively. Since E is the point in the horizontal line AE at which BG subtends a maximum angle, it can easily be proved that AE touches the circle passing round the tri- angle GBE. [See Hall and Stevens' Euclid, p. 242.] The centre D of this circle lies in the vertical line through E. Draw DF per- pendicular to BG, then DF bisects BC and also iGDB. By Euc. III. 20, lGDB = 2iGEB = 2a; .-. IGDF= lBDF=a. GB = 2GF=2DF tan a = 2c tan a. Again, lAEB= I EG Bin alternate segment = - Z EDB at centre _1 /7r_ ~2 V2 .*. ^li = c tan ^^^ = c tan IT a\ 4 ~ 2J 192 ELEMENTARY TRIGONOMETRY. [CHAP. 7. A pillar stands on a pedestal. At a distance of 60 feet from the base of the pedestal the pillar subtends its greatest angle 30° : shew that the length of the pillar is 40v'3 feet, and that the pedestal also subtends 30° at the point of observation. 8. A person walking along a canal observes that two objects are in the same line which is inclined at an angle a to the canal. He walks a distance c further and observes that the objects subtend their greatest angle /3 : shew that their distance apart is 2c sin a sin ^ I (cos a + cos j3). 9. A tower with a flagstaff stands on a horizontal plane. Shew that the distances from the base at which the flagstaff subtends the same angle and that at which it subtends the greatest possible angle are in geometrical progression. 10. The line joining two stations A and B subtends equal angles at two other stations C and D : prove that AB sin CBD = CD sin ADB. 11. Two straight lines ABC, DEC meet at C. If L DAE= L DBE^a, and z EAB = ^, l EBG=y, shew that ^^^ /^sin^sin(a+/3) sm (y-^) sm (a+^+y) 12. Two objects P and Q subtend an angle of 30° at A. Lengths of 20 feet and 10 feet are measured from A at right angles to ^P and AQ respectively to points R and S at each of which PQ subtends angles of 30° : find the length of PQ. 13. A ship sailing N.E. is in a line with two beacons which are 5 miles apart, and of which one is due N. of the other. In 3 minutes and also in 21 minutes the beacons are found to subtend a right angle at the ship. Prove that the ship is sailing at the rate of 10 miles an hour, and that the beacons subtend their greatest angle at the ship at the end of 3^/7 minutes. 14. A man walking along a straight road notes when he is in the line of a long straight fence, and observes that 78 yards from this point the fence subtends an angle of 60°, and that 260 yards further on this angle is increased to 120°. When he has walked 260 yards still further, he finds that the fence again subtends an angle of 60°. If a be the angle which the direction of the fence makes with the road, shew that 13 sin a =5. Also shew that the middle point of the fence is 120 yards distant from the road. xvil] MEASUREMENTS IN MORE THAN ONE PLANE. 193 Measurements in more than one plane. 201. In Art. 199 the base line AB was measured directly toivards the object. If this is not possible we may proceed as follows. From A measm-e a base line AB in any convenient direction in the horizontal plane. At A observe the two angles PAB and PAC ; and at B observe the angle PBA. Let LPAB=a, lPAC=^, LPBA=y, AB = a, PC=x. From A P^ (7, x=PABm^. From A PAB, AB sin PBA PA a sm y sin^P^ sin(a+y)' x=a sin /3 sin y cosec (a + y). 202. To shew how to find the distance hetiveen tivo inaccessible objects. Let P and Q be the objects. Take any two convenient stations A and B in the same horizontal plane, and measm-e the distance between them. At A observe the angles PAQ and QAB. Also if AP, AQ, AB are not in the same plane, measure the angle PAB. At B observe the angles ABP and ABQ. In APABy we know lPAB, lPBA, and AB; so that AP may be found. In A QAB, we know L QAB, z. QBA, and AB ; so that A Q may be found. In APAQ, we know AP, AQ, and L PAQ ; so that PQ may be found. H. K. E. T. 13 194 ELEMENTARY TRIGONOMETRY. [chap. Exaviple 1. The angular elevation of a tower CD at a place A due South of it is 30^, and at a place B due West of A the elevation a is 18°. If AB = a, shew that the height of the tower is / ^ • W. ,-£i ^Af Let CI) = x. From the right-angled triangle DC A, AG=x cot 30°. From the right-angled triangle DCB, BG = x cot 18°. But z BAG is a right angle, .'. BG^-AG^ = a^; .: ^2(cot218°-cot2 30°) = a-; .-. a;2(cosec218°-cosec2 30°) = a2; 4 \2 1 ,/5-l .-. a-2{(^5+l)^-4}=a2; .-. x^2x2^5) = a^ which gives the height required. Exami^le 2. A hill of inclination 1 in 5 faces South. Shew that a road on it which takes a N.E. direction has an inclination 1 in 7. Let AD running East and West be the ridge of the hill, and let ABFD be a vertical plane through AD. Let C be a point at the foot of the hill, and ABC a section made by a vertical plane running North and South. Draw CG in a N.E. direction in the horizontal plane and let it meet BF in G ; draw GH parallel to BA ; then if CH is joined it will represent the direction of the road. XVII.] MEASUREMENTS IN MORE THAN ONE PLANE. 195 Since the inclination of CA is 1 in 5, we may take AB = a, and AG=5a, so that BC^ = 2ia\ .-^N.E Since CBG is a right-angled isosceles triangle. Hence in the right-angled triangle CGH, .-. CH=7a = 7GH. Thus the slope of the road is 1 in 7. EXAMPLES. XVII. c. 1. The elevation of a hill at a place F due East of it is 45°, and at a place Q due South of F the elevation is 30°, If the distance from P to Q is 500 yards, find the height of the hill in feet. 2. The elevation of a spire at a point A due West of it is 60°, and at point B due South of A the elevation is 30°. If the spire is 250 feet high, find the distance between A and B. 3. A river flows due North, and a tower stands on its left bank. From a point A up-stream and on the same bank as the tower the elevation of the tower is 60°, and from a point B just opposite on the other bank the elevation is 45°. If the tower is 360 feet high, find the breadth of the river. 4. The elevation of a steeple at a place A due S. of it is 45°, and at a place B due W. of A the elevation is 15°. If AB=2aj shew that the height of the steeple is a (3^ — 3~^). 13—2 196 ELEMENTARY TRIGONOMETRY. [CHAP. 5. A person due S. of a lightlaouse observes that his shadow cast by the light at the top is 24 feet long. On walking 100 yards due E. he finds his shadow to be 30 feet long. Supposing him to be 6 feet high, find the height of the light from the ground- 6. The angles of elevation of a balloon from two stations a mile apart and from a point halfway between them are observed to be 60°, 30°, and 45° respectively. Prove that the height of the balloon is 440^6 yards. [If AD is a median of the triangle ABC, then 2AD^+2BD^=AB^ + AG^ 7. At each end of a base of length 2a, the angular elevation of a mountain is 6, and at the middle point of the base the elevation is (f). Prove that the height of the mountain is a sin 6 sin cf) Vcosec (<^ + 6) cosec {

, JS, F the points of contact ; then IB, IE, IF are perpendicular to the sides. Now A = siim of the areas of the triangles BIC, CIA, AIB = -ar + - br-\--cr^-{a-\-h + c)r -^sr; whence r = - 208. To express the radius of the inscribed circle in terms of one side and the functions of the half-angles. In the figure of the previous article, we know from Euc. iv. 4 that / is the point of intersection of the lines bisecting the angles, so that o ri LlBD = -^. LlCB = 2' 2 ^i> = r cot|, CZ) = r cot?. 2 2 .'. r ('cot- + cot-j = a; . B^C . B . G . r sin — — - = a sm — sin — ; . B . C a sm — sm — Jt A r=> A cos- 202 ELEMENTARY TRIGONOMETRY. [chap. 209. Definition. A circle which touches one side of a triangle and the other two sides produced is said to be an escribed circle of the triangle. Thus the triangle ABC has three escribed circles, one touching BC, and AB, AC produced; a second touching CA, and BC^ BA produced ; a third touching AB^ and CA, CB produced. We shall assume that the student is familiar with the con- struction of the escribed circles. [See Hall and Stevens' Euclid, p. 255.] For shortness, we shall call the circle inscribed in a triangle the In-circle, its centre the In-centre, and its radius the In- radius ; and similarly the escribed circles may be called the Ex-circles, their centres the Ex-centres, and their radii the Eos-radii. 210. To find the radius of an escHhed circle of a triangle. Let /j be the centre of the circle touching the side BC and the two sides AB and AC produced. Let B^, E^, F^ be the points of contact ; then the lines joining I^ to these points are perpen- dicular to the sides. Let r^ be the radius ; then A = area ABC = area ABI^C- area BI^C = area BI-^A +area CI^A — area BIiC 1 Iz, 1 = 2^^1 + 2 ^i~ 2 '^''i = (s - a) 7-1 ; r.=- s — a Similarly, if r^, r^ be the radii of the escribed circles opposite to the angles B and C respectively, ?'o = s-h' ro = s — c XVIII.] RADII OF THE ESCRIBED CIRCLES. 203 211. To find the radii of the escribed circles in te7'ms of one side and the functions of the half -angles. In the figure of the last article, I^ is the point of intersection of the lines bisecting the angles B and C externally ; so that A A . • . BBj^ = ?'! cot ( 90° - -") - r^ tan - , Ci)i = ri cot fdO" - f ) =^1 tan ^; .-. T-iftan - + tan - j=a; . B+C B G .'. 7*1 sm — ;- — =acos— cos— ; B C a cos — cos — ••• '■,= 2 — . cos- Similarly, , C A A B cos — COS — C cos — COS — COS- • COS- 212. By substituting « = 2i2sin^, b = 2RsinB, c = 2RsmGy in the formulae of Art. 208 and Art. 211, we have .J, . A , B . G r^4it sm — sm — sin — , 2 2 2 ri = 4i2sm-cos-cos-. A . B G gsm-cos-. ^2 = 4-S cos — sin — cos AT? -^ ^ ■ ^ ?*o=4itcos— COS — sm -=. •* 2 2 2 204 ELEMENTARY TRIGONOMETRY. [CHAP. Example 1. Shew that -^ — - + -^-^ — = — . ■^ a b r^ The first side =i(^-^)+i(-^-^^ a\s-a sj o\s-o sj A A _ A{2s-a-b) s{s-a) s{s -b) s (s -a){s — b) Ac Ac{s-c) s{s-a){s-b) s{s-a){s- b) (s - c) Ac{s-c) _c{s- c) ' A2 "" A Example 2. If r^ = To + r^ + r, prove that the triangle is right- angled. By transposition, r^ -r = r^ + r^ ; ,^ . A B G ,^ . A . B . G .'. 4E sin — cos — cos 2" - 4E sm — sm — sin — ,„ A . B G .^ A B . G = 4JK cos — sm — cos — + 4 JR cos — cos — sm — ; Z Z Z li 2i 2i . A( B C . B . C .'. sm - (cos - cos - - sm - sm - A f . B G B = cos - ( sm - cos 2 + cos - sm -^ , . A B+G A . B+G .'. sm - cos — ^ = cos - sm ^^^ ; . I, A ty A ■•' sin-'-=cos--; whence ^ = 45°, and 4 = 90°. 213. Many important relations connecting a triangle and its circles may be established by elementary geometry. With the notation of previous articles, since tangents to a circle from the same point are equal, XVIII.] RELATIONS OP A TRIANGLE AND ITS CIRCLES. 205 we have AF=AE, BB=BF, CD = CE; . • . AF+ {BD + CD) = half the sum of the sides .-. AF-\-a = s. .'. AF==s-a=AK Similarly, BB = BF= s-b, CB= CE= s-c. A Also A A r = ^i^ tan — = (5 - a) tan — B C Similarly, r={s-b) tan — , r={s~c) tan — . A A Again, AF^=^AE^, BF^ = BD^, CE^ = CD^', .'. 2AF^ = AFj^ + AEi={AB-{-BI){) + {AC+CDj) = sum of the sides ; .-. AF^=s=AE^. .-. BD^=BF^ = s-c, CD^ = CE^==s-b. A A Also T-i ■= A F-i tan ^ = s tan — . 2 2 Similarly, ^2 = ^ *^^ "o 5 ^'s = ^ ^^^ "o • 206 ELEMENTARY TRIGONOMETRY. [cHAP. EXAMPLES. XVIILa. 1. Two sides of a triangle are 300 ft. and 120 ft., and the included angle is 150°; find the area. 2. Find the area of the triangle whose sides are 171, 204, 195. 3. Find the sine of the greatest angle of a triangle whose sides are 70, 147, and 119. 4. If the sides of a triangle are 39, 40, 25, find the lengths of the three perpendiculars from the angular points on the opposite sides. 5. One side of a triangle is 30 ft. and the adjacent angles are 22^° and 112^°, find the area. 6. Find the area of a parallelogram two of whose adjacent sides are 42 and 32 ft., and include an angle of 30°. 7. The area of a rhombus is 648 sq. yds. and one of the angles is 150° : find the length of each side. 8. In a triangle if a = 13, 6 = 14, c=15, find r and R. 9. Find 7\, r^, r^ in the case of a triangle whose sides are 17, 10, 21. 10. If the area of a triangle is 96, and the radii of the escribed circles are 8, 12, 24, find the sides. Prove the following formulse A 11. slrr^r.^r^ = A. 12. s{s-a) tan — = A. 13. rrjcot — = A. 14. 4:Rrs = ahc. B G 15. ^i^2^3 = rs^. 16. r cot — cot „ = ^i • 17. Rr (sin A + sin B + sin C) = A. ^4 18. r-^r^-\-rr^ = ah. 19. cos — ^/6c (s - 6) (s - c) = A. A n 20. ri+r2=ccot2- 21. {r^-T){r^-\'r^^a?: XVIII.] RELATIONS OF A TRIANGLE AND ITS CIRCLES. 207 22. ,, cot I =., cot |=.3 cot :^=.cot i cot f cot i ^2^2-^2 222 23. ^ + ^ + ^=^. 24. r,T,+T,i,+i,T,=s'. ^1 ^2 J^3 r 25. ri+r2 + r3-r = 4R. 26. r+ri + r2-r3 = 4)XcosC. 27. b^ sill 2 6'+ c2 gin 25 = 4A. C A — B 28. 4i^cos- = (a + 6)sec— — -. 29. a2_52_27^csin(^4_^)_ -- B?-V- sin A sin B . 2 sm (A - B) 31. If the perpendiculars from A^ B, C to the opposite sides are jt?i, jOg, p^ respectively, prove that /-.N 1 1 1 1 /ON 1111 (1) - + - + -=-; 2) - + =-. Pi P2 Pz ^ Pi P2 Pz '^3 Prove the following identities : 32. {r^-T){r.,^-r){r^-r)^'iRr^. 1 1\ /I 1\ /I 1\ 472 33. r r,/ \r 7*0 / \r r-i r^s /}«2c>2 34. 4A(cot^+cot5 + cot(7) = a2 + &Hc2. 35. ^%^-^^ + «^ = 0. ^1 ^'2 ^3 36. a22,2c2(gin2.4+siu2i? + sin2(7) = 32Al 37. acosA+bcosB+ccosC=4RsinAsinBsinC. 38. acotA+bcotB+ccotC=2(R+r). ABC 39. (6+c) tan^ + (c+a) tan — + ((X + 6)tan — ^ .2 ^ = 4^ (cos J. + cos 5 + cos (7). 40. r (sin J. + sin 5 + sin C) = IR sin .4 sin B sin (7. 41. C0S2 — +C0s2-+C0s2- = 2 + — 7i. 2 2 2 2/t 208 ELEMENTARY TRIGONOMETRY. [chap. Inscribed and circumscribed Polygons. 214. To find the perimeter and area of a regular polygon of n sides inscribed in a circle. Let r be the radius of the circle, and AB a side of the polygon. Join OA^ OB, and draw OD bisecting lAOB; then AB is bisected at right angles in D. And L A OB = - (four right angles) n 277 n A^ -^B Perimeter of polygon = nAB = '2.nAD = '2nOA sin A OD :27zrsin Area of polygon = ?i (area of triangle ^1 OB) 1 , . 27r = - ?iH sm • — . 2 n 215. To find the perimeter and area of a regular polygon of n sides circumscribed about a given cw'cle. Let r be the radius of the circle, and AB a side of the polygon. Let AB touch the circle at B. Join OA, OB, OB ; then OB bisects AB at right angles, and also bisects lAOB. Perimeter of j)olygon = nAB = ^nAD = ^nOD tan A OD = 2?ir tan - . « n ^ Area of polygon = n (area of triangle A OB) = nOD.AD =nr^ tan - . n XVIII.] INSCRIBED AND CIRCUMSCRIBED POLYGONS. 209 216. There is no need to burden the memory with the formulge of the last two articles, as in any particular instance they are very readily obtained. Exaviple 1. The side of a regular dodecagon is 2 ft., find the radius of the circumscribed circle. Let r be the required radius. In the adjoining figure we have AB = 2, aAOB~. AB = 2AD = 2rsm^; .-. 2r sin 15°:= 2; sm Thus the radius is ^JQ + f^'2 feet. Example 2. A regular pentagon and a regular decagon have the same perimeter, prove that their areas are as 2 to ^/5. Let AB be one of the n sides of a regular polygon, the centre of the circumscribed circle, OD perpendicular to AB. Then UAB = a, area of polygon =?i.4D . OD = HAD . AD cot - n na^ , IT — ~-r~ cot — . 4 n Denote the perimeter of the pentagon and decagon by 10c. Then each side of the pen- tagon is 2c, and its area is 5c^ cot — . o Each side of the decagon is c, and its area is -c^cot r^ • Area of pentagon 2 cot 36° 2 cos 36° sin 18° _ 2 cos 36" Area of decagon cot 18° 2 cos 36° 1 + cos 36° 2(^5 + 1) _ 5 + ^0 sin 36° cos 18" 2 2cosn8" sio + r (^-^-^) H. K. E. T. 14 210 ELEMENTARY TRIGONOMETRY. [chap. 217. To find the area of a circle. Let r be the radius of the circle, and let a regular polygon of n sides be described about it. Then from the ad- joining figure, we have area of polygon = n (area of triangle A OB) = n(^AB.OD = \0D.nAB A T = - X perimeter of polygon. A By increasing the number of sides without limit, the area and the perimeter of the polygon may be made to differ as little as we please from the area and the circumference of the circle. Hence r area of a circle = - x circumference =^ X 27rr [Art. 59.] = 77?"=. 218. To find the area of the sector of a circle. Let 6 be the circular measure of the angle of the sector ; then by Euc. vi. 33, area of sector 6 area of circle Stt ' . • . area of sector = ^r- x irr'^ = ^ r^B. An A EXAMPLES. XVIII. b. I7i this Exercise take 7r=-p^ . 1. Find the area of a regular decagon inscribed in a circle whose radius is 3 feet ; given sin 36° = '588. XVIII.] INSCRIBED AND CIRCUMSCRIBED POLYGONS. 211 2. Find the perimeter and area of a regular quindecagon described about a circle whose diameter is 3 yards ; given tan 12° = -213. 