The person charging this material is re¬ sponsible for its return on or before the Latest Date stamped below. Theft, mutilation, and underlining of books are reasons for disciplinary action and may result in dismissal from the University. University of Illinois Library Digitized by the Internet Archive in 2018 with funding from University of Illinois Urbana-Champaign https://archive.org/details/planetrigonometr01lone . By the same Author. A TREATISE ON ELEMENTARY DYNAMICS. Crown 8vo. Second Edition, Reprinted. 7s. 6d. The Athenseum says :—This is one of the best and most complete treatises on elementary dynamics which have come before us for a long time. The principles of the subject are stated with great clear¬ ness, and the illustrative examples are very appropriate. We have been particularly struck with the manner in which the author com¬ bines perspicuity with brevity in his short chapter on “units and dimensions,” a subject which, though apparently very simple, often presents considerable difficulties to a beginner. Special praise is also due to his chapter on the “hodograpli” and “normal accelerations.” The Cambridge Review says :—Mr Loney may be congratulated on the production of a most valuable text-book, at once simple and com¬ plete. The earlier chapters on uniformly accelerated motion and the laws of motion are treated with extreme simplicity....A chapter of wrought-out examples clears up many difficulties which the beginner experiences in tackling problems, but which could scarcely be explained in the text. We are glad to see that the method of the hodograph is used in treating of normal acceleration, and that cycloidal motion is considered as a case of simple harmonic motion. SOLUTIONS TO THE EXAMPLES IN THE ELEMENTARY DYNAMICS. Crown 8vo. 7s. 6d. THE ELEMENTS OF STATICS AND DYNA¬ MICS. Part I. Statics. Third Edition, Revised. Ex. Fcap. 8vo. 4s. 6d. Part II. Elements of Dynamics. Second Edition, Revised and Enlarged. 3s. 6d. The two parts bound in one volume 7s. 6d. SOLUTIONS TO THE EXAMPLES IN THE ELEMENTS OF STATICS AND DYNAMICS. Ex. Fcap. 8vo. 7s. 6d. MECHANICS AND HYDROSTATICS FOR BEGINNERS. Ex. Fcap. 8vo. 4s. 6d. Nature says;—The same high standard of excellence is main¬ tained and the author must again be congratulated on his efforts to place in the hands of a beginner a book which will give him correct ideas of the laws and principles which are included in a study of mechanics....If the reader should fail to understand the chapter on the laws of motion, he must attribute it either to his want of ability or the nature of the subject, for we fail to see how the author could improve his remarks on this part of the subject. PLANE TRIGONOMETRY. Crown 8vo. 7s. 6 d. JLonDon : C. J. CLAY and SONS, CAMBRIDGE UNIVERSITY PRESS WAREHOUSE, AVE MARIA LANE. PLANE TEIPONOMETEY. JLontion: C. J. CLAY and SONS, CAMBRIDGE UNIVERSITY PRESS WAREHOUSE, AYE MARIA LANE. Camfrritige: DEIGHTON, BELL AND CO. Ectp^tg: E. A. BROCKHAUS. . lork: MACMILLAN AND CO. SPECIMEN COPY. PLANE TRIGONOMETRY BY S. L. LONEY, M.A. LATE FELLOW OF SIDNEY SUSSEX COLLEGE, CAMBRIDGE, PROFESSOR AT THE ROYAL HOLLOWAY COLLEGE. PART I. [An Elementary Course, excluding the use of Imaginary Quantities ] CAMBRIDGE: AT THE UNIVERSITY PRESS. 1893 [All Rights reserved.] (Eamimtjge: PRINTED BY C. J. CLAY, M. A. AND SONS, AT THE UNIVERSITY PRESS. GRAINGER 514.5 V - . KU’-k v, I , WTHFMATJCS LIBRARY O CO a CD 111 PREFACE. T 11HIS book consists of the more elementary portions of a text-book on Trigonometry. It covers the whole course usually read in Schools, and excludes De Moivre’s Theorem and the higher Analytical Trigonometry. As Trigonometry consists largely of formulae and the applications thereof, I have prefixed (on pages ix to xiii) a list of the principal formulae which the student should commit to memory. These more important formulae are distinguished in the text by the use of thick type. Other formulae are subsidiary and of less importance. The number of examples is very large. A selection only should be solved by the student on a first reading. On a first reading also the articles marked with an asterisk should be omitted. Considerable attention has been paid to the printing of the book and I am under great obligation to the Syndics of the Press for their liberality in this matter, VI PREFACE. and to the officers and workmen of the Press for the trouble they have taken. I am indebted to Mr W. J. Dobbs, B.A., late Scholar of St John’s College, for his kindness in reading and correcting the proof-sheets and for many valuable sug¬ gestions. For any corrections and suggestions for improvement I shall be thankful. S. L. LONEY. Royal Holloway College, Egham, Surrey. August 23 , 1893 . CONTENTS. CHAP. PAGE I. Measurement of angles. Sexagesimal and Centesimal Measure ......... 1 Circular, or Radian, Measure ..... 5 II. Trigonometrical Ratios for angles less than a right angle.19 Values for angles of 45°, 30°, 60°, 90° and 0° . 32 III. Simple problems in Heights and Distances . . 40 IV. Applications of algebraic signs to Trigonometry . 47 Tracing the changes in the ratios . . . . 52 V. Trigonometrical ratios of angles of any size. Ratios for - < 9 , 90° - <9, 90° + 6 . 64 VI. General expressions for all angles having a given trigonometrical ratio . . . . . . 76 VII. Ratios of the sum and difference of two angles . 87 Product Formulae.93 VIII. Ratios of multiple and submultiple angles . . 105 Explanation of ambiguities.117 Angles of 18°, 36°, and 9°.126 IX. Identities and trigonometrical equations . . .131 Vlll CONTENTS. CHAP. PAGE X. Logarithms. 146 Tables of logarithms ...... 152 XI. Principle of Proportional Parts .... 159 XII. Sides and Angles of a triangle .... 173 XIII. Solution of triangles. 188 Given two sides and the included angle 194 Ambiguous Case. 201 XIV. Heights and Distances. 209 XV. Properties of a triangle. 226 The circles connected with a triangle . 228 Orthocentre and Pedal triangle .... 236 Centroid and Medians. 239 XVI. Quadrilaterals ....... 249 Regular Polygons ...... 255 XVII. Trigonometrical ratios of small angles. sin 0< d = * J‘ ( )~ when 0 is very small. (Art. 228.) Area of a circle = 7 rr 2 . (Art. 233.) THE PRINCIPAL FORMULAE IN TRIGONOMETRY. Xlll sin a cos a + sin (a + (3) + sin (a + 2/?)+... to n terms ViZl a\ n P_ —. (Art. 241.) sin { a + —-— /3 [ sin 7 • £ sin V + cos (a + /3) + cos (a + 2/3) + ... to n terms n — 1 w/2 I -. (Art. 242.) COS i a + /? j sin . P sm- CHAPTER I. MEASUREMENT OF ANGLES, SEXAGESIMAL, CENTESIMAL, AND CIRCULAR MEASURE. 1. In geometry angles are measured in terms of a right angle. This, however, is an inconvenient unit of measurement on account of its size. 2. In the Sexagesimal system of measurement a right angle is divided into 90 equal parts called Degrees. Each degree is divided into 60 equal parts called Minutes, and each minute into 60 equal parts called Seconds. The symbols 1°, 1', and 1" are used to denote a degree, a minute, and a second respectively. Thus 60 Seconds (60") make One Minute (1'), 60 Minutes (60') „ „ Degree (1°), and 90 Degrees (90°) „ „ Right Angle. This system is well established and is always used in the practical applications of Trigonometry. It is not however very convenient on account of the multipliers 60 and 90. L. T. 1 tv 2 TRIGONOMETRY. 3. On this account another system of measurement called the Centesimal, or French, system has been proposed. In this system the right angle is divided into 100 equal parts, called Grades; each grade is subdivided into 100 Minutes, and each minute into 100 Seconds. The symbols l g , 1', and 1" are used to denote a Grade, a Minute, and a Second respectively. Thus 100 Seconds (100") make One Minute (I s ), 100 Minutes (100') „ „ Grade, (l g ), 100 Grades (100 g ) „ „ Right angle. 4. This system would be much more convenient to use than the ordinary Sexagesimal System. As a preliminary, however, to its practical adoption, a large number of tables would have to be recalculated. For this reason the system has in practice never been used. 5 . To convert Sexagesimal into Centesimal Measure , and vice versa. Since a right angle is equal to 90 c and also to 100 g , we have 90 ° = 100 g . 10 g “9" ’ and l g 9° 10 * Hence, to change degrees into grades, add on one- ninth ; to change grades into degrees, subtract one-tenth. Ex. 36°= |^36 + and 64* = ^64 - ^ x 64^° = (64 - 6*4)° = 57*6°. If the angle do not contain an integral number of degrees, we may reduce it to a fraction of a degree and then change to grades. 1 \ g g x 36 J = 40*, MEASUREMENT OF ANGLES. 3 In practice it is generally found more convenient to reduce any angle to a fraction of a right angle. The method will be seen in the following examples; Ex. 1 . Reduce 63° 14' 51" to Centesimal Measure. We have and 17' si''— — 01 ” 20 " 80 ’ 14.QS° 14' 51"= 14*85' = 6 q =*2475°, AQ.047S 63° 14' 51" = 63*2475° = - rt. angle = *70275 rt. angle = 70*2758 = 70» 27-5' = 70s 27' 50' . Ex. 2. Reduce 94^23' 87" to Sexagesimal Measure. 94s 23' 87" = *942387 right angle _90 84-81483 degrees 60 48-8898 minutes _60 53*3880 seconds .*. 94s 23' 87" = 84° 48' 53-388". 6. Angles of any size. Suppose AOA' &nd BOB' to be two fixed lines meeting at right angles in 0, and suppose a revolving line OP (turning about a fixed point at 0) to start from OA and revolve in a direction opposite to that of the hands of a watch. For any position of the re¬ volving line between OA and OB , such as 0P 1 , it will have turned through an angle A0P ly which is less than a right angle. 1—2 4 TRIGONOMETRY. For any position between OB and OA', such as 0P 2 , the angle AOP 2 through which it has turned is greater than a right angle. For any position 0P 3 , between OA' and OB', the angle traced out is AOP 3 , i.e. AOB -{-BOA' + A'OP 3 ,i.e. 2 right angles + A'0P 3 , so that the angle described is greater than two right angles. For any position 0P 4 , between OB' and OA, the angle turned through is similarly greater than three right angles. When the revolving line has made a complete revo¬ lution, so that it coincides once more with OA, the angle through which it has turned is 4 right angles. If the line OP still continue to revolve, the angle through which it has turned, when it is for the second time in the position OP Y , is not AOP 1 but 4 right angles + AOP 1 . Similarly when the revolving line, having made two complete revolutions, is once more in the position 0P 2 , the angle it has traced out is 8 right angles 4- A 0P. 2 . 7. If the revolving line OP be between OA and OB it is said to be in the first quadrant; if it be between OB and OA' it is in the second quadrant; if between OA' and OB' it is in the third quadrant; if it is between OB' and OA it is in the fourth quadrant. 8. Ex. What is the position of the revolving line when it has turned through (1) 225°, (2) 480°, and (3) 1050°? (1) Since 225° =180°+ 45°, the revolving line has turned through 45° more than two right angles and is therefore halfway between OA' and OB'. (2) Since 480° = 360° +120°, the revolving line has turned through 120° more than one complete revolution, and is therefore between OB and OA', and makes an angle of 30° with OB. CIRCULAR MEASURE. 5 (3) Since 1050° —11 x 90° + 60°, the revolving line has turned through 60° more than eleven right angles and is therefore between OB' and OA and makes 60° with OB'. EXAMPLES. I. Express in terms of a right angle the angles 1. 60°. 2. 75° 15'. 3. 63° 17'25". 4. 130° 30'. 5. 210° 30'30". 6. 370° 20'48". Express in grades, minutes, and seconds the angles 7. 30°. 8. 81°. 9. 138° 30'. 10. 35° 47'15". 11. 235° 12'36". 12. 475° 13'48". Express in terms of right angles and also in degrees, minutes, and seconds the angles 13. 120s. 14. 45s 35' 24' '. 15. 39s 45' 36". 16. 255 g 48' 81". 17. 759 g 45' 60". Mark the position of the revolving line when it has traced out the following angles: 4 18. ^ right angle. 19. 3J right angles. 20. 13i right angles. 21. 120°. 22. 315°. 23. 745°. 24. 1185°. 25. 150 g . 26. 420s. 27. 875 g . 28. How many degrees, minutes and seconds are respectively passed over in 11£ minutes by the hour and minute hands of a watch ? 29. The number of degrees in one acute angle of a right-angled triangle is equal to the number of grades in the other; express both the angles in degrees. 30. Prove that the number of Sexagesimal minutes in any angle is to the number of Centesimal minutes in the same angle as 27 : 50. 31. Divide 44° 8' into two parts such that the number of Sexagesimal seconds in one part may be equal to the number of Centesimal seconds in the other part. Circular Measure. 9. A third system of measurement of angles has been devised, and it is this system which is used in all the higher branches of Mathematics. G TRIGONOMETRY. B The unit used is obtained thus; Take any circle APBB', whose centre is 0, and from any point A measure off an arc AP whose length is equal to the radius of the circle. Join OA and OP. The angle AOP is the angle which is taken as the unit of cir¬ cular measurement, i.e. it is the angle in terms of which in this system we measure all others. This angle is called A Radian and is often denoted by l c . 10 . It is clearly essential to the proper choice of a unit that it should be a constant quantity; hence we must shew that the Radian is a constant angle. This we shall do in the following articles. 11. Theorem. The length of the circumference of a circle always hears a constant ratio to its diameter. Take any two circles whose common centre is 0. In the large circle inscribe a regular polygon of n sides, ABCD.... Let OA, OB, OC,... meet the smaller circle in the points ct, b, c, d... and join ah, he, cd,.... Then, by Euc. VI. 2, abed ... is a regular polygon of n sides in¬ scribed in the smaller circle. Since Oa = Ob, and OA = OB, THE RADIAN. 7 the lines ab and AB must be parallel, and hence AB OA Hb = Oa' ' Also the polygon ABGD... being regular, its perimeter, i.e. the sum of its sides, is equal to n . AB . Similarly for the inner polygon. Hence we have Perimeter of the outer polygon n.AB AB OA Perimeter of the inner polygon n . ab ab Oa .ax This relation exists whatever be the number of sides in the polygons. Let then the number of sides be indefinitely increased (i.e. let n become inconceivably great) so that finally the perimeter of the outer polygon will be the same as the circumference of the outer circle, and the perimeter of the inner polygon the same as the circumference of the inner circle. The relation (1) will then become Circumference of outer circle OA Circumference of inner circle Oa Radius of outer circle Hence Radius of inner circle * Circumference of outer circle Radius of outer circle Circumference of inner circle Radius of inner circle Since there was no restriction whatever as to the sizes of the two circles, it follows that the quantity Circumference of a circle Radius of the circle is the same for all circles. 8 TRIGONOMETRY. Hence the ratio of the circumference of a circle to its radius, and therefore also to its diameter, is a constant quantity. 12. In the previous article we have shewn that the . Circumference. « ni . , mi . ratio —^-is the same for all circles. I he value Diameter of this constant ratio is always denoted by the Greek letter 7 r (pronounced Pi), so that ir is a number. TT Circumference A , J . Hence —^--- = the constant number it. Diameter We have therefore the following theorem; The cir¬ cumference of a circle is always equal to 7 r times its diameter or 2 tt times its radius. 13. Unfortunately the value of 77 is not a whole number, nor can it be expressed in the form of a vulgar fraction, and hence not in the form of a decimal fraction, terminating or recurring. The number 7 r is an incommensurable magnitude, i.e. a magnitude whose value cannot be exactly expressed as the ratio of two whole numbers. Its value, correct to 8 places of decimals, is 3T4159265.... 22 . The fraction gives the value of 7 r correctly for the 22 first two decimal places ; for = 3T4285.... 355 The fraction —^ is a more accurate value of 7 r being 1 lo 355 correct to 6 places of decimals; for — = 3T4159203.... JL JL o THE RADIAN. 9 [N.B. The fraction — may be remembered thus; write down the IIo first three odd numbers repeating each twice, thus 113355; divide the number thus obtained into portions and let the first part be divided into the second, thus 113) 355(. The quotient is the value of tt to 6 places of decimals.] To sum up. An approximate value of nr, correct 22 to 2 places of decimals, is the fraction —- ; a more accurate value is 3 a 14159.... By division we can shew that 1 7 T •3183098862.... 14. Ex. 1. The diameter of a tricycle wheel is 28 inches; through ivhat distance does its centre move during one revolution of the icheel? The radius r is here 14 inches. The circumference therefore = 2.7r . 14 = 287 t inches. . 22 22 Taking ir = —, the circumference = 28 x — inches = 7 ft. 4 inches ap¬ proximately. Giving 7 r the more accurate value 3•14159265... the circumference = 28x3*14159265... inches =7 ft. 3*96459... inches. Ex. 2. What must he the radius of a circular running path, round which an athlete must run 5 times in order to describe one mile ? The circumference must be ^ x 1760, i.e. 352, yards. o Hence, if r be the radius of the path in yards, we have 27 t/*=352, 176 i.e. r= — yards. 7T 22 176 x 7 Taking 7r= —, we have r =——— =56 yards nearly. I Taking the more accurate value - =*31831, we have 7T r=176 x *31831= 56*02256 yards. 10 TRIGONOMETRY. EXAMPLES. II. 1. If the radius of the earth be 4000 miles, what is the length of its circumference? 2. The wheel of a railway carriage is 3 feet in diameter and makes 3 revolutions in a second; how fast is the train going? 3. A mill sail whose length is 18 feet makes 10 revolutions per minute. What distance does its end travel in an hour? 4. The diameter of a halfpenny is an inch; what is the length of a piece of string which would just surround its curved edge? 5. Assuming that the earth describes in one year a circle, of 92500000 miles radius, whose centre is the sun, how many miles does the earth travel in a year ? 6. The radius of a carriage wheel is 1 ft. 9 ins., and it turns through 80° in ^th of a second; how many miles does the wheel travel in one hour? 15. Theorem. The radian is a constant angle . Take the figure of Art. 9. Let the arc AB be a quadrant of the circle, i.e. one quarter of the circum¬ ference. By Art. 12, the length of AB is therefore 7TV T , where r is the radius of the circle. By Euc. Yl. 33, we know that angles at the centre of any circle are to one another as the arcs on which they stand. Hence ZAOP arc AP _ r _ 2 Z A OB arc AB n r it ’ — r A AOP = - . zAOB. 7r But we defined the angle AOP to be a Radian. THE RADIAN. 11 2 Hence a Radian = —. Z AOB 7 T = — x a right angle. 7T Since a right angle is a constant angle and since we have shewn (Art. 12 ) that it is a constant quantity, it follows that a Radian is a constant angle, and is therefore the same whatever be the circle from which it is derived. 16. Magnitude of a Radian. By the previous article a radian 2 . , . , 180 = — x a right angle = — 7T 77 180 ° 3 14159265... = 57-2957795 = 57° l7 / 44*8 // nearly. 17. Since a Radian = - x a right angle, 7T therefore a right angle = —. radians, £ so that 180 = 2 right angles = tt radians, and 360 c = 4 right angles = 27 t radians. Hence when the revolving line (Art. 6 ) has made a complete revolution it has described an angle equal to 27r radians; when it has made three complete revolutions it has described an angle of 67 t radians; when it has made n revolutions it has described an angle of 2utt radians. 18. In practice the symbol “ c 55 is generally omitted and instead of “ an angle 7 r c ” we find written “ an angle 7 r.” 12 TRIGONOMETRY. The student must notice this point carefully. If the unit, in terms of which the angle is measured, be not mentioned, he must mentally supply the word “ radians.” Otherwise he will easily fall into the mistake of supposing that 7 r stands for 180°. It is true that ir radians ( 7 r c ) is the same as 180°, but ir itself is a number, and a number only. 19. To convert circular measure into sexagesimal measure or centesimal measure and vice versa . The student should remember the relations, Two right angles = 180° = 200 g = tt radians. The conversion is then merely Arithmetic. Ex. (1) -457t c = -45x180° = 81°= 90s. (2) S c = - x tt c = - x 180°= -x 200?. 7T 7T 7T (3) 40° 15' 36" = 40° 15f = 40*26° = 40 *26 x -A = -22367T radians. 180 (4) 40s 15' 36" = 401536s = 401536 x radians zuu = *200768^ radians. 20. Ex. 1. The angles of a triangle are in a. p. and the number of grades in the least is to the number of radians in the greatest as 40 : tt ; find the angles in degrees . Let the angles be (x-y)°, x°, and (# + ?/) 0 . Since the sum of the three angles of a triangle is 180°, we have 180 = x - y + x + x + y = Sx, so that # = 60. The required angles are therefore (60-?/)°, 60°, and (60 + ?/)°. Now (60-»)°=“x(«0-»)*, (60 + y)° = x (60 + y) radians. 180 and THE RADIAN 1 *2 iO Hence ^ (60 -y) : ^ (60 + ?/) :: 40 : w. 200 60-?/ _ 40 7 r 60 + y 7r i.e. 5 (60-7/) = 60 + ?/, ?#. ?/ = 40. The angles are therefore 20°, 60°, and 100°. Ex. 2. Express in the 3 systems of angular measurement the magni¬ tude of the angle of a regular decagon. The corollary to Euc. I. 32 states that all the interior angles of any rectilinear figure together with four right angles are equal to twice as many right angles as the figure has sides. Let the angle of a decagon contain x right angles, so that all the angles are together equal to 10# right angles. The corollary therefore states that 10#+ 4 = 20, g so that x = - right angles. 5 But one right angle = 90° = 100s = ** radians. Hence the required angle = 144° = 160s = ^ radians. EXAMPLES. III. Express in degrees, minutes, and seconds the angles, 7T C 47r c 1. 2. -5-. 3. 107T c . 4. l c . o o Express in grades, minutes, and seconds the angles, 6 . 47r c IT* 8. 10tt c . Express in radians the following angles: 9. 60°. 10. 110° 30'. 11. 175° 45'. 13. 395°. 14. 60s. 15. 110 s 30'. 5 . 8 C . 12. 47° 25' 36". 16. 345s 25 36' . TRIGONOMETRY. 14 [Exs. III.] 17 Trlt. difference between che two acute angles of a right-angled 9 "to trio~ 6 xfc in - 7 r radians; express the angles in degrees. D 2 3 18. One angle of a triangle is -x grades and another is degrees, o • 2 7T.£ whilst the third is — radians ; express them all in degrees. 19. The circular measure of two angles of a triangle are respectively ^ and \; what is the number of degrees in the third angle ? 2 o 20. The angles of a triangle are in a. p. and the number of degrees in the least is to the number of radians in the greatest as 60 to 7r; find the angles in degrees. 21. The angles of a triangle are in a. p. and the number of radians in the least angle is to the number of degrees in the mean angle as 1:120. Find the angles in radians. 22. Find the magnitude, in radians and degrees, of the interior angle of (1) a regular pentagon, (2) a regular heptagon, (3) a regular octagon, (4) a regular duodecagon, and (5) a regular polygon of 17 sides. 23. The angle in one regular polygon is to that in another as 3 : 2 ; also the number of sides in the first is twice that in the second; how many sides have the polygons? 24. The number of sides in two regular polygons are as 5 : 4, and the difference between their angles is 9°; find the number of sides in the polygons. 25. Find two regular polygons such that the number of their sides may be as 3 to 4 and the number of degrees in an angle of the first to the number of grades in an angle of the second as 4 to 5. 