3. Shew that the areas of the inscribed and circumscribed circles of a regular hexagon are in the ratio of 3 to 4. 4. Find the area of a circle inscribed in a regular pentagon whose area is 250 sq. ft.; given cot 36° = 1*376. 5. Find the perimeter of a regular octagon inscribed in a circle whose area is 1386 sq. inches ; given sin 22° 30' = '382. 6. Find the perimeter of a regular pentagon described about a circle whose area is 616 sq. ft.; given tan 36° = '727. 7. Find the diameter of the circle circumscribing a regular quindecagon, whose inscribed circle has an area of 2464 sq. ft. ; given sec 12° = 1-022. 8. Find the area of a regular dodecagon inscribed in a circle whose regular inscribed pentagon has an area of 50 sq. ft. 9. A regular pentagon and a regular decagon have the same area, prove that the ratio of their perimeters is J'2, : 4/5. 10. Two regular polygons of n sides and 2^1 sides have the same perimeter ; shew that the ratio of their areas is 2 cos - : 1 + cos - . n n 11. If 2a be the side of a regular polygon of n sides, R and r the radii of the circumscribed and inscribed circles, prove that R + r=acot^r-. 12. Prove that the square of the side of a regular pentagon inscribed in a circle is equal to the sum of the squares of the sides of a regular hexagon and decagon inscribed in the same circle. 13. With reference to a given circle, A-^ and B^ are the areas of the inscribed and circumscribed regular polygons of n sides, A^ and B^, are corresponding quantities for regular polygons of 27i sides : prove that (1) A^ is a geometric mean between A^ and B^ ; (2) ^2 is a harmonic mean between A,^ and B^. 14—2 212 ELEMENTARY TRIGONOMETRY. [chap. The Ex-central Triangle. *219. Let ABC be a triangle, 7^, I^, I^ its ex-centres ; then /j/g/g is called the Ex-central triangle of ABC, Let / be the in-centre ; then from the construction for finding the positions of the in-centre and ex-centres, it follows that : (i) The points /, I^ lie on the line bisecting the angle BAC; the points /, I2 lie on the line bisecting the angle ABC; the j)oints 7, 73 lie on the line bisecting the angle A CB. (ii) The points I2, I^ lie on the line bisecting the angle BAG externally ; the points I^, I^ lie on the line bisecting the angle ^5(7 externally ; the points 7^, I^ lie on the line bisecting the angle A CB externally. (iii) The line AI^ is perpendicular to 7373 ; the line BI^ is perpendicular to 737^ ; the line GI^ is perpendicular to 7^72- Thus the triangle ABC \& the Pedal triangle of its ex-central triangle 7i724 [See Art. 223.] (iv) The angles 757^ and ICI-^ are right angles ; hence the points j5, 7, C, I^ are concyclic. Similarly, the points C^ I, A, I^,, and the points A, I, B, I^ are concyclic. (v) The lines AI^, BI^^ GI^ meet at the in-centre 7, which is therefore the Orthocentre of the ex-central triangle IiI^I?,- (vi) Each of the four points 7, 7^, 72, I^ is the orthocentre of the triangle formed by joining the other three points. xvtil] the ex-central triangle. ^13 *220. To find the distances hetioeen the m-centre and ex- centres. Witli the figure of the last article, //, =AL — AI=i\ cosec — — r cosec -— = (n — r) cosec — ix J. 2 2 2 ,T^f.A B C . A . B . C\ A = 4:K sm — cos — cos — — sin — sm — sm — ) cosec — \222 22 2/ 2 .„ B+C .„ . A = 4R cos — - — = 4R sin — . 2 2 Thus the distances are 4R sin — , 4i2 sin — , 4R sin — . A A Jj *221, To find the sides and angles of the ex-central triangle. With the figure of Art. 219, lBI^C=lBIJ+ lCIJ = L BCI+ L CBl [Euc. in. 21] ~2+2~^" 2" Thus the angles are Again, y^ 2' 2' 2* I^I^=I^C+I^G=r^ cosec Z'gO" -^V^aCosec ^90° f . A B C A . B C\ sm — COS — cos — + cos — sm — cos — j s \ 2 2 2 2 2 2/ .p . A + B .p C = 4:R sm — - — = 4:R cos — . 2 2 Thus the sides are 4i2 cos — , AR cos — , 4R cos — . A Ji ^ 214 ELEMENTARY TRIGONOMETRY. [chap. *222. To find the area a7id circum-radius of the ex-central triangle. The area=- (product of two sides) x (sine of included angle) 1 B C = - X 4tR cos— X 4:R cos — x sin ( 90° - ^ -i) ^j,, A B C = 8ii^ cos — cos — cos — . 2 2 2 The circum-radius : Vs 4R cos — A ■ 2R. The Pedal Triangle. *223. Let (7, 5", A" be the feet of the perpendiculars from the an- gular points on the opposite sides of the triangle ABC ; then GHK is called the Pedal triangle of ABC. The three perpendiculars Jl 6^, BH, CK meet in a point which is called the Orthocentre of the triangle ABC. *224. To find the sides and angles of the 'pedal triangle. In the figure of the last article, the points K, 0, G, B are concyclic ; .-. lOGK=lOBK=^0''-A. Also the points H, 0, G^ C are concyclic ; ... lOGH=lOCH=^0°-A', .'. lKGB:=180°-2A. Thus the angles of the pedal triangle are 180° -2^ 180° -2^, 180° -2a Again, L AKH= 180° - z. BKH= L BCH, since the points B, A", H, C are concychc ; .-. iAEH=C. XVIII.] THE PEDAL TRIANGLE. 215 HK _ sin A _ sin A _a •'• AH ~ sin AKH^ smC ~ c ' .", HK=- . AH =- .c co^ A = a co^ A. c c Thus the sides of the pedal triangle are a cos J., b cos B, c cos G. In terms of R, the equivalent forms become R sin 2 A , i2 sin 2^, i2 sin 2 C. If the angle ACB of the given triangle is obtuse, the ex- pressions 180° — 2 C and ccos C are both negative, and the values we have obtained require some modification. We leave the student to shew that in this case the angles are 2 A, 2^, 2(7— 180°, and the sides a cos A, b cos B^ —c cos C. *225. To find the area and circum-radius of the pedal tri- angle. The area = - (product of two sides) x (sine of included angle) = ii2sin2i?.i2sin2C.sin(180 -2.4) = - i22 sin 2.1 sin 25 sin 26'. ^^ . ^. HK i2sin2^ R The circum-radms = ^—, — rfrnr = o • /lono — ettt = "5" • 2 sm HGK 2 sm (180 - 2A) 2 Note. The circum-circle of the pedal triangle is the nine points circle of the triangle ABC. Thus the radius of the nine points circle of the triangle ABG is - . [See Hall and Stevens' Euclid, p. 281.] *226. In Art. 224, we have proved that OG, OH^ OK bisect the angles HGK, KHG, GKH respectively, so that is the in-centre of the triangle GHK. Thus the orthocentre of a tri- angle is the in-centre of the pedal triangle. Again, the line CGB which is at right angles to OG bisects lHGK externally. Similarly the lines AHC and BKA bisect lKHG and lGKH externally, so that ABG is the ex-central triangle of its pedal triangle GHK. 216 ELEMENTARY TRIGONOMETRY. [chap. ■^227. In Art. 219, we have seen that ABC is the pedal tri- angle of its ex -central triangle I^IJl^. Certain theorems depend- ing on this connection are more evident from the adjoining figure, in which the fact that ABC is the pedal triangle of IxI^Jz is brought more prominently into view. For instance, the circum-circle of the triangle ABC is the nine points circle of the triangle I^IJ^o,-! and passes through the middle points of //j, ZZ2, i/g and of /^/g, /g/^, *228. To find the distance between the in-centre and circum- centre. Let >S' be the circum-centre and / the in-centre. Produce A I to meet the circum-circle in H ; join 6'^ and CI. Draw IE perpendicular to AC. Produce HS to meet the circumference in L, and join CL. Then lHIC=lIAC-\-lICA 2 2' lHCI= LICB+ lBCR ^^+ lBAH _G A ~ 2"^2 ' Also .-. LHCI=LmC; ,'. Hl=-EC^^2R^\xi^. AI= IE 0,0^0,0, — =r cosec — 2 2 AI.IH=%Rr. XVIII.] DISTANCES FROM CIRCUM-CENTRB. 217 Produce >S'/ to meet the circumference in M and N. By Euc. III. 35, AI. IH= MI . IX = {R + SI) {R - SI) ; .-. 2Rr=R^~SP', that is, Sr- = R^--2Rr. *229. To find the distance of an ex-centre from the circum- centre. Let >S' be the circum-centre, and / the in-centre ; then A I produced passes through the ex-centre I^. Let AI^ meet the circum-circle in H) join CI, BI, CH, BH, CI^, BI^. Draw I^E-^ perpendicular to AC. Produce HS to meet the circum- ference in jv, and join CL. The angles IBI^^ and ICI^ are right angles ; hence the circle on 11^ as diameter passes through B and C. The chords BH and CH of the circum-circle subtend equal angles at A, and are therefore equal. But from the last article, IIC= III\ .-. HB = HC=HI; hence H is the centre of the circle round IBI^C. .-. JII^=HC=2Rsin^. Now SI-^ — ^2 _ square of tangent from i^ = I^H.I^A = 2R sm — . r^ cosec — = 2Rr,. .'. SI^=E^-\-2Ri\. 218 ELEMENTARY TRIGONOMETRY. [chap. *230. To find the distance of the orthocentre from the circum- centre. With the usual notation, we have SO'^ = SA'^+AO'^- 2SA . AO cos SAO. ^ovfAS=R; A0= AH coBQc G = c cos A cosec C = 2R sin C cos A cosec C = 2R cos A ; lSAO=lSAC- lOAC = (90°-i?)-(90°-(7) = C-B. .'. JSO^ ^ R^ + AR^cos^ A -4R^ cos A cos {C-£) = R^-4Jl^ cos A {cos {B + C) + cos{C-B)} = R^- 8i22 cos A cos B cos (7. The student may apply a similar method to establish the results of the last two articles. ^EXAMPLES. XVni. c. 1. Shew that the distance of the in-centre from A is 4Ji sin — sm — . 2. Shew that the distances of the ex-centre /^ from the angular points A, Bj C are ^_ B C ,j, . A C ,j, . A B 4/t cos — cos — , 4/c sm — cos — , 4/t sm — cos — . 2 2 2 2 2 2 3. Prove that the area of the ex-central triangle is equal to ABC — cosec — cosec — 2t Jt Ji (1) ^Rs ; (2) - A cosec — cosec — cosec 4. Shew that r.Il^. 11^ . Il^=4Jt . lA . IB . la 5. Shew that the perimeter and in-radius of the pedal triangle are respectively 4i2 sin A sin BsinC and 2R cos A cos B cos G. XVIII.] THE EX-CENTRAL AND PEDAL TRIANGLES. 219 6. If g^ A, h denote the sides of the pedal triangle, prove that (2) (^^-^g- , { c'-a')h (a2-62)^ ^^ a^ 52 ^2 • 7. Prove that the ex-radii of the pedal triangle are 2R sin A cos B cos C, 2R cos ^ sin B cos (7, 2^ cos A cos 5 sin C. 8. Prove that any formula which connects the sides and angles of a triangle holds if we replace (1) a, h, c by «cos^, bcosB, ccosC, 3^nd A, B,C by 180°-2.4, 180°-25, 180°-2(7; (2) a, 6, c by a cosec — , b cosec — , c cosec — , and A,B,C by 90° - 1 , 90° - 1 , 90° -^. 9. Prove that the radius of the circum-circle is never less than the diameter of the in -circle. 10. If R — 2rj shew that the triangle is equilateral. 11. Prove that 12. Prove that (1) a.AP-\-b.BP + c.CP = abo; (2) a.AIi^-b.BI^^-c.CIj^ = abc. 13. If GHK be the pedal triangle, and the orthocentre, prove that OG OE OK ^^ AG'^ BH^ CK~ ' (2) ^^ I ^-^ ■ ^^^ 1 ^^ (9(y4-acot^"^OZr-l-6cot^'^0^-f-ccot(7 14. If GHK be the pedal triangle, shew that the sum of the circum-radii of the triangles AHK, BKG, CGH is equal to R + r. 220 ELEMENTARY TRIGONOMETRY. [chap. 15. If AiB^Ci is the ex-central triangle of ABC, and A^B^O^ the ex-central triangle of AiB^C^, and A^B^C^ the ex-central tri- angle of .dg^a^sij ^^^ so ^^ '• fi^^ ^^® angles of the triangle A^B.^C^, and prove that when n is indefinitely increased the triangle becomes equilateral. 16. Prove that (1) 0S^=9R^-a'^-h^-c'^; (2) (9/2 = 2r2 - 4R^ cos A cos BcosO; (3) 0/i2 = 2ri2 - 4i?2 cos A cos i? cos a 17. If/, ^, A denote the distances of the circum-centre of the pedal triangle from the angular points of the original triangle, shew that 4:{f+g^- + h^) = llE'^ + 8R'^cosA cos 5 cos (7. Quadrilaterals. *231. 'To prove that the area of a quadrilateral is equal to - {product of the diagonals) x {sine of included angle). Let the diagonals AC, BD inter- sect at P, and let L DP A = a, and let S denote the area of the quadri- lateral. Al)AC=hAPI) + ACPI) =1 DP. AP sin a + \DP.PC&m{'jr-a) =li)P(^P + Pa)sina =iz>P.J^Csina. Similarly AABC=^^BP,ACsYiia. 2 .-. >S'=^(i)P + ^P)^Csina A =-DB.AC^\na. 2t XVIII.] QUADRILATERALS. 221 *232. To find the area of a quadrilateral in terms of the sides and the sum of two opposite angles. Let ABCD be the quadrilateral, and let a, &, c, d be the lengths of its sides, S its area. By equating the two values of BD^ found from the triangles BAD, BCD, we have a^ + c?^ — 2ad cos J. = 6^ -f c^ — 26c cos C ; .-. a'^ + d^—h'^ — c'^ = 2ad co^ A- '2hc cos C ••(!)• Also /S^sum of areas of triangles BAD, BCD =- a(i sin J. +- 6c sin C ; .-. 4AS'=2ac?sin.4+26csin (7 (2). Square (2) and add to the square of (1) ; .-. 16/S'2 + (a2 + ^2-62-c2)2 = 4a2c^2 + 462c2_8a6cc^cos(^ + C). Let A + G=^a; then cos {A + C) = cos 2a = 2 cos2 a - 1 ; .-. 1652=,4(^^^.5c)2-(a2^^2_52_c2)2_16(^5cC^COs2a. But the first two terms on the right = (2ac? + 26c + «^H c^2 - 62 _ c2) (2ac? + 26c - a2 _ f^2 + 52 + ^2) = {(a+<^)2-(6-c)2}{(6 + c)2-(a-(^)2} = {a-{-d^-h - c) {a+d -h + c) {h + e + a- d) {h + - a^- d) = (2(r - 2c) (2o- - 26) (2a- - 2c?) (2o- - 2a), where a + b + c+d=2a; = 16 (o- - a) (o- - 6) (o- - c) (o- - cT). Thus 152 = (o- - a) (a- -h){(r- c) {(r-d)- abed cos2 a, where 2 a- denotes the sum of the sides, 2a the sum of either pair of opposite angles. *233. In the case of a cycliG quadrilateral, A-\-C=\S(f, so that a ='90°; hence S=^J{(T-a){(T-h){a■-c){,^-d). This formula may be obtained directly as in the last article 222 ELEMENTARY TRIGONOMETRY. [CHAP. by making use of the condition A + C=180° during the course of the work. In this case cos C= - cos A, and sin (7= sin A, so that the expressions (1) and (2) become a^+d^-b^-c^ = 2{ad+bc) cos A, and 4>S'= 2 {ad + be) sin A ; whence by eliminating A we obtain 1 6^2 + ^^2 + (^2 _ 52 _ ^2)2 = 4(^ad+ hcf. *234. To find the diagonals and the circum-radius of a cyclic quadrilateral. If ABCD is a cyclic quadrilateral, we have just proved that 2{ad+bc)co& A = a^-\-d'^-b^-c\ l^ow BI)^ = a^ + d^-2ad cos A ^a2 I ^2 ad{a^+d^-b^-c^) ad+bc _ bc{a^+d^) + ad(b^ + c'^) ad+bc _{ab + cd){ac+bd) ad-Ybc Similarly, we may prove that „2 {ad+bc){ac + bd) AC = :f y— . ao + cd and Thus AC. BI)=ac + bd, [Compare Euc. yi. D.] AC ad+bc BD ab+cd' The circle passing round the quadrilateral circumscribes the triangle ABD ; hence the circum-radius = ;r—. — j 2 sm^ ■ (a d + be) BD _ {ad+ be) BD ~ '2, {ad+ be) sin A~' 4.S =Y^'\/{ab-\-ed){ac + bd){ad+be). XVIII.] QUADRILATERALS. 223 Example. A quadrilateral ABGD is such that one circle can be inscribed in it and another circle circumscribed about it ; shew that ^^"^ 2-ad- If a circle can be inscribed in a quadrilateral, the sum of one pair of the opposite sides is equal to that of the other pair ; .-. a + c = b + d. Since the quadrilateral is cyclic, cos^= _ , , — r^— . [Art. 233.] 2 {ad + he) -^ But a-d = b~c, so that a'^-2ad + d^ = b^-2bc + e^; .'. a2 + d2 _ ^,2 _ c2 = 2 (ad - 6c) ; ad- be :. cos A = ad+bc' A _1- cos A _ be tan — — z r — — -. . 2 1 + cos^ ad ♦EXAMPLES. XVIII. d. 1. If a circle can be inscribed in a quadrilateral, shew that its radius is S/a- where S is the area and 2cr the sum of the sides of the quadrilateral. 2. If the sides of a cyclic quadrilateral be 3, 3, 4, 4, shew that a circle can be inscribed in it, and find the radii of the inscribed and circumscribed circles. 3. If the sides of a cyclic quadrilateral be 1, 2, 4, 3, shew 5 that the cosine of the angle between the two greatest sides is - , and that the radius of the inscribed circle is "98 nearly. 4. The sides of a cyclic quadrilateral are 60, 25, 52, 39 : shew that two of the angles are right angles, and find the diagonals and the area. 5. The sides of a quadrilateral are 4, 5, 8, 9, and one diagonal is 9 : find the area. 6. If a circle can be inscribed in a cyclic quadrilateral, shew that the area of the quadrilateral is \/abcd, and that the radius of the circle is 2 '\/abcd/{a + b+c+d). 224 ELEMENTARY TRIGONOMETRY. [CHAP. 7. If the sides of a quadrilateral are given, shew that the area is a maximum when the quadrilateral can be inscribed in a circle. 8. If the sides of a quadrilateral are 23, 29, 37, 41 inches, prove that the maximum area is 7 sq. ft. 9. If ABCD is a cycUc quadrilateral, prove that B { o-a){^-h) 10. If /, ff denote the diagonals of a quadrilateral and /3 the angle between them, prove that 2fg cos /3 = (a2 + ^2) ^ (&2 + d^y 11. If /3 is the angle between the diagonals of any quadri- lateral, prove that the area is i{(a2+c2)~(62 + c^2)j.tan/3. 12. Prove that the area of a quadrilateral in which a circle can be inscribed is A + C \j abed sm 13. If a circle can be inscribed in a quadrilateral whose diagonals are /and g^ prove that 14. If /3 is the angle between the diagonals of a cyclic quadrilateral, prove that (1) (ac + hd) sin /3 = {ad + 6c) sin A ; (a2+c2)-(62 + c^2)^ (2) cos/3 = 2(ac+6c^) (d) tan 2-(^_a)((r-c) """^ (,, - 6) (o- c^) ' 15. If/, ^ are the diagonals of a quadrilateral, shew that ^=i V4/2^2 - (^2 + C2 - 62 _ e^2)2, 16. In a cyclic quadrilateral, prove that the product of the segments of a diagonal is abed {ac + hd)l{ah -|- cd) {ad + he). XVIII.] MISCELLANEOUS EXAilPLES ON TRIANGLES. 225 235. The following exercise consists of miscellaneous ques- tions involving the properties of triangles. EXAMPLES. XVIII. e. 1. If the sides of a triangle are 242, 1212, 1450 yards, shew that the area is 6 acres. 2. One of the sides of a triangle is 200 yards and the ad- jacent angles are 22*5° and 67'5° : find the area. 3. If ri = %'2 = 2r3 , shew that 3a = 46. 4. If a, b, c are in a. p., shew that r^, ^g, ^3 are in h. p. 5. Find the area of a triangle whose sides are y z z cc XII - + -, - + -, - + ^. z X X y y z 6. If sin A : sin €= sin {A-B): sin {B - C\ shew that a^, b% + l)'r-a, (2^-2)7r + a,J "^ t(2^-l)7r-a, and it will be noticed that even multiples of tt are followed by -|-a, and odd multiples of tt by - a. Thus all angle^ equi-sinal with a are included in the formula ?i7r + (-l)**a, where n is zero, or any integer positive or negative. This is also the formula for all angles which have the same cosecant as a. Example 1. Write down the general solution of sin ^=^ . The least value of 6 which satisfies the equation is - ; therefore 3 ^ the general solution is wtt + { - 1)" ^ . Example 2. Find the general solution of sin^d=3m^a. This equation gives either sin^= +sina (1), or sin^= -sina = sin(-a) (2). XIX.] GENERAL VALUE OP EQUI-COSINAL ANGLES. 233 From (1), ^ = n7r + (-l)"a; and from (2), ^ = W7r + (-!)«(- a). Both values are included in the formula d=mr^a. 239. To find a formula for all the angles which have a given cosine. Let a be the smallest positive angle which has a given cosine. Draw OP and OP' bounding the angles a and 27r — a; then the required angles are those coter- minal with OP and OF, The positive angles are 2p7r + a and 2j07r + (27r - a), where p is zero, or any positive integer. The negative angles are -a and -(27r — a), and those which may be obtained from them by the addition of any negative multiple of 27r ; that is, angles denoted by '^qn - a and '2,qn — {^n — a), where q is zero, or any negative integer. The angles may be grouped as follows : ^^''+-'1 and l^f +,'^-- 2^7r-a, J ((25'-2)7r + a, and it will be noticed that the multiples of tt are always even, but may be followed by + a or by — a. Thus all angles equi-cosinal with a are included in the formula 2?i7r±a, where n is zero, or any integer positive or negative. This is also the formula for all angles which have the same secant as a. Example 1. Find the general solution of cos ^= - - . _ o_. The least value of ^ is tt - - , or -^ ; hence the general solution o o • o 27r IS 2%7r±— . D 234 ELEMENTARY TRIGONOMETRY. [chap. 240. To find a formula for all the angles ivhich have a given tangent. Let a be the smallest positive angle which has a given tangent. Draw OP and OP' bounding the angles a and tt + a ; then the required angles are those coter- minal with OP and OP. The positive angles are 2^77 + a and 2j97r + (7r+a), where p is zero, or any positive integer. The negative angles are — (tt — a) and —(277 — a), and those which may be obtained from them by the addition of any negative multiple of 27r ; that is, angles denoted by '2qiT-{7T — a) and 2qir — {^tt - a)^ where q is zero, or any negative integer. The angles may be grouped as follows : (2^- ^f + «'} and j^,f + ,\^^+"' and it will be noticed that whether the multiple of n is even or odd, it is always followed by + a. Thus all angles equi-tangential with a are included in the formula 6 = n7T-\-a. This is also the formula for all the ansrles which have the O same cotangent as a. Example. Solve the equation cot id = cot 6. The general solution is id^mr + d; mr whence Sd = 7nr, or 6 = 241. All angles which are both equi-sinal and equi-cosinal ■with a are included in the formida 2n7r+a. All angles equi-cosinal with a are included in the formula 27i7r + a ; so that the multiple of tt is even. But in the formula 7i7r+( — l)'*a, which includes all angles equi-sinal with a, when the multiple of tt is even, a must be preceded by the + sign. Thus the formula is 2n7r + a. XIX.] GENERAL SOLUTION OF EQUATIONS. 235 242. In the solution of equations, the general value of the angle should always be given. Example. Solve the equation cos 90= cos 56 - cos d. By transposition, (cos 96 + cos 6) -cos 56 = 0; .-. 2 cos 50 cos 40 -cos 50 = 0; .-. cos50(2cos40-l) = O; ,-. either cos 50=0, or 2cos40-l = O. TT (47i i 1) TT From the first equation, 5d = 2mr±-^ , or 0= -^' ; ^ TT . (6n±l)7r and from the second, 40=2w7r±g , or 0= — — . EXAMPLES. XIX. a. Find the general solution of the equations : 1. sin^=-. 2. sin^=--^. 3. cos (9 =2- 4. tan(9=V3. 5. cot^=-^/3. 6. sec^=-V2. 7. cos2(9 = ^. 8. tan2^=i 9. cosec2(9=^. A o o 10. cos B = cos a. 11. tan^ 6 = tan^ a. 12. sec2^=sec2a. 13. tan 2(9= tan (9. 14. cosec 3d = cosec 3a. 15. cos 3^ = cos 2d. 16. sin 5^ + sin ^ = sin 3^. 17. cos (9 -cos 7^= sin 4^. 18. sin 4^ - sin 3^ + sin 2^- sin ^ = 0. 19. cos ^ + cos 3^ + cos 5(9 + cos 7(9 = 0. 20. sin 5^ cos B = sin 6B cos 2B. 21. sin 11^ sin 4^ + sin 5^ sin 2^ = 0. 22. v/2 cos 3^ -cos (9 = cos 5^. 23. sin 7(9 - ^3 cos 4B = sin B. 24. l + cos^ = 2sin2^. 25. tan2^ + sec^ = l. . 26. cot2^-l=cosec(9. 27. cot (9 -tan (9 = 2. 28. If 2 cos <9 = - 1 and 2 sin B = >J3, find B. 29. If sec B = x/2 and tan ^ = - 1, find B. 236 ELEMENTARY TRIGONOMETRY. [CHAP. 243. In the following examples, the solution is simplified by the use of some particular artifice. Example 1. Solve the equation cosm^ = sinw^. Here cos mO = cos f ^ - nd \ ; where k is zero, or any integer. By transposition, we obtain (m + n) e=( 2k +-JTr, or {m - n) = (2k - ~\ ir. This equation may also be solved through the medium of the sine. For we have sm l--mdj =smnd; .'. ~-md=pTr + {-l)Pnd, where p is zero or any integer ; .-. {m+{-l)Pw}^=Q-_p^^. Note. The general solution can frequently be obtained in several ways. The various forms which the result takes are merely different modes of expressing the same series of angles. Example 2. Solve ;.^3 cos^ + sin ^=1. Multiply every term by - , then Ycos^ + 2sm^ = -, .*. cos ^ COS ^ + sin ^ sin ^ =- ; •. cos I d _7r\_l 6y""2' 1? TT TT TT d = 2n7r + ^ or 2n7r--. A XIX.] GENERAL SOLUTION OF EQUATIONS. 237 Note. In examples of this type, it is a common mistake to square the equation; but this process is objectionable, because it introduces solutions which do not belong to the given equation. Thus in the present instance, ^y3cos^ = l-sin^; by squaring, 3 cos- ^ = (1 - sin 6)^. But the solutions of this equation include the solutions of - ^3 cos 6=1 -sin 6, as well as those of the given equation. Examples. Solve cos 2^ = cos ^ + sin ^. From this equation , cos^ ^ - sin^ 6 = cos 6 + sin 6 ; .•. (cos 6 + sin ^)(cos d - sin 6) = cos ^ + sin 6 ; .-. either eos0 + sin^ = O (1), or cos ^- sin ^ = 1 (2). From (1), tan^=-l, "■ .'. 6 = mr--r. 4 From (2), -j^cos e--j^sme =-j^; T . „ . IT 1 .•. cos 6 cos -r - sm ^ sm -T = -7^ . 4 4 ^2 cos ('+i)=^2' .-. ^ + -=2n7rij; .-. 6 = 2mr or 2mr - — . EXAMPLES. XIX. b. Find the general solution of the equations : 1. tanjo^=cot 2'^. 2. sinwi^+cos?i^=0. 3. cos^-x/3sin^ = l. 4. sin ^-^3 cos ^=1. 5. cos^=v'3(l-sin^). 6. sin ^ + V3cos ^=v/2. 238 ELEMENTARY TRIGONOMETRY. [CHAP. Find the general solution of the equations : 7. cos^-sin^ = -^. 8. cos^ + sin^+^/2 = 0. 9. cosec^ + cot^=v/3. 10. cot ^- cot 2^ = 2. 11. 2 sin ^ sin 3^ = 1. 12. sin .3^ = 8 sin^ /9. 13. tan ^+tan 3^ = 2 tan 2(9. 14. cos ^ - sin ^= cos 2^. 15. cosec ^ + sec ^ = 2 ^2. 16. sec 6 - cosec ^ == 2 ^2. 17. sec 4^ -sec 2^ = 2. 18. cos3^ + 8cos3^=0. 19. 1 + V3 tan2 ^ = (l + ^3) tan d. 20. tanS $ + cot^ = 8 cosec^ 2^ + 12. 21. sin 6 = v/2 sin cf), 4^/3 cos 6 = ^/2 cos 0. 22. cosec d = sj3 cosec (f), cot 6 = 3 cot (j). 23. sec (f) = s,'2 sec 6, cot 6 = a,/3 cot 0. 24. Explain why the same two series of angles are given by the equations 25. Shew that the formulae Un^^^n + a and L-^ -rr +{-lY {^-cS comprise the same angles, and illustrate by a figure. Inverse Circular Functions. 244. If sin 6 = s, we know that 6 may be any angle whose sine is s. It is often convenient to express this statement inversely by writing 6 = bui~^s. In this inverse notation 6 stands alone on one side of the equation, and may be regarded as an angle whose value is only known through the medium of its sine. Similarly, tan-^V^S indicates in a concise form any one of the angles whose tangent is ^3. But all these angles are given by the formula '/itt+q- Thus <9+J=7i7r + (-l)'^J and ^-^ = 2?^7^±^. 4 6 4 3 6 = ta,n~'^j3 and 6 = n77+~ o = T>d,n '■ s,i o ana t) = niT-\-~ are equivalent statements expressed in different forms. XIX.] IIS^ERSE CIRCULAR FUNCTIONS. 239 245. Expressions of the form cos~^^, sin-^a, tan~^& are called Inverse Circular Functions. It must be remembered that these expressions denote angles, and that — 1 is 7iot an algebraical index ; that is, sin~i^ is not the same as (sin .37) "^ or — — . sm X 246. From Art. 244, we see that an inverse function has an infinite number of values. If /denote any one of the circular functions, and/~i {x) = A^ the principal value oi f~^{x) is the smallest numerical value of ^. Thus the principal values of cos-ii, sin-ir--V cos-i(^--^j, tan-i(-l) are 60°, -30°, 135°, -45°. Hence \i x\)Q positive, the principal values of sin~^ x^ cos~i.r, tan~i X all lie between and 90°. If X be negative, the principal values of sin~i^ and tan~i^ lie between and - 90°, and the principal value of cos~ ^ x lies between 90° and 180°. In numerical instances we shall usually suppose that the principal value is selected. 247. If sin 6=Xj we have cos 6 = ^/l — x^. Expressed in the inverse notation, these equations become ^ = sin~i.r, 6 = coB~'^ ^/l-x^. In each of these two statements, 6 has an infinite number of values ; but, as the formulae for the general values of the sine and cosine are not identical, we cannot assert that the equation sin~ 1 .■?; = COS" 1 Vl — ^2 is identically true. This will be seen more clearly from a 1 / a/3 numerical instance. If x = -, thenvl-'^^= "2^ • Here sin~ ^ x may be any one of the angles 30°, 150°, 390°, 510°, ...; and cos"* ^ s/l—x^ may be any one of the angles 30°, 330°, 390°, 690°,..., 240 ELEMENTARY TRIGONOMETRY. [CHAP. 248. From the relations established in the previous chapters, we may deduce corresponding relations connecting the inverse functions. Thus in the identity .. l-tan2^ cos 2^=- — - — ^. , l+tan^^' let tan ^ = 0^, so that ^ = tan~i a ; then 1-^2 cos (2 tan~i a) = 2tan~ia=cos~i 1-^2 Similarly, the formula cos 3^=4 cos^ 6-Z cos 6 when expressed in the inverse notation becomes 3 cos ~ 1 a = cos " 1 (4a3 _ 3c)^). 249. To prove that tan~l.^•^-tan~l y = ts,n~'^- — ~ . \—xy Let tan~i.a7=a, so that tana = ^; and tan~iy=/3, so that tan/3=y. We require a + /3 in the form of an inverse tangent. tan a + tan 3 Now tan(a+/3) = 1 - tan a tan /3 .'. a+i3=tan-i— ^^^; \-xy CO "^ v that is, tan~i.2; + tan~i ?/=tan~i :; — —. ' ^ 1-^y By putting y—x, we obtain 2tan~iA'=tan~i Note. It is useful to remember that tan (tan~ia;4-tan~i?/) =:^^ — a;?/ XIX.] INVERSE CIRCULAR FUNCTIONS. 241 Example 1. Prove that 7 TT tan~i 5 - tan~i 3 + tan~i - = ?i7r + r • 9 4 5 — 3 7 The first side= tan-i - — z-^+ tan-i - i + lo y = tan~i-+ tan~i- o y 8 + 5 =: tan~i „ = tan~i 1 IT = nTT + x • 4 Note. The value of n cannot be assigned until we have selected 7 some particular values for the angles tan~i5, tan-i3, tan i - . If we choose the principal values, then n = 0. Example 2. Prove that sin~^ - + cos~^ 12 l3 il6_ + ^^^65 = 2- We may write this identity in the form sm "i + ^"^"i3 12 TT - sm" 16 65 - = cos" 16 65 4 . 4 Let a = sin~i -= , so that sin a = _ ; o o 12 12 and /S=cos-1y^, so that cos/3 = — We have to express a + /3 as an inverse cosine. Now cos(a + /3) = cosacosj3-sina sin^; whence by reading off the values of the func- tions from the figures in the margin, we have , o^ 3 12 4 5 cos(a + i3) = ^.^-5.^ 16 65' .-. a + p = cos-i-^ 16 65* H. K. E. T. 242 ELEMENTARY TRIGONOMETRY. [chap. It is sometimes convenient to work entirely in terms of the tangent or cotangent. Example 3. Prove that ^ . .n i3 ,125 2 cot~i 7 + cos~-^ - = cosec"-^ ^— _ • 72 — 1 3 The first side=cot-i-Tr — -+ cot-i-r 2x7 4 ^ T 24 ^ , 3 7 4 I = cot-i 24 3 , 7 4 24 3 T + 4 , 1 44 _i 125 =cot-i^^ = coseci^. EXAMPLES. XIX. c. .-ii!. 2. cosec ^— -=tan i-—. 8 15 Prove the following statements 1. mn-i^ = cot-i^. 3. sec(tan-i^) = Vl+^^- 4. 2 tan~i- = tan-i-. 4 1 5. tan'"i--tan-il = tan~i = . o / 2 24 1 6. tan-i— - + cot-i— =tan-i-. 117 ^ X ,4 ^ ,15 ,_,84 7. cot-i--cot-i-g-=cot 'jg. 1 1 32 8. 2tan-i-+tan-i- = tan-i — . 5 4 4o 115 1 9. tan-i-+tan-i-=tan~i7^4-tan~i7Y. 2 o o 11 10. tan-i-+tan-i3+tan-i— = cot-i3. 7 o lo XIX.] INVERSE CIRCULAR FUNCTIONS. 243 Q 3 27 11. tan-i-+sin-i - = tan~^:pr. 5 5 ii 5 40 , 8 . , 240 12. 2cot-i^=tan-iy. 13. 2 tan-i — =sin-i ^g^. 14. sin (2 sin - 1 ;r) = 2^ V 1 — •^^• 15. cos-i.r=2sin-i ^-g-. 16. 2tan-i /- = cos-i — 17. 2tan-i^+tan-i^ + 2tan-i- = ^. 18. sin-ia-cos-i6 = cos-i{&Vl-«^+«^Vl-6^}' 19. «in-i^ + cos-iA = cot-iA. fi^ 1 3 20. cos-i — + 2tan-i- = sin-i -. bo o 3 , 1 —mn 21. tan-im+tan-i7i=cos-i -—======. V(l+wi^)(l+^) ,20 ^ ,16 _il596 23. cos-i^--cos 1-^^=6- 24. tan (2 tan-i ^) = 2 tan (tan-i.37 + tan-i^). 25. tan-ia=tan-i:^^^+tan-i— P|-4-tan-ic. l + ao l + oc 26. If tan-i:r+tan-iy+tan-is=7r, prove that 27. If w = cot - 1 a/coso - tan - 1 a/cos a, prove that sinw=tan2-. 2 16—2 244 ELEMENTARY TRIGONOMETRY. [CHAP. 250. We shall now shew how to solve equations expressed in the inverse notation. Stt Example 1. Solve tan-i2a; + tan~i3a; = ?i7r + -j . „, , , ,2x + dx Sir We have tan-^ z — ^-v=W7r + -r- ; 1 - ox^ 4 ,.___ = tan (^nT+-^j=-l; .-. 6a;2-5a;-l = 0, or (6a; + l)(a;- 1) = 0; - 1 .-. a;=l, or -g. Example 2. Solve sin~^a; + sin~i(l-a:) = cos~^a;. By transposition, sin~i {l-x) = cos~^ x - sirr'^x. Let cos~'^x = a, and sin~ia; = /3; then sin~i(l-a;) = a-j8; .'. l-a; = sin(a-/3) = sinacos/3-cos asin/3. But cos a = x, and therefore sin a = ^^^1 - a;^ ; also sin ^=x, and therefore cos ^— sjl-x^\ :. l-a;=(l-a;2)-a;2 = l-2a;2; .-. 2a;2-a: = 0; whence a; = 0, or -. EXAMPLES. XIX. d. Solve the equations : L sin-i^=cos~i.27. 2. tan~i^=cot~^^'. 3. tan-i (^ + 1) - tan-i (^- l) = cot-i 2. 4. cot~i^ + cot~i2^=— r-. 4 5. sin-i^-cos~i.a7=sin~i(3^-2). 6. cos~i^ — sin~i.:»7=cos~^^V3- XIX.] INVERSE CIRCULAR FUNCTIONS. 245 _ . . x—1 . . a)+l IT 7. tan-i -+tan-i— — = -7. ^-2 x-^^ 4 3 8. 2cot~i2 + cos~i- = cosec~i.27. 5 9. tan~i^'+tan~i(l-^) = 2taii~i\/^-^2, ,1 — a^ ,1 — 62^, . 10. cos-i-—— „-cos-i_— ^=2tan-i.2;. 1 + a^ 1 + 6^^ • 1 2a , ,2^ ,1-62 11. sm 1— — -2+tan-i.j 2=^°^" rT'A2- 12- cot-i-2^+tan ^^23i + y=0- 13. Shew that we can express . , 2a6 . . , 2cc? . ,, - . _i 2^y sin-i-2-rX2+si°~ -2-r:72 m the form sm ^^^3^2 where x and y are rational functions of «, 6, c, c^. 14. If sin [2 cos " 1 {cot (2 tan - 1 ^)}] = 0, find x. 15. If 2 tan - 1 (cos 6) = tan " 1 (2 cosec 6\ find ^. 16. If sin (tt cos 6) = cos (tt sin 6), shew that 2^= + sin~i-. ~ 4 17. If sin (tt cot ^) = cos (tt tan 6), and n is any integer, shew 4*1 + 1 that either cot 2^ or cosec 2^ is of the form — ;. — . 4 18. If tan (tt cot 6) = cot (tt tan 6), and n is any integer, shew that ^ ^ 2?i+l \/4w2 + 47i-15 tan ^ = — 7— + . 4 "~ 4 19. Find all the positive integral solutions of tan~i^+cot~iy = tan'"i 3. 246 ELEMENTARY TRIGONOMETRY. MISCELLANEOUS EXAMPLES. G. 1. If the sines of the angles of a triangle are in the ratio of 4:5:6, shew that the cosines are in the ratio of 12 : 9 : 2. 2. Solve the equations : (1) 2cos3^ + sin2^-l=0; (2) sec3^-2tan2^=2. 3. If tan j3 = 2 sin a sin y cosec (a + y), prove that cot a, cot ^, cot y are in arithmetical progression. 4. In a triangle shew that 4r {7\ + rg + rg) = 2{bc-{-ca + ab) - {a^ + h^ + c^). 5. Prove that 1 11 3 (1) tan-i-- tan-i-+tan-is=tan-i— ; " 7 11 /ci\ • 1 3 , . , 8 . , 36 TT (2) sin-Xg + sm-ij^+sm-i-=-. 6. Find the greatest angle of the triangle whose sides are 185, 222, 259 ; given log 6 = -7781513, Xcos 39° 14'= 9-8890644, diff. for l' = 1032. 7. If tan (a + 6)=^n tan (a - 6\ prove that . ^ = — ^ . • ^ ^' ^ sm 2a ^+1 8. If in a triangle QR^=a'^-\-li^-\-c^, prove that one of the angles is a right angle. 9. The area of a regular polygon of n sides inscribed in a circle is three-fourths of the area of the circumscribed regular polygon with the same number of sides : find n. 10. ABCD is a straight sea-wall. From B the straight lines drawn to two boats are each inclined at 45° to the direction of the wall, and from C the angles of inclination are 15° and 75°. If j5(7= 400 yards, find the distance between the boats, and the distance of each from the sea-wall. CHAPTER XX. FUNCTIONS OF SUBMULTIPLE ANGLES. 