26. The angles of a quadrilateral are in a. p. and the greatest s double the least; express the least angle in radians. 27. Find in radians, degrees, and grades the angle between the hour-hand and the minute-hand of a clock at (1) half-past three, (2) twenty minutes to six, (3) a quarter past eleven. 21. Theorem. The number of radians in any angle whatever is equal to a fraction , whose numerator is the arc which the angle subtends at the centre of any circle , and whose denominator is the radius of that circle. MEASUREMENT OF ANY ANGLE IN RADIANS. 15 Let AOP be the angle which has been described by a line starting from OA and revolv¬ ing into the position OP. With centre 0 and any radius describe a circle cutting OA and OP in the points A and P. JL Let the angle A OB be a radian, so that the arc AB is equal to the radius OA. By Euc. VI. 33, we have so that Z AOP __ Z AOP arc AP _ arc AP A Radian Z AOB arc AB Radius . . , ^ arc AP Z AOP = ^^— x A Radian. Radius Hence the theorem is proved. 22. Ex. 1. Find the angle subtended at the centre of a circle of radius 3 feet by an arc of length 1 foot. arc 1 The number of radians in the angle = Hence the angle radius 3 = J radian = i . — right angle = ~ x 90° = — = IOAt 0 * O O 7T D7T 7r 22 taking ir equal to —. Ex. 2. In a circle of 5 feet radius what is the length of the arc which subtends an angle of 33° 15' at the centre ? If x feet be the required length, we have X - = number of radians in 33° 15' 5 = 33J 180 133 = irr-r 7T. 7 r (Art. 19). 720 133 . 133 22 , , . X = 144 W Cet = 144 X T feet near] y = 2ff feet nearly. 16 TRIGONOMETRY. Ex. 3. Assuming the average distance of the earth from the sun to he 92500000 miles , and the angle subtended by the sun at the eye of a person on the earth to be 32', find the sun's diameter. Let D be the diameter of the sun in miles. The angle subtended by the sun being very small, its diameter is very approximately equal to a small arc of a circle whose centre is the eye of the observer. Also the sun subtends an angle of 32' at the centre of this circle. Hence, by Art. 21, we have D 92500000 = the number of radians in 32' = the number of radians in 8 ^ 15 8 7r 27r 15 X 180 ~ 675 * 185000000 “675“ 7 r miles _185000000 675 22 y miles approximately = about 862000 miles. Ex. 4. Assuming that a person of normal sight can read print at such a distance that the letters subtend an angle of 5' at his eye , find what is the height of the letters that he can read at a distance (1) of 12 feet , and (2) of a quarter of a mile. Let x be the required height in feet. In the first case, x is very nearly equal to the arc of a circle, of radius 12 feet, which subtends an angle of 5' at its centre. Hence x 12 = number of radians in 5' 1 7T X 12 180 * x= i¥o feet = llo x T feet nearly = ^ x ^ inches = about * inch. MEASUREMENT OF ANY ANGLE IN RADIANS. 17 In the second case the height y is given by so that y 440x3 = number of radians in 5 1 7T = L2 X 180’ y=l s t ~ 18 x Y feet nearly = about 23 inches. EXAMPLES. IV. 1. Find the number of degrees subtended at the centre of a circle by an arc whose length is -357 times the radius, taking tt — 3*1416. 2. Express in radians and degrees the angle subtended at the centre of a circle by an arc whose length is 15 feet, the radius of the circle being 25 feet. 3. The value of the divisions on the outer rim of a graduated circle is 5' and the distance between successive graduations is *1 inch. Find the radius of the circle. 4. The diameter of a graduated circle is 6 feet and the graduations on its rim are 5' apart; find the distance from one graduation to another. 5. Find the radius of a globe which is such that the distance between two places on the same meridian whose latitude differs by 1° 10' may be half-an-inch. 6. Taking the radius of the earth as 4000 miles find the difference in latitude of two places, one of which is 100 miles north of the other. 7. Assuming the earth to be a sphere and the distance between two parallels of latitude, which subtends an angle of 1° at the earth’s centre, to be 69^- miles, find the radius of the earth. 8. The radius of a certain circle is 3 feet; find approximately the length of an arc of this circle, if the length of the chord of the arc be 3 feet also. 9. What is the ratio of the radii of two circles at the centre of which two arcs of the same length subtend angles of 60° and 75° ? 10. If an arc, of length 10 feet, on a circle of 8 feet diameter subtend at the centre an angle of 143° 14' 22"; find the value of 7 r to 4 places of decimals. L. T. 2 18 TRIGONOMETRY. [Exs. IV.] 11 . If the circumference of a circle be divided into 5 parts which are in a. p., and if the greatest part be 6 times the least, find in radians the magnitudes of the angles that the parts subtend at the centre of the circle. 12. The perimeter of a certain sector of a circle is equal to the length of the arc of a semicircle having the same radius; express the angle of the sector in degrees, minutes, and seconds. 13. At what distance does a man, whose height is 6 feet, subtend an angle of 10' ? 14. Find the length which at a distance of one mile will subtend an angle of 1' at the eye. 15. Find approximately the distance at which a globe, 5J inches in diameter, will subtend an angle of 6'. 16. Find approximately the distance of a tower whose height is 51 feet and which subtends at the eye an angle of 5 T 5 T '. 17. A church spire, whose height is known to be 45 feet, subtends an angle of 9' at the eye; find approximately its distance. 18. Find approximately in minutes the inclination to the horizon of an incline which rises 3J feet in 210 yards. 19. The radius of the earth being taken to be 3960 miles, and the distance of the moon from the earth being 60 times the radius of the earth, find approximately the radius of the moon which subtends at the earth an angle of 16'. 20. When the moon is setting ,at any given place the angle that is subtended at its centre by the radius of the earth passing through the given place is 57'. If the earth’s radius be 3960 miles, find approximately the distance of the moon. 21. Prove that the distance of the sun is about 81 million geo¬ graphical miles, assuming that the angle which the earth’s radius subtends at the distance of the sun is 8-76", and that a geographical mile subtends 1' at the earth’s centre. Find also the circumference and diameter of the earth in geographical miles. 22. The radius of the earth’s orbit, which is about 92700000 miles, subtends at the star Sirius an angle of about *4"; find roughly the distance of Sirius. CHAPTER II. TRIGONOMETRICAL RATIOS FOR ANGLES LESS THAN A RIGHT ANGLE. 23. In the present chapter we shall only consider angles which are less than a right angle. Let a revolving line OP start from OA and revolve into the position OP, thus tracing out the angle AOP . In the revolving line take any point P and draw PM perpendicular to the initial line OA. In the triangle MOP , OP is the hypothenuse, PM is the perpendicular, and OM is the base. The trigonometrical ratios, or functions, of the angle A OP are defined as follows; MP . Perp. . OP ’ % ' e ' Hyp. ’ 1S OM . Base OP’ l ' e • Hyp.’ MP . Perp. OM ’ l£ ' Base ’ OM . Base MP ’ P^rj). ’ — ie MP ’ e ' Perp. ’ OP . Hyp. OM ’ Base ’ called the Sine of the angle AOP ; Cosine >> )> Tangent Cotangent >> >> >> „ Cosecant >} „ „ Secant 20 TRIGONOMETRY. The quantity by which the cosine falls short of unity, i.e. 1 — cos AOP, is called the Versed Sine of A OP ; also the quantity 1 — sin A OP , by which the sine falls short of unity, is called the Coversed Sine of AOP. 24. It will be noted that the trigonometrical ratios are all numbers. The names of these eight ratios are written, for brevity, sin AOP , cos AOP , tan AOP, cot AOP, cosec AOP , sec AOP, vers AOP, and covers AOP respectively. The two latter ratios are seldom used. 25. It will be noticed, from the definitions, that the cosecant is the reciprocal of the sine, so that cosec A OP = —— \ A . sm AOP So the secant is the reciprocal of the cosine, i.e. sec AOP = - vvytd > cos A OP and the cotangent is the reciprocal of the tangent, i.e. 1 cot A OP = tan A OP ’ 26. To shew that the trigonometrical ratios are always the same for the same angle. We have to shew that if in the revolving line OP any other point P' be taken and PAT be drawn perpendicular to OA, the ratios derived from the triangle TRIGONOMETRICAL RATIOS. 21 OP'M' are the same as those derived from the triangle OPM ; In the two triangles the angle at 0 is common, and the angles at M and M' are both right angles and there¬ fore equal. Hence the two triangles are equiangular and therefore, by Euc. VI. 4, we have MP M'P' OP ” OP' i.e. the sine of the angle AOP is the same whatever point we take on the revolving line. Since, by the same proposition, we have OM _ OM' , MP _ M'P' OP ~ OP' and OM ~ OM ’ it follows that the cosine and tangent are the same whatever point be taken on the revolving line. Similarly for the other ratios. If OA be considered as the revolving line and in it be taken any point P" and P"M" be drawn perpendicular to OP, the functions as derived from the triangle OP"M" will have the same values as before. For, since in the two triangles OPM and OP"M", the two angles P"OM" and OM"P" are respectively equal to POM and OMP , these two triangles are equiangular and therefore similar, and we have M"P" _ MP OM" OM ~OP" “OP ’ and OP" ~ OP * 27. Fundamental relations betiveen the trigonometrical ratios of an angle. We shall find that if one of the trigonometrical ratios of an angle be known, the numerical magnitude of each of the others is known also. o o Let the angle AOP (Fig., Art. 23) be denoted by 6. In the triangle AOP we have, by Euc. I. 47, MP 2 + OM 2 = OP 2 ( 1 ). 22 TRIGONOMETRY. Hence, dividing by OP' 2 , we have (MPy /OMy _ V OP) + [op) - 1> i.e. (sin 0 ) 2 4 - (cos 0) 2 = 1. The quantity (sin 0) 2 is always written sin 2 0 and so for the other ratios. Hence this relation is sin 2 $+cos 2 0 = 1.(2). Again, dividing both sides of equation (1) by OM 2 , we have MP y m) i.e . (tan 0) 2 + 1 = (sec 0 ) 2 , so that sec 2 0 = 1 + tan 2 6 . Again, dividing equations (1) by MP 2 we have 1 + i.e. 1 + (cot 0) 2 = (cosec 0) 2 , so that cosec 2 6 = 1 + cot 2 6 Also, since sin 0 = and cos 0 = , we have Hence sin 0 MP OM MP cos 0 OP * OP tan 6 = OM sin 6 = tan 0. cot 6 = COS0 cos 6 sin 6 (0). Similarly ( 6 ). TRIGONOMETRICAL RATIOS. 23 28. Ex. 1. Prove that VI - cos A + cos A = cosec A — cot A. We have . A ~ cos A - . /QlZ V 1 + C0s4~ V 1- - cos A ) 2 1 - cos A J 1 - cos 2 A by relation (1) of the last article, 1 cos A cos 2 A 1 - cos A sin A * sin A sin A = cosec A - cot A . Ex. 2. Prove that Jsec 2 A + cosec 1 A = tan A + cot A. We have seen that sec 2 A = 1 + tan 2 A , and cosec 2 A = 1 + cot 2 A . sec 2 A + cosec 2 A = tan 2 A + 2 + cot 2 A = tan 2 A + 2 tan A cot A + cot 2 A so that = (tan A + cot A) 2 , yj sec 2 A + cosec 2 A = tan A + cot A , Ex. 3. Prove that (cosec A - sin A) (sec A - cos A) (tan A + cot A) = 1. The given expression - sin A cos A )(- J \cc sin A cos A sin A .. J \cosA ““ J VcosA ' sin A 1 - sin 2 A 1 - cos 2 A sin 2 A + cos 2 A sin A cos A sin A cos A cos 2 A sin 2 A sin A ’ cos A * sin A cos A = 1. 24 TRIGONOMETRY. EXAMPLES. V. Prove the following statements. 1 . 2 . 3. cos 4 A - sin 4 A +1 = 2 cos 2 ^4. (sin A 4 - cos A) (1 - sin A cos A) = sin 3 A 4 - cos 3 A. sin A 14-cosJ. _ 7 -i 4-;— j— = 2 cosec A. 1 4 - cos A sin A 4. cos 6 A 4 - sin 6 A = 1 - 3 sin 2 A cos ' 2 A. 5. 6 . 7. 9. 10 . 11 . /l - sin A _ \ 1 4 - sin A cosec A 4- sec A - tan A. cosec A cosec ^4-1 cosec ^4 4-1 cosec A = cos A. — 2 sec 2 A. cot A 4 - tan A 8. (sec A 4 - cos A) (sec A - cos ^4) = tan 2 y4 4 -sin 2 ^4. 1 cot A 4 - tan A 1 sec A - tan A = sin A cos A. = sec A 4 - tan A. 1 - tan A cot ^4-1 1 4 - tan A cot ^4 4-1* 14- tan 2 A _ sin 2 A 1 4 - cot 2 A ~ cos 2 A * , 0 sec A - tan A 9 13. -—i-=1 - 2 sec A tan A 4-2 tan~y4. sec A 4- tan A tan A 1 - cot A cot A . . . 4- ^—t -7 = sec A cosec ^14-1 1 - tan A 15. cos A 1 - tan A + sm A . z - — = sm A 4- cos A . 1 - cot A 16. (sin A + cos A) (cot A + tan A) =sec A + cosec A. 17. sec 4 A - sec 2 A = tan 4 A 4- tan 2 A. [Exs. V.] TRIGONOMETRICAL RATIOS. 25 18. cot 4 A + cot 2 A =cosec 4 A - cosec 2 A . 19. ij cosec 2 A — 1 = cos A cosec A. 20. sec 2 A cosec 2 A — tan 2 A + cot 2 A+2. 21 . tan 2 A - sin 2 A = sin 4 A sec 2 A. 22. (l + cot4 - cosec A) (1 + tan A + sec A) = 2. i ii 1 23. 24. 25. 26. 27. 28. 29. 30. 31. 32. 33. 34. 35. cosec A - cot A sin A sin A cosec A + cot A cot A cos A cot A - cos A cot A + cos A cot A cos A cot A + tan B cot B + tan A 1 ( = cot A tan B. 1 + sec 2 a - cos 2 a ' cosec* a - sin* a cos 2 a sin 2 a = 1 - cos 2 a sin 2 a 2 + cos 2 a sin 2 a * sin 8 A - cos 8 A = (sin 2 A - cos 2 ^4) (1 — 2 sin 2 A cos 2 A). cos A cosec A - sin A sec A - -;— -= cosec A - sec A. cos A + sin A tan A + sec A - 1 _ 1 + sin A tan A - sec A + l cos A (tan a + cosecj3) 2 - (cot/3 - sec a) 2 =2 tana cot/S (coseca + sec/3). 2 sec 2 a - sec 4 a - 2 cosec 2 a + cosec 4 a = cot 4 a - tan 4 a. 1 - sin A 1 + sin A = 1 + 2 tan A (tan A - sec A). (cosec A + cot A) covers A - (sec A + tan A) vers A = (cosec A - sec A) (2 - vers A covers A). sec A cosec A (1 + cot4 + tan4) (sin A -cos A)=- -„ .-. v ' x cosec 2 A sec 2 A 2 versin A + cos 2 A = 1 + versin 2 A. 29. Limits to the values of the trigonometrical ratios. From equation (2) of Art. 27 we have sin 2 # + cos 2 # = 1. 26 TRIGONOMETRY. Now sin 2 # and cos 2 #, being both squares, are both necessarily positive. Hence, since their sum is unity, neither of them can be greater than unity. [For if one of them, say sin 2 6, were greater than unity, the other, cos 2 0, would have to be negative, which is impossible.] Hence neither the sine nor the cosine can be numeri¬ cally greater than unity. Since sin # cannot be greater than unity therefore cosec #, which equals > cannot be numerically less than unitv. «/ So sec #, which equals —— 7 ., cannot be numerically ^ cos # J less than unity. 30. The foregoing results follow easily from the figure of Art. 23. For, whatever be the value of the angle AOP , neither the side OM nor the side MP is ever greater than OP. MP . Since MP is never greater than OP the ratio ^ is never greater than unity, so that the sine of an angle is never greater than unity. Also since OM is never greater than OP, the ratio OM OP is never greater than unity, i.e. the cosine is never greater than unity. 31. We can express the trigonometrical ratios of an angle in terms of any one of them. TRIGONOMETRICAL RATIOS. 27 The simplest method of procedure is best shewn by examples. Ex. 1 . To express all the trigono¬ metrical ratios in terms of the sine . Let A OP be any angle 0. Let the length OP be unity and let the corresponding length of MP be s. By Euc. i. 47, OM = VOP 2 - MP 2 = Vl^7 2 . Hence sin 6 = MP OP cos 6 = tan 6 = = Vl — s 2 = V1 — sin' 2 0, itfP 5 sin # OM V1 — 5 2 Vl — sin 2 # ? and cot 6 = cosec 0 = sec 0 = MP~ OP _ MP ~ OP __ 0M~ Vl — s 2 _ Vl — sin 2 6 s sin 6 y 1 _ 1 5 sin 6 ’ _1 1 Vl — s 2 Vl — sin 2 #* The last five equations give what is required. Ex. 2. To express all the trigonometrical relations in terms of the cotangent. Taking the usual figure let the length MP be unity, and let the corre¬ sponding value of OM be x. By Euc. i. 47, op= voiip+ mp 2 =vr+^r 28 TRIGONOMETRY. Hence and n OM x COt ^ ~ MP ~i~ X) sin 6 = MP OP Vl+a; 2 Vl + cot 2 6 ’ cos 6 = tan 6 — sec 6 = cosec 6 = OM x cot 6 OP \f\+x 2 Vl-fcot 2 0 5 MP 1 1 OM x cot 6 ’ OP 1+# 2 V1 + cot 2 6 OM x cot 6 OP _ Vl + MP ~ 1 x- = Vl + cot 2 ^. The last live equations give what is required. It will be noticed that, in each case, the denominator of the fraction which defines the trigonometrical ratio was taken equal to unity. For example, the sine is MP OP , and hence in Ex. 1 the denominator OP is taken equal to unity. The cotangent is OM MP and hence in Ex. 2 the side MP is taken equal to unity. Similarly suppose we had to express the other ratios in terms of the cosine, we should, since the cosine is equal OM to -Qp , put OP equal to unity and OM equal to c. The working would then be similar to that of Exs. 1 and 2. In the following examples the sides have numerical values. TRIGONOMETRICAL RATIOS. 29 Ex. 3. If cos 9 equal - , find the values of the other ratios. Along the initial line OA take OBI equal to 3, and erect a perpen¬ dicular MP. Let a line OP, of length 5, revolve round 0 until its other end meets this perpendicular in the point P. Then AOP is the angle 0. By Euc. i. 47, MP= JOF*-OM?= 3 2 =4. Hence clearly 4 4 3 5 5 sin0 = ^, tan 0 = -, cot 9 — -, cosec 0 = - and sec 0 = -. 5 3 4 4 3 Ex. 4. Supposing 9 to be an angle whose sine is - , to find the numeri¬ cal magnitude of the other trigonometrical ratios. Here sin 0 = -, so that the relation (2) of Art. 27 gives O + cos 2 0=1, i.e. i.e. cos 2 0 = 1 - 1 _ 8 9 “ 9 ’ cos 0 = 2^/2 3 * Hence tan 9 = cot 0 = sin 9 _ 1 _ cos 9 2 n /2 4 1 tan 0 ~ 2 ' /2 ’ cosec 6 =-—= 3, sin 9 Q 1 3 3^2 S6C cos 9 2^2 4 2^/2 vers 0=1- cos 0=1 — covers 0=1- sin 0 = 1-- O 2 3* and 32. In the following table is given the result of expressing each trigonometrical ratio in terms of each of the others. 30 TRIGONOMETRY. <3b 0 0 co O 0 Qb 0 a) co O 0 T“H i—1 rH | 1 | ^b ^b Qb ^b o 0 0 rH 0 Cl o 0 0 0 CO 0 0 CO CO O CO CO O O 0 O O 0 0 0 0 > > > I ^b ci 0 0 co O 0 ^b 0 o co O 0 rH rH rH 1 <3b ^b 1 1 <3b <3b o 0 rH 0 ^b rH Qb Cl 0 0 0 0 0 0 0 0 CO CO 0 0 CO CO CO CO > > > 0 0 co o 1 0 I CO ! I ^b o 0 co !> ^b ci 4^ O ^b Cl 4-o C Qb • <3b Cl 40) O <3b 4-0 O 0 rH 0 + rH -4-0 o 0 0 + rH rH 4-o O 0 -4-0 o 0 0 + rH 4-0 O 0 > > > ^b ci -+o> O 0 + I r—I > ^b rH ^b ^b Cl d d + Qb ^b Cl d d Cl d d d 4-0 rH d rH d 40 40 d 4-0 + d d d 4-5 d 4-0 + + ' rH 40 M rH > > > > ^b d d db d d ho <3b Qb > > <3b w co O o I 1—I > ^b <3b ^b ^b Cl Cl Cl Cl <3b d • rH ^b d • rH d • rH Qb a • rH d CO d co 1 co 1 d rH co 1 • rH GO 1 • rH co • rH CO rH rH rH rH > > > > <3b Qb <3b ^b d in 3 ■+3 0 • r-4 0 d O 0 CO 0 40 1 1 c > CO <3b • rH co d *00 O 0 CO o O TRIGONOMETRICAL RATIOS. 31 EXAMPLES. VI. 1. Express all the other trigonometrical ratios in terms of the cosine. 2. Express all the ratios in terms of the tangent. 3. Express all the ratios in terms of the cosecant. 4. Express all the ratios in terms of the secant. 5. The sine of a certain angle is ^ ; find the numerical values of the other trigonometrical ratios of this angle. 12 6. If sin 0 = —, find tan 6 and versin 6. It) 7. If sin A = ^i find tan A, cos A, and sec A. ol 4 8. If cos 6 = -=, find sin 6 and cot 6. 5 9 9. If cos A = —, find tan A and cosec A. 3 10. If tan 0 = -, find the sine, cosine, versine and cosecant of 6. ,, T n , 1 ,io i p cosec 2 6 - sec 2 6 11. If tan 6 = -j- , find the value of -- rh • V7 cosec 2 6 + sec 2 6 15 12. If cot 6=—, find cos 6 and cosec 6. o 13. If secA = -, find tan A and cosec A. 14. If 2 sin 6 = 2 - cos 0, find sin 6. 15. If 8 sin 6 = 4: + cos 6, find sin 6. 16. If tan d-r sec 6= 1*5, find sin 6. 17. If cot 6 + cosec 6 = 5, find cos 6. 18. If 3 sec 4 6 + 8 = 10 sec 2 6, find the values of tan 6. 19. If tan 2 6 + sec 6 = 5, find cos 6. 20. If tan 6 + cot 0 = 2, find sin 6. 21. If sec 2 6 = 2 + 2 tan 6, find tan 6. 22. If tan Q — ^- x i XJr ^ s j n q an( j cos 2x +1 32 TRIGONOMETRY. Values of the trigonometrical ratios in some useful cases. 33. Angle of 45°. Let the angle AOP traced out be 45°. Then, since the three angles of a triangle are together equal to two right angles, Z OPM = 180°— z POM- z PMO = 180° - 45° - 90° = 45° = Z POM. . *. OM = MP = a (say), and OP = V OM 2 -f MP 2 = f 2. a. . 0 MP a 1 .*. sm 4o = cos 45 = OP f2.a *J2’ OM a 1 and OP V2. a V2’ tan 45° = 1. 34. Angle of 30°. Let the angle AOP traced out be 30°. Produce PM to P' making MP' equal to PM. The two triangles OMP and OMP' have their sides OM and MP' equal to OM and MP and also the contained angles equal. Therefore OP' = OP, and Z OP'P = z OPP' = 60°, so that the triangle P'OP is equilateral. TRIGONOMETRICAL RATIOS. 33 A Hence OP 2 = PP' 2 = 4 PM* = 4 OP 2 - 4a 2 , where OM equals a. 3OP 2 = 4a 2 , so that OP = ^, and ifP = iOP= 4 V-3 V3 sin 30° = JfP_l OP ~ 2 ’ of\o _OM_ ' 2 a_f3 COS 30 - 0 p-^ . ^3 2 > and tan 30° = sin 30 c cos 30° \/3 * 35. Angle of 60°. Let the angle AOP traced out be 60°. Take a point N on OA, so that MN = OM = a (say). The two triangles OMP and NMP have now the sides OM and MP equal to NM and MP respectively, and the included angles equal, so that the triangles are equal. . PN = OP, and Z PNM = Z POM = 60°. The triangle OPN is therefore equilateral, and hence 0P = 0N = 20M=2a. MP = *J0P 2 — OM 2 = — a 2 = . a. P L. T. 3 34 TRIGONOMETRY. Hence and sin 60° = MP V3a V3 OP “ 2a - 2 j cos 60 c = Oif OP (X 2a 1 2 5 tan sin 60° cos 60°‘ 36. Angle of 0°. Let the revolving line OP have turned through a very small angle, so that the angle MOP is very small. P The magnitude of MP is q" - A then very small and initially, before OP had turned through an angle big enough to be perceived, the quantity MP was smaller than any quantity we could assign, i.e. was what we denote by 0. Also, in this case, the two points M and P very nearly coincide, and the smaller the angle AOP the more nearly do they coincide. Hence, when the angle AOP is actually zero, the two lengths OM and OP are equal and MP is zero. Hence . c,e_MP_ 0 sm 0 yp Qp 0, cos 0° = OM OP and OP OP tan 0° = y = 0. OM Also cot 0° = the value of ~p when M and P coincide = the ratio of a finite quantity to something infinitely small = a quantity which is infinitely great. Such a quantity is usually denoted by the symbol x . TRIGONOMETRICAL RATIOS. Hence Similarly And cot 0° = co . A o OP 1 cosec 0 = XF 7 ) — 20 also. MP n o_0P _ sec 0 — 1. 37. Angle of 90°. Let the angle A OP be very nearly, but not quite, a right angle. When OP has actually described a right angle the point M coincides with 0, so that then 0M is zero and OP and MP are equal. Hence MP sin 90° = "p = cos 90° = o OM OP tan MP a finite quantity OM an infinitely small quantity’ A = a number infinitely large = co . cot 90° = OM 0 MP MP = 0 , OP sec 90° = 77 T 7 = oo , as in the case of the tangent, and OM cosec 90° = OP _0P MP ~ OP = 1 . 38. Complementary Angles. Def. Two angles are said to be complementary when their sum is equal to a right angle. Thus any angle 0 and the angle 90° — 6 are complementary. 3—2 36 TRIGONOMETRY. 39. To find the relations between the trigonometrical ratios of ttvo complementary angles . Let the revolving line, starting from OA, trace out any acute angle A OP, equal to 6. From any point P on it draw PM perpendicular to OA. Since the three angles of a triangle are together equal to two right angles, and since OMP is a right angle, the sum of the two angles MOP and OPM is a right angle. They are therefore complementary and Z OPM = 90° - e. [When the angle OPM is considered, the line PM is the “ base ” and MO is the “ perpendicular.”] We then have sin (90 — 6) = sin MPO = = cos A OP = cos 6, cos (90° — 6 ) = cos MPO = — sin A OP = sin 6 , tan (90° — 6) = tan MPO = cot A OP = cot 0, 7 PM PM cot (90° — 6) = cot MPO — jyj-Q — tan A OP = tan 0 , PO cosec (90° — 0) = cosec MPO = sec A OP = sec 0, PO and sec (90° — 0) = sec MPO = = cosec A OP = cosec 0. TRIGONOMETRICAL RATIOS. 37 Hence we observe that the Sine of any angle = the Cosine of its complement, the Tangent of any angle = the Cotangent of its comple¬ ment, and the Secant of an angle = the Cosecant of its comple¬ ment. From this is apparent what is the derivation of the names Cosine, Cotangent, and Cosecant. 