251. Trigonometrical ratios of 22^° or | . From the identity 2sin2 22|°=:l-cos45°, we have 4sin2 22j° = 2-2cos 45° = 2- ^2; .-. 2sin22|° = \/2^2 (1). In like manner from 2cos2 22^° = l+cos45°, we obtain 2cos22^° = V2+V2 (2). In each of these cases the positive sign must be taken before the radical, since 22|° is an acute angle. 1 _ cos 45° Again, tan 22i° = — . ,,, = cosec 45° - cot 45° ; ^ sm45 .-. tan22|° = ^2-l. 252. We have seen that 2 cos - = \/2 + v/2 ; o but , . 4cos2:^ = 2+2cosf ; io o .-. 4cos2^ = 2 + V'2+x/2; lb .-. 2cos^=v/2 + V2 + V2. Similarly, 2 cos ^ = ^ 2-\-J^ + \/2 + j2 ; and so on. 248 ELEMENTARY TRIGONOMETRY. [cHAP. 253. Suppose that cos A = - and that it is required to find . A sm-. . A sm — = 2 V 2 ~V 2(^"2)-V 4~-2 This case differs from those of the two previous articles in that the datum is less precise. All we know of the angle A is contained in the statement that its cosine is equal to -, and without some further knowledge respecting A we cannot remove A the ambiguity of sign in the value found for sin — . We now proceed to a more general discussion. A A 254. Given cos A to find sin — and cos — and to explain the -presence of the two values in each ca^e. From the identities A A 2sin2 — =1 — cos^, and 2 cos^ — =l+cos^, 2 2 we have ±\ . A sm- = A -cos J. 2 ' -.yi , A /I + cos A and cos — Thus corresponding to one value of cos A^ there are two values for sin — , and two values for cos — . The presence of these two values may be explained as follows. If cos J. is given and nothing further is stated about the angle -4, all we know is that A belongs to a certain group of equi- cosinaZ angles. Let a be the smallest positive angle belonging A to this group, then A'='2mr±a. Thus in finding sin— and A cos- we are really finding the values of sin ■- (2n7r + a) and cos - (27i7r ± a). XX.] FUNCTIONS OF SUBMULTIPLE ANGLES, Now sm-(27?7r±a) = sin f W7r±- a , .a = sm rnr cos - + cos nir sm - = ±sm-, — 2' for sin nir = and cos titt = + 1. Again, cos - (2mr ± a) == cos titt cos - + sin rnr sin - 249 = ±cos-. . A Thus there are two values for sin — and two vahies for A cos -- when cos A is given and nothing further is known re- Ji specting A. 255. Geometrical Illustration. Let a be the smallest posi- tive angle which has the same cosine as A ; then ^ = 2?i7r±a, and we have to find the sine A and cosine of ^ , that is of A Each of the angles denoted by this formula is bounded by one of the lines OP^, OP^, OP^, OP^. Now sin XOP2 = sin - , sin XOP^ = - sin - , sin XOP^ = - sin - , AAA cos XOP2 = — COS - , COS XOP^ = - cos - , COS XOP^ = cos - . A A ^ Thus the values of sin— are +sin-, and the values of cos — A A ^ are ±cos-. A 250 ELEMENTARY TRIGONOMETRY. [CHAP. 256. If cos^ is given, and^ lies between certain known limits, the ambiguities of sign in the formulae of Art, 254 may be removed. 7 Example. If cos A = -^, and A lies between 450° and 540°, find 25 A A sin — and cos ^ . Ji it . A /l-co^A /l/, 7\ /16 ,4 ^^^2 = \/"~2— ^Vn'' 25) = V 25=^^-5' A /I + COB A /rr, 7\ / 9 3 ^°'2 = V 2 — =\/n 25J = \/25='^5- A A A Now — lies between 225° and 270°, so that sin — and cos ^ are both negative ; A 4: ^3 /. sin-=-g, andcos-=-^. A A 257. To find sin — and cos — in terms o/sin A and to explain the 'presence of four values in ea/ih case. A A We have sin^ — -|- cos^ -^ = 1, and 2sm— cos — =sm^. 2 2 / A A\^ By addition, ( sin — + cos "^ ) = 1 + ^i^ ^ j / A A\^ by subtraction, (sin — — cos — j = 1 - sin J.. .-. sin — + COS — = ±\/l+sin^ (1), A A . and sin — — cos — = +vl -sin J[ ..(2). A A A A By addition and subtraction, we obtain sin — and cos — ; and A A since there is a double sign before each radical, there are four XX.] FUNCTIONS OF SUBMULTIPLE ANGLES. 251 .A A values for sin — , and four values for cos — corresponding to one value of sin ^. The presence of these four values may be explained as follows. If sin A is given and nothing else is stated about the angle A all we know is that A belongs to a certain group of equi-sinal angles. Let a be the smallest positive angle belonging to this A A group, then A=mr + {-\Ya. Thus in finding sin^ and cos — we are really finding 1 1 sin-{7i7r+(-l)"a}, and cos - {7^7^ + (- l)"a}. First suppose n even and equal to 2m ; then sin- {?i7r + (-l)"a} = sin iimr-\-^\ a . a = sm nvrr cos - + cos mrr sm - A A 2 since sin Tmr = 0, and cos ^mr = ± 1. Next suppose n odd and equal to 2m + 1 ; then sin -{nn + i- 1)" a} = sin (niTr +^-fj /tt a\ , . /tt a = sm mrr cos [^ - ^ 1 +cos mn sm ( - - - A Thus we have four values for sin — when sin A is given and nothing further is known respecting A. In like manner it may be shewn that cos — has the four values + cos - , ± cos ( o — « j • 252 ELEMENTARY TRIGONOMETRY. [chap. 258. Geometrical Illustration. Let a be the smallest positive angle which has the same sine as A ; then -4=W7r+(-l)"a, A and we have to find the sine and cosine of — , that is of -{wrr + i-lYa}. If n is even and equal to 2m, this expression becomes mir + - . A If n is odd and equal to 2m + 1, the expression becomes The angles denoted by the formula m7r+- are bounded by one of the lines OP-^ or OP^ ; and those denoted by the formula m-n + ( ^ "" 9 ) ^^® bounded by one of the lines OP, or OP,. Now sin XOP^ = sin 2' sin XOP^= - sin XOPj = - sin 5 ; sin XOP^= sin ( ? - S sin XOP^ = - sin XOP^ = - sin Thus the values of sin — are A + sin - and + sin i tz — 7: 2 ~ \2 2 Similarly the values of cos — are + cos | and + cos f - - - j . XX.] FtTNCTIONS OF SUBMULTIPLE ANGLES. 253 259. If in addition to the value of sin A we know that A lies between certain limits, the ambiguities of sign in equations (1) and (2) of Art. 257 may be removed. A A Example 1. Find sin— and cos-^ in terms of sin A when A lies between 450° and 630°. In this case — lies between 225° and 315°. From the adjoining figure it is evident that between these limits sin— is greater than cos — and is negative. • ^ ^ r^ — = — 7 .-. sm — + C0S — = - >/l + sm^, and sm- A - cos ^ = - sj\ - sin A. 225 A . 2 sin — = - ^/l 4- sin ^ - >/l - sin A, and 2 cos — = -^l + sin^+ >/l-sin4. Example 2. Determine the limits between which A must lie in order that 2 cos A= - /Jl + sin 2 A - ^1 - sin 2 A. The given relation is obtained by combining sin^ + cos^= - ;^i + sin24 (1), and sin A-co3A= + ^1- sin 2A (2). From (1), we see that of sin A and cos A the numencally greater is negative. From (2), we see that the cosine is the greater. Hence we have to choose limits between which cos A is numerically greater than sin A and is negative. From the figure we see that A lies between 2mr+-r and 2mr-r~, ■ 4 4 254 ELEMENTARY TRIGONOMETRY. [CHAP. Example 3. Trace the changes of cos 6 - sin 6 in sign and mag- nitude as d increases from to 27r. cos ^- sin ^ = ,^2 f — cos ^ — th sin j = ,^2 ( cos 6 cos J - sin ^ sin J J =^2cosr6'+|j . As 6 increases from to j , the expression is positive and decreases from 1 to 0. As d increases from j to "x ' ^^^ expression is negative and in- creases numerically from to - J2. As 6 increases from -r- to -j- 4 4 decreases numerically from - sj^ to As d increases fro increases from to sj'^- As 6 increases from — to 27r, the expression is positive and decreases from ^2 to 1. 260. To find the sine and cosine of 9°. Since cos 9°>sin 9° and is positive, we have sin 9° -f cos 9° = + Vl + sin 18°, and sin 9° - cos 9° = - a/I - sin 18°. .-. sin9°-fcos9°==-fy^l+^^^5_ll=+l,^3:^, sin9°-cos9°=-y^l-'^^^ = -iV5^V5. As 6 increases from x *° "I"' *^^ expression is negative and As increases from "x *° X' *^® expression is positive and and .-. sin9° = i{\/3-i-V5-V5-V5}» and cos 9° = - {Vs+VS + V^+v^}- XX.] FUNCTIONS OF BUBMULTIPLE ANGLES. 255 EXAMPLES. XX. a. 1. When A lies between - 270° and — 360°, prove that — cos A .A /l-c( ^^^2 = -V~^ 119 A A 2. If cos A = 3— r , find sin — and cos — when A lies be- 169 2 2 tween 270° and 360'. 3. If cos A= - ^^ , find sin — and cos — when A lies be- tween 540° and 630°. A A 4. Find sin — and cos — in terms of sin A when A lies Ji A between 270° and 450°. 5. Find sin— and cos— in terms of sin^ when — lies between 225° and 315°. A A 6. Find sin— and cos— in terms of sin^ when A lies 2 2 between -450° and -630°. 24 A A 7. If sin A=Tr-s find sin — and cos — when A lies between 25 2 2 90° and 180°. 8. If sin A— — TTT- , find sin — and cos — when A lies be- 289 2 2 tween 270° and 360°. 9. Determine the limits between which A must lie in order that (1) 2 sinyl = Vl + sin 2^ -- Vl - sin ^A \ (2) 2cos^ = - Vl+sin2^ + \/l-sin2^; (3) 2 sin ^ = - \/l -|-sirr2Z-}-\/l - sin 2^. 256 ELEMENTARY TRIGONOMETRY. [cHAP. 10. If yl = 240°, is the following statement correct ? 2 sin — =Vl 4-sin J. — Vl — sin A. If not, how must it be modified ? 11. Prove that (1) tan7i° = v/6-V3+V2-2; (2) cotl42i° = V2+N/3-2-V6. 12. Shew that sin 9° lies between '156 and '157. 13. Prove that (1) 2sinll° 15' = x/2-\/2TV2; (2) tanll°15' = \/4 + 2V2-(^/2 + l). 14. When 6 varies from to 27r trace the changes in sign and magnitude of (1) cos^+sin^; (2) sin ^-^3 cos ^. 15. When 6 varies from to tt, trace the changes in sign and magnitude of . tan^+cot^ ^ 2sin^-sin2^ ^ ' tan^-cot^' ^ ' 2sin^ + sin2(9" 261. To find tan — when tan A is given and to explain the presence of the two values. Denote tan ^1 by ^ ; then 2 tan — t = tan A = ^^ l-tan2:^ A A .'. ?tan2- + 2tan--jJ=0; A A ^ A -2 + \/4 + 4p -l + VrH2 .*. tan — = =^— = "^"^ . 2 2i t The presence of these two values may be explained as follows. XX.] FUNCTIONS OF BUBMULTIPLE ANGLES. 257 If a be the smallest positive angle wtiich has the given tangent, then J=?i7r + a, and we are really finding the value of tan-(n7r + a). (1) Let n be even and equal to 2m ; then tan -(n7r+ a) = tan ( mTT +- j=tan-. (2) Let n be odd and equal to 2m + 1 ; then tan-(w7r + a) = tan(??i7r+| + ^y=tan (2 + 2)* A a ■ Thus tan — has the two values tan - and tan ( ~ 2 2 (i-i)- T, , ^ TP . ,„^n .1 i^ X ^ -1- Jl + taji^A Example 1. If A = 170°, prove that tan — = -^ — . ^ tan A A A Here — is an acute angle, so that tan ^ must be positive. Hence in the formula — the numerator must have the same tan 4 sign as the denominator. But when ^ = 170°, tan>4 is negative, and therefore we must choose the sign which will make the numerator ^. ^, ^ A -1-Vl + tanM negative : thus tan — = ^ : ■ • 2 tan A Example 2. Given cos A = -6, find tan - , and explain the double answer. ^A _1- cos ^ _ '4 _ 1 ^^"^ 2-l + cosA~l^Q~V . A 1 .". tan — = ziz — . 2 2 Here all we know of the angle A is that it must be one of a group of equi-cosinal angles. Let a be the smallest positive angle of this group ; then A = 2mr ± a. .-. tan ^ = tan ( ?i7r ± - J = tan f ± - j = ± tan - . Thus we have two values differing only in sign. H. K. E. T. 17 258 ELEMENTARY TRIGONOMETRY. [chap. 262. When any one of the functions of an acute angle A is given, we may in some cases conveniently obtain the functions of — 5 as in the following example. b . A Example. Given cos A =- , to find the functions of — . Make a right-angled triangle VQB, in which the hypotenuse PQ = a, and base QB = b', then cos PQR =-^-^ = - = cos^; PQ a :. iPQR=A. Produce RQ to S making QS=QP; .: iPSQ= iSPqJ-^lPQR=^, Now SR = a + b, and PR = Ja^-b% .-. PS^ = {a + bf + {a^-b'^) = 2a^ + 2ab; .: PS=^2a{a + b). The functions of ^ may now be written down in terms of the sides of the triangle PRS. 263. From Art. 125, we have cos A — ^ cos3 A 3 cos 3* Thus it appears that if cos^l be given we have a cubic A A equation to find cos — ; so that cos — has three values. Similarly, from the equation sin^ 3 sin ^ — 4 sin^ — it appears that corresponding to one value of sin A there are three values of sin — . It will be a useful exercise to prove these two statements analytically as in Arts. 254 and 257. In the next article we shall give a geometrical explanation for the case of the cosine. XX.] FUNCTIONS OF SUBMULTIPLE ANGLES. 259 A 264. Given cos A to find cos — , a7id to explain the presence of o the three values. Let a be the smallest positive angle with the given cosine ; then A = 27i7r±a, and we have to find all the values of cos-{27i7r±a). Consider the angles denoted by the formula -(27i7r±a), and ascribe to n in succession the values 0, 1, 2, 3, When 71 = 0, the angles are +- , bounded by OF^ and OQ^ ; when ?i = 1, the angles are -^ ± ^ , bounded by OPg ^^^ OQ^ when ?i=2, the angles are -^ ± - , bounded by OP^ and OQ^. By giving to n the values 3, 4, 5, ... we obtain a series of angles coterminal with those indicated in the figure. Thus OPi, 0^1, OP^, OQ^, OP^, OQs bound all the angles included in the formula x(27i7r + a). Now cos XOQi = cos XOP^ = cos - ; cos cos XOP^=cosXOQ^=cos (^ - ^) ; XOQ^ = cos XOP^ = cos (^ + ^) • Thus the values of cos — are cos - , oos — - — , cos — - — , o o o o 17—2 260 ELEMENTARY TRIGONOMETRY. EXAMPLES. XX. b. 1. If ^ = 320°, prove that A _ -l + Vl + tanM 2 tan J. 2. Shew that . . l + Vl+ta n2 2l ^„ tanJ.= -^ — -— — ^r-^ when -4 = 110. tan ^^ 12 3. Find tan J. when cos2J. = — and A lies between 180° and 225°. A 4 4. Find cot— when cosJ.= -^ and A lies between 180° 2 5 and 270°. 5. If cot 2d = cot 2a, shew that cot 6 has the two vahies cot a and -tana. 6. Given that sin ^=sin a, shew that the values of sin ^ are 3 TT + a sm-, sm-g-, -sm 7. If tan ^= tan a, shew that the values of tan - are «5 . a , TT + a , TT — a tan ^ , tan — — - , — tan — — . 8. Given that cos 3^ = cos 3a, shew that the values of sin^ are + sina, — sinf- + aj, sinl — ±a 9. Given that sin 3^= sin 3a, shew that the values of cos^ are 4- nns n p.ns I — \- n \ . pos i + COSa, COS(^±aj, cos(— ±a). A T CHAPTER XXI. LIMITS AND APPROXIMATIONS. 265. If 6 he the radian measure of an angle less than a right angle^ to shew that sin B, 6, tan 6 are in ascending order of magnitude. Let the angle 6 be represented by A OP. "With centre and radius OA de- scribe a circle. Draw FT at right angles to OP to meet OA produced in T, and join PA. Let r be the radius of the circle. Area of A^ 0P= I AO .OP sin AOP=l r^ sin ; area of sector AOP=~ r^ ; area of AOPT= I OP . PT= Ir.r tan e = lr'^ tan 0. 2 2 2 But the areas of the triangle A OP, the sector AOP, and the triangle OTP are in ascending order of magnitude ; that is, - r2 sin 6, - rH, - r^ tan B are in ascending order of magnitude ; .'. sin B, B, tan B are in ascending order of magnitude. 262 ELEMENTARY TRIGONOMETRY. [CHAP. 266. When 6 is indefinitely diminished, to pi^ove that — -^ and — 2 — ^^^^ have unity for their limit. 6 In the last article, we have proved that sin ^, 6, tan 6 are in ascending order of magnitude. Divide each of these quantities by sin 6 ; then 1, -; — ;: , ;7 are iu ascending order of magnitude ; ' sm^' cos^ ^ ^ ' a that is, — — ;: lies between 1 and sec 6. sm^ But when 6 is indefinitely diminished, the limit of sec^ is 1 : hence the limit of —. — ^ is 1 ; that is, the limit of — .— ' sm^ ' ' 6 is unity. Again, by dividing each of the quantities sin (9, 6, tan $ by tan 6, we find that cos 6, - — ^ , 1 are in ascending order of mag- tan o nitude. Hence the limit of — ^— is unity. These results are often written concisely in the forms ''tan ff" LtJ^^\ = l, Lt.(^- = 1. ■ a Example. Find the limit of n sin - when w = oo . n . e . n .6 n sm - = d . - . sm - : nan I sm - -7- - ) ; \ n nj n ft R but since - is indefinitely small, the limit of sin - -4- - is unity ; 71 it 71 .'. Lt. f n sin n=co Similarly Lt. n/ i n tan -] = d. XXI.] LIMITS AND APPROXIMATIONS. 263 267. It is important to remember that the conclusions of the foregoing articles only hold when the angle is expressed in radian measure. If any other system of measurement is used, the results will require modification. Example. Find the value of Lt. i ) . n=0\ n J Let 6 be the number of radians in n° ; then n 6 ^ 180^ , . o . ^ zTKR = - > and n = ; also sm n° = sm 6 ; loU IT IT sin n° IT sin d ir sin 6 " n 180^ 180' d ' When n is indefinitely small, d is indefinitely small ; ^^/sinn^\ jr j;^^/sin^\. n=Q\ n J 180* 6=Q\ e J' n=o\ n J 180 268. When 6 is the radian measure of a very small angle, we have shewn that sin d ^ , , tan , -^=1, cos^=l, "^- = 1; . that is, sin^=^, cos^ = l, tan^ = ^. Hence r tan B=rd, and therefore in the figure of Art. 265, the tangent PT is equal to the arc FA, when /. A OF is very small. In Art. 270, it will be shewn that these results hold so long as B is so small that its square may be neglected. When this is the case, we have sin (a + 6) = sin a cos 6 + cos a sin 6 = sina + ^cosa; cos (a + 6) = cos a cos 3 — sin a sin 6 = cosa — ^sina. 264 ELEMENTARY TRIGONOMETRY. [chap. Exaviple 1. The inclination of a railway to tlie horizontal plane is 52' 30", find how many feet it rises in a mile. Let OA be the horizontal plane, and OP a mile of the railway. Draw PA'' perpendicular to OA. het FN =x feet, iPON=0; then PN Yyp = sin 6 = 6 approximately. But ^=: radian measure of 52' 30" = -~ x -^_ = b x -^ ; bO loO o loO X 7 22 1760 X 3 1760 X 3 X 22 8 ^ 7 "^ 180 ' 8x180 242 = ^ = 801 Thus the rise is 80| feet. Example 2. A pole 6 ft. long stands on the top of a tower 54 ft. high : find the angle subtended by the pole at a point on the ground which is at a distance of 180 yds. from the foot of the tower. Let A be the point on the ground, BC the tower, CD the pole. Let ABAC = a, lCAD=6; ., . PC 54 1 then tan a= - - = = — ; AB 540 10' . , ^, BD 60 1 tan (a + 6) = ■ — = = - ^^ ^ AB 540 9 But tan(a + ^) = tan a + tan 6 tan a + 1 — tan a tan 6 1 10 1 + 1-6 tan a 1 + log . lO-^* ' approximately ; 10 whence 6 = - that is, the angle is — of a radian, and therefore 91' , . 1 180 , contams — x — degrees. 91 IT On reduction, we find that the angle is 37' 46" nearly. XXI.] LIMITS AND APPROXIMATIONS. 265 269. If 6 he the number of radians in an acute angle, to prove that cos ^ > 1 — -^ , and sin 6>6— ~. Since cos ^ = 1 — 2 sin^ - , and sin - < - ; 2 2 2 .-. cos^>l-2(^0'; that is, cos ^ > 1 — — . . 6 6 6 6 Again, sin ^ = 2 sin - cos „ = 2 tan - cos^ - ; but tan 2 > 2 ; 6 6 .'. sin ^>2 -cos^- ; 2 2 .-. sin^><9(l-sin-- 6 6 But sin 7: < TT , and therefore .'. sm6>6--r- 4 270. From the propositions established in this chapter, it follows that if 6 is an acute angle, 2 cos 6 lies between 1 and 1 — „ , 6^ and sin 6 lies between 6 and ^ — -r • 4 Thus cos 6 = 1 -kS^ and sin ^ = ^ — ^•'^^, where k and k' are proper fractions less than ^ and j respectively. 266 ELEMENTARY TRIGONOMETRY. [CHAP. Hence if B be so small that its square can be neglected, cos^ = l, sin^=^. Example. Find the approximate value of sin 10". The circular measure of 10" is — — — -- — -- or 180x60x60 64800' sin 10" < TTTT^^TTT and > ■ -f— -V. 4V64800; 64800 64800 4 - 61^00 = ^^^^^- =— ■ TT < 64800 •00005 and (g^^) < -000000000000125; •'• ^^°^^"< 64^00 ^^^ >64^- I (-000000000000125). Hence to 12 places of decimals, sin 10"= xtIt^t.^ -000048481368... . 64800 271. To shew that when n is an indefinitely large integer, the ,. ., . 6 6 6 6 mn6 limit of cos - cos - cos - . .. cos ^— = — -— . •^ 2 4 8 2'' 6 We have sin 6 — 2 sin - cos - 2 2 = 2^ sm - cos - cos - 4 4 2 ...6666 = 2"^ sin - cos - cos - cos - 8 8 4 2 on • ^ ^ 6 6 6 =2" sm - cos- ... cos - cos- cos^. 6 6 6 6 sin 6 cos - COS - COS - . . . COS 2 4 8'" 2" ^ . ^ ■ 2"sm- a But the limit of 2" sin— is 6, and thus the proposition is established. [See Art. 266.] XXI.] LIMITS AND APPROXIMATIONS. 267 272. To shew that - continually decreases from \ to - as 6 continually increases from to — . We shall first shew that the fraction sin (9 sin(^ + /0 . -0 ^:^ IS positive, h denoting the radian measure of a small positive angle. ^, (6 + h)sm6-d (sin d cos h + cos 6 sin h) This fraction = ^^ ~-r- — r\ d {a + n) 6 sin ^ (1 — cos h) + {hsmB-0 cos 6 sin h) Now tan 6>6, that is sin d>$ cos 0, and h > sin h ; .•. A sin^>^cos^ sinA. Also 1 — cos h is positive ; hence the numerator is positive, and therefore the fraction is positive ; sin {d + h) sin 6 •*• e+h '^~r'' .'. — ^— continually decreases as 6 continually increases. 6 wi. z) A sin^ ^ , , .IT sin^ 2 W hen ^ = 0, — -r- = 1 ; and when 6 = ~ ^ — -x- = - . A t) TT Thus the proposition is established. EXAMPLES. XXI. a. In this Exercise take tt = -=- . 1. A tower 44 feet high subtends an angle of 35' at a point A on the ground : find the distance of A from the tower. 2. From the top of a wall 7 ft. 4 in. high the angle of depression of an object on the ground is 24' 30" : find its distance from the wall. 268 ELEMENTARY TRIGONOMETRY. [CHAP. 3. Find the height of an object whose angle of elevation at a distance of 840 yards is 1° 30'. 4. Find the angle subtended by a pole 10 ft. 1 in. high at a distance of a mile. 5. Find the angle subtended by a circular target 4 feet in diameter at a distance of 1000 yards. 6. Taking the diameter of a penny as 1'25 inches, find at what distance it must be held from the eye so as just to hide the moon, supposing the diameter of the moon to be half a degree. 7. Find the distance at which a globe 1 1 inches in diameter subtends an angle of 5'. 8. Two places on the same meridian are 11 miles apart: find the difference in their latitudes, taking the radius of the earth as 3960 miles. 9. A man 6 ft. high stands on a tower whose height is 120 ft. : shew that at a point 24 ft. from the tower the man subtends an angle of 31 '5' nearly. 10. A flagstaff standing on the top of a cliff 490 feet high subtends an angle of "04 radians at a point 980 feet from the base of the cliff : find the height of the flagstaff. 11. When 71 = 0, find the limit of sinn' . sinn" n 71 1 27r 12. When ?i = oo , find the limit of - nr^ sin — . . 2 % When ^=0, find the limit of 1-cos^ -. m sin m6 — n sin nd 13 . 14 ~~ 6 sin d ' ' tan mO + tan 016 15. If ^ = "01 of a radian, calculate cos (— + ^ ] . 16. Find the value of sin 30° 10' 30". 17. Given cos ( q + ^ ) ~ '^^y ^^^ ^^® sexagesimal value of ^, ' XXI.] DISTANCE AND DIP OF THE HORIZON. 269 Distance and Dip of the Visible Horizon. 273. Let A be a point above the earth's surface, BCD a section of the earth by a plane passing through its centre £J and A. Let AB cut the circumference in B and J). From A draw AC to touch the circle BCD in C, and join EC. Draw AF at right angles to AD ; then L.FAC is called the dip of the horizon as seen from A. Thus the dip of the horizon is the angle of depression of any point on the horizon visible from A. 274. To find the distance of the horizon. In the figure of the last article, let AB=h, EB = ED = r, AC=x', then by Euc. iii. 36, AC^ = AB.AD', that is, x^ = h {^r + A) = 2Ar + h\ \ For ordinary altitudes K^ is very small in comparison with %hr ; hence approximately x^^2hr and x=sj'^hr. In this formula, suppose the measurements are made in miles^ and let a be the number of feet in AB ; then a = 1760x3x/^. By taking r= 3960, we have 2 X 3960 X g _ 3a 1760x3 ~T* Thus we have the following rule : Twice the square of the distance of the horizon measured in miles is equal to three times the height of the place of observation measured in feet. Hence a man whose eye is 6 feet from the ground can see to a distance of 3 miles on a horizontal plane. ^^=' 270 ELEMENTARY TRIGONOMETRY. [chap. Example. The top of a ship's mast is 66f ft. above the sea-level, and from it the lamp of a lighthouse can just be seen. After the ship has sailed directly towards the lighthouse for half-an-hour the lamp can be seen from the deck, which is 24 ft. above the sea. Pind the rate at which the ship is sailing. B C Let L denote the lamp, D and E the two positions of the ship, B the top of the mast, G the point on the deck from which the lamp is seen; then LCB is a tangent to the earth's surface at A. [In problems like this some of the hnes must necessarily be greatly out of proportion.] Let AB and ^C be expressed in miles; then since DB = Q&^ feet and EC = 24: feet, we have by the rule 4^2 = 1 X 661=100; .-. ^5=10 miles. AC^ = lx24: = S&; :. yiC=6 miles. But the angles subtended by AB and AG at the centre of the earth are very small ; .-. arc AD = AB, and arc AG=AE. [Art. 268.] .-. arc DE = AD - AE = AB- AC = 4: mUes. Thus the ship sails 4 miles in half-an-hour, or 8 miles per hour. 275. Let $ be the number of radians in the dip of the horizon ; then with the figure of Art. 273, we have . EC r f,h\-^ h h? 2sin^- = ^4-, h SB Since B and - are small, we may replace sin- by - and neglect the terms on the right after the first. XXI.] DISTANCE AND DIP OF THE HORIZON. 271 Thus -77 = -, or 6 = 2 r 1 2k V r' Let N be the number of degrees in 6 radians ; then 180^ _ 180 /2A ^ ~~ IT ~ TV Si r ' Now v'?'=63 nearly ; hence we have ap^jroximately 180 X 7 X sl^Ji N= 22x63 10 or N=^^2h, a formula connecting the dip of the horizon in degrees and the height of the place of observation in miles. [ EXAMPLES. XXI. b. 22 . . ' Here tt = -^ , and radius of earth = 3960 miles. 1. Find the greatest distance at which the lamp of a lighthouse can be seen, the light being 96 feet above the sea- level. 2. If the lamp of a lighthouse begins to be seen at a distance of 15 miles, find its height above the sea-level. 3. The tops of the masts of two ships are 32 ft. 8 in. and 42 ft. 8 in. above the sea-level : find the greatest distance at which one mast can be seen from the other. 4. Find the height of a ship's mast which is just visible at a distance of 20 miles from a point on the mast of another ship which is 54 ft. above the sea-level. 5. From the mast of a ship 73 ft. 6 in. high the lamp of a lighthouse is just visible at a distance of 28 miles : find the height of the lamp. 6. Find the sexagesimal measure of the dip of the horizon from a hill 2640 feet high. 272 ELEMENTARY TRIGONOMETRY. 7. Along a straight coast there are lighthouses at intervals of 24 miles : find at what height the lamp must be placed so that the light of one at least may be visible at a distance of 3^ miles from any point of the coast. 8. From the top of a mountain the dip of the horizon is l'8i° : find its height in feet. 9. The distance of the horizon as seen from the top of a hill is 30'25 miles : find the height of the hill and the dip of the horizon. 10. If X miles be the distance of the visible horizon and JV degrees the dip, shew that V n ,^_ ^ /lO When ^=0, find the limit of sin 4^ cot ^ -- 1-cos^+sin^ vers 26 cot^ 26' * 1 — cos 6 — sin 6 ' 13. AVhen 6 = a, find the limit of . . sin ^ — sin a , cos ^- cos a ^^^ ' 6-a ' ^^^ I^a • 14. Two sides of a triangle are 31 and 32, and they include a right angle : find the other angles. 15. A person walks directly towards a distant object P, and observes that at the three points A, B, C, the elevations of P are a, 2a, 3a respectively : shew that AB = ZBC nearly. 16. Shew that — ^ — continually increases from 1 to oo as ^ 6 continually increases from to ^ . CHAPTER XXII. GEOMETRICAL PROOFS. 276. To find the expansion o/tan (^4 -\-B) geometrically. Let LLOM=A,Bnd lMON=B; then lLON=A+B. In ON take any point P, and draw PQ and PR perpendicular to OL and OM respectively. Also draw RS and RT perpendicular to OL and PQ respectively. , ,,^j,,PQ_RS-\-PT RS PT OS ^ OS 'OS RS PT OS "^ OS 1- TR TP' Now TP' OS RS ^ , , TR . , -Yo=tan^, and -^^=tanJ.; also the triangles ROS and TPR are similar, and therefore TP_PR_, „ 'OS~OR~^^'^^' , , „, tan^ + tan^ H. K. E. T. 18 274 ELEMENTARY TRIGONOMETRY. [chap. Ill like manner, with the help of the figure on page 95, we may obtain the expansion of tan {A — B) geometrically. 277. To prove geometrically the formulm for transformation of sums into products. Let /-EOF he denoted by A, and lEOG by B. With centre and any radius describe an arc of a circle meeting OG in H and OF in K. Bisect L KOH by OL ; then OL bisects HK at right angles. Draw KP, HQ., LR perpendicular to OF, and through L draw MLN parallel to OE meeting KP in M and QH in N. It is easy to prove that the triangles MKL and NHL are equal in all respects, so that KM=NII, ML = LN, PR = RQ. Also L. GOF—A - B, and therefore lHOL=lKOL = A-B lEOL=B^^ A-B A+B {KM+LR) + {LR-NH) _^ LR OK ~ OK' XXII.] TRANSFORMATION OF SUMS INTO PRODUCTS. 275 . • . sill .4 + sin 5 = 2 7:;-^ . 7^-^= 2 sin ROL cos KOL ^ . A+B A-B = 2 sm — - — cos — - — . cos^+cos^=^. + ^^=-^^- _ {OR-PR) + {OR + RQ ) ^^OR OK ''OK = 2y-^. ^.= 2 cos ROL cos KOL OL OK ^ A+B A-B = 2co3 -— cos ^ -- • . , . „ KP JIQ KF-SQ sm A - sm B=-^~^. - ^= — O^T" { KM+LR) - (LR - NH) _^KM OK OK = 2^^. ^.=2 cos LKM sin KOL KL OK A+B . A-B = 2cos— g— sm— g- , A+B since L XZif=comp' of lKLM= L ML0= lLOE^—^- . ^ . OQ OF OQ-OF cos5-cos^ = ^-^-^ = — ^^^ _ {OR + RQ)-{OR-FR) _ ^PR_^ML - OK " 0K~ OK = 2 ^^ . ^= 2 sin Z^Jf sin ^OZ KL UK „ . ^+^ . A-B = 2 sin — - — sm — r — . 18—2 276 ELEMENTARY TRIGONOMETRY. [chap. 278. Oeoiiietrical proof of the 2 A formulce, Let BPB be a semicircle, BB the diameter, C the centre. On the circumference, take any point P, and join PB, PC, PD. Draw PN perpendicular to BD, Let lPBD=A, then B lPCD = 2A. And z.iyPi) = comp' of lPDN= lPBD = A. . PN _2PN^_2PN^ PN BP_ sm 2.4-^p - 2^p - ^jy -^ ^p • j^jy = 2 sin PBN cos PBD = 2 sin J. cos J.. CN 2CN CN+CN CP ~ BD~ BD _ {BN-BC) + {CD-NJ)) _ B N-NB ~ BD ~ BD _BN BP ND PD ~BP ' BD PD' BD — cos A . cos A — sin A . sin ^1 =008^-4 — sin^^l. ,_CN _ CD-DN _ DN 2DN cos2A-^p- ^p -1 ^p-l ^^ , ^DN DP ^ ^ . , . . = l-2^p.-^-^=l-2smA.smA N D cos 2 A = 1-2 sin2 A. CN_BN-BC cos zA — -^jp — -^p : BN 2BN CP BD' = 2 BN BP BP' BD — 2cos'^ A — 1. — 1 = 2 cos A . cos A — 1 XXII.] VALUE OF SIN 18° 277 tan 2 A = ^^r^ = FJV 2PN 2PN CN 2GN PN ^ BN BN-ND ^ BN 1- JS'D BN 2 tan A 1- ND PN PN ' BN 1 — tan xi 2 tan>4 ^l-tan2.1 tan^ 279. To find the value of sin 18° geometrically Let ABD be an isosceles triangle in which each angle at the base BD is double the ver- tical angle A ; then ^ + 2^ + 2^ = 180°, and therefore J. = 36°. Bisect lBAD by AE \ then AE bisects BD at right angles ; .'. L BAE =18°. sml8 =— ^ = -, AB a Thus where AB=a, and BE=x. From the construction given in Euc. iv. 10, AC=BD = 2BE=2x, and AB.BC=AC^; .-. a (a -2^) = (2^)2; . • , 4tx^ -f- 2ax -a^ = 0; -2a±\/20a2 -l±^d x = 8 The upper sign must be taken, since x is positive. 1 no V ^ ~ •*- Sin 18 = . Thus 278 ELEMENTARY TRIGONOMETRY. CHAP. Proofs by Projection. 280. Definition. If from any two points A and B, lines AC and BD are drawn perpendicular to OX, then the intercept CD is called the projection of AB upon OX. Through A draw AE parallel to OX; then CI)=AE=^ABcosBAE; that is, CB^AB cos a, where a is the angle of inclination of the lines AB and OX. 281. To skew that the projection of a straight line is equal to the projection of an equal and parallel straight line drawn from a fixed point. Let AB be any straight line, a fixed point, which we shall call the origin, OP a straight line equal and parallel to AB. Let CD and OM be the pro- jections of AB and OP upon any straight line OX drawn through the origin. Draw A E parallel to OX. The two triangles AEB and OMP are identically equal ; .-. OM=AE=CD', that is, projection of OP = projection of AB. 282. In the figure of the last article, two straight lines OP and OQ can be drawn from equal and parallel to AB ; it is therefore necessary to have some means of fixing the direction in which the line from is to be drawn. Accordingly it is agreed to consider that the direction of a line is fixed hy the order of the letters. Thus AB denotes a line drawn from A to B, and BA denotes a line drawn from ^ to ^. D X XXII.] PROOFS BY PROJECTION. 279 Hence OP denotes a line drawn from the origin parallel to AB^ and OQ denotes a line drawn from the origin parallel to BA. Similarly the direction of a projected line is fixed by the order of the letters. Thus CD is drawn to the right from C to D and is positive, while DC 19, drawn to the left from D to C and is negative. Hence in sign as well as in magnitude OM=CD, and ON=^DC; that is, projection of OP = projection of AB, and projection of (9^= projection of BA. Thus the projection of a straight line can be represented both in sign and magnitude by the projection of an equal and parallel straight line draion from the origin. P Y 283. Whatever be the direction of AB, the line OP will fall within one of the four quadrants. Also from the definitions given in Art. 75, we have OM ^^ p ^ = cos ZOP, that is, OM=OPq,o^XOP, whatever be the magnitude of the angle XOP. We shall always sup- pose, unless the contrary is stated, that the angles are measured in the positive direction. 284. Let be the origin, P and Q any two points. Join OP, OQ, PQ, and draw PM and QN perpendicular to OX. We have OM=^ON^NM, since the line NM is to be regarded as O negative ; that is, N X the projection of OP=projection of 0^+projection of QP. 280 ELEMENTARY TRIGONOMETRY. [chap. Hence, the projection of one side of a triangle is equal to the sum of the projections of the other two sides taken in order. Thus projection of (9 $= projection of OP + projection of P^; projection of §P = projection of $0 + projection of OP, General Proof of the Addition Formulae. 