40. The student is advised before proceeding any further to make himself quite familiar with the following table. [For an extension of this table, see Art. 76.] Angle 0° 30° 45° 60° 90° Sine 0 1 2 1 s/2 s/3 2 1 Cosine 1 v/3 2 1 s/2 1 2 0 Tangent 0 1 s/3 1 v/3 GO Cotangent GO s/3 1 1 s/3 0 Cosecant GO 2 s/2 2 s/3 1 Secant 1 2 s/3 s/2 2 GO If the student commits accurately to memory the portion of the above table included between the thick lines, he should be able to easily reproduce the rest. 38 TRIGONOMETRY. For (1) the sines of 60 and 90 are respectively the cosines of 30° and 0°. (Art. 39.) (2) the cosines of 60° and 90 are respectively the sines of 30° and 0°. (Art. 39.) Hence the second and third lines are known. (3) The tangent of any angle is the result of dividing the sine by the cosine. Hence any quantity in the fourth line is obtained by dividing the corresponding quantity in the second line by the corresponding quantity in the third line. (4) The cotangent of any angle is the reciprocal of the tangent, so that the quantities in the fifth row are the reciprocals of the quantities in the fourth row. (5) Since cosec 0 = . - , the sixth row is obtained v sin 0 by inverting the corresponding quantities in the second row. (6) Since sec 0 = — the seventh row is similarly v cos 0 obtained from the third row. EXAMPLES. VII. 1. If ^ = 30°, verify that (1) cos 2A = cos 2 ^4 - sin 2 yl = 2 cos 2 A - 1, (2) sin 2^1 =2 sin A cos A, (3) cos 3^ = 4 cos 3 ;! - 3 cos A, (4) sin 3A = S sin A - 4 sin 3 ^4, _ / . , 2 tan A and ( 5 ) tan2d =i -— 2i . [Exs. VII.] TRIGONOMETRICAL RATIOS. 39 2. If A =45°, verify that (1) sin 2A = 2 sin A cos A, (2) cos 2A = 1 - 2 sin 2 A, , ^ . 2 tan A and (3) tan 2d Verify that 3. sin 2 30° + sin 2 45° -f- sin 2 60° = ?. 4. tan 2 30° +tan 2 45° +tan 2 60° = 4J. 5. sin 30° cos 60° + cos 30° sin 60°= 1. cos 45° cos 60° - sin 45° sin 60° = v /3 -1 2 v /2 ’ 6 . CHAPTER III. SIMPLE PROBLEMS IN HEIGHTS AND DISTANCES. 41. One of the objects of Trigonometry is to find the distances between points, or the heights of objects, without actually measuring these distances or these heights. 42. Suppose 0 and P to be two points, P being at a higher level than 0. Let OM be a horizontal line drawn through 0 to meet in M the vertical line drawm through P. The angle MOP is called the Angle of Elevation of the point P as seen from 0. Draw PN parallel to MO , so that PN is the hori¬ zontal line passing through P. The angle NPO is the Angle of Depression of the point 0 as seen from P. 43 . Two of the instruments used in practical work are the Theodo¬ lite and the Sextant. The Theodolite is used to measure angles in a vertical plane. The Theodolite, in its simple form, consists of a telescope attached to a flat piece of wood. This piece of wood is supported by three legs and can be arranged so as to be accurately horizontal. HEIGHTS AND DISTANCES. 41 This table being at 0 and horizontal and the telescope being initially pointing in the direction Oil/, the latter can be made to rotate in a vertical plane until it points accurately towards P. A graduated scale shews the angle through which it has been turned from the horizontal, i.e. gives us the angle of elevation MOP. Similarly, if the instrument were at P, the angle NPO through which the telescope would have to be turned, downward from the horizontal, would give us the angle NPO. The instrument can also be used to measure angles in a horizontal plane. 44 . The Sextant is used to find the angle subtended by any two points D and E at a third point F. It is an instrument much used on board ships. Its construction and application are too complicated to be here considered. 45. We shall now solve a few simple examples i:i heights and distances. Ex. 1. A vertical flagstaff stands on a horizontal plane; from a point ■distant 150 feet from its foot the angle of elevation of its top is found to he 30°; find the height of the flagstaff. Let MP (Fig. Art. 42) represent the flagstaff and 0 the point from which the angle of elevation is taken. Then OM =150 feet, and aMOP = 30°. Since PMO is a right angle, we have MP 1 —- = tan i¥OP= tan 30 6 =-~ (Art. 33). OM JS v ' MP ™ V 5 v 5 o Now, by extraction of the square root, we have V3 = 1*73205.... Hence MP= 50 x 1*73205... feet = 86*6025... feet. Ex. 2. A man ivislies to find the height of a church spire which stands on a horizontal plane; at a point on this plane he finds the angle of elevation of the top of the spire to he 45°; on walking 100 feet toward the tower he finds the corresponding angle of elevation to he 60°; deduce the height of the tower and also his original distance from the foot of the spire. 42 TRIGONOMETRY. Let P be the top of the spire and A and B the two points at which the angles of elevation are taken. Draw PM perpendicular to AB produced and let MP be x . We are given dP = 100 feet, ZMAP= 45°, and z MBP = 60°. We then have AM x — cot 45°, and — = cot 60 ° = T X sjd Hence x AM—x , and BM — —j . 100 = AM-BM = x- ^- = x 1 \/0 = 50 [3 +1-73205...] = 236*6... feet. Also AM=x, so that both of the required distances are equal to 236-6... feet. Ex. 3. From the top of a cliff , 200 feet high, the angles of depression of the top and bottom of a tower are observed to be 30° and 60° ; find the height of the tower . Let A be the point of observation and BA the height of the cliff and let CD be the tower. Draw AE horizontally, so that z EAC= 30° and Z PAD — 60°. Let x feet be the height of the tower and produce DC to meet AE in E, so that CE = AB - x = 200 - x. Since lADB= lDAE = C) 0° (Euc. i. 29), *. DB = AB cot ADD = 200 cot 60° = ^^. \'o Also 200 - x DB GE = tan 30°=--. EA J 3 HEIGHTS ANI) DISTANCES. 43 so that DB 200 200 - x = —— = n/3 x — 200 - 200 IT = 1331 feet. Ex. 4. A man observes that at a point due south of a certain tower its angle of elevation is 60°; he then walks 300 feet due west on a horizontal plane and finds that the angle of elevation is then 30°; find the height of the tower and his original distance from it. Let P be the top, and PM the height, of the tower, A the point due south of the tower and B the point due west of A. The angles PM A, PMB , and MAB are therefore all right angles. For simplicity, since the triangles PAM , PBM , and ABM are in different planes, they are reproduced in the second, third, and fourth figures and drawn to scale. We are given AB = 300 feet, z PAM= 60°, and z PBM= 30°. Let the height of the tower be x feet. From the second figure so that From the third figure — 3 »- BM = cot S0° = s /S, BM = s /3 . x, so that x 44 TRIGONOMETRY. From the last figure we have BM 2 = AM 2 + AB 2 , i.e. 3x 2 = 1 a: 2 + 300 2 . o .-. 8x 2 = 3 x 300 2 . X = 3 2 ^ 2 3 = 15 °- ? 2 ' = 75x ^ 6 = 75x2-44949... = 183-71... feet. Also his original distance from the tower = a;cot 60° = 4; = 75 x J2 v 3 = 75 x (1-4142.,.) = 106-065... feet. EXAMPLES. VIII. 1. A person standing on the bank of a river observes that the angle subtended by a tree on the opposite bank is 60°; when he retires 40 feet from the bank he finds the angle to be 30°; find the height of the tree and the breadth of the river. 2. At a certain point the angle of elevation of a tower is found to be 3 such that its cotangent is —; on walking 32 feet directly toward the tower D 2 its angle of elevation is an angle whose cotangent is -. Find the height 5 of the tower. 3. At a point A the angle of elevation of a tower is found to be such that its tangent is — ; on walking 240 feet nearer the tower the tangent _L Z 3 of the angle of elevation is found to be - ; what is the height of the tower ? 4. Find the height of a chimney when it is found that on walking towards it 100 feet in a horizontal line through its base, the angular elevation of its top changes from 30° to 45°. 5. An observer on the top of a cliff, 200 feet above the sea-level, observes the angles of depression of two ships at anchor to be 45° and 30° respectively ; find the distances between the ships if the line joining them points to the base of the cliff. [EXS. VIII.] HEIGHTS AND DISTANCES. 45 6. From the top of a cliff an observer finds that the angles of depression of two buoys in the sea are 39° and 26° respectively; the buoys are 300 yards apart and the line joining them points straight at the foot of the cliff; find the height of the cliff and the distance of the nearest buoy from the foot of the cliff, given that cot 26° = 2-0503, and cot 39°= 1-2349. 7. The upper part of a tree broken over by the wind makes an angle of 30° with the ground, and the distance from the root to the point where the top of the tree touches the ground is 50 feet; what was the height of the tree ? 8. The horizontal distance between two towers is 60 feet and the angular depression of the top of the first as seen from the top of the second which is 150 feet high is 30°; find the height of the first. 9. The angle of elevation of the top of an unfinished tower from a point distant 120 feet from its base is 45°; how much higher must the tower be raised so that its angle of elevation at the same point may be 60° ? 10 . Two pillars of equal height stand on either side of a roadway which is 100 feet wide; at a point in the roadway between the pillars the elevations of the tops of the pillars are 60° and 30° ; find their height and the position of the point. 11 . The angle of elevation of the top of a tower is observed to be 60°; at a point 40 feet above the first point of observation the elevation is found to be 45°; find the height of the tower and its horizontal distance from the points of observation. 12. At the foot of a mountain the elevation of its summit is found to be 45°; after ascending one mile up a slope of 30° inclination the elevation is found to be 60°. Find the height of the mountain. 13. What is the angle of elevation of the sun when the length of its shadow is ^3 times its height ? 14. The shadow of a tower standing on a level plane is found to be 60 feet longer when the sun’s altitude is 30° than when it is 45°. Prove that the height of the tower is 30 (1 + ^/3) feet. 15. On a straight coast there are three objects A, B, and C such that AB = BC = 2 miles. A vessel approaches B in a line perpendicular to the coast and at a certain point AC is found to subtend an angle of 60°; after sailing in the same direction for ten minutes AC is found to subtend 120°; find the rate at which the ship is going. 46 TRIGONOMETRY. [Exs. VIII.] 16. Two flagstaffs stand on a horizontal plane. A and B are two points on the line joining the bases of the flagstaffs and between them. The angles of elevation of the tops of the flagstaffs as seen from A are 30° and 60° and, as seen from B, they are 60° and 45°. If the length AB be 30 feet, find the heights of the flagstaffs and the distance between them. 17. P is the top and Q the foot of a tower standing on a horizontal plane. A and B are two points on this plane such that AB is 2 32 feet and QAB is a right angle. It is found that coi PAQ — ~ and 0 cot PBQ = ?; find the height of the tower. 18. A square tower stands upon a horizontal plane. From a point in this plane from which three of its upper corners are visible their angular elevations are respectively 45°, 60°, and 45°. Shew that the height of the tower is to the breadth of one of its sides as >^6(^/5+ 1) to 4. 19. A lighthouse, facing north, sends out a fan-shaped beam of light extending from north-east to north-west. A steamer sailing due west first sees the lighthouse when it is 5 miles away from the lighthouse and continues to see it for 30^/2 minutes. What is the speed of the steamer ? 20. A man stands at a point X on the bank XY of a river with straight and parallel banks and observes that the line joining X to a point Z on the opposite bank makes an angle of 30° with XY. He then goes along the bank a distance of 200 yards to Y and finds that the angle ZYX is 60°. Find the breadth of the river. 21. A man, walking due north, observes that the elevation of a balloon, which is due east of him and is sailing toward the north-west, is then 60° ; after he has walked 400 yards the balloon is vertically over his head; find its height supposing it to have always remained the same. CHAPTER IV. APPLICATION OF ALGEBRAIC SIGNS TO TRIGONOMETRY. 46. Positive and Negative Angles. In Art. 6 in treating of angles of any size we spoke of the revolving line as if it always revolved in a direction opposite to that in which the hands of a watch revolve, when the watch is held with its face uppermost. This direction is called counter-clockwise. When the revolving line turns in this manner it is said to revolve in the positive direction and to trace out a positive angle. When the line OP revolves in the opposite direction, i.e. in the same direction as the hands of the watch, it is said to revolve in the negative direction and to trace out a negative angle. This negative direction is clockwise. 47. Let the revolving line start from OA and revolve until it reaches a position OP which lies between OA' and OB' and which bisects the angle A'OB'. If it has revolved in the positive a direction it has traced out the positive angle whose measure is + 225°. 48 TRIGONOMETRY. If it has revolved in the negative direction it has traced out the negative angle — 135°. Again, suppose we only know that the revolving line is in the above position. It may have made one, two, three ... complete revolutions and then have described the positive angle 4- 225°. Or again it may have made one, two, three... complete revolutions in the negative direction and then have described the negative angle - 135°. In the first case the angle it has described is either 225°, or 360° + 225°, or 2 x 360° + 225°, or 3 x 360° + 225° . i.e. 225°, or 585°, or 945°, or 1305°.... In the second case the angle it has described is — 135°, or - 360° - 135°, or - 2 x 360° - 135°, or - 3 x 360° - 135° . i.e. — 135°, or —495°, or —855°, or — 1215 .... 48. Positive and Negative Lines. Suppose that a man is told to start from a given milestone on a straight road and to walk 1000 yards along the road and then to stop. Unless we are told the direction in which he started we do not know his position when he stops. All we know is that he is either at a distance 1000 yards on one side of the milestone or at the same distance on the other side. In measuring distances along a straight line it is therefore convenient to have a standard direction; this direction is called the positive direction and all distances measured along it are said to be positive. The opposite direction is the negative direction and all distances measured along it are said to be negative. The standard or positive directions for horizontal lines is towards the right. POSITIVE AND NEGATIVE LINES. 49 The length OA is in the positive direction. The length OA' is in the © A' negative direction. If -g--g- A the magnitude of the distance OA or OA' be a, the point A is at a distance -I- a from 0 and the point A' is at a distance — a from 0. All lines measured to the right have then the positive sign prefixed; all lines to the left have the negative sign prefixed. If a point start from 0 and describe a positive distance OA and then a distance AB back again toward 0, equal numerically to b, the total distance it has described measured in the positive direction is OA + AB i.e. + a + (— b ), i.e. a — b. 49. For lines at right angles to A A' the positive direction is from 0 towards the top of the page, i.e. the direction of OB (Fig. Art. 47). All lines measured from 0 towards the foot of the page, i.e. in the direction OB ', are negative. 50. Trigonometrical ratios for an angle of any magni¬ tude. Let OA be the initial line (drawn in the positive direction) and let OA' be drawn in the opposite direction to OA. Let BOB' be a line at right angles to OA, its positive direction being OB. Let a revolving line OP start from OA and revolving in either direction, positive or negative, trace L. T. 50 TRIGONOMETRY. out an angle of any magnitude whatever. From a point P in the revolving line draw PM perpendicular to AO A'. [Four positions of the revolving line are given in the figure, one in each of the four quadrants, and the suffixes 1, 2, 3 and 4 are attached to P for the purpose of distinction.] We then have the following definitions, which are the same as those given in Art. 23 for the simple case of an acute angle: MP OP OM OP MP OM OM MP OP OM OP MP is called the Sine of the angle A OP, JJ J) Cosine j? >> )} ?> Tangent 5 > >5 >) )) Cotangent Secant 5 ) ?> i) » Cosecant )! 5 J The quantities 1 — cos A OP , and 1 — sin A OP are respectively called the Versed Sine and the Coversed Sine of A OP. 51. In exactly the same manner as in Art. 27 it may be shewn that, for all values of the angle AOP (= 0), we have ANGLES OF ANY MAGNITUDE. 51 sin 2 0 + cos 2 0 = 1, sin 6 . & - - = tan 0, COS (7 sec 2 0 = 1 + tan 2 0, and cosec 2 # = 1 + cot 2 #. 52. Signs of the trigonometrical ratios. First quadrant . Let the revolving line be in the first quadrant, as 0P 3 . This revolving line is always positive. Here 0M 1 and M 1 P 1 are both positive, so that all the trigonometrical ratios are then positive. Second quadrant. Let the revolving line be in the second quadrant, as 0P 2 > Here M 2 P 2 is positive and 0M 2 is negative. The sine, being equal to the ratio of a positive quantity to a positive quantity, is therefore positive. The cosine, being equal to the ratio of a negative quantity to a positive quantity, is therefore negative. The tangent, being equal to the ratio of a positive quantity to a negative quantity, is therefore negative. The cotangent is negative. The cosecant is positive. The secant is negative. Third quadrant. If the revolving line be, as 0P 3 , in the third quadrant, we have both M 3 P 3 and 0M 3 negative. The sine is therefore negative. The cosine is negative. The tangent is positive. The cotangent is positive. The cosecant is negative. The secant is negative. UNIVERSITY OF ItLINO/S library 4—2 52 TRIGONOMETRY. Fourth quadrant . Let the revolving line be in the fourth quadrant, as 0P 4 . Here ilf 4 P 4 is negative and OM a is positive. The sine is therefore negative. The cosine is positive. The tangent is negative. The cotangent is negative. The cosecant is negative. The secant is positive. The annexed table shews the signs of the trigono¬ metrical ratios according to the quadrant in which lies the revolving line, which bounds the angle considered. B sin + cos - tan cot - cosec + sec - sin + cos + tan + cot + cosec + sec + > O A sin - sin - cos ~ cos + tan + tan - cot + cot cosec - cosec - sec sec + B' 53. Tracing of the changes in the sign and magnitude of the trigonometrical ratios of an angle , as the angle increases from 0° to 360 . Let the revolving line OP be of constant length a. CHANGES IN THE TRIGONOMETRICAL RATIOS. 53 When it coincides with OA the length 0M X is equal to a and, when it coincides with OB , the point M 1 coincides with 0 and OMj vanishes. Also, as the revolving line turns from OA to OB , the dis¬ tance 0M X decreases from a to zero. Whilst the revolving line is in the second quadrant and is revolving from OB to OA ', the distance 0M 2 is negative and increases numerically from 0 to a [• i.e . it decreases algebraically from 0 to — a]. In the third quadrant the distance 0M 3 increases algebraically from -a to 0, and in the fourth quadrant the distance 0M 4 increases from 0 to a. In the first quadrant the length M 1 P 1 increases from 0 to a ; in the second quadrant M 2 P 2 decreases from a to 0; in the third quadrant M 3 P 3 decreases algebraically from 0 to — a ; whilst in the fourth quadrant Af 4 P 4 increases algebraically from — a to 0. 54. Sine. In the first quadrant as the angle in- MP . 0 creases from 0 to 90 u , the sine, i.e. — - —- 1 , increases from - a a to -, i.e. from 0 to 1. a In the second quadrant as the angle increases from a 0 90° to 180°, the sine decreases from - to i.e. from 1 to 0. a a In the third quadrant as the angle increases from 180° to 270°, the sine decreases from - to —- , i.e. from 0 to — 1. a a 54 TRIGONOMETRY. In the fourth quadrant as the angle increases from 270 to 360°, the sine increases from —- to-, i.e. from a a — 1 to 0. 55. Cosine. In the first quadrant the cosine, which is equal to — — , decreases from - to -, i.e. from 1 to 0. a a a In the second quadrant it decreases from - to —-, i.e, a a from 0 to — 1. In the third quadrant it increases from —- to i.e. a a from — 1 to 0. In the fourth quadrant it increases from — to i.e. a a from 0 to 1. 56. Tangent. In the first quadrant M 1 P 1 increases M,P i from 0 to a and 0M 1 decreases from a to 0, so that OM\ continually increases (for its numerator continually in¬ creases and its numerator continually decreases). When 0P 1 coincides with OA, the tangent is 0; when the revolving line has turned through an angle which is slightly less than a right angle, so that OP l nearly coincides with OB , then M 1 P l is very nearly equal to M l P 1 . a and 0M Y is very small. The ratio is therefore very large, and the nearer 0P l gets to OB the larger does the ratio become, so that, by taking the revolving line near enough to OB, we can make the tangent as large as we please. This is expressed by saying that when the angle is equal to 90° its tangent is infinite. CHANGES IN THE TRIGONOMETRICAL RATIOS. 55 The symbol oo is used to denote an infinitely great quantity. Hence in the first quadrant the tangent increases from 0 to oo. In the second quadrant when the revolving line has described an angle AOP 2 slightly greater than a right angle, M 2 P 2 is very nearly equal to a and 0M 2 is very small and negative, so that the corresponding tangent is very large and negative. Also, as the revolving line turns from OB to OA ', M 1 P 1 decreases from a to 0 and 0M 2 is negative and decreases from 0 to — a, so that when the revolving line coincides with OA' the tangent is zero. Hence in the second quadrant the tangent increases from — x to 0. In the third quadrant both M 3 P 3 and OM s are negative, and hence their ratio is positive. Also, when the revolving line coincides with OB\ the tangent is infinite. Hence in the third quadrant the tangent increases from 0 to x . In the fourth quadrant df 4 P 4 is negative and OM± is positive, so that their ratio is negative. Also, as the revolving line passes through OB' the tangent changes from + x to - x [just as in passing through OB], Hence in the fourth quadrant the tangent increases from — x to 0. 57. Cotangent. When the revolving line coincides with OA , M 1 P l is very small and OM Y is very nearly equal to a, so that the cotangent, i.e. the ratio , is infinite to start with. Also, as the revolving line rotates 56 TRIGONOMETRY. from OA to OB, the quantity M 1 P 1 increases from 0 to a and 0M 1 decreases from a to 0. Hence in the first quadrant the cotangent decreases from x to 0. » In the second quadrant M 2 P 2 is positive and 0M 2 negative, so that the cotangent decreases from 0 to , i.e. from 0 to — oo . In the third quadrant it is positive and decreases from x to 0 [for as the revolving line crosses OB' the cotangent changes from — x to x ]. In the fourth quadrant it is negative and decreases from 0 to — x . 58. Secant. When the revolving line coincides with OA the value of 0M 1 is a, so that the value of the secant is then unity. As the revolving line turns from OA to OB, OM l decreases from a to 0, and when the revolving line CL coincides with OB the value of the secant is ^ , i.e. x. Hence in the first quadrant the secant increases from 1 to x . In the second quadrant 0M 2 is negative and decreases from 0 to — a. Hence in this quadrant the secant in¬ creases from — x to — 1 [for as the revolving line crosses OB the quantity OM x changes sign and therefore the secant changes from + x to — x ]. In the third quadrant 0M 3 is always negative and increases from — a to 0; therefore the secant decreases from — 1 to — x . In the fourth quadrant 0M l is always positive and increases from 0 to a. Hence in this quad¬ rant the secant decreases from x to + 1. CHANGES IN THE TRIGONOMETRICAL RATIOS. 57 59. Cosecant. The change in the cosecant may be traced in a similar manner to that in the secant. In the first quadrant it decreases from go to +1. In the second quadrant it increases from +1 to + oo. In the third quadrant it increases from — x to — 1. In the fourth quadrant it decreases from — 1 to — x . 60. The foregoing results are collected in the annexed table. In the second quadrant the sine decreases from 1 to 0 cosine decreases from Oto-1 tangent increases from - oo to 0 cotangent decreases from Oto-oo secant increases from-oo to-1 cosecant increases from 1 to oo A' O In the third quadrant the sine decreases from 0 to-1 cosine increases from- 1 to 0 tangent increases from 0 to oo cotangent decreases from oo to 0 secant decreases from - 1 to-oo cosecant increases from - oo to-1 In the first quadrant the sine increases from 0 to 1 cosine decreases from 1 to 0 tangent increases from 0 to oo cotangent decreases from oo to 0 secant increases from 1 to oo cosecant decreases from oo to 1 A In the fourth quadrant the sine increases from - 1 to 0 cosine increases from 0 to 1 tangent increases from-oo to 0 cotangent decreases from 0 to - oo secant decreases from oo to 1 cosecant decreases from - 1 to - oo 61. Periods of the trigonometrical functions. As an angle increases from 0 to 2tt radians i.e whilst the revolving line makes a complete revolution its sine first increases from 0 to 1, then decreases from 1 to — 1, and finally increases from — 1 to 0, and thus the sine goes through all its changes returning to its original value. 58 TRIGONOMETRY. Similarly as the angle increases from 2ir radians to 47r radians, the sine goes through the same series of changes. Also the sines of any two angles which differ by four- right angles, i.e. 27r radians, are the same. This is expressed by saying that the period of the sine is 2 tt . Similarly the cosine, secant, and cosecant go through all their changes as the angle increases by 2ir. The tangent, however, goes through all its changes as the angle increases from 0 to it radians, i.e. whilst the revolving line turns through two right angles. Similarly for the cotangent. The period of the sine, cosine, secant and cosecant is therefore 2ir radians; the period of the tangent and cotangent is ir radians. Since the values of the trigonometrical functions repeat over and over again as the angle increases, they are called periodic functions. #62. The variations in the values of the trigono¬ metrical ratios may be graphically represented to the eye by means of curves constructed in the following manner. Sine-Curve. Let OX and OF be two straight lines at right angles . SINE-CURVE. 59 and let the magnitudes of angles be represented by lengths measured along OX. Let R ly R 2 , be points such that the distances 0R ly RiR 2 , R 2 R 3 ,". are equal. If then the distance OR x represent a right angle, the distances 0R 2 , 0R 3) 0R 4 ,... must represent two, three, four,... right angles. Also if P be any point on the line OX, then OP represents an angle which bears the same ratio to a right angle that OP bears to OR x . [For example, if OP be equal to i OR x then OP would represent one- O third of a right angle; if P bisected P 3 P 4 then OP would represent 3J right angles.] Let also 0R X be so chosen that one unit of length represents one radian; since 0R 2 represents two right angles, i.e. 7 r radians, the length 0R 2 must be 7 r units of length, i.e. about units of length. In a similar manner negative angles are represented by distances 0R X , 0R 2 ',... measured from 0 in a negative direction. At each point P erect a perpendicular PQ to represent the sine of the angle which is represented by OP ; if the sine be positive the perpendicular is to be drawn parallel to 0 Y in the positive direction; if the sine be negative the line is to be drawn in the negative direction. [For example, since OB 1 represents a right angle, the sine of which is 1, we erect a perpendicular B X B X equal to one unit of length; since OP 2 represents an angle equal to two right angles, the sine of which is zero, we erect a perpendicular of length zero; since OP 3 represents three right angles, the sine of which is -1, we erect a perpendicular equal to -1, i.e. we draw P 3 P 3 downward and equal to a unit of length; if OP were equal to one-third of OB x it would represent ^ of a right angle, i.e. 30°, O 60 TRIGONOMETRY. the sine of which is - , and so we should erect a perpendicular PQ equal to one-half the unit of length.] The ends of all these lines, thus drawn, would be found to lie on a curve similar to the one drawn above. It would be found that the curve consisted of portions, similar to 0B 1 R 2 B 3 R 4 , placed side by side. This corre¬ sponds to the fact that each time the angle increases by 27 t, the sine repeats the same value. # 63. Cosine-Curve. B r 3 ' r 2 ' Y ... Q i \ R i r 2 r 3 / ^ % % % V \ V V • % « % • •» ~ ~ -— / R/ O P \ r 4 X _ + The Cosine-Curve is obtained in the same manner as the Sine-Curve, except that in this case the perpendicular PQ represents the cosine of the angle represented by OP. The curve obtained is the same as that of Art. 62 if in that curve we move 0 to R x and let OF be drawn along R l B 1 . * 64. Tangent-Curve. In this case, since the tangent of a right angle is infinite and since 0R 1 represents a right angle, the per¬ pendicular drawn at R 1 must be of infinite length and the dotted curve will only meet the line R Y L at an infinite distance. TANGENT-CURVE. 61 Since the tangent of an angle slightly greater than a right angle is negative and almost infinitely great, the t 1 1 1 1 « $ 1 I 1 0 0 0 0 0 0 0 9 0 0 9 0 0 ,* 0 0 0 / Y O 1 / 1 / • • L / « f # t 0 0 0 0 9 0 0 0 4 4 0 4 0 4 4 j «3 / R 2 / / 0 0 0 0 0 0 0 0 9 # 0 i i i % P Ri /R 2 / 0 0 0 9 4 9 9 4 9 4 1 / 9 1 1 f 4 f t 1 L' X CO CC dotted curve immediately beyond LRJJ commences at an infinite distance on the negative side, i.e. below, OX. The Tangent-Curve will clearly consist of an infinite number of similar but disconnected portions, all ranged parallel to one another. Such a curve is called a Discon¬ tinuous Curve. Both the Sine-Curve and the Cosine- Curve are, on the other hand, Continuous Curves. #65. Cotangent-Curve. If the curve to represent the cotangent be drawn in a similar manner, it will be found to meet 0 Y at an infinite distance above 0 ; it will pass through the point R 1 and touch the vertical line through R 2 at an infinite distance on the negative side of OX. Just beyond R 2 it will start at an infinite distance above R 2) and proceed as before. The curve is therefore discontinuous and will consist of an infinite number of portions all ranged side by side. 62 TRIGONOMETRY. *66. Cosecant-Curve. Y R 2 ' Ri' \ i 1 i \ 1 1 • * • t 1 \ v Bi R3 O / \ / \ / \ 1 . < ( 1 | 1 , 1 * 1 * 1 1 1 Ri r 2 /* p''v / / » ! 1 \ ! \ 1 1 / 1 ; • ; 1 1 1 1 f 1 1 1 « 1 1 • 1 1 When the angle is zero the sine is zero, and the cosecant is therefore infinite. Hence the curve meets OY at infinity. When the angle is a right angle the cosecant is unity, and hence R 1 B 1 is equal to the unit of length. When the angle is equal to two right angles its cosecant is infinity, so that the curve meets the perpen¬ dicular through R 2 at an infinite distance. Again, as the angle increases from slightly less to slightly greater than two right angles, the cosecant changes from + x to - x . Hence just beyond R 2 the curve commences at an infinite distance on the negative side of, i.e. below, OX. *67. Secant-Curve. If, similarly, the Secant-Curve be traced it will be found to be the same as the Cosecant- Curve would be if we moved OY to R^B^ ANGLES OF ANY SIZE. 63 MISCELLANEOUS EXAMPLES. IX. 1. In a triangle one angle contains as many grades as another con¬ tains degrees, and the third contains as many centesimal seconds as there are sexagesimal seconds in the sum of the other two; find the number of radians in each angle. 2 . Find the number of degrees in the angle at the centre of a circle whose radius is 5 feet which is subtended by an arc of length 6 feet. 3. To turn radians into seconds prove that we must multiply by 206265 nearly, and to turn seconds into radians the multiplier must be •0000048. 4 . 5 . If sin 6 equal find the values of cos 6 and cot 6. If . . m 2 + 2 mn sin 6 = —— -,, m 2 + 2 mn + 2u“ prove that 6. If prove that 7 . Prove that . m 2 4 - 2 mn tan 6 — - -— . 2 mil + 2 Ji- cos - sin 6 = x /2 sin 6, cos 6 + sin 6 = J2 cos 6 . cosec 6 a - cot 6 a = 3 cosec 2 a cot 2 a +1. 8. Express 2 sec’ 2 A - sec 4 A- 2 cosec 2 A + cosec 4 A in terms of tan A . 9 . Solve the equation 3 cosec 2 6 = 2 sec 6. 10. A man on a cliff observes a boat at an angle of depression of 30°, which is making for the shore immediately beneath him. Three minutes later the angle of depression of the boat is 60°. How soon will it reach the shore ? 11. Prove that the equation sin 6 = x + - is impossible if x be real. X 12 . Shew that the equation sec 2 6 = — ,y . is only possible when (x + iyY x = y. CHAPTER V. TRIGONOMETRICAL FUNCTIONS OF ANGLES OF ANY SIZE AND SIGN. [On a first reading of the subject, the student is recommended to confine his attention to the first of the four figures given in Arts. 68, 69 and 72.] 68. To find the trigonometrical ratios of an angle { — 6) in terms of those of 6, for all values of 9. ANGLES OF ANY SIZE AND SIGN. 65 Let the revolving line, starting from OA, revolve through any angle 9 and stop in the position OP. Draw PM perpendicular to OA (or OA produced) and produce it to P\ so that the lengths of PM and MP' are equal. In the geometrical triangles MOP and MOP' we have the two sides OM and MP equal to the two OM and MP\ and the included angles OMP and OMP' are right angles. Hence (Euc. I. 4), the magnitudes of the angles MOP and MOP' are the same and OP is equal to OP'. In each of the four figures, the magnitudes of the angle AOP (measured counter-clockwise) and of the angle AOP' (measured clockwise) are the same. Hence the angle A OP' (measured clockwise) is denoted by — 9. Also MP and MP' are equal in magnitude but are opposite in sign. (Art. 49.) We have therefore sin ( - 6 ) = and tan ( — 9) = cot ( — 9) = cosec ( — 9) = sec ( — 9 ) = MP' -MP OP' ~ OP ~ OM OM 0P'~ 0P~ MP' -MP 0M~ OM OM OM MP' ~ — MP ~ OP’ OM MP'~ — MP~ OP' OP 0M~ 0M~ = — sin 6 , = cos 6 , = — tan 9, = — cot 9 , — cosec (7, = sec 9. L. T. 5 66 TRIGONOMETRY. Exs. and sin (- 30°) = - sin 30°= - 1 2 ’ tan (- 60°) = - tan 60°= - ^/3, cos (- 45°) = cos 45°=— . 69. To find the trigonometrical ratios of the angle (90° — 0) in terms of those of 9, for all values of 9. The relations have already been discussed in Art. 39, for values of 8 less than a right angle. Let the revolving line, starting from OA, trace out any angle A OP denoted by 6. To obtain the angle 90° — 9 } let the revolving line rotate to B and then rotate from B in the opposite ANGLES OF ANY SIZE AND SIGN. 67 direction through the angle 6, and let the position of the revolving line be then OP'. The angle AOP' is then 90° — 6. Take OP' equal to OP and draw P'M' and PM per¬ pendicular to OA, produced if necessary. Also draw P'N' perpendicular to OB, produced if necessary. In each figure the angles AOP and BOP' are numeri¬ cally equal, by construction. Hence in each figure Z MOP = z N'OP' = Z OP'M', since ON' and M'P' are parallel. Hence the triangles MOP and M'P’O are equal in all respects, and therefore OM = M'P' numerically, and OM' = MP numerically. Also in each figure OM and M'P' are of the same sign, and so also are MP and OM', ie. OM = + M'P', and OM = + MP. Hence and sin (90 — 6 ) = sin AOP' = cos (90 — &) = cos AOP' = tan (90° — 6) = tan A OP' — cot (90° — 6) = cot A OP' = sec (90° — 0) = sec A OP' = cosec (90° — 6) = cosec A OP' = M'P' 0M n 0P r - OP ~ cos 0M 1 MP . * OP’ ~ OP ~ sin MP'_0M_ 0M’ ~ MP ~ cot d ’ 0M'_MP_ M'P' ~ 0M ~ tan OP’ OP a 0M’ ~ MP~ C0SeC 6 OP' OP a M'P’~ 0M~ sec6 ’ 5—2 68 TRIGONOMETRY. 70. To find the trigonometrical ratios of the angle (90 + 6) in terms of those of 0, for all values of 6. Let the revolving line, starting from OA, trace out any angle 6 and let OP be the position of the revolving line then, so that the angle AOP is 6. Let the revolving line turn through a right angle from OP in the positive direction to the position OP\ so that the angle AOP' is (90 c + 6 ). Take OP' equal to OP and draw PM and PM' perpendicular to AO A' . In each figure, since POP' is a right angle, the sum of the angles MOP and P'OM' is always a right angle. Hence Z MOP = 90° - Z P'OM' = Z 0PM'. ANGLES OF ANY SIZE AND SIGN. 69 The two triangles MOP and MP'0 are therefore equal in all respects. Hence OM and M'P' are numerically equal, as also MP and OM are numerically equal. In each figure OM and M'P' have the same sign, whilst MP and OM' have the opposite, so that M'P' = + OM , and OM' = - MP. We therefore have sin (90° + 6 )== sin A OP' = cos (90 + 6 ) = cos A OP' = M'P' OM OP' ~ OP = cos 6 , OM __ - MP OP' ~ OP MP' 0 M tan (90° + 0) = tan A OP' = = fgp = — sin 6 , = — cot 9, OM' - MP cot (90° + 6) = cot A OP' = jMP, = = - tan 0, O P' C) P sec (90° + 0) = sec A OP' = p = — cosec 9 , OP' OP and cosec (90 4- 6) = cosec A OP' = = sec <9. Exs. and sin 150°= sin (90° + 60°) = cos 60° = \, cos 135° = cos (90° + 45°) = - sin 45° = - A , V 2 tan 120°= tan (90°+ 30°)= - cot 30°= - v /3. 71. Supplementary Angles. Two angles are said to be supplementary when their sum is equal to two right angles, i.e. the supplement of any angle 6 is 180° — 6. 70 TRIGONOMETRY. Exs. The supplement of 30°= 180° - 30°= 150°. The supplement of 120°= 180°- 120°= 60°. The supplement of 275° = 180° - 275°= -95°. The supplement of - 126°= 180° - (- 126°) = 306°. 72. To find the values of the trigonometrical ratios of the angle (180" — 6) in terms of those of the angle 6, for all values of 6. A A Let the revolving line start from OA and describe any angle AOP (= 6). To obtain the angle 180° — 6, let the revolving line start from OA and, after revolving through two right angles (i.e. into the position OA'), then revolve back through an angle 6 into the position OP’, so that the angle A'OP' is equal in magnitude but opposite in sign to the angle AOP. The angle AOP' is then 180 — 0. ANGLES OF ANY SIZE AND SIGN. 71 Take OP' equal to OP and draw PM' and PM perpendicular to AOA\ The angles MOP and M'OP' are equal and hence the triangles MOP and M'OP' are equal in all respects. Hence OM and OM' are equal in magnitude and so also are MP and M'P '. In each figure OM and OM' are drawn in opposite directions, whilst MP and M'P' are drawn in the same direction, so that OM' = - OM , and M'P' = + MP. Hence we have M'P' Mp sin (180 — 6) = sin A OP' = -jjp = gp = sin 0, cos (180 — 0) = cos A OP = jyp, = -Qp- = - cos 6, M'P' M P tan (180° — 0) = tan A OP' = qJjt = — j = — tan 0 , OM' -OM cot (180° — 0) = cot A OP' = jp-p, = ~ cot 0, f)P' r\p sec (180° - 6) = sec AOP =~ = = - sec 6, OP’ OP and cosec (180° - 0) = cosec A OP' = = pp = cosec 6. Exs. sin 120° = sin (180° - 60°) = sin 60° = n/3 2 t cos 185°=cos (180° -45°)= - cos 45°= - tan 150°=tan (180°-30°)= - tan 30°= - and 72 TRIGONOMETRY. 73. To find the trigonometrical ratios of (180 + 0) in terms of those of 0, for all values of 6. The required relations may be obtained geometrically, as in the previous articles. The figures for this propo¬ sition are easily obtained and are left as an example for the student. They may also be deduced from the results of Art. 70, which have been proved true for all angles. For putting 90 + 6 = B, we have sin (180 + 6) = sin (90 + B) = cos B (Art. 70) = cos (90 + 8) — — sin 0, (Art. 70) and cos (180 + 6) = cos (90° + B) — — sin B (Art. 70) = — sin (90 + 6) = — cos 6. (Art. 70). % So tan (180° + 6) = tan (90 c + B) = — cot B = — cot (90° + 6) — tan 0, and similarly cot (180° + 0) — cot 0 , sec (180° + 8) = — sec 0 , and cosec (180° -f 0) = — cosec 8. 74. To find the trigonometrical ratios of an angle (360° + 8) in terms of those of 8, for all values of 0. In whatever position the revolving line may be when it has described any angle 0 ) it will be in exactly the same position when it has made one more complete revolution in the positive direction, i.e. when it has described an angle 360° + 0. ANGLES OF ANY SIZE AND SIGN. 73 Hence the trigonometrical ratios for an angle 360° 4- 0 are the same as those for 6. It follows that the addition or subtraction of 360°, or any multiple of 360°, to or from any angle does not alter its trigonometrical ratios. 75. From the theorems of this chapter it follows that the trigonometrical ratios of any angle whatever can be reduced to the determination of the trigonometrical ratios of an angle which lies between 0° and 45°. For example, sin 1765° = sin [4 x 360° + 325°] = sin 325° = sin (180° + 145°) = - sin 145° = - sin (180° - 35°) = - sin 35° tan 1190° = tan (3 x 360° + 110°) = tan 110° = tan (90° -f 20°) = — cot 20° and cosec ( — 1465°) = — cosec 1465° = — cosec (4 x 360° + 25°) = — cosec 25° (Art. 74) (Art. 73) (Art. 72); (Art. 74) (Art. 70); (Art. 68) m (Art. 74). Similarly any other such large angles may be treated. First, multiples of 360° should be subtracted until the angle lies between 0° and 360°; if it be then greater than 180° it should be reduced by 180°; if then greater than 90 the formulae of Art. 70 should be used, and finally, if necessary, the formulae of Art. 69 applied. 74 TRIGONOMETRY. 76. The table of Art. 40 may now be extended to some important angles greater than a right angle. Angle 0° 30° 45° 60° o o Gi 120° 135° 150° 180° ' Sine 0 1 2 1 V2 V3 2 1 v/3 2 1 J2 1 2 0 Cosine 1 \/3 2 1 V2 1 2 0 1 ~ 2 1 n /2 v/3 2 -1 Tangent 0 1 /Q V 1 v /3 oo — \/3 -1 1 v/3 0 1 Cotangent 00 v /3 1 1 V3 0 1 v/3 -1 -v/3 1 OC Cosecant 00 2 2 V3 1 2 73 v/2 2 oc Secant 1 2 v/3 V2 2 00 -2 -v/2 2 v/3 -1 EXAMPLES. X. Prove that 1. sin 420° cos 390° + cos (- 300°) sin ( - 330°) = 1. 2. cos 570° sin 510° - sin 330° cos 390°=0. and 3. tan 225° cot 405° + tan 765° cot 675° = 0. What are the values of cos A - sin A and tan A + cot A when A ha& the values 7r 6 . 57r T’ 3 ’ 4. 5. 27r 3 ’ 7. j- and 8. [Exs. X.] EXAMPLES. 75 What values between 0° and 360° may A have when 9. sin A = , 10. cos A — 1 ~ 2 * 11. tan A = - 12. CO > 1 II O O 13. sec A = - ± and 14. cosec A = Express in terms of the ratios of a positive angle, which is less than 45°, the quantities 15. sin (-65°). 16. cos (-84°). 17. tan 137°. 18. sin 168°. 19. cos 287°. 20. tan (- 246°). 21. sin 843°. 22. cos (-928°). 23. tan 1145°. 24. cos 1410°. 25. cot (-1054°). 26. sec 1327° and 27. cosec (-756°). What sign has sin A + cos A for the following values of A ? 28. 140°. 29. 278°. 30. - 356° and 31. -1125°. What sign has sin A - cos A for the following values of A? 32. 215°. 33. 825°. 34. -634° and 35. -457°. 36. Find the sines and cosines of all angles in the first four quadrants whose tangents are equal to cos 135°. Prove that 37. sin (270° + A) = -cos A, and tan (270° + A) = -cot A. 38. cos (270° - A) =- sin A, and cot (270° - A) = tan A. CHAPTER VI. GENERAL EXPRESSIONS FOR ALL ANGLES HAVING A GIVEN TRIGONOMETRICAL RATIO. 77. To construct the least positive angle whose sine is equal to a , where a is a proper fraction. Let OA be the initial line and let OB be drawn in the positive direction perpendicular to OA. Measure off along OB a distance ON which is equal to a units of length. [If a be negative the point N will lie in BO produced.] Through N draw NP parallel to OA. With centre 0 and radius equal to the unit of length describe a circle and let it meet NP in P. Then AOP will be the required angle. Draw PM perpendicular to OA , so that sin A OP = MP _ ON 0P~ OP The sine AOP is therefore equal to the given quantity and AOP is therefore the angle required. ANGLES HAVING A GIVEN COSINE. 77 78. To construct the least positive angle whose cosine is equal to h where h is a proper /inaction. Along the initial line measure off a distance OM equal to h and draw MP perpendicular to OA. -^ [If h be negative M will lie on the other side of 0 in the line AO produced.] With centre 0 and radius equal to unity, describe a circle and let it meet MP in P. Then AOP is the angle required. For a mi OM b , cos AOP = gp = - = b. 79. To construct the least positive angle whose tangent is equal to c. Along the initial line measure off OM equal to unity and erect a per¬ pendicular MP. Measure off* MP equal to c. Then a r\T> MP tan AOP = -Qjp=c, so that AOP is the required angle. 80. It is clear from the definition given in Art. 50, that, when an angle is given, so also is its sine. The converse statement is not correct; there is more than one angle having a given sine; for example, the angles 30°, 150°, 390 J , — 210 ,... all have their sine equal to Hence, when the sine of an angle is given, we do not definitely know the angle; all we know is that the angle is one out of a large number of angles. 78 TRIGONOMETRY. Similar statements are true if the cosine, tangent, or any other trigonometrical function of the angle be given. Hence, simply to give one of the trigonometrical functions of an angle does not determine it without ambiguity. 81. Suppose we know that the revolving line OP coincides with the initial line OA. All we know is that the revolving line has made 0, or 1 , or 2 , or 3,... complete revolutions, either positive or negative. But when the revolving line has made one complete revolution the angle it has described is (Art. 17) equal to 2tt radians. Hence when the revolving line OP coincides with the initial line OA , the angle that it has described is 0 , or 1 , or 2, or 3... times 27 t radians, in either the positive or negative directions, i.e. either 0 , or ± 2i r, or + 47 r, or + 67 r... radians. This is expressed by saying that when the revolving line coincides with the initial line the angle it has de¬ scribed is 2n7T, where n is some positive or negative whole number. 82. Theorem. To find a general expression to in¬ clude all angles which have the same sine . Let AOP be the smallest positive angle having the given sine and let it be denoted by a. Draw PM perpendicular to OA and produce MO to M' making MO equal to OM' and draw M'P' parallel and equal to MP. As in Art. 72 the angle AOP' is equal to it — a. ANGLES HAVING THE SAME SINE. 79 When the revolving line is in either of the positions OP or OP'. and in no other position, the sine of the angle traced out is equal to the given sine. When the revolving line is in the position OP it has made a whole number of complete revolutions and then described an angle a, i.e. by the last article it has described an angle equal to 2r7r 4- a.(1) where r is zero or some positive or negative integer. When the revolving line is in the position OP' it has, similarly, described an angle 2rir + AOP\ i.e . an angle 2r7r 4- 7r — a, i.e. (2 r + 1) 7r — a.(2) where r is zero or some positive or negative integer. All these angles will be found to be included in the expression T17T 4- (— l) n a .