285. In Fig. 1, let a line starting from OX revolve until it has traced the angle ^4, taking up the position Oi/, and then let it further revolve until it has traced the angle B^ taking up the final position ON. Thus XON is the angle A^B. Fig. I Fig.2. In ON take any point P, and draw PQ perpendicular to OM], also draw OR equal and parallel to ()P. Projecting upon OX, we have projection of 0P= projection of 0^+ projection of QP = projection of 0^ + projection of OR. .-. OPco^XOP-=OQcosXOQ + ORco^XOR (1) = OP cos B cos XOQ + OP sin B cos XOR ; . • . cos XOP = cos B cos XO ^ + sin 5 cos XOR ; that is, cos {A+B) = cos B cos A + sin B cos (90° + A) = cos A cos B — sin A sin B. Projecting upon 01\ we have only to write Y for A^ in (1); XXII.] PROOFS BY PROJECTION. 281 thus OP COS TOP = OQcos YOQ + OR cos FOE = OP cos B cos rOQ + OP sin B cos FOE ; .• . cos TOP = cos B cos TO^ + sin B cos FOi^ ; that is, cos {A+B - 90°) = cos B cos (.1 - 90°) + sin i? cos ^ ; .*. sin(J.+^) = sin J. cos^+cos J. sin5. In Fig. 2, let a line starting from OX revolve until it has traced the angle A, taking up the position OM, and then let it revolve back again until it has traced the angle B, taking up the final position ON. Thus XON is the angle A-B. In OiV take any point P, and draw PQ perpendicular to MO produced ; also draw OR equal and parallel to QP. Projecting upon OX, we have as in the previous case OP cos XOP = OQ cos XOQ + OR cos XOR = OP cos (180° - B) cos XOQ ■^ OP sin {180° -B) cos XOR; . • . cos XOP =-cosB cos XOQ+ sin B cos XOR ; that is, cos (A-B)=- cos B cos {A - 180°) + sin B cos {A - 90°) = - cos ^ ( — cos ^4) +sin 5 sin ^ = cos A cos B + sin ^4 sin B. Projecting upon OF, we have OP cos FOP=OQ cos FOQ+OR cos rO/2 ; = OP cos (180° - 5) cos FOQ + OP sin (180° - B) cos FOi? ; . • . cos FOP =-cosB cos FOQ + sin ^ cos FOB ; that is, cos {A-B- 90°)= - cos B cos (^ - 270°) + sin B cos (.4 - 180°) ; .*. sin (J - B)= -cos5(-sin J.) + sini5(-cos J.) = sin A cos B - cos A sin B. 282 ELEMENTARY TRIGONOMETRY. [CHAP. 236. The above method of proof is appHcable to every case, and therefore the Addition Formulae are imiversally estabUshed. The universal truth of the Addition Formulae may also be deduced from the special geometrical investigations of Arts. 110 and 111 by analysis, as in the next article. 287. When each of the angles A, B, A+B is less than 90°, we have shewn that cos(^ + ^) = cos^ cos^-sin^sin^ (1). But cos {A +^)=sin {A +^ + 90°) = sin (A +90° + B) ; also cos ^ = sin {A + 90°), and - sin ^ = cos (A + 90°). [Art. 98.] Hence by substitution in (1), we have sin {A + 90° + B) = sin (A + 90°) cos B + cos {A + 90°) sin B. In like manner, it may be proved that cos (^+90° + B) = cos {A + 90°) cos 5 - sin (4 + 90°) sin B. Thus the formulae for the sine and cosine of ^ +^ hold when A is increased by 90°. Similarly we may shew that they hold when B is increased by 90°. By repeated applications of the same process it may be proved that the formulae are true when either or both of the angles A and B is increased by any multiple of 90°. Again, cos(^+5)=cos^cos jB — sin^ sin^ (1). But cos (A+B)=- sin (Z+^- 90°) = - sin (A-90° + B) ; also cos ^ = - sin (^1 - 90°), and sin A =cos (A - 90°). [Arts. 99 and 102.] Hence by substitution in (1), we have sin (^-90° + B) = sin {A - 90°) cos B + cos {A - 90°) sin B. Similarly we may shew that cos {A-90° + B) = cos {A - 90°) cos B - sin {A - 90°) sin B. Thus the formulae for the sine and cosine of A+B hold when .1 is diminished by 90°. In like manner we may prove that they are true when B is diminished by 90°. XXII.] MISCELLANEOUS EXAMPLES. H. 283 By repeated applications of the same process it may be shewn that the formulae hold when either or both of the angles A and B is diminished by any multiple of 90°. Further, it will be seen that the formulae are true if either of the angles ^ or 5 is increased by a multiple of 90° and the other is diminished by a multiple of 90°. Thus sin {P+Q) = sin FcosQ + cos F sin Q, and cos(P+$) = cos Pcos ^ — sin Psin ^, where P=J±m. 90°, and Q=B±n. 90°, m and n being any positive integers, and A and B any acute angles. Thus the Addition Formulae are true for the algebraical sum of any two angles. MISCELLANEOUS EXAMPLES. H. 1. If the sides of a right-angled triangle are cos 2a + cos 2/3+ 2 cos (a +/3) and sin 2a + sin 2/3 + sin 2 (a +/3), shew that the hypotenuse is 4 cos^ — ~— . Ji 2. If the in-centre and circum-centre be at equal distances from BG, prove that cos ^ + cos (7=1. 3. The shadow of a tower is observed to be half the known height of the tower, and some time afterwards to be equal to the height : how much will the sun have gone down in the interval ? Given log 2, Ztan 63° 26' = 10-3009994, diff. for r = 3159. 4. If (l+sin«)(l+sini3)(l+siny) = (1 - sin a) (1 - sin /3) (1 - sin y), shew that each expression is equal to +cos a cos/3 cos y. 5. Two parallel chords of a circle lying on the same side of the centre subtend 72° and 144° at the centre : prove that the distance between them is one-half of the radius. Also shew that the sum of the squares of the chords is equal to five times the square of the radius. 284 ELEMENTARY TRIGONOMETRY. [CHAP. 6. Two straight railways are inclined at an angle of 60°. From their point of intersection two trains A and B start at the same time, one along each line. A travels at the rate of 48 miles per hour, at what rate must B travel so that after one hour they shall be 43 miles apart 1 7. If a = cos-i- + cos-if , a shew that sin^ a = -^ ^ cos a+%i,. 8. If ^, q, r denote the sides of the ex-central triangle, prove that ^2 je c2 2ahc , 9. A tower is situated within the angle formed by two straight roads OA and OB, and subtends angles a and /3 at the points A and B where the roads are nearest to it. If OA = a, and OB = b, shew that the height of the tower is Va^ - i>^ sin a sin j3 / Vsin (a + /3) sin (a - /3). 10. In a triangle, shew that r^+rj^+r^^+r^^=l6R^-a^-b'--cl 11. If AD be a median of the triangle ABC, shew that (1) cot^.4Z) = 2cot^+cot^; (2) 2 cot ADC= cot B - cot C. 12. If p, q, r are the distances of the orthocentre from the sides, prove that fa h c\ _ /a b c\ /b . c d\ ( c a b \V q r)~\j) q Graphical Representation of the Circular Functions. 288. Definition. Let / {x) be a function of x which has a single value for all values of x, and let the values of x be repre- sented by lines measured from along OX or 0X\ and the values of f{x) by lines drawn perpendicular to XX'. Then with the figure of the next article, if OM represent any value of .r, and MP the corresponding value off{x), the curve traced out by the point F is called the Graph of / (x). r\ b c a\ ( c a [- + -■ --) - + - r) \q r V \r p XXII.] GEAPHS OF SIN 6 AND COS 6. 285 Graphs of sinO and cos 6. 289. Suppose that the unit of length is chosen to represent a radian ; then any angle of 6 radians will be represented by a line OM which contains 6 units of length. Graph of sin^. Let MP^ drawn perpendicular to OX, represent the value of sin 9 corresponding to the value OM of 6 ; then the curve traced out by the point P represents the graph of sin 6. As OJf or B increases from to - , MP or sin 6 increases from to 1, which is its greatest value. As Oif increases from - to tt, MP decreases from to 1. As OM increases from tt to — , 3IP increases numerically from to — 1. As OM increases from — to 27r, MP decreases numerically from — 1 to 0. As OM increases from 27r to 477, from 47r to Gtt, from Gtt to 877, , MP passes through the same series of values as when OM increases from to 2rr. Since sin ( - ^) = - sin 6, the values of MP lying to the left of are equal in magnitude but are of opj)osite sign to values of MP lying at an equal distance to the right of 0. Thus the graph of sin ^ is a continuous waving line extending to an infinite distance on each side of 0. The graph of cos 6 is the same as that of sin ^, the origin being at the point marked - in the figure. 286 ELEMENTARY TRIGONOMETRY, [chap. Graphs of tan and cot 9. 290. As before, suppose that the unit of length is chosen to represent a radian ; then any angle of radians will be repre- sented by a line OM which contains 3 units of length. Let J/P, drawn perpendicular to OX, represent the value of tan 6 corresponding to the value OM of ; then the curve traced out by the point F represents the graph of tan d. By tracing the changes in the value of tan 6 a,s 6 varies from to Stt, from £77 to 47r, , it will be seen that the graph of tan consists of an infinite number of discontinuous equal branches as represented in the figure below. The part of each branch be- neath JiX' is convex towards XX', and the part of each branch above XX' is also convex towards XX' ; hence at the point where any branch cuts XX' there is what is called a poi7it of injieonon, where the direction of curvature changes. The proof of these statements is however beyond the range of the present work. The various branches touch the dotted lines passing through the points marked + ?, ^-^'^ ■2' -^2' at an infinite distance from XX'. Graph of tan^. The student should draw the graph of cot^, which is very similar to that of tan 6. XXII.] GRAPHS OP SEC 6 AND COSEC 6. 287 Graphs of sec 6 and cosec 6. 291. The graph of sec^ is represented in the figure below. It consists of an infinite number of equal festoons lying alter- nately above and below XX'^ the vertex of each being at the unit of distance from XX'. The various festoons touch the dotted lines passing through the points marked -2' -T' -T' ' at an infinite distance from XX'. The graph of cosec 6 is the same as that of sec ^, the origin TT being at the point marked - ^ in the figure. CHAPTER XXIII. SUMMATION OF FINITE SERIES. 292. An expression in which the successive terms are formed by some regular law is called a series. If the series ends at some assigned term it is called a finite series ; if the number of terms is unlimited it is called an infinite series. A series may be denoted by an expression of the form where Un + i, the (n + 1)* term, is obtained from u^, the ii"* term, by replacing nhy n+1. Thus if %^ = cos{a+n^), then w,J^.l = cos{a-^-(?^+l)/3} ; and if it^ = cot 2"'"i a, then u,^ + ^ = cot 2** a. 293. If the r"* term of a series can be expressed as the difference of two quantities one of which is the same function of r that the other is of r+1, the siun of the series may be readily found. For let the series be denoted by u^+U2 + u^ + +^ln, and its sum by S, and suppose that any term then S= (^2 - Vj) + (Vg - V.2) + (^4 - Vg) -f . . . + (v„ - V^_j) + («^n + 1 - Vn) Example. Find the sum of the series cosec a + cosee 2a + cosec 4a + + cosec 2""^ a. sm - sin ' 1 "^"2 V cosec a = — ^^ -i) sin a . a . . a . sm ^ sm a sm - sm a SUMMATION OF FINITE SERIES. 289 Hence cosec a = cot ^ - cot a. If we replace a by 2a, we obtain cosec 2a = cot a - cot 2a. Similarly, cosec 4a = cot 2a - cot 4a, cosec 2"-i a = cot 2"-2 a - cot 2"'~i a. By addition , ^S = cot ^ - cot 2"'-i a. 294. 3^0 ^nS' sm I = cos ( a - 1 ) - cos ( a + -^— /3 i / n-l («+-2- = 2sin(a + ^-/3)siny; -f 2 . / ?i-l ^ , .•.^=^_sin(a+-2— /3 sin 2 H. K. E. T. 19 290 ELEMENTARY TRIGONOMETRY. [CHAP. 295. In like manner we may shew that the sum of the cosine-series cos a + cos (a 4- /3) + cos (a + 2/3) -{- . . . 4- cos {a + (ti - 1) ^} sin| 296. The formulae of the two last articles may be expressed verbally as follows. The sum of the sines of a series ofn angles in a. p. n diff. sm 2 . first angle + last angle . diff. 2 The sum of the cosines of a series ofn angles in a. p. n diff. sm- . diff. sm-f- 2 first angle + last angle Example. Pind the sum of the series cos a + cos 3a + cos 5a + + cos (2n - 1) a. Here the common difference of the angles is 2a; „ sin na a + (2n - 1) a .• S:=—. cos ^— TT — -sm a 2 _ sin na cos na _ sin 27ia "" sin a ~ 2 sin a nQ 297. If sin -^ = 0, each of the expressions found in Arts. 294 and 295 for the sum vanishes. In this case —z=kir^ or B= — - , where h is any integer. ^ n Hence the sum of the sines and the sum of the cosines of n angles in arithmetical progression are each equal to zero, when the common difference of the angles is an even midtiple of - . XXin] SUMMATION OF FINITE SERIES. 291 298. Some series may be brought under the rule of Art. 296 by a simple transformation. Example 1. Eind the sum of n terms of the series cos a - cos (a + /3) + cos (a + 2j8) - cos (a + 3/3) + This series is equal to cosa + cos(a+j3 + 7r) + cos(a + 2/3 + 27r)+cos(a + 3)3 + 37r)+ , a series in which the common difference of the angles is jS + tt, and the last angle is a + (w-l) (jS + tt). „ 2 j , (n-l)(3 + 7r) ] sm^ Example 2. Find the sum of n terms of the series sina + cos(a+/3)-sin(a + 2j3)-cos(a + 3/3) + sin(a + 4|8) + This series is equal to sina + sin fa + j3 + | J + sin(a + 2/3 + 7r) + sin(a + 3j3+-^) + , TT a series in which the common difference of the angles is j8 + - . . n(2/3 + 7r) o 4 . ( (n-l)(2^ + 7r)) 2/3 + 7r 4 ( (n-l)(2^ + 7r) j ^ EXAMPLES. XXIII. a. Sum each of the following series to n terms : 1. sin a + sin 3a 4- sin 5a 4- 2. cos a + cos (a -/3) + cos (a — 2/3) + 3. sina + sinf a — )+sin( a W J, TT , 27r , Stt 4. cos -7 4- cos -^+ cos -7-+ 19—2 292 ELEMENTARY TRIGONOMETRY. [CHAP. Find the sum of each of the following series : 5. cos-+cos^+cos-+ + cos^. 27r , 477 , 67r , , 207r _ 6. C0S- + C0S-+C0S-+ +COS— . 7. sin-+sin hsin 1- to ?i - 1 terms. n n n 8. COS- + C0S — hcos — 1- to 2?2< - 1 terms. - n 71 n 9. smna + sm{n—l)a+sm{n-2)a-\- to 2?i terms. Sum each of the following series to n terms : 10. sin <9- sin 2^ + sin 3^ -sin 4^ + 11. cosa-cos(a-/3)^-cos(a-2/3)-cos(a-3^) + 12. cos a - sin (a -iS)- cos (a -2/3)+ sin (a -3/3) + 13. sin 2^ sin ^ + sin 3^ sin 2^ + sin 4^ sin 3^ + 14. sinacos3a + sin3acos5a + sin5aCOs7a+., 15. sec a sec 2a + sec 2a sec 3a + sec 3a sec 4a + 16. cosec 6 cosec 3^ +cosec 3^ cosec 56 + cosec 5^ cosec 7^ + 17. tan-seca+tan^sec-+tan-3sec^+ 18. cos 2a cosec 3a + cos 6a cosec 9a + cos 1 8a cosec 27a + 19. sin a sec 3a + sin 3a sec 9a + sin 9asec27a+ 20. The circumference of a semicircle of radius a is divided into n equal arcs. Shew that the sum of the distances of the several points of section from either extremity of the diameter is a I cot (-*^-0- 21. From the angular points of a regular polygon, perpen- diculars are drawn to XX' and FY' the horizontal and vertical diameter of the circumscribing circle : shew that the algebraical sums of each of the two sets of perpendiculars are equal to zero. XXIII.] SUMMATION OF FINITE SERIES. 293 299. By means of the identities 2sin2a = l— cos2a, 2 cos2a=l + cos2aj 4 sin^ a = 3 sin a - sin 3a, 4 cos^ a=3 cos a + cos 3a, we can find the sum of the squares and cubes of the sines and cosines of a series of angles in arithmetical progression. Example 1. Find the sum of n terms of the series sin2a + sin2(a + /3) + sin2(a + 2/3) + 25f={l-cos2a} + {l-cos(2a + 2/3)} + {l-cos(2a + 4j8)} + =w-{cos2a + cos(2a + 2/3) + cos(2a + 4j3)+ }; _ sin n^ 2a + {2a + (n- 1)2^3} ^ — W ; — cos ^ J sm /3 2 2 2sm^ * "^ Example 2. Find the sum of the series cos3a + cos^3a + cos3 5a+ +cos3(2n-l)a. 4/Sf = (3 cos a + cos 3a) + (3 cos 3a + cos 9a) + (3 cos 5a + cos 15a) + = 3(cosa + cos3a + cos5a+ ) + (cos 3a + cos 9a + cos 15a + ) 3sinna (a+(2n-l)a] sin 3na f3a + (2n- 1) 3a] = — . cos ■{ — -!-— '—]■ + -^--— cos ] ^—^ — V ; sm a ( 2 j sm 8a ( 2 j „ 3 sin na cos na sin 3wa cos 3na 4 sm a 4 sm 3a 300. The following further examples illustrate the principle of Art. 293. Example 1. Find the sum of the series tan~iq — z — x — s+tan-iq — ^r—^ — ^+ +tan~i^; =-:— o- l + 1.2.a;2 l + 2.3.a;2 l-^n{n + l)x^ As in Art. 249, we have cc tan~i ; — — 5= tan-i (r+V)x- tan~i rx ; 1 + r (r + 1) a;2 ^ ' :. /S=tan~i(?i + l)a;-tan-ia;. 294 ELEMENTARY TRIGONOMETRY. [CHAP. Example 2. Find the sum of n terms of the series l^al.al.a, ta.na + -^tSLn-^ + ^^tan-, + ^^t&n-^+ We have tan a = cot a - 2 cot 2a. Eeplacing a by ^ and dividing by 2, we obtain 1, o 1 , a , 2*^^2'^2 2~ • Sunilarly, -^ *an ^i = 22 ^°* 2^ ~ 2 2 ' l.a 1 . a la 2n— 1 *a° 2^ = 2^ °°* 2^i^ - 2^i=2 cot 2^2 Byaddition, ..J^ cot ^,-2 cot 2a. EXAMPLES. XXIII. b. Sum each of the following series to n terms ; 1. cos2<9+cos2 3^ + cos2 5(9 + 2. sin2a4-sin2fa + yJ + sin2U+-^j + 3. cos2a + cos2U--j +cos2 ^a- — j+ 4. sin3^ + sin3 2^ + sin3 3^ + 5. sin^a + sin^i aH ) H-sin^ { a-\ ) + « / 27r\ , / 47r\ 6. cos^a + cos^ I a + COS'^ [a ) + \ nj \ n) 7. tan ^ + 2 tan 2(9 + 22 tan 22^ + , 8. 