(3), where n is zero or a positive or negative integer. For, when n = 2r, since (— l) 2r = 4-1, the expression (3) gives 2r7r 4- a, which is the same as the expression (1). Also, when n = 2r 4- 1, since (— l) 2r+1 = — 1, the expres¬ sion (3) gives (2r4-l)7r —a, which is the same as the expression (2). Cor. Since all angles which have the same sine have also the same cosecant, the expression (3) includes all angles which have the same cosecant as a. 83. Theorem. To find a general expression to in¬ clude all angles which have the same cosine. Let AOP be the smallest angle having the given cosine and let it be denoted by a. 80 TRIGONOMETRY. Draw PM perpendicular to OA and pro¬ duce it to P\ making PM equal to MP'. When the revolving line is in the position OP or OP', and in no other position, then, as in Art. 78, the cosine of the angle traced out is equal to the given cosine. When the revolving line is in the position OP it has made a whole number of complete revolutions and then described an angle a, i.e. it has described an angle 2nir + a, where n is zero or some positive or negative integer. When the revolving line is in the position OP' it has made a whole number of complete revolutions and then described an angle — a, i.e. it has described an angle 2??7r—a. All these angles are included in the expression 2n7T ± o . (1) where n is some positive or negative number. Cor. The expression (1) includes all angles having the same secant as a. 84. Theorem. To find a general expression for all angles which have the same tangent. Let A OP be the smallest angle having the given tangent, and let it be denoted by a. Produce PO to P' making OP' equal to OP and draw P'M' per¬ pendicular to OM'. As in Art. 78 the angles AOP and A OP' have the same tangent; also the angle AOP' — i r + a. When the revolving line is in ANGLES HAVING THE SAME TANGENT. 81 the position OP, it has described a whole number of complete revolutions and then turned through an angle a, i.e. it has described an angle 2r7r + a . .( 1 ), where r is zero or some positive or negative integer. When the revolving line is in the position 0P\ it has similarly described an angle 2r7 r -f {it + a), 1.6. (2.V + 1) 7T -+■ CL .(2). All these angles are included in the expression n7r + a .(3). For, when n is even, (= 2r say), the expression (3) gives the same angles as the expression (1). Also, when n is odd, (= 2r +1 say), it gives the same angles as the expression (2). Cor. The expression (3) includes all angles which have the same cotangent as a. 85. Ex. 1. Write down the general expression for all angles , /3 (i) whose sine is equal to —-, (2) whose cosine is equal to - ^, 1 and (3) whose tangent is equal to n / 3 - (1) The smallest angle, whose sine is is 60°, i.e. J. u O Hence, by Art. 82, the general expression for all the angles which have this sine is W7T+(-l) n ^. (2) The smallest positive angle, whose cosine is - h is 120°, i.e. ~. £ o L. T. 6 82 TRIGONOMETRY. Hence, by Art. 83, the general expression for all the angles which have this cosine is 2?r 2mr ± — . O (3) The smallest positive angle, whose tangent is , is 30°, i.e. g . Hence, by Art. 84, the general expression for all the angles which have this tangent is m r + 7r 6 ’ Ex. 2. What is the most general value of 0 satisfying the equation sin 2 0 = 7 ? 4 Here we have sin 0 = ± - . Taking the upper sign, . . 1 .7 r sm 0 = - = sm t; 2 o 7r .*. 0 — ?17T + ( - l) W g • Taking the lower sign, sin 0 =-^ = sin^-g 0 = mr + (- l) n ( - Putting both solutions together we have 7 T sr 7T 0 = wtt±(- IV 1 - b or, what is the same expression, 0 = ?Z7T± 7T 6* Ex. 3. What is the most general value of 6 ivhich satisfies both of the equations sin 0 = - i and tan 0— ? 2 sj o Considering only angles between 0° and 360° the only angles, whose sine is - ^, are 210° and 330°. Similarly the only angles, whose tan- gent is , are 30° and 210°. \/o EXAMPLES. 83 The only angle, between 0° and 360°, satisfying both conditions is 7tt therefore 210 °, i.e. 6 * The most general value is hence obtained by adding any multiple 1 7T of four right angles to this angle, and hence is 2mr + — where n is any positive or negative integer. EXAMPLES. XI. What are the most general values of 0 which satisfy the equations 1. sin0=^. 2. sin d = - . 3. sin 0 = -^. 4. cos 0 = . 5. . cosd=-~. Li 6. „ 1 cos0 = - —- . s! 2 7. tan 0 = ^/3. 8 . tan 6 = - 1. 9. • t-H II o o 10. seed = 2. 11. „ 2 cosec 0— -7-. Vd 12. sin 2 0 = 1. 13. cos 2 <9 = -.-. 4 14. tan 2 0 = - . o 15. 4 sin 2 0 = 3. 16. 2 cot 2 6 = cosec 2 0. 17. sec 2 0 = f 9 3* 18. What is the most general value of 0 that satisfies both of the equations 1 cos 0 — - V2 and tan 0 = l? 19. What is the most general value of 0 that satisfies both of the equations cot 6 — -*J 3 and cosec 0 = - 2 ? 20. If cos(A-2?)=i, and sin (A + B) = ^ f find the smallest positive Li Li values of A and B and also their most general values. 21. If tan (A- B) = 1, and sec (A + B) = -j -, find the smallest positive v ® values of A and B and also their most general values. 6—2 84 TRIGONOMETRY. [Exs. XI.] 22. Find the angles between 0° and 360° which have respectively (1) /3 i their sines equal to , (2) their cosines equal to- - , and (3) their tan- gents equal to . 23. Taking into consideration only angles less than 180°, how many 5 1 4 values of x are there if (1) sin# = -, (2) cos#=-, (3) cos#= - (4) i o o 2 tan.'rzr-, and (5) cot x— - 7 ? O 24. Given the angle x construct the angle y if (1) sin y = 2 sin x, (2) tan y = 3 tan x, (3) cos y = | cos x , and (4) sec y = cosec x. 25. Shew that the same angles are indicated by the two following 7T 7T 7r formulae : (1) (2 n - 1) ^ + ( - l)’ 1 g , and (2) 2 h 7 t ± - , n being any integer. 26. Prove that the two formulae (1) (^2n + ^'j 7r± a and (2) n7r+ (- l) n denote the same angles, n being any integer. Illustrate by a figure. 27. If 0- a = mr+(-l) n (3 prove that 6 = 2imr + a + (3 or else that 0 — (2m +1) 7r -f a - where m and n are any integers. 28. If cospO + cosqd = 0, prove that the different values of 6 form two arithmetical progressions in which the common differences are 2ir p~q 2tt p + q and respectively. 29. Construct the angle whose sine is --. 86. An equation involving the trigonometrical ratios of an unknown angle is called a trigonometrical equation. The equation is not completely solved unless we obtain an expression for all the angles which satisfy it. Some elementary types of equations are solved in the following article. EQUATIONS. 85 87. Ex. 1. Solve the equation 2 sin 2 #+ ^3 cos x +1 = 0. The equation may be written 2-2 cos 2 x + ^3 cos x +1 = 0, i.e. 2 cos 2 x - ^/3 cos x - 3 = 0, i.e. (cos x - y/3) (2 cos x 4-^/3) = 0. The equation is therefore satisfied by cos # = ^3, or cos# = - V3 2 * There is no angle whose cosine is ^/S, so that the first factor gives no solution. The smallest positive angle, whose cosine is - is 150°, i.e. ^. ^ D Hence the most general value of the angle, whose cosine is - V3 Q7T is 2/i7r± —. (Art. 83.) This is the general solution of the given equation. Ex. 2. Solve the equation tan 50 = cot 20. The equation may be written -20). Now the most general value of the angle, that has the same tangent as “20, is, by Art. 84, ?i7r + ^-20, where n is any positive or negative integer. The most general solution of the equation is therefore 5d = mr + 20 £ where n is any integer. 7r tan 56 = tan ( - EXAMPLES. XII. Solve the equations 1. cos 2 6 — sin 0—^ = 0. 2. 2 sin 2 0 + 3 cos 0 = 0. 3. 2y/3 cos 2 0 = sin 6. 4. cos 6 4- cos 2 0 = 1. 86 TRIGONOMETRY. [Ess. XII.] 5. 4 cos 6 - 3 sec 6= 2 tan 6. 6. sin 2 0-2 cos 0 + t = 4 :0. 7. tan 2 6 - (l + N /3) tan 6 + *j3 = 0. 8. cot-0+(v 3 + v O | cot 0 + 1 = 0. 9. cot 6 - ah tan 6 = a- ■b. 10. tan 2 0 +cot 2 0 = 2. 11. sec 6 - 1 = ( x /2 - 1) tan 6. 12. sin 50= -t^. • v /2 13. sin 90 = sin 6 . 14. sin 30 = sin 20. 15. cos??i0 = cos n6. • 16. sin 20 = cos 30. 17. cos 56 = cos 46. 18. cosm0 = sin n6. 19. cot 6 = tan 86. 20. cot 0 = tan n6. 21. 2 tan 20 = tan-. 0 22. tan 20 tan 0 = 1. 23. tan 2 30 = cot 2 a. 24. tan 30 = cot 0. 25. tan 2 30 = tan 2 a. 26. 3 tan 2 0 = 1. 27. tan mx + cot nx = 0. 28. tan (7rcot 0) = cot (tt tan 0). 29. sin(0-0) = i and LJ COS (0 + 0) = 1 2* 30. cos (2x + 3y) = -, cos (3x + 2y) = V 3 2 * 31. Find all the angles between 0° and 90° which satisfy the equation sec 2 0 cosec 2 0 + 2 cosec 2 6 = 8. 5 32. If tan 2 0 = -, find versin 6 and explain the double result. 33. If the coversin of an angle be -, find its cosine and cotangent. O CHAPTER VII. TRIGONOMETRICAL RATIOS OF THE SUM AND DIFFERENCE OF TWO ANGLES. 88. Theorem. To prove that sin (A + jB) = sin A cos B 4- cos A sin B , and cos (A + B) = cos A cos B — sin A sin B. Let the revolving line start from OA and trace out the angle A OB (=A), and then trace out the further angle BOO (= B). In the final position of the revolving line take any point P, and draw PM and PJS T perpendicular to OA and OB respectively; through N draw NR parallel to AO to meet MP in R and draw NQ perpendicular to OA. 88 TRIGONOMETRY. The angle RPN =90° - z PNR = z RNO = z NOQ = A. tt • , , m ■ a rxn MP MR + RP Hence sin ( A + B) = sm A OP = gp = - OP QN RP _QN ON , RP NP 0P + OP ~ ON 0P + NP~0P = sin A cos B + cos RPN sin B. ;. sin (A + B) = sin A cos B + cos A sin B. Again cos (A + B) = cos A OP = ~ _ OQ RN_ 0Q ON RNNP OP OP ON OP NP OP = cos A cos B — sin RPN sin B. cos (A + B) = cos A cos B — sin A sin B. 89. The figures in the last article have been drawn only for the case in which A and B are acute angles. The same proof will be found to apply to angles of any size, due attention being paid to the signs of the quantities involved. The results may however be shewn to be true of all angles, without drawing any more figures, as follows. Let A and B be acute angles, so that, by Art. 88, we know that the theorem is true for A and B. Let ^4 1 = 90° + A, so that, by Art. 70, we have sin A x = cos A, and cos A 1 = - sin A. Then sin (A 1 + 5) = sin {90°+(A 4-L')} =cos (A+ 2^), by Art. 70, = cos A cos B - sin A sin B — sin A 1 cos B + cos A 2 sin2>. Also cos (A 1 + B) = cos [90° + (A + B)] = - sin (A + B) = - sin A cos B - cos A sin B = cos A 1 cos B - sin A 1 sin B. Similarly, we may proceed if B be increased by 90°. ADDITION AND SUBTRACTION FORMULAS. 89 Hence the formulae of Art. 88 are true if either A or B be increased by 90°, i.e. they are true if the component angles lie between 0° and 180°. Similarly, by putting A 2 = 90 o + A 1 , we can prove the truth of the theorems when either or both of the component angles have values between 0° and 270°. By proceeding in this way we see that the theorems are true uni¬ versally. 90. Theorem. To prove that sin ( A — B) = sin A cos B — cos A sin J3, and cos (A — B) = cos A cos B + sin A sin B. Let the revolving line starting from the initial line OA trace out the angle ° D A OB (= A) and then re¬ volving in the opposite di¬ rection, trace out the angle BOC, whose magnitude is B. The angle A OC is there¬ fore A — B. Take a point P in the final position of the revolv¬ ing line, and draw PM and PN perpendicular to OA and OB respectively; from N draw NQ and NR perpendicular to OA and MP respectively. The angle RPN= 90°- Z PNR = Z RNB= Z Q0N=A . Hence sm {A — B) — sm A00 = Qp~= - Qp — = -— OP OP QN ON PR PN ON OP PN OP = sin A cos B — cos RPN sin B, sin (A — B) = sin A cos B — cos A sin B. so that 90 TRIGONOMETRY. Ai / a t)\ OM OQ + QM OQ NR Also cos (A B ) Qp OR OB~^ OB OQ ON NR NB . D . , rDD . _ + t=ttv = cos A cos B + sm NBR sin B, ON OB ' NB OB so that cos (A — B) = cos A cos B + sin A sin B. 91. The proofs of the previous article will be found to apply to angles of any size, provided that due attention be paid to the signs of the quantities involved. Assuming the truth of the formulae for acute angles, we can shew them to be true universally without drawing any more figures. For, putting ^4 1 = 90° + yf, we have, (since sin A x = cos A, and cos A 1 = - sin A), sin (A 1 -B) = sin [90° + (A - B)] = cos (A - B) (Art. 70) , =cos A cos i^ + sin A sin B = sin A 1 cos B - cos A 1 sin B. Also cos (^fj — Z>) = cos [90° + (A — Z>)]= — sin (A — B) (Art. 70) = - sin A cos B + cos A sin B = cos ^ cos I? + sin A 1 sini>. Similarly we may proceed if B be increased by 90°. Hence the theorem is true for all angles which are not greater than two right angles. So, by putting ^2 = 90° + ^, we may shew the theorems to be true for all angles less than three right angles, and so on. Hence, by proceeding in this manner, we may shew that the theorems are true for all angles whatever. 92. The theorems of Arts. 88 and 90 which give respectively the trigonometrical functions of the sum and differences of two angles in terms of the functions of the angles themselves are often called the Addition and Sub¬ traction Theorems. ADDITION AND SUBTRACTION FORMULAS. 91 93 . Ex. 1 - Find the values of sin 75° and cos 15°. sin 75° = sin (45° + 30°) = sin 45° cos 30° + cos 45° sin 30° 1 */3 J l 1_/ S /3 + 1 — N /2 ’ 2 + *72 2 ~ 2 n /2 ’ and cos 75° = cos (45° + 30°) = cos 45° cos 30° - sin 45° sin 30° 1 V3 1 1 V3-1 ~J2 2 J2 2~ 2s]2 * Ex. 2. Assuming the formulae for sin(# + y) and cos(.r + y), deduce the formulae for sin (x - y) and cos (x - y). We have sin x = sin {(x- y) + y} = sin(x-y) cos y + cos (x - y) sin y .(1), and cos x = cos {(x - y) + y} = cos (x «- y) cosy - sin (a; - y) sin y .(2). Multiplying (1) by cosy and (2) by sin y and subtracting, we have sin x cos y - cos x sin y = sin (x - y) {cos 2 y + sin 2 y} = sin ( x -- y). Multiplying (1) by sin y and (2) by cosy and adding, we have sin x sin y + cos x cos y = cos [x — y) {cos 2 y + sin 2 y} = cos (x - y ). Hence the two formulae required are proved. These two formulae are true for all values of the angles since the formulae from which they are derived are true for all values. EXAMPLES. XIII. 3 9 1. If sin a = and cos (3 = find the value of sin (a - (3) and cos (a + /3). 45 33 2. If sina = ^ and sin/3=gg, find the values of sin (a-/3) and sin (a + /3). 15 12 3. If sin a = — and cos £ = jg , find the values of sin (a + j8), cos (a - (3), and tan («+/*)• Prove that 4. sin (A +B) sin (A - B) = sin 2 A - sin 2 B. 5. cos (A + B) cos (A - B) = cos 2 A - sin 2 B. 6. cos (45° - A) cos (45° - B) - sin (45° - A) sin (45° - B) = sin (A + B). 92 TRIGONOMETRY. [Exs. XIII.] 7. sin (45° 4 .4) cos (45° - B) + cos (45° 4 A) sin (45° - B) = cos (A-B). 8 sin (A-B ) + s in ( B - C) sin (C - A ) _ Q cos A cos B cos B cos C cos C cos A ~ 9. sin 105° 4 cos 105° = cos 45°. 10. sin 75° - sin 15° = cos 105° 4 cos 15°. 11 . cos a cos (7 - a) - sin a sin (7 - a) = cos a. 12. cos (a + /3) cos y - cos (/3 + 7) cos a = sin (3 sin (7 - a). 13. sin (k + 1) A sin (n — 1) A + cos (n +1) A cos (n— 1) A = cos 2 A. 14. sin (?i4l) A sin (w + 2) A 4 cos (n 4 I) A cos (n 4 2) A = cos A. 94. From Arts. 88 and 90, we have, for all values of A and B, sin (A + B) = sin A cos B + cos A sin B , and sin (A — B) = sin A cos B — cos A sin B. Hence, by addition and subtraction, we have sin (A + B) + sin (A — B) = 2 sin A cos B .(1), and sin (A + B) — sin (A — B) = 2 cos A sin B .(2). From the same articles we have, for all values of A and B , cos (A + jB) = cos A cos B — sin A sin B, and cos (A — B) = cos A cos B + sin A sin B . Hence, by addition and subtraction, we have cos (4+5) + cos (A-B) = 2 cos A cos B .(3), and cos (A — B) — cos (A + B) = 2 sin A sin B . (4). Put A + B = (7, and A — B = D, so that C + D 2 and B = G-D ) PRODUCT FORMULAE. 93 On making these substitutions the relations (1) to (4) become, for all values of C and D, _ „ . C + D C-D sin C + sin D = 2 sin —_— cos —— C+D C-D sin C - sin D = 2 cos —_ sin —-— C-D • • • • • II. 2 C-D C + D cos C + cos D — 2 cos —_— cos C + D and cos D — cos C = 2 sin — 0 — sin — - jL ^ [The student should carefully notice that the left-hand member of IV is cos D — cos C and not cos C — cos 1).] Ill ... IV. 95. These relations I to IV are extremely important and should be very carefully committed to memory. On account of their great importance we give a geo¬ metrical proof for the case when C and D are acute angles. Let AOC be the angle C and ADD the angle D. Bisect the angle COD by the straight line OE. On OE take a point P and draw QPR perpendicular to OP to meet OC and OD in Q and R respectively. Draw PL, QM and RN perpendicular to OA, and through R draw RST perpendicular to PL or QM to meet them in S and T respectively. Since the angle DOC is C — D, each of the angles DOE and EOC is—-—, and also 2 Z AOE= Z AOD + z DOE = D + D ■ 2 2 Since the two triangles POR and POQ are equal in all respects we have OQ = OR, and PR = PQ, so that RQ = 2 RP. ■94 TRIGONOMETRY. Hence QT — 2PS, and RT = 2RS, i.e. MN — 2ML. Therefore MQ + NR = TQ + 2ZS = 2 SP + 2 LS = 2LP. Also OM + ON = 20M+ MN = 20M + 2ML = 20L. Hence sin C + sin D = MQ NR _ MQ + NR OQ + OR ~ OR a . C+D C-D — 2 sm —_ — cos —„— . 2 2 . . . _ . MQ NR MQ - NR Again sm U — sin 1) = ^ - - - OR qp qp pp = 2^4 = 2^. ™ = 2 cos SPR sin ROP (y xt XtJL \J JAj _ (7+1) . <7-D = 2 cos —=— sm 2 for Z SPR = 90°-zSPO = z LOP = C+D' „ ri 01/ OiY Oil/+OiV Also cos (7 + cos D = + OQ OR OL OP OR 2°i- 2 _ 0R~ OP OR o rnp pdp o (7 4-D C — D = 2 cos LOP cos i Oii = 2 cos —~ cos — s 2 2 PRODUCT FORMULAE. 95 Finally cos D — cos G = ON _0M OR OQ ON- OM OR _ MN SR _ 2SR PR ~ OR OR " PR OR = 2 sin SPR . sin POR = 2 sin C+D 2 sin G-D 2 96. The student is strongly urged to make himself perfectly familiar with the formulae of the last article and to carefully practise himself in their application; perfect familiarity with these formulae will considerably facilitate his further progress. The formulae are very useful because they change sums and differences of certain quantities into products of certain other quantities, and products of quantities are, as the student probably knows from Algebra, easily dealt with by the help of logarithms. We subjoin a few examples of their use. Ex. 1. Ex. 2. . .. . .. . . 60 + 40 60-40 sm 60 + sin 40 = 2 sin — -— cos —— ™ n . 30 + 70 . 70-30 cos 30 - cos 70 = 2 sm —-— sm —-— = 2 sin 50 cos 0. — 2 sin 50 sin 20. Ex. 3. sin 75° - sin 15° cos 75° +cos 15° 75°+ 15° . 75° -15° 2 cos-- sm--- Z 2 75°+15° 75°-15° 2 cos---cos--- 2 cos 45° sin 30° _ _ n 1 - o- T~o -ono = tan 30 = -7- = 2 cos 4o° cos 30° J 3 = •57735 [This is an example of the simplification given by these formulae; it would be a very long and tiresome process to look out from the tables the values of sin 75°, sin 15°, cos 75 c , and cos 15°, and then to perform the division of one long decimal fraction by another.] 96 TRIGONOMETRY. Ex. 4. Simplify the expression (cos 0 - cos 30) (sin 80 + sin 20) (sin 50 - sin 0) (cos 40 - cos 60) * On applying the formulae of Art. 94 this expression . 0 + 30 . 30- 0 . 80 + 20 80-20 2 sin --- sin —-— x 2 sin —-— cos —-— 0 50 + 0 . 50-0 . 40 + 60 . 60-40 2 cos —-— sin —-— x 2 sin — —- sin — — 4 . sin 20 sin 0 . sin 50 cos 30 4 . cos 30 sin 20 . sin 50 sin 0 EXAMPLES. XIV. Prove that sin 70 - sin 50 cos 70 + cos 50 sin A + sin 3A 1 . 3. 4. 5. 6 . 7. 9. 10 . 11 . 12 . 13. = tan 0. 2 . cos 60 - cos 40 sin 60 + sin 40 cos A + cos 3A sin 7A - sin A sin 8A - sin 2A = tan 2 A. — cos 4 A sec 5A. cos 2B + cos 2A m . n - ir7 =cot(i4+B) cot (A-B), cos 2B - cos 2A sin 2A + sin 2 B _ tan (A + B) sin 2A - sin 2 B ~ tan (A-B)' sm A + sm 2A A — ^ 7 = cot —. cosA-cos2+ 2 cos 2 B - cos 2 A , . . . rtT) - 7 —=tan ( A-B sm 2 B + sm 2A 8 . sin 5A - sin 3 A cos 3 A + cos 5A cos (+ + P) + sin (A -B) — 2 sin (45° +A) cos (45°+ P). cos 3 A - cos A cos 2 A - cos 4A + sin A sin 3^4 - sin A sin 4 A - sin 2A cos 2A cos 3 A * sin (4 A - 2B) + sin (4i? - 2A) cos (4A - 2 B) + cos (4 B - 2.4) tan 50 + tan 30 = tan (A + B). tan 50 - tan 30 — 4 cos 20 cos 40. - tan 0. tan A . [Exs. XIV.] 14. 15. 16. 17. 18. 19. 20 . 21 . 22 . 23. 24. PRODUCT FORMULAE. cos 36 + 2 cos 56 + cos 76 . 0i3 , o. — = cos 26 - sin 26 tan 66. 97 = tan 4 A. cos 6 + 2 cos 3 6 + cos 56 sin A + sin 3 A + sin 5A + sin 7A cos A + cos 3 A + cos 5A + cos 7A sin {6 + (p) - 2 sin 0 + sin (6 - (p) — -- ~ tcin v. cos (6 + (p) - 2 cos 6 + cos (6 - i.e. by 9 — — a 4- any odd multiple of 7r.(1), and 0 = a + any even multiple of ir .(2), i.e. 6 must = (—l) n a + mr, where n is any positive or negative integer. For when n is odd this expression agrees with (1), and when n is even it agrees with (2). 103. Find the general value of all angles having the same cosine. The equation we have now to solve is cos 9 = cos a, i.e. cos a — cos 9 — 0, i.e. . 9 + a . 9 - a 2 sm —— sm —^ = 0, and it is therefore satisfied by i.e. by . 9 + a . 9 — a A sm —jr- — 0, and by sm —= 0, ~~^~ a = any multiple of n r, 6 a = any multiple of i r, and by 2 104 TRIGONOMETRY. i.e. by 6 = — a + any multiple of 27 r, and by 6 = a + any multiple of 27r. Both these sets of values are included in the solution 6 = 2?i7T + a, where n is any positive or negative integer. 104. Find the general value of all angles having the same tangent. The equation we have now to solve is tan 6 — tan a = 0, i.e. sin 6 cos a — cos 6 sin a = 0, i.e. sin(0 —a) = 0. 6 — ol = any multiple of ir = n7r, where n is any positive or negative integer, so that the most general solution is 6 = nnr -f a. CHAPTER VIII. THE TRIGONOMETRICAL RATIOS OF MULTIPLE AND SUBMULTIPLE ANGLES. 105. To find the trigonometrical ratios of an angle 2A in terms of those of the angle A. If in the formulae of Art. 88 we put B = A, we have sin 2A = sin A cos A + cos A sin A = 2 sin A cos A, cos 2 A = cos A cos A — sin A sin A = cos 2 A — sin 2 A = (1 — sin 2 A) — sin 2 A — 1 — 2 sin 2 A, and also and = cos 2 A — (1 — cos 2 A) = 2 cos 2 A — 1 ; tan 2A = tan A + tan A 1 — tan A . tan A 2 tan A 1 — tan 2 A * Now the formulae of Art. 88 are true for all values of A and B ; hence any formulae derived from them are true for all values of the angles. In particular the above formulae are true for all values of A. 106 TRIGONOMETRY. 106. An independent geometrical proof of the formulae of the preceding article may be given, for values of A which are less than a right angle. Let QCP be the angle 2A. With centre G and radius CP describe a circle and let QG meet it again in 0. Join OP and PQ , and draw PN jDerpendicular to OQ. By Euc. Til. 20, the angle Q0P=^QCP = A, and the angle i \ T PQ = Z QOP = A. Hence sin 2A = NP GP 2NP _ 0 NP _ 0 NP OP 2 CQ “ 0Q~ OP ‘ OQ = 2 sin NOP cos POQ, since OPQ is a right angle, also cos = 2 sin A cos A ; CN _2CN _{0C+CN)-(0C CP ~ OQ ~ OQ ON-NQ ON OP NQPQ -CN) OQ OP OQ PQ OQ = cos 2 A — sin 2 A and tan 2 A = 2 NP NP GN ON-NQ 2 tan A NP ON 1 - NQ PN PN ON 1 — tan 2 A ‘ MULTIPLE ANGLES. 107 Ex. To find the values of sin 15° and cos 15°. Let the angle 2A be 30°, so that A is 15°. Let the radius CP be 2a, so that we have CN — 2a cos 30° = u n /3, and NP = 2a sin 30° = a. Hence ON=OC+ GN= a (2 + ^3), and NQ = CQ-CN=a( 2 - N /3). OP 2 = ON.OQ = a(2 + > J-d) x4a so that OP = aJ2 (V3 + 1), and PQ 2 = QN.QO-a( 2 - N /3) x 4a, so that PQ = aJ-2 U/3-1). Hence and sin 15° = PQ _ V2 (V 3 - 1) _ V 3 - 1 OQ~ 4 “ 2J2 ’ cos 15° = OP OQ V*2 (V 3 + 1) V 3 +1 4 ” 2y/2 * (Euc. vi. 8), 107. To find the trigonometrical f unctions of 3 A in terms of those of A. By Art. 88, putting B equal to 2A, we have sin 3 A = sin ( A + 2 A) = sin A cos 2 A + cos A sin 2A = sin A (1 — 2 sin 2 A) + cos A . 2 sin A cos A = sin A ( 1 — 2 sin 2 A) -f 2 sin A (1 — sin 2 A). Hence sin 3 A = 3 sin A — 4 sin 3 A.(1). So cos 3 A = cos (A + 2A) = cos A cos 2 A — sin A sin 2A = cos A (2 cos 2 A — 1) — sin A . 2 sin A cos A = cos A (2 cos 2 A — 1) — 2 cos ^(i — cos 2 A). Hence cos 3A = 4 cos 3 A — 3 cos A.(2). 108 TRIGONOMETRY. Also tan 3 A = tan (A + 2A) = tan A + tan 2 A 1 — tan A tan 2 A tan A 4- 2 tan A 1 — tan 2 A tan A (1 — tan 2 A) 4- 2 tan A 1 — tan A . 2 tan A 1 — tan 2 A (1 — tan 2 A) — 2 tan 2 A Hence tan 3A 3 tan A — tan 3 A 1 — 3 tan 2 A 108. By a process similar to that of the last article, the trigonometrical ratios of any higher multiples of 0 may be expressed in terms of those of 0. The method is however long and tedious. In a later chapter better methods will be pointed out. As an example let us express cos 50 in terms of cos 6. We have cos 5 6 = cos (3# + 26) = cos 3# cos 26 — sin 3# sin 26 = (4 cos 3 6 — 3 cos 6) (2 cos 2 0—1) — (3 sin 6 — 4 sin 3 6) . 2 sin 6 cos 6 = (8 cos 5 0 — 10 cos 3 0 4- 3 cos 0) — 2 cos 0. sin 2 0 (3 — 4 sin 2 0) = (8 cos 5 0 — 10 cos 3 0 4- 3 cos 0) — 2 cos 0(1 — cos 2 0) (4 cos 2 0 — 1) = (8 cos 5 0 — 10 cos 3 0 4- 3 cos 0) — 2 cos 0 (5 cos 2 0 — 4 cos 4 0—1) = 16 cos 5 0— 20 cos 3 04-o cos 0. MULTIPLE ANGLES. 109 EXAMPLES. XVII. 1, Find the value of sin 2 a when 3 . 12 16 (1) cos a — —, (2) sina = — and (3) tan a = — . r> lo x ' nK 63 2. Find the value of cos 2a, when 15 (1) cos a = —, (2) sina = -, and (3) tana = 1 / r> 3. If tan 8 = - , find the value of a cos 2 8 + b sin 28. a 12 * Prove that . sin 2 A 4. 6 . 1 + cos 2A 1 - cos 2 A = tan A. = tan 2 A . = cot A. 7. 9, 1 + cos 2A 8. tan A -cot A— -2 cot 2A. 1 - cos A + cos B - cos (A + B) . A B n ^— tan — cot — . 1 + cos A - cos B - cos (A + B) 2 2 sin 2A 1 - cos 2A tan A + cot A—2 cosec2 A, cosec 2A + cot 2A = cot A . 10 11 r ^L=tan(45° ± ^). 1 =f sin A \ 2 12 . sec 8A - 1 _ tan 8 A sec 4A - 1 — tan 2 A 13. l + tan 2 (45° -A) 1 - tan 2 (45°-A) = cosec 2A. 14. 15. sin (a + j8) _ 2 ^ ” tan a ;/ sin 2 A — sin 2 B sin A cos A - sin B cos B = tan (A +B). 16. tan ^ + 0^ “tan 8^ = cos A + sin A cos A - sin A 17. cos A - sin A cos A + sin A 2 tan 28. — 2 tan 2 A. _ _ iti i rm . / j -i kox 4 COS 2A 18. cot (A +15°) - tan (A -15°) = — 2 ^ 110 TRIGONOMETRY. [Exs. XVII.] 19. 21 . 22 . 23. 24. 25. 26. 27. 28. 29. 30. 31. 32. 33. 35. 36. 37. 38. 39. 40. sin 6 + sin 2 6 = tan 6. = tan l + cos# + cos2# sin (n +1) A - sin (n -1) A cos (ft+ 1)^4 + 2cosft^ +cos(ft- 1)A sin (ft +1) A + 2 sin nA + sin (n-1) A , A --- - --7—- 7 --t——-—= cot — . cos (n - 1) A - cos (?i+ 1) A 2 sin (2ft +1) A sin A — sin 2 ( n + 1) A - sin 2 nA. sin (A + 3B) + sin (3^4 +B) _ ^ l+sin0-cos# . 20. -v——- = tan 1 + sin 0 + cos0 A sin 2A + sin 2 B 2 cos (A + B). A SA sin 3A + sin 2^4 - sin A —A sin A cos — cos — tan 2^4 = (sec 2^4 + 1) J sec 2 ^4-1. cos 3 20 + 3 cos 26 = 4 (cos 6 6 - sin 6 6). 1 + cos 2 26 — 2 (cos 4 6 + sin 4 6). sec 2 ^4(1 + sec 2A) = 2 sec 2^4. cosec A — 2 cot 2 A cos A = 2 sin A . cot A = i ^cot ^ - tan . sin a sin (60° - a) sin (60° + a) = ^ sin 3a. cos a cos (60 - a) cos (60° + a) = - cos 3a. cot a + cot (60 + a) - cot (60° - a) = 3 cot 3a. cos 20° cos 40° cos 60° cos 80° = . lb sin 20° sin 40° sin 60° sin 80° = 16 cos 4a = 1 - 8 cos 2 a + 8 cos 4 a. sin 4^4 = 4 sin A cos 3 ^4-4 cos A sin 3 A . cos 6a = 32 cos 6 a - 48 cos 4 a + 18 cos 2 a - 1. tan 3 A tan 2.4 tan A = tan 3^4 - tan 2.4 -- tan A . 2 cos 2 n 6 +1 2 cos 6 + 1 = (2 cos 6 - 1) (2 cos 26 - 1) (2 cos 2 2 6 - 1) (2 cos 2 7l_1 6-1), 41. to I <3> SUBMULTIPLE ANGLES. Ill Submultiple angles. 109. Since the relations of Art. 105 are true for all values of the angle A, they will be true if instead of A we substitute —, and therefore if instead of 2 A we put La 2 . —, i.e. A. 2 Hence we have the relations sin A = 2 sin — cos —.(1), 2 2 n A . 0 A cos A = cos 2 — — sin- — 2 2 = 2 cos 2 ^-1=1—2 sin 2 ^. ( 2 ), 2 2 04- A 2 tan — and tan A = ————- .(3). 1 - tan 2 *= From (1) we also have .A A 2 sm — cos — • ^ 2 2 sm A ----- A . 2 A cos 2 — + sm — 2 tan ^ =--, by dividing numera- 1 + tan 2 ~ 2 tor and denominator by cos 2 — . InS I 112 TRIGONOMETRY. 2 A . 2 A cos -g — sin — So cos A - ---. cos 2 —+ sm 2 — 1 — tan 2 ~ 2 ~7a' 1 + tan — V 110. To express the trigonometrical ratios of the angl in terms of cos A. From equation (2) of the last article we have cos A = 1 — 2 sin 2 —, so that 2 sin 2 — = 1 — cos A , 2 and therefore sin ^ = ± . (1). A Again, cos A = 2 cos 2 9 — 1, LJ A so that 2 cos 2 — = 1 + cos A, 2 and therefore cos ~ = ± ^ J * — . (2). . A A Sill — --7 Hence, tan — -- = 4- a /-—^— A - .(3). 2 A ~ V 1 + cos A v 7 COS — 2 RATIOS OF 4 IN TERMS OF COS A. 113 ^ * 111. In each of the preceding formulae it will be noted that there is an ambiguous sign. In any particular case the proper sign can be determined as the following examples will shew. . 1. Given cos 45°= , find the values of sin 22 and cos 22J°. The equation (1) of the last article gives, by putting A equal to 45°, sin 22J°= ± - cos 45° — i i- •j 2 = *\/ 2-^2 = ±2 n/ 2 ->/ 2 - Now sin 22J° is necessarily positive, so that the upper sign must be taken. Hence So cos 22 J sin 22\° = 1 j2-^2. also cos 22J° is positive; V2+V 2 /. cos22J° = 2 2 v/3 Ex. 2. Given cos 330° = ^— , find the values of sin 165° and cos 165°. u The equation (1) gives sin 165 0=± \/ r - cos 330 c = ± 1- V3 »W 4 -2^3 Also JS-1 ~ 2V2 * cos 165° -v 5 + cos 330° — ± 2 > 2 v/3+1 -V 4 + V3 = ± 2^2 ' L. T. 8 114 TRIGONOMETRY. Now 165° lies between 90° and 180°, so that, by Art. 52, its sine is positive and its cosine is negative. Hence sin 165° = sf 3-1 2^2 ’ and cos 165° = \/3 +1 2J2 * From the above examples it will be seen that, when the angle A and its cosine are given, the ratios for the angle — may be determined without Z any ambiguity of sign. When however only cos A is given, there is an ambiguity in finding A A sin — and cos —. The explanation of this ambiguity is given in the next z z article. 112. To explain why there is ambiguity when cos £ and sin — are found from the value of cos A. Zd We know that, if n be any integer, cos A = cos ( 2n7r + A) =k (say). A 2 Hence any formula which gives us cos — in terms of k } should give us also the cosine of 2n7r 4- A xt 2utt + A JN ow cos---= cos A 2 A _ . .A A = cos nir cos — + sm nir sin — = cos nir cos — 2 2 2 A = ± cos —, according as n is even or odd. 115 RATIOS OF - IN TERMS OF SIN A. u A Similarly any formula giving us sin — in terms of k , should give us also the sine of 2n7r + A ,, . 2nir ± A . ( A' Also sm ---= sm ( nir ± A . A = sm nir cos -=r + cos nir sm — 2 “ 2 = + cos nir sin A . A = + sm — . 2 Hence in each case we should expect to obtain two A A values for cos — and sin — , and this is the number which A A the formulae of Art. 110 give. 113. To express the trigonometrical ratios of the angle A — in terms of sin A. From equation (1) of Art. 109 we have a . A A . 2 sm — cos — = sm A _ A A Also .A o A . sm 2 — -f cos 2 — = 1, always (i> ( 2 ). First adding these equations, and then subtracting them, we have • 9 A _ . A A A . . sm- — + 2 sm — cos — + cos 2 — = 1 + sm A, i . 0 A _ . A A A ^ . A and sm- — — 2 smcos ^ + cos 2 - = l- smi; A A A A 8—2 116 TRIGONOMETRY. /. A , Ay , . . (sin — + cos J = 1 + sin J., -A \ 2 and (sin — — cos = 1 — sin A ; A Jf __ so that sin — + cos = + Vl + sin A .(3), a __ and sin — cos = ± Vl — sin A .(4). By adding, and then subtracting, we have A 2 sin —= + Vl +sin A ± Vl - sin A. (5), ^ __ _ and 2 cos — = ± V 1 + sin A + V 1 — sin A .(6). The other ratios of ^ are then easily obtained. 114. In each of the formulae (5) and (6) there are two ambiguous signs. In the following examples it is shewn how to determine the ambiguity in any particular case. Ex. 1 . Given that sin 30° is find the values of sin 15° and cos 15°. Putting ^4 = 30°, we have from relations (3) and (4), sin 15° +cos 15°= ± ^/l + sin 30°= ± ^ , v ^ sin 15°-cos 15°= ± >/l-sin 30°= Now sin 15° and cos 15° are both positive and cos 15° is greater than sin 15°. Hence the expressions sin 15° + cos 15° and sin 15° - cos 15° are respectively positive and negative. 117 RATIOS OF - IN TERMS OF SIN A. A Hence the above two relations should be sin 15° + cos 15° = + , v and sin 15° - cos 15°= - s/*' Hence sinl5 ° = ^/2 1 ’ and C0Sl5 °^ij¥- Ex. 2. Given that sin 570° is equal to - ~ , find the values of sin 285° and cos 285°. Putting A equal to 570°, we have sin 285° + cos 285° = ± J 1 + sin 570° = ± -ir, v z and sin 285° - cos 285°= ± *Jl - sin 570°= ± Now sin 285° is negative, cos 285° is positive, and the former is numerically greater than the latter, as may be seen by a figure. Hence sin 285° + cos 285° is negative and sin 285° - cos 285° is also negative. and /. sin 285° + cos 285°= sin 285° - cos 285° = 1 V2’ V3 n/2- Hence and ^^115. To explain why there is ambiguity when sin — £ and cos — are found from the value of sin A. We know that, if n be any integer, sin {n7r 4- (— l) n A) = sin A = h (say). (Art. 82.) 118 TRIGONOMETRY. i Hence any formula which gives us sin — in terms of k, should give us also the sine of U7r ~ ^ ^ ^ . z First, let n be even and equal to 2m. Then . UTT + i-l ) 71 A . ( A sin--= sm ( m7r + ~2 A .A .A = sin mir cos — + cos mir sin — = cos mrr sin £ A 2 = + sm A 2 J according as m is even or odd. Secondly , let n be odd and equal to 2^ + 1. Then . mr + (— IV 2, A . 2n7r + 7r — A sm-—--= sm —---= sm 2 2 P7T + 7T — A 7T — A 7T — A A — smpir cos —-—- + cos pir sm —-— = cos pir cos — z z z a - m = ± COS —, according as p is even or odd. Hence any formula which gives us sin — in terms of 2 j sin A should be expected to give us, in addition, the values of . A A j A — sm — , cos — and — cos — , 2 2 2 i.e. 4 values in all. This is the number of values which we get from the formulae of Art. 113, by giving all possible values to the ambiguities. RATIOS OF =■ IN TERMS OF SIN A. 119 £j A In a similar manner it may be shewn that when cos — is found from sin A, we should expect 4 values. 116. In any general case we can shew how the ambiguities in relations (3) and (4) of Art. 113 may be found. We have . A A .. / 1 . sm _ + c ° s _ = V 2 si A 1 A' Sm 2 + ^ 2°° s 2 • 4 IT j. 7 r / 77* sin - 7 T cos 2 - 7 + COS 4 2 sm 4_ = \/2 sm (i + 2 ) = V2 The right-hand member of this equation is positive if 7T A ^ + -g lie between 2nir and 2mr 4- 7 r, A . 7T 37T i.e. if — lie between 2nir — — and 2 w 7 t + — . 2 4 4 A A A Hence sin — + cos is positive if lie between 2 /i 7 r — 7 and 2 ?i 7 r + ; 4 4 it is negative otherwise. Similarly we can prove that sm _-c OS - = V 2 sm^- 5 j. A A Therefore sin — — cos — is positive if Li Li lie between 2nir and 2nir + it, i.e. if — lie between 2nir + 7 and 2utt + ^. 2 4 4 120 TRIGONOMETRY. It is negative otherwise. The results of this article are shewn graphically in the following figure. OA is the initial line and OP, OQ, OR and OS bisect the angles in the first, second, third and fourth quadrants respectively. Numerical Example. Within what limits must — lie if j __ _ 2 sin -y = - J 1 + sin A - J 1 - sin A. In this case the formulae of Art. 113 must clearly be sin ^ -f- cos - Jl-\- sin A .(1), A A _ and sin — - cos - = - sjl - sin A .(2). A Jj For the addition of these two formulae gives the given formula. From (1) it follows that the revolving line which bounds the angle must be between OQ and OR or else between OR and OS. to | tu RATIOS OF - IN TERMS OF TAN A. A 121 From (2) it follows that the revolving line must lie between OR and OS or else between OS and OP. Both these conditions are satisfied only when the revolving line lies between OR and OS, and therefore the angle — lies between 2?i 7 r - AL and 2mr - ^ . 4 4 117. To express the trigonometrical ratios of — in terms of tan A. From equation (3) of Art. 109 we have A 2 tan tan A = 1 — tan 2 A * Hence - ^ , A 2 x A .*. 1 — tan 2 — = -- T tan -77 . 2 tan A 2 A tan 2 -77 + 7 A 2 tan A 2 x A tan — + = 1 + 2 tan 2 A tan 2 A 1 + tan 2 A tan 2 A , A .*. tan — + = + V 1 + tan 2 A 2 tan A tan A A ± V 1 + tan 2 A — 1 .* * tan ^7 =- 7 - 1 - 2 tan A 118. The ambiguous sign in equation (1) can only be determined when we know something of the magnitude of A. Given tan 15° = 2 - */3, find tan 7J°. Putting A = 15° we have, from equation (1), of the last article, 1Q _ ±Vi + (2-n/3) 2 -1_ tan 7i°= 2-^3 2-^3 (!)■ 122 TRIGONOMETRY. Now tan 7J° is positive so that we must take the upper sign. Hence tan 7£° = —- = (V6-\/2-l) (2 + n /3)= n / 6- v /3 + V2-2 = (V3 - s /2)(V2-1). 15° Since tan 15° = tan 195°, the equation which gives us tan — in terms 195° of tan 15° may be expected to give us tan —- in terms of tan 195°. In fact the value obtained from (1) by taking the negative sign before the , . . 195° radical is tan ——. Hence tan 195° _ - s/8 - y 3 - 1 _ - (s/6 -s/2) -1 so that 2 2-y3 2-^/3 = (- s/6 + s/2 -1) (2 + s/3) = - (s/3 + s/2) (s/2 +1), -cot7J°=tan97J°=-(s/3+s/2) (s/2 + 1). * *119. To explain why there is ambiguity when tan y is found from the value of tan A. We know, by Art. 84, that, if n be any integer, tan (nir + A ) = tan A —k (say). £ Hence any equation which gives us tan — in terms of k Z may be expected to give us tan n7r . j : also. First , let n be even and equal to 2m. Then mr + A 2mir + A ( A' tan --— = tan - - -= tan I ■ mm + -y = tan — , as in Art. 84. z Secondly , let n be odd and equal to 2 p + 1 . RATIOS OF 4 IN TERMS OF SIN A. a 123 rri j_ U7T + A (2p + 1) 7T + A Then tan ^— = ^ an 2 - = tan (pir +* = tan 17 ^ — (Art. 84) A = — cot y • (Art. 70.) A Hence the formula which gives us the value of tan — should be expected to give us also the value of — cot • An illustration of this is seen in the example of the last article. EXAMPLES. XVIII. 1. If sin 0 = - and sin0 = -, find the values of sin (0 + 0) and a O sin (20 + 20). 2. The tangent of an angle is 2*4. Find its cosecant, the cosecant of half the angle and the cosecant of the supplement of double the angle. 3. If cosa = ^i and sinfi = f, find the values of sin 2 a ■ r and ol 5 A cos 2 , the angles a and /3 being positive acute angles. 3 4 ct — B 4 . If cosa = ^ and cos /3=^, find the value of cos the angles 5 O a a and p being positive acute angles. 5. Given sec 0 = 1J, find tan - and tan 0. a A 6. If cos A — *28, find the value of tan — , and explain the resulting ambiguity. 7. Find the values of (1) sin7J°, (2) cos7J°, (3) tan22|°, and (4) tan 11|°. 124 TRIGONOMETRY. [Exs. XVIII.] 8. If sin 0 + sin

/i - sin ^4, find within what limits — must lie when ( 1 ) the two positive signs are taken, ( 2 ) the two negative ,, ,, ,, and (3) the first sign is negative and the second positive. 31. Prove that the sine is algebraically less than the cosine for any , , , ~ 37T _ angle between 2mr - — and 2mr 4- : where n is any integer. *x TC 126 TRIGONOMETRY. [Exs. XVIII.] ^4 32. If sin — be determined from the equation 3 sin A — 3 sin 4 - 4 sin 3 4 , O Jj prove that we should expect to obtain also the values of . 7 t — A . 7 r + A sin —-— and - sin —- . 33 . If cos — be found from the equation O cos .4 = 4 cos 3 4 - 3 cos 4 o 3 prove that we should expect to obtain also the values of 2tt - A , 2tt + A cos —-— and cos —-— 120. By the use of the formulae of the present chapter we can now find the trigonometrical ratios of some important angles. To find the trigonometrical functions of an angle of 18°. Let 6 stand for 18°, so that 20 is 36° and 3# is 54°. Hence 20 = 90° — 30 , and therefore sin 20 = sin (90° — 30) = cos 30. 2 sin 0 cos 0 = 4} cos 3 0 — 3 cos 0 (Arts. 105 and 107). Hence, either cos 0 = 0, which gives 0 = 90°, or 2 sin 0 = 4< cos 2 $ — 8 = 1 — 4 sin 2 0. .*. 4 sin 2 0 -f 2 sin 0 = 1. By solving this quadratic equation, we have 4 ANGLES OF 18° AND 36°. 127 In our case sin 0 is necessarily a positive quantity. Hence we take the upper sign, and have . _ —- o 5 1 sin 18 = -—-— . 4 Hence cos 18° — Vl — sin 2 18° = 6-2V5 1 - 16 10 + 2V5 16 V10 + 2V5 The remaining trigonometrical ratios of 18° may be now found. Since 72° is the complement of 18°, the values of the ratios for 72° may be obtained by the use of Art. 69. 121 . To find the trigonometrical functions of an angle of 36°. Since cos 20 = 1 — 2 sin 2 0 , (Art. 105), cos 36° = 1 - 2 sin 2 18° = 1 - 2 = 1 - 3-V5 4 so that cos 3 6 Hence sin 36° = Vl - cos 2 36° = V5 + 1 1 - 6 + 2 a/ 5 _ VlO -2 Vo 16 “ 4 The remaining trigonometrical functions of 36° may now be found. Also, since 54° is the complement of 36°, the values of the functions for 54° may be found by the help of Art. 69. 122 . The value of sin 18° and cos 36° may also be found geometrically as follows. 128 TRIGONOMETRY. Let ABC be a triangle constructed, as in Euc. iv. 10, so that each of the angles B and G is double of the angle A. Then 180° = A + B + C = A + 2A + 2A, so that A = 86°. Hence, if AD be drawn perpendicu¬ lar to BG, we have ZBAD = 18°. By Euclid’s construction we know that BG is equal to AX where X is a point on AB, such that AB.BX=AX\ Let AB = a , and AX = x. This relation then gives a (a — x) — x 1 2 , i.e. x 2 + ax = a 2 , V5-1 i.e. x = a —-— . Hence sin 18° = sin BAD = BD BA 1 BG 2 BA 1 x \/5 — 1 2 a 4 Again (by Euc. iv. 10) we know that AX and XG are equal; hence if XL be perpendicular to AG, then L bisects AG. Hence cos 36° = AL a ii = r i Vo — 1 V5 + 1 _V5 + 1 (V5 — 1)(V5 + 1) ~~ 4 ANGLES OF 9° AND 81°. 129 123. To find the trigonometrical f unctions for an angle «/9°. Since sin 9° and cos 9° are both positive the relation (3) of Art. 113 gives • ./—— 7 —Too A , V5-1 V3 + V5 sin 9 + cos 9 = V1 + sin 18 = \^/ 1 H- j — =-^- Also, since cos 9° is greater than sin 9° (Art. 53), the quantity sin 9° — cos 9° is negative. Hence the relation (4) of Art. 113 gives sin 9° — cos 9° — — Vl — sin 18° = — 1 — — Vo - do =- 2 ^. (2) * By adding (1) and (2), we have . V3 +V5 - V5-V5 sm 9 =---> 4 and, by subtracting (2) from (1), we have 0 o’ V3 + V5 + V5 — Vo cos 9 =--—-- 4 The remaining functions for 9° may now be found. Also, since 81° is the complement of 9°, the values of the functions for 81° may be obtained by the use of Art. 69. EXAMPLES. XIX. Prove that 1. sin 2 72° - sin 2 60°=AA . o 2. cos 2 48° - sin 2 12° = - ^ * . L. T. 9 130 TRIGONOMETRY. [Exs. XIX.] 3. cos 12° + cos 60° + cos 84° = cos 24° + cos 48°. 4 . 7T.27r.37r.47r 5 sin - sin -=* sin — sin — = — . 5 5 o 5 lb .7r . 137T 1 5. sin — + sm y 0 — - g • _ 7T . 137T 6 - sin 10 sm 10 4' 7. tan 6° tan 42° tan 66° tan 78° = 1. 7T 27T 37T 47T 57T 67T 7 7T 1 s. cos Jg cos Jg cos ^ cos - cos n cos ^ cos is = 97 15 15 15 15 9. „ _ 27T 47T 87T 16 cos — cos — cos — cos 15 15 15 147T ~15 = 10. Two parallel chords of a circle, which are on the same side of the centre, subtend angles of 72° and 144° respectively at the centre. Prove that the perpendicular distance between the chords is half the radius of the circle. 11 . In any circle prove that the chord which subtends 108° at the centre is equal to the sum of the two chords which subtend angles of 36° and 60°. 12. Construct the angle whose cosine is equal to its tangent. 13. Solve the equation 4 cos 0-3 sec 0 = 2 tan 0. CHAPTER IX. IDENTITIES AND TRIGONOMETRICAL EQUATIONS. 124 . The formulae of Arts. 88 and 90 can be used to obtain the trigonometrical ratios of the sum of more than two angles. For example sin (A + R + G) = sin (A + R) cos G + cos (A + B ) sin G = [sin A cos B + cos A sin R] cos G + [cos A cos B — sin A sin R] x sin C = sin A cos R cos C + cos A sin R cos C + cos A cos R sin C — sin A sin R sin G. So oos (A 4 * R + G) = cos (A -f R) cos C — sin (A + R) sin G = (cos A cos R — sin A sin R) cos C — (sin A cos R + cos A sin R) sin C = cos A cos R cos G — cos A sin R sin (7 — sin A cos R sin C — sin A sin R cos (7. 9—2 132 TRIGONOMETRY. Also tan (A 4 B 4 C) = tan (i+5) + tan G 1 — tan (A 4 B) tan G tan A -f tan B ri z --- -7— -4- tan U 1 — tan A tan B tan A 4 tan B 1 — -— - . - tan L tan A 4 * tan B 4 - tan C — tan A tan B tan C 1 — tan B tan G — tan G tan A — tan A tan B' 125. The last formula of the previous article is a particular case of a very general theorem which gives the tangent of the sum of any number of angles in terms of the tangents of the angles themselves. The theorem is tcLii (Aj 4 A 2 4 Ao 4" • • • 4“ A n ) — Sl ~~ S s s 5 ~ s 7 + ... ^ 1 —s 2 4- s 4 —s 0 + ..^ ’ where s x = tan A 1 + tan A 2 4 ... 4 tan A n = the sum of the tangents of the separate angles, So — tan A 1 tan A 2 4 tan A x tan A 3 4 = the sum of the tangents taken two at a time, s 3 = tan Aj tan A 2 tan A 3 4 tan A 2 tan A s tan A 4 4 = the sum of the tangents taken three at a time, and so on. Assume the relation (1) to hold for n angles and add on another angle A n+1 . Then tan ( A 1 4 A 2 4 ... 4- A w+1 ) = tan [(Aj 4 A 2 4 ... 4* A n ) 4 A n+i] __ tan (A 4 4 A 2 4 ... 4- A n ) 4 tan A n+1 1 — tan (Aj 4 Ao 4 ... 4 A n ) . tan A n+1 TANGENT OF THE SUM OF ANGLES. 133 #3 + #5 ~ S 7 + .»» 1 — #3 + . . . + tan A n+i 1 S x — S% + S 5 ... . 1 — --tan A n+1 1 — #2 + #4 ... Let tan A x , tan A 2 , ... tan A n+l be respectively called t\) 4 >* " ^n-\- 1 * Then tan (A x + A 2 + ... + A n+1 ) r _ (#1 “‘ S 3 + $5 * * • ) + t n + x (1 “ S 2 + $4 • • •) (1 $2 “b &4 • • •) ('^i £3 + ^5 • • •) ^n +1 _ _ ($1 + t n+1 ) ~ ( S s + S 2 t)l+l) + (#5 + ^4 t n + 1 ) • » • 1 ( S 2 + #1 tn+l) + ($4 + S 3 ^71+1) “ (#6 + ^5 tn+i) • • • But Si + t n +Y = (t\ + 4 + • • • t n ) "1" t n +i = the sum of the (n + 1 ) tangents, .9,+ tn+i — (tit 2 + t. 2 t s + ...) + (t x + t 2 +• •. + tn) t n +i « = the sum, two at a time, of the(?i+ 1 ) tangents. £3 + ^2 t)i- j_i — (tit 2^3 + t^t^ti + •. .) + (^1^2 + ^ 2^3 + • • •) tn+i = the sum three at a time of the (n + 1) tangents and so on. Hence we see that the same rule holds for (n +1) angles as for n angles. Hence, if the theorem be true for n angles, it is true for (n + 1) angles. But, by Arts. 98 and 124, it is true for 2 and 3 angles. Hence the theorem is true for 4 angles; hence for 5 angles .... Hence it is true universally. Cor. If the angles be all equal and there be n of them and each equal to #, then s x = n. tan 0 ; s 2 = n C 2 tan 2 # ; s 3 = n G z tan 3 #. 134 TRIGONOMETRY. Ex. Write down the value of tan 40. 4 tan 0 — 4 (7 3 tan 3 0 1 - 4 C 2 tan 2 0 + 4 C 4 tan 4 0 4 tan 0-4 tan 3 0 1-6 tan 2 0 + tan 4 0 * Ex. Prove that tan oO = 5 tan 0-10 tan 3 0 + tan 5 0 1-10 tan 2 0 + 5 tan 4 0 126. By a method similar to that of the last article it may be shewn that sin (A 1 + = cos A 1 cos J-o... cos A n ($i — s 3 4- s 5 — . • .)> and that cos (A 1 + A 2 + ... + A n ) = cos A 1 cos -4 2 ... cos A n (1 — s 2 + s 4 — ...), where s lf s 2y s 3) ... have the same values as in that article. 127. Identities holding between the trigono¬ metrical ratios of the angles of a triangle. When three angles A , B and G , are such that their sum is 180°, many identical relations are found to hold between their trigonometrical ratios. The method of proof is best seen from the following examples. Ex. 1. If A + B + C= 180°, to prove that sin 2A + sin 2 B + sin 2C = 4 sin A sin B sin C. sin 2^4 + sin 2 B + sin 2C = 2 sin (A+B) cos (A - B) + 2 sin C cos C. o Since A + B + (7 = 180°, A +R = 180° - C, sin (A + B) = sin C, cos (A + B)~ - cos G. we have and therefore and (Art. 72) IDENTITIES. 135 Hence the expression = 2 sin C cos [A - B) + 2 sin C cos C = 2 sin C [cos (A - B) + cos G ] = 2 sin C [cos (A - B) - cos (A + B)] — 2 sin C . 2 sin A sin B = 4 sin A sin B sin C. Ex. 2. 7/ j:+£+< 7=180 o , , < -r\ t i A B . C prove that cos A + cos - cos C= -1 + 4 cos — cos — sin — . Z Z Z The expression Now so that and therefore and = cos A + (cos B - cos C) o 2 A i , o • B + c • C ~ B = 2 cos- — - 1 + 2 sin —-— sin —-— 2 2 2 B + C— 180° ~ A f ^= 90 °-^- . B + C A sm - =cos—, B + G . A cos 2 = sm 2 ' Hence the expression 0 n A - _ A . C — B = 2 cos 2 — - 1 + 2 cos — sin ——- z z z 0 ^ r A , • 2 cos — I cos —+ sin—- P 1- . ^ r • B + G • c-B-i 2cos 2 L Sm_ 2“ + Sm ^ _ J O A O * C B 1 = 2 cos — . 2 sin — cos — - 1 z z z 1 . 4 B . <7 -1 + 4 cos — cos — sin — . z z z Ex. 3. If A+B + G = 180°, prove that sin 2 A + sin 2 B + sin 2 (7 = 2 + 2 cos A cos B cos G. 136 TRIGONOMETRY. Let S =sin 2 A + sin 2 B + sin 2 C, so that 2&=2sin 2 _4 + l-cos2B + l-cos2(7 = 2 sin 2 A + 2-2 cos (B + C) cos (B - C) = 2-2 cos 2 A + 2-2 cos (B + C) cos (B - C). S = 2 + cos A [cos (B - C) + cos (B+C)], since cos A — cos {180° - (B + C)} = - cos (B+C). £ = 2 + cosA. 2 cos B cos C. = 2 + 2 cos A cos B cos C. Ex. 4. If A+B+C= 180°, prove that tan A + tan B + tan C =tan A tan B tan C. By the third formula of Art. 124, we have tan A + tan B + tan C - tan A tan B tan C an ( + + — i _ (t an b tan C + tan C tan A + tan A tan B) * But tan (A + B + C)- tan 180° = 0. Hence 0 = tan A + tan B + tan C - tan A tan B tan C , i.e. tan A + tan B + tan C = tan A tan B tan C. This may also be proved independently. For tan (A + B) = tan (180° - C) = - tan C. tan A + tan B , „ ^—i- 7 T-- = - tan C. _ 1 - tan A tan B tan A + tan B = - tan C + tan A tan B tan C, i.e. tan A + tan B + tan C = tan A tan B tan C. Ex. 5. If x + y + z=xyz, prove that 2x 2 y 2 z _ 2x 2 y 2 z 1 - x 2 1 - y 2 1 - z 2 1 - x 2 * 1 - y 2 ’ 1 z 2 * Put x = tan A, y = tanB, and z — tan (7, so that we have tan A + tan B + tan C = tan A tan B tan C. tan A + tan B , - • -= — tan C ' 1 - tan A tan B tan (A + B) = tan ( 7 r- C). so that [Art. 72.] Hence 2x 1-x 2 ^ \ -y 2 ' 1-z 2 1 - tan 2 A 1 - tan 2 B ' 1 - tan 2 C = tan 2A -f tan 2B 4- tan 2(7= tan 2 A tan 2 B tan 2(7, (by a proof similar to that of the last example) 2 y 4- IDENTITIES. A 4- B + C — nir + ir, 2 z 2 tan A 2 tan B 2 tan G 4* - 1 — o + 2x 2 y 2 z 1 - x 2 * 1 - y 2 * 1 - z 2 ' EXAMPLES. XX. If A 4- B 4- <7 =180°, prove that 1. sin 2A + sin 2 B - sin 2(7 —4 cos A cos B sin C. 2. cos 2 A 4 - cos 2 B 4- cos 2 C— - 1 - 4 cos A cos B cos (7. 3. cos 2A 4- cos 2 B - cos 2(7= 1 - 4 sin A sin B cos (7. * ABC 4. sin A 4- sin B + sin (7=4 cos cos cos jr . Li Li Li A JB C 5. sin A 4-sin B - sin (7 = 4 sin — sin — cos —. Z Z Z ABC 6. cos A 4- cos B 4- cos (7 =14-4 sin — sin — sin — . Z Z Z 7. sin 2 A + sin 2 B - sin 2 (7 = 2 sin A sin B cos C. 8. cos 2 A 4- cos 2 B 4- cos 2 (7 = 1-2 cos A cos B cos (7. 9. cos 2 A 4- cos 2 B - cos 2 (7=1-2 sin A sin B cos (7. 10. sin 2 ^ 4- sin 2 ~ 4- sin 2 ^= 1 - 2 sin ^ sin ^ sin ^. ii • 9 -^ • *> B • i o ^ B . G 11. sin- - + sm 2 — -sm-— = 1 -2 cos ^cos-^ sin —. Z Z Z Z Z Z _ n . A. B . B, G , C. A , 12. tan — tan — + tan — tan - 4- tan — tan — = 1. Li Li Z Li Z Li io A , B , c . A . B 13. cot — 4-cot— +cot— = cot — cot g cot 2 . 14. cot B cot G 4- cot (7 cot A + cot A cot B = 1. 137 138 TRIGONOMETRY. [Exs. XX.] 15. sin (B + 2 C) + sin (C + 2A) + sin (A + 2 B) . . B-C . C-A . A-B = 4 — 2 “ sm -y- sin . 16. 17. 18. . A . B . C ^ . . 7T — A . 7T — B . 7 T—C sin — + sin — + sin --1 = 4 sin —-— sin —— sin —-— . 2 2 2 4 4 4 sin 2A + sin 2B + sin 2(7 _ . A . B . C —;—.-=—--:—— = 8 sin — sin - sin-. sin A + sin B + sm G 2 2 2 sin (B + C - A) + sin (C + A - B) + sin (A + B - C) = 4 sin A sin B sin C. If A + B + G— 2S prove that 19. sin (S - A) sin (S - B) + sin S sin (S - C) = sin A sin B. 20. sin S sin (S - A) sin (S - B) sin (S - G) = 1 - cos 2 A - cos 2 B - cos 2 C + 2 cos A cos B cos C. 21. sin (S - A) + sin (S - B) + sin (S - G) - sin S = 4 sin A . B . C 2 sm 2 Sm 2 - 22. cos 2 S +cos 2 (S - A) + cos 2 (S - B) + cos 2 (S-C) = 2 + 2 cos A cos B cos C. 23. cos 2 A + cos 2 B + cos 2 G + 2 cos A cos B cos C = 1 + 4 cos S cos (S -- A) cos (S - B) cos (S-C). 24. If a + /3 + 7 + 5 = 27r, prove that a + B a + y a + 5 cos a + cos p + cos 7 + cos 5 + 4 cos — ~ cos — Q cos ——- = 0, Ld Li a • • o. • . * . a+B . a +7 a+5 and sm a - sm p + sm 7 - sm 5 + 4 cos —sm —cos —= 0. Li Ll Ll 25. If the sum of four angles be 180° prove that the sum of their cosines taken two and two together is equal to the sum of their sines taken similarly. Prove 26. 1 - COS 2 6 - COS 2 (p - COS 2 Xp+2 COS 6 COS