1 1 1 1 ^ I cos a + cos 3a cos a + cos 5a cos a + cos 7a XXIII.] SUMMATION OF FINITE SERIES. 295 9. sin2 e sin 2^ +- sin2 2^ sin ^B + - sin^ 4^ sin 8<9 + A 6 6 6 6 10. 2 cos 6 sin2 - + 22 cos - sin^ 22+23 cos —^ sin^ -3+ 14. tan-' p-^|_^-p+tan-' ^ _^^^_^^, + tan-i--_-A_.,4-...,. 15. From any point on the circumference of a circle of radius r, chords are drawn to the angular points of the regular inscribed polygon of n sides : shew that the sum of the squares of the chords is '2.nr^. 16. From a point P within a regular polygon of ^n sides, perpendiculars P^^, P^2) -^^3) •••^^2n ^^^ drawn to the sides : shew that where r is the radius of the inscribed circle. 17. If J.i^2^3--^2rt + i is a regular polygon and P a point on the circumscribed circle lying on the arc A^A^n + i^ shew that P^ + P^2+-+^^2n + l = ^^2 + ^^4+-+^^ 2n* 18. From any point on the circumference of a circle, perpen- diculars are drawn to the sides of the regular circumscribing polygon of n sides : shew that (1) the sum of the squares of the perpendiculars is —^ ; (2) the sum of the cubes of the perpendiculars is -— . CHAPTER XXIV. MISCELLANEOUS TRANSFORMATIONS AND IDENTITIES. Symmetrical Expressions. 301. An expression is said to be symmetrical with respect to certain of the letters it contains, if the value of the expression remains unaltered when any pair of these letters are interchanged. Thus cos a + cos /3 + cos y, sin a sin ^ sin y, tan (a - ^) + tan O - <9) +tan (y - <9), are expressions which are symmetrical with respect to the letters a, /?, y. 302. A symmetrical expression involving the sum of a number of quantities may be concisely denoted by writing down one of the terms and prefixing the symbol S. Thus 2 cos a stands for the sum of all the terms of which cos a is the type, S sin a sin ^ stands for the sum of all the terms of which sin a sin /3 is the type ; and so on. For instance, if the expression is symmetrical with respect to the three letters a, /3, y, % cos /3 cos y = COS /3 cos y + COS y COS a-f cos a COS /3 ; 2 sin (a- ^)=sin (a- ^) + sin (j3- ^) + sin (y- ^). 303. A symmetrical expression involving the 'product of a number of quantities may be denoted by writing down one of the factors and prefixing the symbol 11. Thus 11 sin a stands for the product of all the factors of which sin a is the type. For instance, if the expression is symmetrical with respect to the three letters a, /3, y, n tan (a + ^) = tan (a + ^) tan (/3 + ^) tan (y + ^) ; n (cos/3 + COS y) = (cos /3+COSy) (cos y + cos a) (cos a+COS /3). SYMMETRICAL EXPRESSIONS. 297 304. With the notation just explained, certain theorems in Chap. XII. involving the three angles A, B, C, which are con- nected by the relation A+B + C=180°, may be written more concisely. For instance ^ sin 2 J. = 411 sin J. ; 3 sin A = 411 cos — ; 2 Stan J. = n tan J.; Stan — tan — = 1. Example 1. Find the ratios of a : & : c from the equatione acos0 + 6 sin ^=c and acos^ + bsm-e = 0; whence by cross multiplication a _ & _ c sin - sin ^ ~ cos d - cos ~ sin cos 6 - cos sin ^ ' a he ' d) + e . d)-d ^ . d) + d . (b-d Sin(0-^) 2 cos ---— sm ^-jr— 2 sm ^—^r— sm ^^-^— J a Dividing each denominator by 2 sin , we have c 6 + d) . e + d) d-a2(a2_62_^^2). .*. a;2>a2- b^. Thus the minimum value is ^Ja^ - 62. Example 3. li a, b, c, k are constant quantities and a, /3, 7 variable quantities subject to the relation a tan a + 6 tan /S + c tan 7= fc, find the minimum value of tan2 a + tan2 p + tan2 7. By multiplying out and re-arranging the terms, we have (a2 + 62 + c2^ ^tan2 a + tan2 j3 + tan2 7) - (a tan a + 6 tan /3 + tan 7)2 = (6 tan 7 - c tan /3)2 + (c tan a - a. tan 7)2+ (a tan jS - 6 tan a)2. But the minimum value of the right side of this equation is zero; hence the minimum value of (a2 + 62 + c2) (tan2 a + tan2 /3 + tan2 7) - /c2 = ; that is, the minimum value of tan2a + tan2/3 + tan2 7= — — — — ^. a^ + b^-\-c^ 318 ELEMENTARY TRIGONOMETRY. [CHAP. i EXAMPLES. XXV. a. When 6 is variable find the minimum value of the following expressions : I. jocot^ + 2'tan^. 2. 4sin2^+cosec2^. 3. 8sec2^+18cos2(9. 4. S-Scos^+cos^^ Prove the following inequalities : 5. tan2a+tan2/3 + tan2'y>tan/3tany + tan y tan a + tan a tan j3. 6. sin2a + sin2/3>2(sina + sin/3-l). When B is variable, find the maximum value of 7. sin^+cos^. 8. cos^+v'Ssin <9. 9. acos(a + ^) + 6sin^. 10. p cos ^ + g' sin (a + ^). If (r=a + /3, where a and /3 are two angles each lying between and ^ , and a is constant, find the maximum or minimum value of II. sin a + sin /3. 12. sin a sin /3. 13. tan a + tan /3. 14. coseca+cosec/3. li A, B^ C are the angles of a triangle, find the maximum or ^nimum value of 15. cos A cos B cos C. 16. cot A -f cot B + cot C. ABC 17. sin^ — + sin''* — + sin2 — . 18. sec ^1+ sec 5+ sec (7. A Jt Ji 19. tan2^ + tan2|+tan2|. rf7-5e2tan|tan^=1.1 20. cot2^+cot2 5 + cot2a {Use '% cot B 0010=1.] 21. If 62<4ac, find the maximum and minimum values of a sin2 ^ + 6 sin ^ cos 6-\-c cos2 6. XXV.] ELIMINATION. 319 22. If a, /3j y lie between and - , shew that sin a + sin ^ + sin •y>sin (a + /S + 7). 23. Vl a and 6 are two positive quantities of which a is the greater, shew that a cosec ^>6 cot ^ + sj o? - W. 24. Shew that — 5- r — : — -^ hes between 3 and - . sec2^ + tan^ 3 oc T7- J 4-1, • 1 »tan2^-cot2^ + l 25. 1 md the maximum value of 7 — n-^- — 70-7 — :r . tan "T cot^ a — i 26. If a, 6, c, h are constant positive quantities, and a, /3, y variable quantities subject to the relation a cos a + 6 cos /3+ccosy = ^, find the minimum value of cos^a+cos^/S+cos^y and of a cos^ a + 5 cos^ /3 + c cos^ y. Elimination. 322. No general rules can be given for the elimination of some assigned quantity or quantities from two or more trigono- metrical equations. The form of the equations will often suggest special methods, and in addition to the usual algebraical artifices we shall always have at our disposal the identical relations sub- sisting between the trigonometrical fimctions. Thus suppose it is required to eUminate 6 from the equations a?cos^=a, ^ cot 6=h. Here sec^=-, and tan^=v; a but for all values of ^, we have sec2^-tan2^=l. .•.by substitution, '^^~¥ From this example we see that since 6 satisfies two equations (either of which is sufficient to determine 6) there is a relation, independent of 6j which subsists between the coefficients and 320 ELEMENTARY TRIGONOMETRY. [CHAP. constants of the equations. To determine this relation we eliminate d, and the result is called the eliminant of the given equations. 323. The following examples will illustrate some useful methods of elimination. Example 1. Eliminate 6 between the equations Zoos ^ + w sin ^ + w=0 and ^cos d + qQm.d + r=0. From the given equations, we have by cross multiplication cos ^ _ sin ^ _ 1 mr- nq ~ up - Ir ~ Iq- mp ' „ mr-nq . . ^ np-lr .: cos 6 = , and sm 6= ^ : Iq - mp Iq - mp whence by squaring, adding, and clearing of fractions, we obtain {mr —nqf + {np - lr)-= {Iq - mp)^. The particular instance in which q = I and p= -m is of frequent occurrence in Analytical Geometry. In this case the eliminant may be written down at once ; for we have I cos 6 + m sind= -n, and lsia6-m cos d= -r; whence by squaring and adding, we obtain P + m^ = n^+r''^. Example 2. Eliminate 6 between the equations <^ by o -. , . ^ -. — ^^=c2 and lta,nd=:m. cos d sm From the second equation, we have sin ^ _ cos ^ _ ^ysm^T+cos^ _ 1 . m ,*. smg=- . , and cos^= Jm^ + l^ s/m^ + P By substituting in the first equation, we obtain ax by _ c2 XXV.] ELIMINATION. 321 Example 3. Eliminate 6 between the equations a; = cot^ + tan^ and 7/ = sec ^ - cos 0. From the given equations, we have and U/ • tan^ tan (9 sec2^ tan^' y = sec ^- 1 sec- 6 - 1 sec 6 seed tan2^ sec, shew that {a^-b- Vf=a^ (2 - a\ 15. If cos 6 - sin B = b, and cos 3^ + sin 3^ = of, shew that a =36 -263. 16. Eliminate B from the equations : a cos ^ - 6 sin ^ = c, 2a6 cos 2(9 + (a^ - 6^) sin 2^ = 2c2. 17. If ,2;= a cos ^ + 6 cos 2^, and y = asin^+6sin2^, shew that a^ {{x^bf-\-y^\ = {x^^y'^-b'^f. XXV.] ELIMINATION. 325 -- -r„tan((9 + a) a+b , -. . r oa 18. it , = — ,- , and a cos 2a -^-o cob 26 =c, tan(^-a) a -6' ' shew that a^ + c^— 2ao cos 2a = 6^. 19. If a;=a (sin 3^ — sin 0), and y = a (cos ^ - cos 3^), shew that {x^+y^) {2a^ — x^- y^f = 4(2%^. Eliminate 6 from the equations : 20. ^'cos^— y sin^=acos 2^, .r sin ^ 4- 3/ cos ^ = 2a sin 2^. «^ . >, yi /-o 5 cos2(9 . sin2^ 1 21. ^sm^-vcos^=V^^+y^ — — + ,0 = g . » ■ ^cos^ ysin^ ^ . . . / ., . „ - — ^ 5-7, 22. V—i — = 1, ^sm^-'ycos^=V«^sm2^ + 62cos2^. a 23. If cos (a -Z6) = m cos^ ^, and sin (a - 3^) = m sin^ 6, shew that m^ 4- m cos a = 2. Ehminate 6 and from the equations : 24. tan^+tan0 = ^, cot (9 + cot 0=3/, 6 + (f) = a. 25. sin ^+ sin ^ = a, cos ^ + cos = 5, ^ — = a. 26. asin^ ^-)-6 cos^ ^= shew that -5 + ttj = sec^ ^ , a-^ 0^ 2 x^ y^ „ a 28. If tan ^+ tan = a, cot ^ + cot (^ = 6, 6 — (^ — a^ shew that a6 (a6 — 4) = (a + 6)^ tan^ a. Eliminate 6 and (jf) between the equations : 29. « cos2 B-\-h sin^ ^ = m cos^ 0, a sin^ ^ + 6 cos^ ^ = % sin^ 0, m tan^ ^ — ?z. tan^ <^ = 0. 30. A'cos^+ysin^ = 2a-v^3, ^cos(^ + 0)+3/sin(^+0) = 4a, X cos (^ — 0) +y sin {B-(j)) = 2a. 31. csin^=asin(^+0), (X sin (/> = 6 sin ^, cos ^ — cos = 2m. 326 ELEMENTARY TRIGONOMETRY. [CHAP. Application of Trigonometry to the Theory of Equations. 326. In the Theory of Equations it is shewn that the solu- tion of any cubic equation may be made to depend on the solution of a cubic equation of the form ^■\-ax-\-h = h. In certain cases the solution is very conveniently obtained by Trigonometry. 327. Consider the equation cfi — q^x — r=^^ (1), in which each of the letters c[ and r represents a positive quantity. From the identity cos 3^=4 cos^ ^ - 3 cos ^, o ., 3 , cos Z6 ^ we have cos^^ — -cos^ ;; — =0 (2). 4 4 ^ ' Let x=y cos ^, where y is a positive quantity; then from (1), cos3(9--^cos^-^=0 (3). If the equations (2) and (3) are identical, we have 2 ~ I ? ^^ ^^^* y y= + ^ / -^ , since y is positive ; and cos 3^ _ r _ /27r2 ^ 4 ~2^3-\/64^3 5 /27r2 whence cos 3^ = W XT • Hence the values of 6 are real if '^^r^<^(f ; that is, if ©'<(!)'• /27r2 Let a be the smallest angle whose cosine is equal to W -j-3- ; then cos 3^ = cos a ; whence 3^ = 2?^7^ ± a. Thus the values of cos 6 are a 27r+a %r — a ^-c* * , /« n cos - , cos , cos -^r — , [See Art. 264.J XXV.] APPLICATION TO THE THEORY OF EQUATIONS. 327 But ^=y COS 6= f. -^ COS 3 and therefore the roots oi x^ —qx — r—0 are Stt — a -COS-—, 328. Following the method explained in the preceding article, we may use the identity . „ , 3 . . sin 3^ _ sm^^-- sm 6-{ : — =0 4 4 to obtain the solution of the equation x^—qx-\-r = 0, each of the quantities represented by q and r being positive. Example. Solve the equation x^ - 12x + 8 = 0. fTT 1 • o ^ 3 . „ sin 3^ . We have sin3^--7Sin^H -. — =0. 4 4 In the given equation put x=y sin 6, where y is positive ; then 12 8 sin3^- — sin ^ + — = 0. y. yi 3 12 •*• i = "p' whence y = 4; ^ sin 3^ 8 1 , • Q. 1 and — i — = -- = -; whence sin 3^ = - . 4 2/ S 2 Suppose that d is estimated in sexagesimal measure ; then 36' = 7i.l80° + (-l)"30°. By ascribing to n the values 0, 1, 2, 3, 4 we obtain ^ = 10°, ^ = 50°, d = nQ°, 6 = 110°, ^ = 250°; and by further ascribing to n the values 5, 6, 7, ... it will easily be seen that the values of sin d are equal to some one of the three quantities sin 10°, sin 50°, - sin 70°. But x=y sin ^=4 sin ^, and therefore the roots are 4 sin 10°, 4 sin 50°, -4 sin 70°, 328 ELEMENTARY TRIGONOMETRY. [CHAP. Application of the Theory of Equations to Trigonometry. 329. In the Theory of Equations it is shewn that the equation whose roots are a^, «2, Og, , a^is, {x - a-^ {x — a,^ {x — a^ (.r-a„)=0, or .^" - >S'i.r"-i + >S'2:r«-2 - /S'g.r"-^ + + ( - 1)">S'„= 0, where S-^ = sum of the roots ; >S'2 = sum of the products of the roots taken two at a time ; >S'3 = sum of the products of the roots taken three at a time; >S'„= product of the roots. [See Hall and Knight's Higher Algebra, Art. 538 and Art. 539.] Example 1. If a, ^, y are the values of 6 which satisfy the equation at&n^e + {2a-x)t&ne + rj = (1), shew that (i) if tan a + tan /3= 7t, then ah^ + {2a -x)h = y; (ii) if tan a tan ^ = A-, then y^ + (2a - x) ak^ = a^k^. (i) From the theory of equations, we have from (1), tan a + tan /3 + tan 7=:0 ; .•. 7?+tan7 = 0, or tan7=-/t. But atan37 + (2a-.'c) tan7 + 2/=:0; .-. aW-\r{2a-x) h-y = 0. (ii) From the theory of equations, we have from (1), V tan a tan /Stan 7= --; •^ a .'. fctan7= --, or tan7= --^. 'a ak Substituting in a tan^ 7 + (2a - x) tan 7 + ?/ = 0, we have .-. 2/2 + (2a - x) ak^ - a^k^=0. XXV.] APPLICATION OF THE THEORY OF EQUATIONS. 329 Example 2. Shew that ./27r \ ,/27r \ 3 eos2 a + cos^ l "o" + " ) + ^^^ I T ~ "^ ) ~ •? ' Suppose that cos3^=/!;j then 4cos3^-3cos^ = cos30=fc; 3 k .'. COS^ ^ - - COS ^ - - = 0. 4 4 The roots of this cubic in cos 6 are /27r \ , /27r \ cos a, cos I "v + a ) » aiid cos I — -a \ , where a is any angle which satisfies the equation cos3a = fc. For shortness, denote the roots by a, b, c; then a^ + h~ + c'^ = {a + b + c)^-2{bc + ca + ah) .•. cos^ « + cos2^^ + a)+cos2(^^-a^=|. 330. If 56 = 2n7r, where n is any integer, we have 3d=2mr-2d', .-. sin 3^= -sin 2(9. The values of sin d found from this equation are ^ . Stt . 47r . Qtt . Stt 0, sin — , sm — , sin — , sm — , being obtained by giving to n the values 0, 1, 2, 3, 4. It will easily be seen that no new values of sin 6 are obtained by ascribing to n the values 5, 6, 7, -—= — sm — - = — sm - 5 5 5 But sin -^= — sin -^ = — sin - , , . Stt . 27r and sm — = — sm — ; 5 D hence rejecting the zero solution, the values of sin B found from the equation sin 3^= — sin 2^ are + sm - , and + sin -;r . ~ 5 ~ o 330 ELEMENTARY TRIGONOMETRY. [CHAP. If we put sin 6= a;, the equation sin 3^= - sin 2^ becomes 3x — 427^ = — '2aj a/I — •^^• Dividing by x, and thus removing the solution ^=0, we have or 16^-20^24.5 = 0. This is a quadratic in x'^, and as we have just seen the values of ^2 are 77 J . „ 27r sm-^ - and siU'* - . 5 5 From the theory of quadratic equations, we have . „7r . oStt 20 5 . „7r . „27r 5 sm2-sm2- = -. Example. Shew that . 2ir . 47r . Stt 1 -_ sm _ + sm-y + sm y = -- ^7, If 16=2mr, where n is any integer, we have sin 4^= -sin 3^. The values of sin 6 found from this equation are . 27r . 47r . . Stt 0, ±sm— -, ±siny, ±smy, . Gtt . Sir since sm -=- = - sm y- . If sin d = x, the equation sin 4:6= - sin Bd becomes Ax (1 - 2x^) jr^= 4a;3 - 3a; ; whence rejecting the solution a; = 0, we obtain 16 (1 - 4:x^ + 4:x^) (1 - a;2) = Ux^ - 24:X^ + 9, or 64x6- 112a;H56a;2- 7 = (1). The values of x^ found from this equation are . „27r . „47r . ^Stt sm2y, sm2y, sm-y; XXV.] APPLICATION OF THE THEORY OF EQUATIONS. 331 , . ,27r . „47r . „87r 112 7 hence sm^ — + sm^ — + sm^ — = -— - = - . Ill Uu: rx. T, ^ . 27r . 47r . 27r . Stt . ^tt . Stt But sm -7=- sm — + sin— sin — + sm — sm — 1 if 2x GttN / Gtt IOttX / 47r 127r COS — = 0. . 27r , . 47r , . SttV • o27r , . ,47r . „87r 7 siny + smy+smyj =sm2— + sin2— + s:u2— = -; . 27r . 47r . Stt 1 ,„ .-. smy + siny + smy =-V7. 331. If ^d = 27i7r, where n is any integer, we have 4<9=2»7r-3^; .'. cos 4^= cos 3^. By giving to n the values 0, 1,2, 3, the values of cos B obtained from this equation are _ 27r 4:77 Gtt 1, cos -=- , cos --- , cos — - . ' 7 ' 7 ' 7 It will easily be seen that no new values of cos 6 are found by ascribing to n the values 4, 5, 6, 7, ; for Stt Gtt IOtt 47r -=- = cos -Tz- , cos — — = cos — - . 7 7 7 7 ' Now cos4^=8cos*^-8cos2^ + l, and therefore if ;r = cos 6, the equation cos 46 = cos ZO becomes 8^4 _ 8^2 _{. 1 = 4:^3 -3^, or 8x^-4a!^-8x^+3x + l=0. Eemoving the factor x-1, which corresponds to the root cos 6=1, we obtain 8^3+4^^-4^^-1=0, the roots of which equation are 277 47r Gtt cos — , cos -;=- , cos -^ . 332 ELEMENTARY TRIGONOMETRY. [CHAP. Example 1. Find the values of tan^ - + tan^ — + tan^ — and tan - tan — tan -^ . If ld=mr, where n is any integer, we have tan 4^= -tan 3^. By writing tan d = t, this equation becomes 4f-4f3 _ U-t^ l_6f2 + ^- l-3f2' or f6_ 21^4 + 35^2 _ 7 = 0. The roots of this cubic in t^ are tan^^, tan^y, tan^^. .-. tan2 ^ + tan2 ^ + tan^ ^ = 21, and tan - tan — tan - = sjl, the positive value of the square root being taken, since each of the factors on the left is positive. Example 2. Shew that . TT . 27r . Stt 13 cos4 - + cos4 — + cos4 y = jg ; -• 1 1* . 27r . OTT . ^ _ and sec^ - + sec* — + sec* — = 416. Let y denote any one of the quantities „7r „27r „ 37r C0S2-, COS2y, C0S2y; then 1'y = l + x, where x denotes one of the quantities 27r 47r Gtt cos -=- , cos -=- , COS -^ . 