, and C such that A+R + C=180°, prove that the theorem is still true if we substitute for A, B , and G respectively the quantities (1) 90°-^, 90 o -|,and90 o -|, or (2) 180° -2A, 180°- 2B, and 180°- 2C. If x + y + z = xyz prove that 34 . 3.c - x s 3 y - y 3 1 - 3x'' J + I-3 y 3 + 3z - z z _ 3x - x z 3 y - y 3 1 - 3z 2 = 1^3x 2 ' 1 - 3y* * 3z - z 5 1 - 3s 2 and 35 . x (1-y 2 ) (1 - z 2 )+ y (1 - z 2 ) (1 - x 2 )+ z (1 - x 2 ) (1 - y 2 ) = 4xyz. 128. The Addition and Subtraction Theorems may be used to solve some kinds of trigonometrical equations. Ex. Solve the equation sin x + sin 5x = sin 3#. 140 TRIGONOMETRY. By the formulae of Art. 94 the equation is 2 sin 3x cos 2x = sin 3x. .*. sin 3x = 0, or 2 cos 2x = 1. If sin 3x = 0, then 3x = mr. If cos 2x = ^, then 2x = 2 mr ± ^ ^ o Hence ?17T 7r # = — , or ?i7r + 77 . o b 129. To sofoe an equation of the form a cos 6 + b sin 6 — c. Divide both sides of the equation by Va 2 + 6 2 , so that it may be written b a V a 2 + b 2 cos 6 + Va 2 + b 2 sin 6 = V a 2 + b 2 Find from the table of tangents the angle whose tangent is — and call it a. 6 a Then tan a = -, so that a sm a = fa 2 + b 2 , and cos a = a Va 2 + b' 1 The equation can then be written cos a cos 6 + sin a sin 0 = Va 2 4- b 2 i.e . cos (0 — a) = Va 2 + 6- GRAPHIC SOLUTION. 141 Next find from the tables, or otherwise, the angle /? whose cosine is . , v a 2 + b 2 so that cos /3 — c Va 2 + b 2 9 [N.B. This can only be done when c is <\/a 2 + b 2 .] The equation is then cos (9 — a) = cos /3. The solution of this is 9 — a — %mr + /3, so that 9 = 2 ? 77 T + ol + /3, where n is any integer. Angles, such as a and /3, which are introduced into trigonometrical work to facilitate computation are called Subsidiary Angles. 130. The above solution may be illustrated graphically as follows; Measure OM along the initial line equal to a, and MP perpen¬ dicular to it, and equal to b. The angle MOP is then the angle whose tangent is -, i.e. a. With centre 0 and radius OP, i.e. Va 2 4 - b 2 , describe a circle and measure ON along the initial line equal to c. Draw QNQ perpendicular to ON to meet the circle in Q and Q f ; the angles NOQ and Q'ON are therefore each equal to /3. The angle QOP is therefore a — /3 and Q'OP is a + /3. 142 TRIGONOMETRY. Hence the solutions of the equation are respectively 2nir + QOP and 2nir 4- Q' OP. The construction clearly fails if c be > Va 2 + b 2 , for then the point N would fall outside the circle. 131. As a numerical example let us solve the equation 5 cos 0 - 2 sin 0 = 2. given that tan 21° 48'= —. Dividing both sides of the equation by J& + 2 2 i-e. V 2 9. we have J29 COS0 ~ &9 sine = ^- Hence cos 0 cos 21° 48' - sin 0 sin 21° 48' = sin 21° 48' = sin (90 - 68° 12') = cos 68° 12'. cos (^+21° 48') = cos 68° 12'. Hence 6 + 21° 48' = 2?itt± 68° 12'. (Art. 83) 0 = 2mr - 21° 48' ± 68° 12' = 2?i7r-^ or 2n7r + 46° 24', where n is any integer. EXAMPLES. XXI. Solve the equations 1. sin 0 + sin 10 = sin 40. 3. cos 0 + cos 30 = 2 cos 20. 5. cos 0 - sin 30 = cos 20. 7 . cos 0 + cos 20 + cos 30 = 0. 9. sin 20 - cos 20 - sin 0 +cos 0 = 0. 2. cos 0-h cos 70 = cos 40. 4. sin 40 - sin 20 = cos 30. 6. sin 70 = sin 0 + sin 30. 8. sin 0 +sin 30 +sin 50 = 0. 10 . sin (30 + a) + sin (30 - a) + sin (a - 0) - sin (a + 0) = cos a. [Exs. XXI.] TRIGONOMETRICAL EQUATIONS. 143 11 . 12 . 14. 16. 18. 20 . 22 . 23. 24. 25. 27. 29. 31. 33. 34. 36. 37. 38. 39. 40. either -.o - n+l. .n -1 . . . 13. sin —0 = sm —— 04 -sin 0. 15. cos md + cos n 6 = 0. 17. sin 30 + cos 20 = 0. 19. sin 0 4-cos 0 = ^2. 21 . sin x 4- cos x = 2 cos A . cos (30 4- a) cos (30 - a) 4- cos (56 4- a) cos (56 - a) = cos 2a. cos n 6 = cos (n - 2) 6 4- sin 6 . sin md + sin n 6 = 0. sin 2 116 - sin 2 (n - 1) 6 = sin 2 6 . V^cos 6 + sin 6 = sin 6 - cos 6 = ^2. 5 sin 6 4- 2 cos 6 = 5 (given tan 21° 48' = *4). 6 cos x 4- 8 sin x = 9 (given tan 53° 8' = 1^). 1 + sin 2 6 = 3 sin 6 cos 6 (given tan 71° 34' = 3). cosec 6 = cot 6 + 26. cosec x = 1 4- cot x. (2 +a/3) cos 6 = 1 - sin 6 . 28. tan 0 +sec 0=*/3. cos 20 = cos 2 6 . 30. 4 cos 0 - 3 sec 0 = tan 0. cos 20+ 3 cos 0 = 0. 32. cos 30+ 2 cos 0 = 0. cos 20= (^2 + 1) ^cos 0— . cot 0-tan 0 = 2. 35. 4 cot 20 = cot 2 0 - tan 2 0. 3 tan (0 - 15°) = tan (0 + 15°). tan 0 + tan 20 4- tan 30 = 0. tan 0 -f tan 20 + ^/3 tan 0 tan 20 = ^3. sin 3a = 4 sin a sin (x 4- a) sin (x - a). Prove that the equation x 3 - 2x 4-1 = 0 is satisfied by putting for x of the values v /2 sin 45°, 2 sin 18°, and 2 sin 234°. 132. Ex. To trace the changes in the sign and magnitude of the expression sin 6 4- cos 6 as 6 increases from 0 to 360°. We have sin 6 4- cos 6 = f2 = *J2 [sin 6 cos 45° + cos 6 sin 45°] = f2 sin (0 + 45°). —7- sin 6 4 —jzr cos 6 'Y Jj 2 144 TRIGONOMETRY. As 0 increases from 0 to 45°, sin (0 + 45°) increases from sin 45° to sin 90°, and hence the expression increases from 1 to V2. As 0 increases from 45° to 135°, 0 + 45° increases from 90° to 180°, and hence the expression is positive and decreases from f2 to 0. As 0 increases from 135° to 225°, the expression changes from f2 sin 180° to f2 sin 270°, i.e. it is negative and decreases from 0 to — f2. As 0 increases from 225° to 315°, the expression changes from f2 sin 270° to f2 sin 360°, i.e. it is negative and increases from — fi2 to 0. As 0 increases from 315° to 360°, the expression changes from V2sin 360° to f2 sin 405°, i.e. it is positive and increases from 0 to 1. 133. Ex. To trace the changes in the sign and magnitude of a cos 0 + b sin 0 , and to find the greatest value of the expression. We have a cos 0 -\-b sin 0 = fa 2 4- b 2 a fa? + b 2 COS 0 -f f a 2 + b‘ sin 0 Let a be the smallest positive angle such that a cos a = . . V a? + b 2 and sin a = . ----- . V a 2 + b 2 The expression therefore = fa 2 + b 2 [cos 0 cos a + sin 0 sin a] = fa 2 + b 2 cos (0 — a). As 0 changes from a to 360° -f- a, the angle 0 — a changes from 0 to 360°, and hence the changes in the sign and magnitude of the expression are easily obtained. MAXIMUM VALUE. 145 Since the greatest value of the quantity cos (0 — a) is unity, i.e. when 6 equals a, the greatest value of the expression is V a 2 -t- b 2 . Also the value of 6 which gives this greatest value is such that its cosine is .. a _ Va 2 + b 2 * EXAMPLES. XXII. As 0 increases from 0 to 360°, trace the changes in the sign and magnitude of 1. sin 0- cos 0, 2. sin 0 + ^3 cos d, j^N. B. sin 0 + ^/3 cos 6 = 2 sin d + ^ cos = 2 sin (0 + 6O°).^j 3 . sin 6 - JS cos 6 . 5. sin 6 cos 6 , 4. cos 2 6 - sin 2 6 . sin 6 + sin 26 cos 6 + cos 26 ' 7 . sin (?r sin 6 ). 8. cos ( 7 r sin 6 ). sin {tt cos 6) 9. cos(7rsin 6 ) * 10 . Trace the changes in the sign and magnitude of angle increases from 0 to 90°. sin 30 cos 20 as the L. T. 10 CHAPTER X. LOGARITHMS. 134. Supposing that we know that 1 O' 2 - 4031205 = 253, 10 2 - 6095944 = 407, and 10 3 - 0127149 = 102971, we can shew that 253 x 407 — 102971 without performing the operation of multiplication. For 253 x 407 = 10 2 - 4031205 x 10 2 - 6095944 _ ^02-4031205+2.6095944 = 10 5 * 0127149 = 102971. Here it will be noticed that the process of multiplica¬ tion has been replaced by the simpler process of addition. Again supposing that we know that 10 4 -9oo4055 = 79507, and that 10 1 - 6334685 = 43, we can easily shew that the cube root of 79507 is 43. For ^79507 = [79507] J = (lO 4 * 9004055 ) 4 _ ^ Q 3 X 4 -9004055 _ 10 1 -6334685 43. Here it will be noticed that the difficult process of extracting the cube root has been replaced by the simpler process of division. LOGARITHMS. 147 135. Logarithm. Def. If a be any number and x and N two other numbers such that a x = N, then x is called the logarithm ofN to the base a and is written log a N. Exs. Since 10 2 =100, therefore 2 = log 10 100. Since 10 5 —100000, therefore 5 = log 10 100000. Since 2 4 =16, therefore 4 = log 2 16. Since 83 = [8i] 2 = 2 2 =4, therefore | = log 8 4. o Since 9"t 11 9 | 3 s 1 27 , therefore N.B. Since a° = 1 always, the logarithm of unity to any base is always zero. 136. In Algebra, if m and n be any real quantities whatever, the following laws, known as the laws of indices, are found to be true: (i) a m x a n — a m+n , (ii) d ,n - 5 - a n = a m ~ n , and (iii) {a m ) n = a mn . Corresponding to these we have three fundamental laws of logarithms, viz. (i) log a (mn) = log a m + log a n, (ii) log a = log a m - log a n, and (iii) log a m n = n log a m. The proofs of these laws are given in the following articles. 10—2 148 TRIGONOMETRY. 137. The logarithm of the product of two quantities is equal to the sum of the logarithms of the quantities to the same base, i.e. . log a (mn) = log a m + log a n. Let x = loga m > so that a x = m, and y = loga n, so that a y = n. Then mn = a x x a y = a xJry . :. loga mn — x-vy (Art. 135, Def.) = loga m + loga n. 138. The logarithm of the quotient of two quantities is equal to the difference of their logarithms , i.e . Let and Then = log a m - log a n. x = loga m, so that a x = m, (Art. 135, Def.) y = loga n , so that a y = n. m - = a x +-a y = a x ~ y . n = x — y (Art. 135, Def.) = loga m ~ loga n. 139. The logarithm of a quantity raised to any power is equal to the logarithm of the quantity multiplied by the index of the power , i.e. log a (m n ) = n log a m. Let x = loga so that a x = m. Then m n = (a x ) n = a nx . lo ga(m n ) = nx (Art. 135, Def) = n loga Wi. LOGARITHMS. 149 140. Common system of logarithms. In the system of logarithms which we practically use the base is always 10, so that, if no base be expressed, the base 10 is always understood. The advantage of using 10 as the base is seen in the three following articles. 141. Characteristic and Mantissa. Def. If the logarithm of any number be partly integral and partly fractional, the integral portion of the logarithm is called its characteristic and the decimal portion is called its mantissa. Thus, supposing that log 795 = 2 9003671, the number 2 is the characteristic and *9003671 is the mantissa. Negative characteristics . Suppose we know that log 2 = *30103. Then log \ = log 1 — log 2 = 0 — log 2 = — *30103, so that log \ is negative. Now it is found convenient, as will be seen in Art. 143, that the mantissse of all logarithms should be kept positive. We therefore instead of —*30103 write - [1 — *69897], so that log i = - (1 - *69897) = - 1 + *69897. For shortness this latter expression is written 1*69897. The horizontal line over the 1 denotes that the integral part is negative; the decimal part however is positive. As another example 3*4771213 stands for -3 +*4771213. 142. The characteristic of the logarithm of any number can always be determined by inspection. 150 TRIGONOMETRY. (i) Let the number be greater than unity. Since 10° = 1, therefore log 1 = 0 ; since 10 1= =10, therefore log 10 =1; since 10 2 = 100, therefore log 100 = 2, and so on. Hence the logarithm of any number lying between 1 and 10 must lie between 0 and 1, that is, it will be a decimal fraction and therefore have 0 as its characteristic. So the logarithm of any number between 10 and 100 must lie between 1 and 2, i.e. it will have a characteristic equal to 1. Similarly the logarithm of any number between 100 and 1000 must lie between 2 and 3, i.e. it will have a characteristic equal to 2. So, if the number lie between 1000 and 10000, the characteristic will be 3. Generally, the characteristic of the logarithm of any number will be one less than the number of digits in its integral part. Exs. The number 296*3457 has 3 figures in its integral part and therefore the characteristic of its logarithm is 2. The characteristic of the logarithm of 29634*57 will be 5-1, i.e. 4. (ii) Let the number be less than unity. Since 10°= 1, therefore log 1 = 0 ; since 10 _1 = —= *1, therefore log *1 = — 1 since 10 -2 = i — ’01, therefore log *01 = —2; 10 2 1 10 since 10 -3 = ^ = *001, therefore log’001 and so on. CHARACTERISTIC OF ANY LOGARITHM. 151 The logarithm of any number between 1 and T there¬ fore lies between 0 and — 1, and so is equal to — 1 + some decimal, i.e. its characteristic is 1. So the logarithm of any number between T and *01 lies between — 1 and — 2, and hence it is equal to — 2 -f some decimal, i.e. its characteristic is 2. Similarly the logarithm of any number between *01 and *001 lies between — 2 and — 3, i.e. its characteristic is 3. Generally, the characteristic of the logarithm of any decimal fraction ivill he negative and numerically will he greater hy unity than the number of cyphers following the decimal point. For any fraction between 1 and ‘I {e.g. *5) has no cypher following the decimal point and we have seen that its characteristic is 1. Any fraction between ‘I and *01 {e.g. *07) has 1 cypher following the decimal point and we have seen that its characteristic is 2. Any fraction between *01 and ‘001 {e.g. *003) has two cyphers following the decimal point and we have seen that its characteristic is 3. Similarly for any fraction. Exs. The characteristic of the logarithm of the number *00835 is 3. The characteristic of the logarithm of the number *0000053 is 6. The characteristic of the logarithm of the number *34567 is 1. 143. The mantissce of the logarithm of all numbers , consisting of the same digits , are the same. This will be made clear by an example. Suppose we are given that log 66818 = 4*8248935. 152 TRIGONOMETRY. Then log 668T8 = log 66818 100 = log 66818 — log 100 (Art. 138) = 4 8248935 - 2 = 2-8248935 ; S log -66818 = log -~T p = log 66818 - log 100000 (Art. 135) = 4-8248935 - 5 = 1-8248935. So log -00066818 = log = log 66818 - log 10 8 = 4-8248935 - 8 = 4-8248935. Now the numbers 66818, 66818, *66818, and *00066818 consist of the same significant figures and only differ in the position of the decimal point. We observe that their * logarithms have the same decimal portion, i.e. the same mantissa, and they only differ in the characteristic. The value of this characteristic is in each case deter¬ mined by the rule of the previous article. It will be noted that the mantissa of a logarithm is always positive. 144. Tables of logarithms. The logarithms of all numbers from 1 to 108000 are given in Chambers’ Tables of Logarithms. Their values are there given correct to seven places of decimals. The student should have access to a copy of the above table of logarithms or to some other suitable table. It will be required for many examples in the course of the next few chapters. On the opposite page is a specimen page selected from Chambers’ Tables. It gives the mantissae of the logarithms of all whole numbers from 52500 to 53000. No. 01234 56789 Diff. 5250 7201593 1676 1758 1841 1924 2007 2089 2172 2255 2337 51 2420 2503 2586 2668 2751 2834 2916 2999 3082 3164 52 3247 3330 3413 3495 3578 3661 3743 3826 3909 3991 53 4074 4157 4239 4322 4405 4487 4570 4653 4735 4818 54 4901 4983 5066 5149 5231 5314 5397 5479 5562 5645 55 5727 5810 5892 5975 6058 6140 6223 6306 6388 6471 56 6554 6636 6719 6801 6884 6967 7049 7132 7215 7297 57 7380 7462 7545 7628 7710 7793 7875 7958 8041 8123 58 8206 8288 8371 8454 8536 8619 8701 8784 8867 8949 59 9032 9114 9197 9279 9362 9445 9527 9610 9692 9775 60 9857 9940 0023 0105 0188 0270 0353 0435 0518 0600 5261 7210683 0766 0848 0931 1013 1096 1178 1261 1343 1426 62 1508 1591 1674 1756 1839 1921 2004 2086 2169 2251 63 2334 2416 2499 2581 2664 2746 2829 2911 2994 3076 64 3159 3241 3324 3406 3489 3571 3654 3736 3819 3901 65 3984 4066 4149 4231 4314 4396 4479 4561 4644 4726 66 4809 4891 4973 5056 5138 5221 5303 5386 5468 5551 67 5633 5716 5798 5881 5963 6045 6128 6210 6293 6375 68 6458 6540 6623 6705 6787 6870 6952 7035 7117 7200 69 7282 7364 7447 7529 7612 7694 7777 7859 7941 8024 70 8106 8189 8271 8353 8436 8518 8601 8683 8765 8848 on 5271 8930 9013 9095 9177 9260 9342 9424 9507 9589 9672 82 72 9754 9836 9919 0001 0084 0166 0248 0331 0413 0495 1 8 *2 Ifi 73 7220578 0660 0742 0825 0907 0990 1072 1154 1237 1319 3 25 74 1401 1484 1566 1648 1731 1813 1895 1978 2060 2142 4 33 75 2225 2307 2389 2472 2554 2636 2719 2801 2883 2966 5 41 76 3048 3130 3212 3295 3377 3459 3542 3624 3706 3789 6 49 *7 7*7 77 3871 3953 4036 4118 4200 4282 4365 4447 4529 4612 7 o 7 78 4694 4776 4858 4941 5023 5105 5188 5270 5352 5431 o OO 9 74 79 5517 5599 5681 5763 5846 5928 6010 6092 6175 6257 80 6339 6421 6504 6586 6668 6750 6833 6915 6997 7079 5281 7162 7244 7326 7408 7491 7573 7655 7737 7820 7902 82 7984 8066 8148 8231 8313 8395 8477 8559 8642 8724 83 8806 8888 8971 9053 9135 9217 9299 9382 9464 9546 84 9628 9710 9792 9875 9957 0039 0121 0203 0286 0368 85 7230450 0532 0614 0696 0779 0861 0943 1025 1107 1189 86 1272 1354 1436 1518 1600 1682 1765 1847 1929 2011 87 2093 2175 2257 2340 2422 2504 2586 2668 2750 2832 88 2914 2997 3079 3161 3243 3325 3407 3489 3571 3654 89 3736 3818 3900 3982 4064 4146 4228 4310 4393 4475 90 4557 4639 4721 4803 4885 4967 5049 5131 5213 5296 5291 5378 5460 5542 5624 5706 5788 5870 5952 6034 6116 92 6198 6280 6362 6445 6527 6609 6691 6773 6855 6937 93 7019 7101 7183 7265 7347 7429 7511 7593 7675 7757 94 7839 7921 8003 8085 8167 8250 8332 8414 8496 8578 95 8660 8742 8824 8906 8988 9070 9152 9234 9316 9398 96 9480 9562 9644 9726 9808 9890 9972 0054 0136 0218 97 7240300 0382 0464 0546 0628 0710 0792 0874 0956 1038 98 1120 1202 1283 1365 1447 1529 1611 1693 1775 1857 99 1939 2021 2103 2185 2267 2349 2431 2513 2595 2677 5300 2759 2841 2923 3005 3086 3168 3250 3332 3414 3496 154 TRIGONOMETRY. 145. To obtain the logarithm of any such number, such as 52687, we proceed as follows. Run the eye down the extreme left-hand column until it arrives at the number w 5268. Then look horizontally until the eye sees the figures 7035 which are vertically beneath the number 7 at the top of the page. The number corresponding to 52687 is there¬ fore 7217035. But this last number consists only of the digits of the mantissa, so that the mantissa required is *7217035. But the characteristic for 52687 is 4. Hence log 52687 = 4-7217035. So log *52687 = 1*7217035, and log *00052687 = 4*7217035. If again the logarithm of 52725 be required the student will find (on running his eye vertically down the extreme left-hand column as far as 5272 and then horizontallv w along the row until he comes to the column under the digit 5) the number 0166. The bar which is placed over these digits denotes that to them must be prefixed not 721 but 722. Hence the mantissa corresponding to the number 52725 is *7220166. Also the characteristic of the logarithm of the number 52725 is 4. Hence log 52725 = 4*7220166. So log *052725 = 2*7220166. We shall now work a few numerical examples to shew the efficiency of the application of logarithms for purposes of calculation. 146. Ex. Find the value of ^/*23*4. Let x = = so that 1 i 5 logz = glog (23-4), by Ait. 139. EXAMPLES OF LOGARITHMS. 155 In tlie table of logarithms we find, opposite the number 234, the logarithm 3692159. Hence log 23*6 = 1-3692159. Therefore log x = = [1-3692159] = -2738432. Again in the table of logarithms we find, corresponding to the logarithm 2738432, the number 187864, so that log 1-87864 = -2738432. .-. £=1-87864. Ex. 2. Find the value of (6-45 ) 3 x ^/-U0034 (9"37 ) 2 x 4/8^ ' Let x be the required value so that, by Arts. 138 and 139, log x = log (6-45 ) 3 + log (-00034)3 - log (9*37 ) 2 - log ^8-93 = 3 log (6-45) + hog (-00034) - 2 log (9-37) - \ log 8-93. o 4 Now in the table of logarithms we find opposite the number 645 the logarithm 8095597, „ ,, ,, 34 „ ,, 5314789, „ „ „ 937 „ „ 9717396, „ „ „ 893 „ „ 9508515. Hence loga = 3x-8095597+ 5 (4*5314789) o - 2 x -9717396 - = x -9508515. 4 But 5 (4-5314789) = 5 [ 6 + 2-5314789] D O = 2+ *8438263. .-. log £=2-4286791 + [2 + -8438263] -1-9434792 - -2377129 = 3-2725054-4-1811921 = 1 + 4-2725054 - 4-1811921 = 1-0913133. 