7 7 7 From Art. 381, the equation whose roots are 1ir 4:17 GtT COS — - , COS -=- , cos -=- 7 7 7 is 8ic3+4a;2-4a;-l = 0: XXV.] APPLICATION OF THE THEORY OP EQUATIONS. 333 whence by substituting x = 2y-l, it follows that C0S2-, C0S2y, C0S2y are the roots of the equation 8 (27/-l)3 + 4(22/- 1)2-4 (2^-1)- 1=0, or 64?/3 - 802/2 + 24'j/ -1 = 0. ^TT ,27r ^Stt 80 5 ••• cos-'- + cos-'- + cos--=g^ = -; and ^cos2-cos2y = — = -. By squaring the first of these equations and subtracting twice the second equation, we have .IT .lir . Stt 13 cos* „ + cos* y + cos* y- = JQ • By putting 2 = - , we see that ., TT ^ 27r „ 37r sec-^, sec-y, sec^ y are the roots of the equation 23 _242H 80^-64 = 0; .-. sec* ^ + sec* y + sec* y = (24)2 - (2 x 80) = 416. 332. To find cos bO and sin 5^, we may proceed as follows : cos 5^+ cos ^ = 2 cos 3^ cos 2^ = (4 cos'^ (9 - 3 cos 6) (4 cos^ ^ - 2) > .-. cos5(9=16cos5^-20cos3^+5cos^. sin 5^ + sin ^ = 2 sin 3^ cos 2^ = (3 sin ^ - 4 sin^ <9) (2 - 4 sin'^ &) ; .'. sin 5^=16 sin^ ^ - 20 sin^ ^+5 sin 6. It is easy to prove that cos6^=32cos6^-48cos*^+18cos2 8-1, and sin QB = cos 6 (32 sin^ ^ - 32 sin^ ^ + 6 sin 6). 334 ELEMENTARY TRIGONOMETRY. [CHAP. EXAMPLES. XXV. c. Solve the following equations : 1. ^-3^^-1=0. 2. ^^3_3_;j;^i=o. 3. ^-3^'-V3 = 0. 4. 8^-6^-+v/2=0. 5. 8a^x^-eax + 2ain3A = 0. 6. ^-3a%- 2^3 cos 3^=0. 7. If sin a and sin /3 are the roots of the equation asin^^-f 6sui^ + c = 0, shew that (1) if sin a + 2sin/3 = l, then a^-\-2b^+dab-\-ac=0j (2) if csina=asin/35 then a + c=+6. 8. If tan a and tan /3 are the roots of the equation a tan^ d — b tan d + c = 0, and if a tan a + b tan /3 = 2&, shew that 62 (2a - 5) + c (a -6)2=0. 9. If tan a, tan /S, tan y are the roots of the equation atan^ ^ + (2a — ^^)tan^+3/=0, and if a (tan^ a ^tan^ /3) = 2^ — 5a, shew that ^ ±3/ = 3a. 10. If cos a, cos /3, cos y are the roots of the equation cos^ ^ + a cos^ ^ + 6 cos ^+c = 0, and if cos a (cos /3 + cos y) = 26, prove that abc-\-2b'^-\-c^=0. Prove the following identities : 11. seca+sec( — + aj+secf — -aj= -3 sec 3a. 12. sm^a+sm^(^+a\+sm^(^ + a}j = ^. 13. cosec a + cosec ( -^ + a ) + cosec ( -^ + a j = 3 cosec 3a. 14. cosec^ - + cosec^ -^^ = 4, 5 o 277 . 47r 1 , 27r 47r 1 15. cos -^ + cos -^ = — ^ , and cos -^ cos -=- = - 7 . o 5 2 o 5 4 XXV.] APPLICATION OF THE THEORY OF EQUATIONS. 335 16. Form the equation whose roots are (1) cos^, cosy , cosy; (2) sm2-, sm--, sm^ — . 17. Form the equation whose roots are 7' 7 ' 7 * « = 3 ^2,7j- 21 '^^^ wtt and shew that 2 sin^ "tt "^ TH ^^^ ^ cosec* -^r- =32. 71=1 7 ib ,1=1 7 18. Form the equation whose roots are ,^. ^TT 47r Gtt Stt (1) cos — , cos—, cos—, cosy; ,^. TT StT StT 777 (2) COS-, COS—, COS — , COS—. 19. Form the equation whose roots are „7r „27r „37r „47r C0S2-, C0S2 — , C0S2 — , COS^ — , and shew that 2 cos^-r- =t^, and ^ sec^ -^- = 1120. n=i y lo n=i y 20. Form the equation whose roots are ■ 9 77 . „ 277 , „ 377 , „ 477 tan^ - , tan^ — , tan-^ — , tan^ — y y y y and shew that cot^ - + cot^ — + cot^ -5- = 9. y y y 21. Form the equation whose roots are 77 277 377 477 577 cos—, cos — , cos — ) cos — , cos — ; and shew that ^ sec^ — - = 60, and 11 sec jj = 32. »=i 11 n=i 11 336 ELEMENTARY TRIGONOMETRY. [CHAP. MISCELLANEOUS EXAMPLES. L 1. If ^1- sin^^ 11. v/2. ,2_„2 p2 + g2 2)"^ + gr2' 2pq P ,2_ 1. 5. 6. H. 3 11. J. 1. 22° 30'. 6. 45° -B. 9. 18°. 13. 30°. 2. li 7. 9. 12. 0. p' + q' IV. a. Page 26. 3. 0. 4. 2 8. 2. 13. IH. d. 2yV 14. 2 15. 6. IV. b. Page 28. 2. 64° 59' 30". 3. 79° 58' 57". 4. 45° + 4. 6. 60° + B. 10. 9°. 14. 15°. 7. 50°. 11. 22° 30'. 30. 1. 8. 60°. 12. 45°. 31. tan A. IV. c. Page 31. 1. 45°. 2. 60°. 3. 60°. 4. 45°. 5. 60°. 6. 30°. 7. 45°. 8. 60°. 9, 45°. 10. 60°. 11. 45°. 12. 60°. 13. 45°. 14. 30°. 15. 45°. 16. 60°. 17. 30°. 18. 60°. 19. 45°. 20. 60°. 21. 30°. 22. 30°. 23. 60°. 24. 45°. 25. 45° or 30°. [(2 sin - -l){ta ne- 1)= = 0.] 26. 60° . 28. lor- ANSWERS. 339 MISCELLANEOUS EXAMPLES. A. Page 32. 1. (1) -2537064; (2) -704. 3. |,g. *• ^' ^. 6. (1) 15° 28' 7-5"; (2)l'37-2". 7. 41, ^,^,|1. 8. (1) possible; (2) impossible; (3) possible. 10. VI+c3^ Vr+coP^. 11. 6. cot a 12 ^^ Jm^ + n^ 20 29 V^^^^+^'' ^ ' '21' 20- 18. 10°. 20. (1) 30°; (2) 45°. 22. 30°. 26. —. 29. (1) 30°; (2) 30°. V. a. Page 37. 1. c = 2, 5 = 60°, C = 30°. 2. a= 6 ^3, ^ = 60°, (7 = 30°. 3. c = 8^3, ^ = 30°, B = 60°. 4. = 30^3, -B = 30°, C= 60°. 5. 6 = 20^2,^ = = 45°. 6. 0=10^/3, ^ = 30°, 5 = 60°. 7. a=2^2, .B=C=45°. 8. a = 9, .4=60°, C = 30°. 9. 5 = 60°, 6 = 27, c = 18V3. 10. C = 60°, 6 = 2, = 2^3. 11. .5 = 30°, a = 4;^3, 6 = 4. 12. 5 = 90°, a = 3^3, c = 3. 13. ^ = 30°, a = 50, c = 50^3. 14. 0=90°, a = 20, c = 40. 15. ^ = 90°, a = 4;^2, 6=4. 16. ^ = 90°, 6 = 4, = 4^3. 17. 700. 18. 31. 19. 86-47. 20. 97*8. 21. 0=54°, a = 73, 6 = 124. 22. 5 = 68° 17', = 21° 43', 6 = 93. 23. 0=50° 36', a = 34-3875, c = 30-435. 24. c = 353, A = 39° 36', B = 50° 24'. V. b. Page 39. 1. 10^3. 2. a=10V2, c = 20. 3. ^5 = 10 ^3 ft., ^0=10 ft., ^D = 5 ^3 ft. 4. 12,4. 6. 24^3. 6. 20(^3-1). 7. 20(3 + ^3). 8. DO =5X> = 100. 22—2 340 ELEMENTARY TRIGONOMETRY. VI. a. Page 42. 1. 173-2 ft. 2. 277-12 ft. 3. 60°. 4. 50 ft.; 100 ft. 5. 22-5 ft.; 38-97 ft. 6. 30 ft. 7. 200 yds. 8. 51 yds., 81 yds. 9. 86-6 yds. 10. 46-19 ft. 11. 273-2 ft. 12. Each = 70-98ft. 13. Smiles. 14. 73-2 ft. 15. 64 ft. 16. 300 ft. 17. 1193 yds. 18. 277-12 yds. VI. b. Page 47. 1. 665-6 yds.; 1131-2 yds. 2. 3-464 miles ; 6 miles. 3. 29 miles. 4. 10 miles per hour. 5. 10 miles; 24-14 miles. 6. 16 miles ; S. 25° W. 7. 9-656 miles. 8. 5*77 miles; 11-54 miles. 9. 295-1 knots. 10. 5-196 miles per hour ; 18 miles. 11. 31 minutes past midnight. 12. 38-97 miles per hour. VII. a. Page 54. TT _ TT lir TT TT 1. J. 2. g. 3. jg. 4. -. 6. jg. 11. 1-4399. 12. 1-1999. 13. 2-7489. 14. '9163. 15. 135°. 16. 28°. 17. 33° 20'. 18. 37° 30'. 19. 22° 30'. 20. 30°. 21. 37° 30'. 22. 165°. 23. -638. 24. 1-332. 25. 2-0262. 26. 2-9979. VII. b. Page 56. 1. |. 2. A^. 3. 4i. 4. -|. 5. 9. 3 TT 27r la. ^^ ^'"' 6. ^. 7. 1. 13. ^, ^. 14. -g-, —. VII. C. Page 60. 1. -. 2. 300 ft. 3. A radian. 5 4. 5-85 yards. 5. 330. 6. tt of a second. 7. 58§. 8. 40 yds. 9. 1-15192 miles. 10. 3-581. 11. 2° 6'. 12. 45 feet. ANSWERS. 341 MISCELLANEOUS EXAMPLES. B. Page 61. 95-26. 221°, I 1. 9°. 2. 6. 30°. 8. -^2 , ^ 10. a = 6 v/3, c = 12, perp. = 8 JS. 14. 120°, 36°, 24°. 15. 17. (1) possible; (2) impossible. 54°. 67i°. 35 8 ' 4. 3438 inches. 12. 17-32 ft. 18. 8-66 miles. 13' 19. 5' 3' 15 • 21. 90. 24. 4 miles per hour, 1-732 : 25. TT 8' 26. (1) 30°; (2) 30°. 27. : 29. 200 yards. 30. 33 feet. VIII. a. Page 69. 1. Second. 2. Third. 3. First. 4. Third. 5. Second. 6. Second. 7. Third. 8. Third. 9. Sine. 10. Cosine. 11. Tangent. 12. Sine. 13. Sine. 14. Tangent. 15. /O Sine. /Q 16. All. 17. Cosine. 18. 60°, Y • 19- 80", ^^ 20. 45°, 1. 21. 45°, ^2. 22. 30°, 2. 23. ''°' Tr 24. 45°, 1. 25. 60°, 2. 26. 60°, V3. VIII. b. Page 72. 1. -x/3. 1 3. 1 2* 12 12 ^ ^ 4 3 4. l3' T" 5- -4. -4. 6. 3' 5" 7. x/3 1 2 ' V3" 8. 1, -^^2. 9. 13 ' 12 IX. Page 79. 1. cot J^ decreases from qo to 0, then increases numerically from to - 00 , then decreases from 00 to 0, then increases numerically from to - 00 . 2. cosec d decreases from co to 1, then increases from 1 to 00 . 3. cos d decreases numerically frora - 1 to 0, then increases from to 1. 4. tan A decreases from 00 to 0, then increases numerically from to - 00 . 5. sec d decreases numerically from - od to - 1, then increases numerically from - 1 to -00. 6. 3. 7. 1. 8. -2. 9. 2. 342 ELEMENTARY TRIGONOMETRY. MISCELLANEOUS EXAMPLES. 0. Page 81. 1. i|. 3. A = &0°,B = SO-,a = ^-. 4. ^. 5. 1313 miles, nearly. 6. 301 feet. 7. 10* 8. 12-003 inches. 10. 200 feet. X. a. Page 87. 1. 1 v/2- .. |. 3. s/S. 4. -s/2. 5. n/3 2 • 6. 1. 7. -1. 8. 1 2' 9. 2. 10. - 1. 11. v/3 2 " 12. -2. 13. -2. ''■ -J2' 15. V3. 16. sin^. 17. tan J. 18. —cos 4. 19. - sec A. 20. -cos^. 21. -tan^. 22. - cos 9. 23. tan 0. 24. - cosec 6 25. 1. 26. 2 sin A. 27. 1. X. b, . Page 91. 1. 1 2* 2. -f • ^• 1 2* 4. 1 2' 5. -1. 6. 1. 7. ;!■ «■ 1 v/3 9. -v/2. -;^- 11. 0. 12. ;^- - -V2. 14. 1 2' 15. -V2 16. 1 "V2" 17. - 1. 18. 2. 19. 1 x/3" 2 21. ±30°, ±330°. 22. 210°, 330°, -30 1°, -150°. 23. 120°, 300°, -60°, -240°. 24. 135°, 315°, -45 °, -225°. 30. 3. 31. cot2^. 32. -1. 34. -4. XI. a. Page 97. ^' 25' ^' 65' 65* 36 ANSWERS. 343 XI. b. Page 100. , n 1 n 12 ^ 278 1 I. 1. 2. -. 3. 0; 3^. 4. -^; -. II. cos A cos B cos G - cos A sin B sin C - sin A cos S sin C - sin jL sin B cos C; sin A cos J5 cos (7 - cos A sin B cos G + cos 4 cos £ sin C + sin A sin B sin C tan .4 - tan B - tan G - tan J. tan B tan G 12. 13. 1 - tan B tan C + tan C tan ,4 + tan A tan J5 * cot A cot J5 cot G - cot ^ - cot B - cot C cot B cot C + cot C cot A + cot ^ cot B-1' XI. d. Page 104. 7 ^17 ,24 3 ^* "§• ^* 25- ^' 25- *• 4* ^ 24 1 1 ^' 25' 25* ^' 3' ''^ T XI. e. Page 106. 23 117 9 27* 125' •" 13* XII. a. Page 112. 1. sin 4(9 + sin 2=nT^^ . «iV« --'- 4, V' — 3 * XIX. d. Page 244. 1. 'i- 2. ±1. 3. ±2. 4. 4 5. 1 1, or -. 6. 0, or ±^. 7. v/2 8. 25 24' 9. 1 2* 10. a~b 1 + ab' 11. 12. v/3. 13. a; = ac — &d, y= be + ad. 14. il, or ±(1 =^v/2). 15. TT mr, or wtt + - . 19. a; = l, y = 2; x = :2, 2/ = 7. MISCELLANEOUS EXAMPLES. G. Page 246. 6. 78° 27' 4". 9. 6. 10. 800 yds, 146-4 yds, 546-4 yds. XX. a. Page 255. ^ . A 5 A 12 . A 15 A 8 2. sm- = -,cos-=--. 3. sm-=--,cos-. = -. 352 ELEMENTAEY TRIGONOMETRY. 4. 2 sm--= - >yi + sin ^+ A^/l-sin 4; 2 cos -= - ^1 + sin ^ - >/l - sin ^. j[ 5. 2sin— = - ^yi + sin^ - A^l -sin J.; A 2cos — = - ^/l^-sin^ + ^yl-sin^. 6. 2sin— = + .v/l + sin^ + >y'l-sin^; 2 cos - = + ^1 + sin ^ - ^/l - sin ^. . A 4. A ^ ^ . A It) A 8 7. sm-=.-, cos- = ^. 8. «m2=i7. cos-=-- 9. (1) 2mr~- and 2TO7r + -r; (2) 2?i7r + -— and 2w7r+-T-: 4 4 4 4 (3) 2w7r + -T- and 2?i7r+— -. 4 4 10. No; 2sin-= Vl+'sirTZ+^n^'sin'I. 14. (1) =V2cos(^^-^^; (2) =2sin(^^-|). 15. (1) =-sec2^; (2) =tan2^. XX. b. ■ Page 260. 1 1 XXI. a. Page 267. 1. 1440 yards. 2. 342f yards. 3. 22 yards. 4. 6' 34". 5. 13' 45". 6. 11 ft. 11 in. 7. 210 yards. 8. 9' 49". 10. 50 ft. 1 1 v/3 13. g. 14. m-n. 15. g " 200 16 l + yV3-.503 17 !1n^^-39.7' ANSWERS. 353 XXL b. Page 271. 1. 12 miles. 2. 150 ft. 3. 15 miles. 4. 80 ft. 8 in. 5. 204 ft. 2 in. 6. 104 ft. 2 in. 7. 54' 33". 8. 10560 ft. 9. 610 ft., |v'110i^iiiutes = 26'13". 11. 8. 12. -1. 13. (1) cos a; (2) -sin a. 14. 45° 54' 33", 44° 5' 27". MISCELLANEOUS EXAMPLES. H. Page 283. 3. 18° 26' 6". 6. 35 miles or 13 miles per hour. » XXIII. a. Page 291. sin 27za ^ . nQ ( '' - 1 o\ / • /^ 1. ^. . 2. sm-^cos( a--^— /3 / sm^. 2 sm a 2 \ 2 ' ) I 2 {"■^^^/'^k- *• mr (n + l)ir / . ir cos I a + — J / sm — . 4. sm ^ cos ^-i^ — / sm — 5. 1 2' 1 „ . TT „ 27r 6. -K- 7. cot -. 8. -cos — . 2 2» n 9. sin na. 10. sm \ hmi 2 ^+ 2 ")/'^"' 2 ' . w(7r-/3) f (?i-l)(7r-/3)] / . 7r-/3 11. sm ^ ^^ cos ja+ ^^ -^ —> / sm —~ 71 cos ^ sin n^ cos (71 + 2) ^ 7r-2i3 sm — ~-, 4 13. 14. 2 2 sin ^ sin 27?a sin 2 (ti + 1) a n sin 2a 2 sin 2a 2 15. coseca{tan(7i + l)a -tana}. 16. cosec 26 {cot ^ - cot {2n + 1) ^ } . 17. tan a - tan — . 18. ^ (cosec a -cosec 3"' a). 19. - (tanS^a-tana). H. K, E. T. 23 354 ELEMENTARY TRIGONOMETRY. XXIII. b. Page 294. n sm4w^ „ n „ ^* 1. - -I . 2. -. 3. -. 2^4sm20 2 2 ^ . nd . {n^l)d . Snd d(n+l)d 3 sm -^ sm -^^ — ~- sm -^ cos — ^— 4. ;; 4 sm - 4 sm — ^ <^ 5. 0. 6. 0. 7. cot d* - 2" cot 2^^^. 1 , , ,. . sin 2^ sin 2"+!^ 8. - coseca{tan(n + l) a-tana}. 9. — x ^z^ — . 10. sin2e-2'*sin2 — . 11. tan-^a;- tan"! 2" n+1 12. tan-i(?i + l)-^. 13. tan-i{lf7i(n+l)} - ^, 14. tan~i?i(n + l). XXIV. a. Page 301. a + 3 I a-8 , . a + 8 / a-/S 2. a; = ticos— 2"^ / COS— 2^, ?/ = 6 sm — ^-^ / cos "2" . 3. a; = a (cosa + sina), ?/ = b (sina-cos a). A • a + i3 + 7 . 3 + y-a. . 7 + a-^ . a + /3-7 13. 4 sm ^ — - sm ^^ — | sm -^^ — - — - sm — ^ — '- . 14. 4 sm — ^ — - sm — ^ — ^ cos "- — | cos -- — ^-^ . X,. _4oos(^±to-^)ncos(^-±p-%^). 22. (1) (a2 + &2)a;2_2&ca;+c2-a2=0; (2) (a2 + &2)2 a;2 - 2 (a^ - 62) (2c^ _ a [Use cos 2a cos 2j8=cos2(a - /3) - sin2(a + |S).] (2) (a2 + &2)2 a;2 - 2 (a^ - 62) (2c2 -a^-b^) x + a'^ + ¥ + 4c4 - 2a262 -4a2c2-462c2 = 0. ANSWERS. ^ 355 XXV. a. Page 318. 1. 2VS- 2. 4. 3. 24. 4. 2. 7. V^. 8. 2. 9. ^a^ - 2a& sin a + 6^. 10. iJp^-{-2yqB\na-\-q^. 11 . Maximum = 2 sin - . 12. Maximum = sin^ - . 13. Minimum = 2 tan ■ 15. Maximum = -. 8 14. Minimum = 2 cosee ^ . 16. Minimum = ^3. 17. Minimum = - . 4 19. Minimum = 1. 18. Minimum = 6. 21. -(a + c)±- v'6^ + (a-c)2. 26. fcV(«' + &' + c2); A;2/(a + & + c). 20. Minimum = 1. 25 5 25. 3. 1. ^+r=2. XXV. b. Page 324. 2. x'^ + y- = d^ + h\ 3. 6^ 3, ^2 (2 _ a^). 4 2 2 4 4. y[x^-l) = 2. 5. (a2-62)2^16„^, 6. xhf-x^y^=l. 222 f ? 6 4 7. a2&2(a2 + 62)_i, 3. a;3 + ^3=a=l 9. xh/-x^y^=a\ X" y , 10. ^ + 75 = 1. 11 22 12. x^ + ^f' = 2. 13. (a; + ?/)^+{a;-?/)3=2. 2 2 2 9 9 16. ft2 + 52^2c2. 20. (a; + ?/)3+(a;-i/)3=2a3. 21. |:2 + ^ = 1. 2 2 22. ^ + |-zz:a + &, or {a(i/2-&2)-?;(a;2-a2)}2=-4a&a;2i/2. 24. xy = {y-x)ia,-n.a. 25. «2^^2_2cos a = 2. 26. a + 6 = 2a6. 29. (a + &)(m + w) = 2mw. 30. a;2 + 2/2 = 16a2. 31. (a - 6) {c2 - (rt + &)2} =iahcm. 356 ELEMENTARY TRI&ONOMETRY. XXV. c. Page 334. 1. 2 cos 20°, -2 cos 40°, -2 cos 80°. 2. 2 sin 10°, 2 sin 50°, -2 sin 70° 3. 2 cos 10°, -2 cos 50°, - 2 cos 70°. 4. sin 15°, sin 45°, -sin 75°. 11 1 5. -sin^, -sin (60°-^), --sin (60°+^). a a a 6. 2acos^, 2acos(120°±^). 16. (1) 8x^-4x^-4cx + l = 0- (2) 64?/3-80?/2+242/-l = 0. 17. 64?/8-112?/2 + 567/-7 = 0. 18. (1) 16x^ + Sx^ - 12a;2 _ 4;c + 1 = ; (2) IQx^ - 8x^ - 12^2 + 4a; + 1 = 0. 19. 256?/^ -4482/3 + 2401/2 -40?/ + 1 = 0. 20. i8 _ 36^6 + i26f4_84«2 + 9^0. 21. 32a;5 - 16^4- 32x3 + 12a;2 + 6x- 1 = 0. CAMBRIDGE: PRINTED BY 0. J. CLAY. M.A. AND SONS, AT THE UNIVERSITY PRESS. MATHEMATICAL WORKS BY Messrs HALL AND KNIGHT, PUBLISHED BY MACMILLAN AND CO. ALGEBRA FOR BEGINNERS. By H. S. Hall, M.A., formerly Scholar of Christ's College, Cambridge, Master of the Military Side, CUfton College; and S. K. Knight, B.A., M.B., Ch.B., formerly Scholar of Trinity College, Cambridge, late Assistant Master at Marlborough College. Globe 8vo. 2s. With Answers. 2s. Qd. Educational Times : — "We pronounce this book the best of its size which we have seen." University Correspondent: — "Those masters who have already adopted Messrs Hall and Knight's Elementai-y Algebra in their schools, will welcome this new work for the use of their junior classes . . . 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The very fa- vourable reception accorded to previous editions leads us to hope that in its present more complete form the work will be found suitable as a first MACMILLAN AND CO., LONDON. Works by H. S. HALL, MA and S. R. KNIGHT, B.A. text-book for every class of student, and amply sufficient for all whose study of Algebra does not extend much beyond the Binomial Theorem. Academy: — "We will not say that this is the best Elementary Algebra for school use that we have come across, but we can say that we do not rememlDer to have seen a better .... It is the outcome of a long experience of school teaching, and so is a thoroughly practical hook. All others that we have in our eye are the works of men who have had considerable experience with senior and junior students at the universities, but have had little, if any, acquaintance with the poor creatures who are just stumbling over the threshold of Algebra .... Buy or borrow the book for yourselves and judge, or write a better .... 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Military and Naval Science Agriculture and Forestry Domestic Economy Hygiene Commerce Technology Geography History Art Divinity PAGK 30 30 30 32 33 33 34 35 35 36 37 38 38 39 39 39 40 40 40 41 41 41 42 44 44 GREEK AND I.ATIN CLASSICS. Elementary Classics ; Classical Series ; Classical Library, (1) Texts, (2) Trans- lations; Grammar, Composition, and PMlology; Antiquities, Ancient History, and Philosopliy. ^ELEMENTARY CLASSICS. 18mo, Eigliteenpence each. The following contain Introductions, Notes, and Vocabularies, and in some cases Exercises. ACCIDENCE, LATIN, AND EXERCISES ARRANGED FOR BEGINNERS.— By W. Welch, M.A., and C. G. Duffield, M.A. AESCHYLUS.— PROMETHEUS VINCTUS. By Rev. H. M. Stephenson, M.A. ARRIAN.— SELECTIONS. With Exercises. By Rev. John Bond, M.A., and Rev. A. S. Walpole, M.A. AULUS GELLIUS, STORIES FROM.— Adapted for Beginners. Witli Exercises. By Rev. G. H. Nall, M.A., Assistant Master at Westminster. OiESAR.— THE HELVETIAN WAR. 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