156 TRIGONOMETRY. In the table of logarithms we find, opposite the number 12340 the logarithm 0913152, so that . . . log *12340=1*0913152. Hence log x = log *12340 nearly, and therefore x = *12340 nearly. When the logarithm of any number does not quite agree with any logarithm in the tables but lies between two consecutive logarithms, it will be shewn in the next chapter how the number may be accurately found. Ex. 3. Having given log 2 = *30103, find the number of digits in 2 67 and the position of the first significant figure in 2 -37 . We have log 2 67 = 67 x log 2 = 67 x *30103 = 20*16901. Since the characteristic of the logarithm of 2 67 is 20 it follows, by Art. 142, that in 2 67 there are 21 digits. Again log2" 37 = - 37log 2= - 37 x *30103 = -11*13811 = 12*86189. Hence by Art. 142, in 2 -37 there are 11 cyphers following the decimal point, i.e. the first significant figure is in the twelfth place of decimals. Ex. 4. Givenlog3 = -4:7n213 i log7 = - 8450980 and log 11 = 1*0413927, solve the equation 3 x x 7 2x +i = 11 x+ 5. Taking logarithms of both sides we have log 3 X + log 7 2x+1 = log ll x + 5 . .*. x log 3 + {2x + 1) log 7= (x + 5) log 11. .*. x [log 3 + 2 log 7 - log 11] = 5 log 11 - log 7. 5 log 11 - log 7 X “ log 3 + 2 log 7 - log 11 5*2069635 - *8450980 “ *4771213 +1 *6901960 - 1*0413927 _ 4^618655 “1*1259246 LOGARITHMS TO DIFFERENT BASES. 157 147. To prove that log a m = log b m x log a b. Let log a m — x, so that a x = m. Also let log& m = y , so that b y = m. /. a x = b y . Hence l°ga (O = lo g a (b y ). .*. x — y log a b (Art. 139). Hence lo g a m = log & m x log a b. By the theorem of the foregoing article we can from the logarithm of any number to a base b find its logarithm to any other base a. It is found convenient, as will appear in a subsequent chapter, not to calculate the logarithms to base 10 directly, but to calculate them first to another base and then to transform them by this theorem. EXAMPLES. XXIII. 1. Given log 4 = *60206 and log 3= *4771213, find the logarithms of •8, *003, *0108, and (*00018)L 2 . Given log 11 = 1*0413927 and log 13 = 1*1139434, find the values of (1) log 1*43, (2) log 133*1, (3) log \/l43 and (4) log 4^00169. 3 . What are the characteristics of the logarithms of 243*7, *0153, 2-8713, -00057, -023, 4/24615, and (24589)5? 4 . Find the 5th root of -003, having given log 3 = -4771213 and log 312936 = 5-4954243. 5 . Find the value of (1) 7^, (2) (84)5 and (3) (-021)v, having given log 2 = -30103, log 3 = -4771213, log 7= -8450980, log 132057 = 5-1207283, log 588453 = 5-7697117 and log 461791 = 5-6644438. 158 TRIGONOMETRY. [Exs. XXIII.] 6 . Having given log 3 = *4771213, find the number of digits in (1) 3 43 , (2) 3 2 7, and (3) 3 62 , and the position of the first significant figure in (4) 3 -13 , (5) 3 -43 , and ( 6 ) 3" 65 . 7 . Given log 2 = *30103, log 3 = *4771213 and log 7 = *8450980, solve the equations 2 X . 3 x+i =7 x , 22as+l . 333+2 _. 743 and 723 _^ 2 *- 4 = 3 3 *- 7 . 8 . From the tables find the seventh root of *000026751. Making use of the tables find the approximate values of 9 . 4/645 -3. 10 . ^82357. „ y/5x4/7 4/8 x 4/9 • 7-2 x 8-3 9 '4 ^ 16-5' 8^ x 11T ^74 x ^62 ’ / CHAPTER XI. TABLES OF LOGARITHMS AND TRIGONOMETRICAL RATIOS. PRINCIPLE OF PROPORTIONAL PARTS. 148. We have pointed out that the logarithms of all numbers from 1 to 108000 may be found in Chambers’ «/ Mathematical Tables, so that, for example, the logarithms of 74583 and 74584 may be obtained directly therefrom. Suppose however we wanted the logarithm of a number lying between these two, e.g. the number 74583*3. To obtain the logarithm of this number we use the Principle of Proportional Parts which states that the increase in the logarithm of a number is proportional to the increase in the number itself. Thus from the tables we find log 74583 = 4*8726398.(1), and log 74584 = 4*8726457 .(2). The quantity log 74583*3 will clearly lie between log 74583 and log 74584. Let then log 74583*3 = log 74583 + x = 4*8726398 + x (3). 160 TRIGONOMETRY. From (1) and (2) we see that for an increase 1 in the number the increase in the logarithm is •0000059. The Theory of Proportional Parts then states that for an increase of - 3 in the number the increase in the logarithm is •3 x 0000059, i.e., ‘00000177. Hence log 74583'3 = 4-8726398 + '00000177 = 4-87264157. 149. As another example we shall find the value of log '0382757 and shall exhibit the working in a more concise form. From the tables we obtain log -038275 = 2-5829152 log-038276 = 2-5829265. Hence difference for •000001 = -0000113. Therefore the difference for •0000007 =-7 x -0000113 = •00000791, log -0382757 = 2 5829152 + -00000791 = 2-58292311. Since we only require logarithms to seven places of decimals we omit the last digit and the answer is 25829231. 150. The converse question is often met with, viz., to find the number whose logarithm is given. If the logarithm be one of those tabulated the required number is easily found. The method to be followed when this is not the case is shewn in the following examples. PROPORTIONAL PARTS. 161 Find the number whose logarithm is 2*6283924. On reference to the tables we find that the logarithm 6283924 is not tabulated, but that the nearest logarithms are 6283889 and 6283991, between which our logarithm lies. We have then log 425*00 = 2*6283889.(1), and log 425*01 = 2*6283991.(2). Let log (425*00 + x) = 2*6283924.(3). From (1) and (2) we see that corresponding to a difference *01 in the number there is a difference *0000102 in the logarithm. From (1) and (3) we see that corresponding to a difference x in the number there is a difference *0000035 in the logarithm. Hence we have x : *01 :: *0000035 : *0000102. 35 *35 /. X =IQ2 X 01 “ 102“ 00343 nearl y* Hence the required number = 425*00 + *00343 = 425*00343. 151. Where logarithms are taken out of the tables the labour of subtracting successive logarithms may be avoided. On reference to page 153 there is found at the extreme right a column headed Diff. The number 82 at the head of the figures in this column gives the difference corresponding to a difference unity in the numbers on that page. This number 82 means *0000082. The rows below the 82 give the differences correspond¬ ing to T, * 2 ,.... Thus the fifth of these rows means that the difference for *5 is *0000041. As an example let us find the logarithm of 52746*74. From page 153 we have log 52746 = 4*7221895 diff. for *7 = *0000057 diff. for *04 (= x diff. for 4^ = *0000003 L. T. log 52746*74 = 4*7221955. 11 162 TRIGONOMETRY. 152. The proof of the Principle of Proportional Parts will not be given at this stage. It is not strictly true without certain limitations. The numbers to which the principle is applied must contain not less than five significant figures, and then we may rely on the result as correct to seven places of decimals. For example, we must not apply the principle to obtain the value of log 2*5 from the values of log 2 and lo g 3. For, if we did, since these logarithms are *30103 and *4771213, the logarithm of 2*5 would be *389075. But from the tables the value of log 2*5 is found to be *3979400. Hence the result which we should obtain would be manifestly quite incorrect. Tables of trigonometrical ratios . 153. In Chambers’ Tables will be found tables giving the values of the trigonometrical ratios of angles between 0 J and 45°, the angles increasing by differences of 1'. It is unnecessary to separately tabulate the ratios for angles between 45 and 90°, since the ratios of angles between 45° and 90° can be reduced to those of angles between 0 C and 45°. (Art. 75.) For example, [sin 76° 11' = sin (90° - 13° 49') = cos 13° 49', and is therefore known]. Such a table is called a table of natural sines, cosines, etc. to distinguish it from the table of logarithmic sines, cosines, etc. PROPORTIONAL PARTS. 163 If we want to find the sine of an angle which contains an integral number of degrees and minutes we can obtain it from the tables. If, however, the angle contain seconds we must use the principle of proportional parts. Ex. 1. Given sin 29° 14'= ’4883674, and sin 29° 15' = *4886212, find the value of sin 29° 14' 32". By subtraction we have difference in the sine for l'= *0002538. QO difference in the sine for 32" = — x *0002538= -00013536, 60 .*. sin 29° 14' 32"= *4883674 + •0 0013536 = •48850276. Since we want our answer only to seven places of decimals we omit the last 6, and, since 76 is nearer to 80 than 70, we write sin 29° 14'32" = *4885028. N.B. When we omit a figure in the eighth place of decimals we add 1 to the figure in the seventh place, if the omitted figure be 5 or a number greater than 5. Ex. 2. Given cos 16° 27' = *9590672, and cos 16° 28' = *9589848, find cos 16° 27' 47". We note that, as was shewn in Art. 55, the cosine decreases as the angle increases. Hence for an increase of 1', i.e. 60", in the angle, there is a decrease of *0000824 in the cosine. Hence for an increase of 47" in the angle there is a decrease of 47 — x *0000824 in the cosine. 60 cos 16° 27' 47" = -9590672 - ^ x -0000824 oU = *9590672 - *0000645 = *9590672 -*0000645 = *9590027. 11 — 2 164 TRIGONOMETRY. 154. The inverse question, to find the angle, when one of its trigonometrical ratios is given, will now be easy. Ex. Find the angle icliose cotangent is 1*4109325, having given cot 35° 19' = 1*4114799, and cot 35° 20' = 1*4106098. Let the required angle be 35° 19' +a;, so that cot (35° 19' + x) = 1*4109325. From these three equations we have For an increase of 60" in the angle a decrease of *0008701 in the cotangent, „ „ x" „ „ „ ,, *0005474 „ „ .*. x : 60 :: 5474 : 8701, so that # = 37*7. Hence the required angle = 35° 19'37*7". 155. In working all questions involving the applica¬ tion of the Principle of Proportional Parts the student must be very careful to note whether the trigonometrical ratios increase or decrease as the angle increases. As a help to his memory he may observe that in the first quadrant the 3 trigonometrical ratios whose names begin with co-, i.e. the cosine, the cotangent, and the cosecant, all decrease as the angle increases. Tables of logarithmic sines , cosines , etc. 156. In many kinds of trigonometric calculation, as in the solution of triangles, we often require the logarithms of trigonometrical ratios. To avoid the inconvenience of first finding the sine of any angle from the tables and then obtaining the logarithm of this sine by a second application of the tables, it has been found desirable to have separate tables giving the logarithms of the various TABLES OF LOGARITHMIC SINES, ETC. 165 trigonometrical functions of angles. As before it is only necessary to construct the tables for angles between 0° and 45°. Since the sine of an angle is always less than unity, the logarithm of its sine is always negative (Art. 142). Again, since the tangent of an angle between 0° and 45° is less than unity its logarithm is negative, whilst the logarithm of the tangent of an angle between 45° and 90° is the logarithm of a number greater than unity and is therefore positive. 157. To avoid the trouble and inconvenience of print¬ ing the proper sign to the logarithms of the trigonometric functions, the logarithms as tabulated are not the true logarithms, but the true logarithms increased by 10. For example, sine 30° = Hence log sin 30° = log \ — — log 2 = — ’30103 = 1*69897. The logarithm tabulated is therefore 10 + log sin 30°, i.e. 9*69897. Again, tan 60° = \/3. Hence log tan 60° = ^ log 3 = \ (*4771213) = *2385606. The logarithm tabulated is therefore 10 + *2385606, i.e. 10*2385606. The symbol L is used to denote these “ tabular logarithms,” i.e. the logarithms as found in the English books of tables. Thus L sin 15° 25' = 10 + log sin 15° 25', and L sec 48° 23' = 10 + log sec 48° 23'. 166 TRIGONOMETRY. 158. If we want to find the tabular logarithm of any function of an angle, which contains an integral number of degrees and minutes, we can obtain it directly from the tables. If, however, the angle contain seconds we must use the principle of proportional parts. The method of procedure is similar to that of Art. 152. We give an example and also one of the inverse question. Ex. 1 . Given L cos^c 32° 21'= 10*2715733, and L cosec 32° 22'= 10*2713740, find L cosec 32° 21' 51". For an increase of 60" in the angle there is a decrease of *0001993 in the logarithm. Hence for an increase of 51" in the angle the corresponding decrease is ^ x -0001993, i.e. -0001694. Hence L cosec 32° 21' 51" = 10-2715733 - -0001694 = 10-2714039. Ex. 2. Find the angle such that the tabular logarithm of its tangent is 9*4417250. From the tables we have L tan 15° 27' = 9*4415145, and L tan 15° 28' = 9*4420062. Let L tan (15° 27' + x") = 9*4417250. We then have x" _ 9*4417250-9*4415145 60" “ 9*4420062 - 9*4415145 _ *0002105 ~ *0004917’ so that x = 60 x 2105 4917 = nearly 26. Hence the required angle is 15° 27' 26". PROPORTIONAL PARTS. 167 Ex. 3. Given L sin 14° 6' = 9 *3867040 find L cosec 14° 6'. Here log sin 14° 6 ' — L sin 14° 6' - 10 = -1 +-3867040. Now log cosec 14° 6' = log —— * sin 14° 6 = - log sin 14° 6' = 1--3867040 =-6132960. Hence L cosec 14° 6' = 10-6132960. The error to be avoided is this; the student sometimes assumes that because log cosec 14° 6' = - log sin 14° 6', he may therefore assume that L cosec 14° 6'= - L sin 14° 6'. This is obviously untrue. EXAMPLES. XXIV. 1. Given log 35705 = 4-5527290 and log 35706 = 4*5527142, find the values of log 35705-7 and log 35*70585. 2. Given log 5*8743 = -7689487 and log 587 *44 = 2*7689561, find the values of log 58743*57 and log *00587432. 3 . Given log 47847 = 4*6798547 and log 47848 = 4*6798638, find the numbers whose logarithms are respectively 2*6798593 and 3*6798617. 168 TRIGONOMETRY. [Exs. XXIV.] 4. Given log 258-36 = 2-412225 3 and log 2*5837= *4122421 find the numbers whose logarithms are •4122378 and 2*4122287. 5. From the table on page 153 find the logarithms of (1) 52538*97, ( 2 ) 527*286, (3) *000529673, and the numbers whose logarithms are (4) 3*7221098, (5) 2*7240075 and ( 6 ) *7210386. 6 . Given sin 43° 23'= -6868761 and sin 43° 24'= -6870875, find the value of sin 43° 23' 47" 7. Find also the angle whose sine is *6870349. 8 . Given cos 32° 16'= *8455726 and cos 32° 17'= *8454172, find the values of cos 32° 16' 24" and of cos 32° 16' 47". 9. Find also the angles whose cosines are *8454832 and *8455176. 10. Given and tan 76° 21'= 4-1177784 tan 76°22' = 4-1230079, find the values of tan 76° 2 T 29" and tan 76° 21 ' 47". 11, Given cosec 13° 8 '= 4-4010616 and cosec 13° 9' = 4-3955817, find the values of cosec 13° 8 ' 19'' and cosec 13° 8 ' 37''. 12. Find also the angle whose cosecant is 4*396789. 13. Given L cos 34° 44' = 9-9147729 and L cos 34° 45'=9*9146852, find the value of L cos 34° 44' 27". 14. Find also the angle 0, where L cos 0 = 9*9147328. PROPORTIONAL PARTS. 169 [Exs. XXIV.] 15. Given L cot 71° 27' = 9*5257779 and L cot 71° 28' = 9*5253589, find the value of L cot 71° 27' 47 and solve the equation L cot 6 = 9*5254782. 16. Given L sec 18° 27' = 10*0229168 and L sec 18° 28'= 10*0229590, find the value of L sec 18° 27' 35". 17. Find also the angle whose L sec is 10*0229285. 18. Find in degrees, minutes, and seconds the angle whose sine is *6, given that log 6 = 7781513, L sin 36° 52' = 9 *7781186 and L sin 36° 53' = 9 *7782870. 159. On the next page is printed a specimen page taken from Chambers’ tables. It gives the tabular log¬ arithms of the ratios of angles between 32 and 33° and also between 57 c and 58°. The first column gives the L sine for each minute between 32° and 33°. In the second column under the word Diff. is found the number 2021. This means that *0002021 is the difference between Zsin32 c O' and L sin 32° V ; this may be verified by subtracting 9*7242097 from 9*7244118. It will also be noted that the figures 2021 are printed half¬ way between the numbers 9*7242097 and 9*7244118, thus clearly shewing between what numbers it is the difference. This same column of Differences also applies to the column on its right-hand side which is headed Cosec. Similarly the fifth column, which is also headed Diff, may be used with the two columns on the right and left of it. LOGARITHMIC SINES, TANGENTS, AND SECANTS. 32 Deg*. t Sine Diff. Cosec. Tang. Diff. Cotang. Secant Diff. Cosine / 0 9*7242097 2021 2020 2018 2018 2015 2015 10-2757903 9-7957892 2811 2810 2809 2S08 2808 2807 10-2042108 10*0715795 790 790 791 791 792 792 9-9284205 60 1 9-7244118 10-2755882 9*7960703 10-2039297 10*0716585 9-9283415 59 2 9*7246138 10*2753862 9*7963513 10-2036487 10-0717375 9-9282625 58 3 9*7248156 10-2751844 9*7966322 10-2033678 10-0718166 9-9281834 57 4 9-7250174 10-2749826 9-7969130 10-2030870 10-0718957 9-9281043 56 5 9*7252189 10-2747811 9-7971938 10-2028062 10-0719749 9-9280251 55 6 9*7254204 2013 2012 2011 2009 2008 10-2745796 9-7974745 2806 2805 2804 2804 2803 10*2025255 10*0720541 793 793 794 794 795 9-9279459 54 7 9-7256217 10-2743783 9"7977551 10-2022449 10-0721334 9*9278666 53 8 9*7258229 10-2741771 9-7980356 10-2019644 10-0722127 9-9277873 52 9 9*7260240 10*2739760 9-7983160 10-2016840 10*0722921 9-9277079 51 10 9-7262249 10*2737751 9-7985964 10-2014036 10*0723715 9-9276285 50 11 9-7264257 2007 2005 2004 2003 2002 10-2735743 9-7988767 2802 2801 2800 2800 2799 10*2011233 10-0724510 795 796 796 797 797 9*9275490 49 12 9-7266264 10-2733736 9-7991569 10-2008431 10*0725305 9*9274695 48 13 9-7268269 10-2731731 9-7994370 10-2005630 10-0726101 9-9273899 47 14 9-7270273 10-2729727 9*7997170 10-2002830 10-0726897 9-9273103 46 15 9*7272276 10-2727724 9*7999970 10*2000030 10-0727694 9-9272306 45 16 9*7274278 2000 1999 1998 1996 1996 10-2725722 9-8002769 2798 2798 2796 2796 2795 10-1997231 10-0728491 798 798 799 800 800 9*9271509 44 17 9-7276278 10-2723722 9-8005567 10-1994433 10-0729289 9*9270711 43 18 9-7278277 10-2721723 9-8008365 10-1991635 10-0730087 9-9269913 42 19 9*7280275 10-2719725 9-8011161 10-1988839 10-0730886 9-9269114 41 20 9-7282271 10-2717729 9-8013957 10-1986043 10-0731686 9-9268314 40 21 9'7284267 1993 1993 1991 1990 1989 10-2715733 9-8016752 2794 2794 2793 2792 2791 10-1983248 10-0732486 800 801 801 802 803 9-9267514 39 22 9-7286260 10-2713740 9-8019546 10-1980454 10-0733286 9-9266714 38 23 9*7288253 10-2711747 9*8022340 10-1977660 10-0734087 9-9265913 37 24 9-7290244 10*2709756 9-8025133 10-1974867 10-0734888 9-9265112 36 25 9-7292234 10-2707766 9-8027925 10-1972075 10-0735690 9-9264310 35 26 9*7294223 1988 1986 1985 1983 1983 10-2705777 9-8030716 2790 2790 2789 2788 2788 10-1969284 10-0736493 803 803 805 804 805 9-9263507 34 27 9-7296211 10*2703789 9.8033506 10-1966494 10-0737296 9-9262704 33 28 9*7298197 10-2701803 9-8036296 10-1963704 10-0738099 9-9261901 32 29 9-7300182 10-2699S18 9-8039085 10-1960915 10*0738904 9*9261096 31 30 9*7302165 10-2697835 9*8041873 10-1958127 10-0739708 9-9260292 30 31 9-7304148 1981 1980 1978 1977 1976 10-2695852 9-8044661 2786 2786 2786 2784 2784 10*1955339 10-0740513 806 806 806 808 807 9*9259487 29 32 9*7306129 10*2693871 9*8047447 10-1952553 10-0741319 9-9258681 28 33 9*7308109 10-2691891 9-8050233 10-1949767 10-0742125 9-9257075 27 34 9-7310087 10-2689913 9*8053019 10*1946981 10-0742931 9-9257069 26 35 9*7313064 10*2687936 9-8055803 10-1944197 10*0743739 9-9256261 25 36 9*7314040 1975 1974 1972 1971 1970 10-2685960 9*8058587 2783 2782 2781 2781 2780 10-1941413 10-0744546 808 809 809 810 810 9-9255454 24 37 9*7316015 10*2683985 9-8061370 10-1938630 10*0745354 9-9254646 23 38 9-7317989 10-2682011 9-8064152 10-1935848 10-0746166 9-9253837 22 39 9-7319961 10*2680039 9-S066933 10-1933067 10-0746972 9-9253028 21 40 9-7321932 10-2678068 9-8069714 10-1930286 10*0747782 9-9252218 20 41 9-7323902 1968 1967 1966 1965 1963 10-2676098 9-8072494 2779 2779 2777 2777 2777 10-1927506 10-0748592 811 811 812 813 812 9-9251408 19 42 9*7325870 10-2674130 9-8075273 10-1924727 10-0749403 9*9250597 18 43 9-7327837 10-2672163 9*8078052 10-1921948 10-0750214 9-9249786 17 44 9-7329803 10-2670197 9-8080829 10-1919171 10-0751026 9-9248974 16 45 9-7331768 10*2668232 9-8083606 10-1916394 10-0751839 9*9248161 15 46 9*7333731 1962 1961 1960 1958 1957 10-2666269 9-8086383 2775 2775 2774 2773 2773 10-1913617 10-0752651 814 814 814 815 815 9-9247349 14 47 9-7335693 10*2664307 9*8089158 10-1910842 10-0753465 9-9246535 13 48 9-7337654 10*2662346 9-8091933 10-1908067 10*0754279 9-9245721 12 49 9-7339614 10-2660386 9-8094707 10*1905293 10-0755093 9-9244907 11 50 9*7341572 10-2658428 9-8097480 10-1902520 10*0755908 9-9244092 10 51 9-7343529 1956 1955 1953 1952 1951 10*2656471 9-8100253 2772 2771 2770 2770 2769 10-1899747 10*0756723 S16 817 817 817 S19 9-9243277 9 52 9-7345485 10-2654515 9-8103025 10-1896975 10*0757539 9-9242461 8 53 9*7347440 10*2652560 9-8105796 10-1894204 10*0758356 9-9241644 7 54 9-7349393 10-2650607 9-8108566 10-1891434 10-0759173 9-9240827 6 55 9-7351345 10*2648655 9-S111336 10*1888664 10*0759990 9-9240010 5 56 9-7353296 1950 1949 1947 1946 10-2646704 9-8114105 2768 2768 2767 2766 10*1885895 10-0760809 818 819 820 820 9-9239191 4 57 9*7355246 10-2644754 9-8116873 10-1883127 10-0761627 9-9238373 3 58 9-7357195 10-2642805 9-8119641 10-1880359 10*0762446 9-9237554 2 59 9*7359142 10-2640858 9-8122408 10-1877592 10-0763266 9"9236734 1 60 9*7361088 10*2638912 9-8125174 10-1874826 10-0764086 9-9235914 0 / Cosine Diff. Secant Cotang. Diff. Tang. Cosec. Diff. Sine / 57 Deg PROPORTIONAL PARTS. 171 160. There is one point to be noticed in using the columns headed Diff. It has been pointed out that 2021 (at the top of the second column) means *0002021. Now the 790 (at the top of the eighth column) means not *000790, but *0000790. The rule is this; the right-hand figure of the Diff. must be placed in the seventh place of decimals and the requisite number of cyphers prefixed. Thus Diff. = 9 Diff. = 74 Diff. = 735 Diff. = 2021 whilst Diff. = 12348 means that the difference is *0000009, )> *0000074, *0000735, • 0002021 , *0012348. 161. Page 170 also gives the tabular logs, of ratios between 57° and 58°. Suppose we wanted L tan 57° 20'. We now start with the line at the bottom of the page and run our eye up the column which has Tang, at its foot. We go up this column until we arrive at the number which is on the same level as the number 20 in the extreme right-hand column. This number we find to be 10T930286, which is therefore the value of L tan 57° 20'. EXAMPLES. XXV. 1. Find d given that cos d= *9725382, cos 13° 27'=*9725733, diff. for l' = 677. 3 2. Find the angle whose sine is -, given sin 22° l' =-3748763, diff. for l' = 2696. 172 TRIGONOMETRY. [Exs. XXV.] 3. Given cosec 65° 24' = 1*0998243, diff. for 1' = 1464, find the value of cosec 65° 24' 37" and the angle whose cosec is 1*0997938. 4. Given L tan 22° 37' = 9*6197205, diff. for l' = 3557, find the value of L tan 22° 37' 22" and the angle whose L tan is 9*6195283. 5. Find the angle whose L cos is 9*993, given L cos 10° 15' = 9*9930131, diff. for l' = 229. 6 . Find the angle whose L sec is 10*15, given L sec 44° 55'= 10*1498843, diff. for l' = 1260. • 7. From the table on page 170 find the values of € (1) L sin 32° 18' 23", (2) L cos 32° 16' 49", (3) L cot 32° 29' 43", (4) L sec 32° 52' 27", (5) L tan 57° 45' 28", ( 6 ) L cosec 57° 48' 21", and (7) L cos 57° 58' 29". 8 . With the help of the same page solve the equations (1) L tan 6 = 10*1959261, (2) L cosec 0 = 10*0738125, (3) L cos 0 = 9*9259283, and (4) L sin 0 = 9 ’9241352. 9. Take out of the tables L tan 16° 6 ' 23" and calculate the value of the square root of the tangent. 10. Change into a form more convenient for logarithmic computation (i.e. express in the form of products of quantities) the quantities (i) 14 -tan .r tan y, ( 2 ) 1 - tan x tan y, (3) cot.r+ tan ?/, (4) cot x -tan?/, (5) 1 - cos 2x and ( 6 ) tan x 4 - tan y cot x 4 - cot y 1 4-cos 2x ’ CHAPTER XII. RELATIONS BETWEEN THE SIDES AND THE TRIGONOMETRICAL RATIOS OF THE ANGLES OF ANY TRIANGLE. 162. In any triangle ABC , the side BC, opposite to the angle A, is denoted by ~bT +{a - h ) ir b 2 + c 2 - a 2 , ox c 2 + a 2 - b 2 -i[r- c 2 ) 2abc + (g2 ~ a2) 2ai C ~ +(a2 ~ 62) a 2 + b 2 -c 2 2 abc ] 2abck = 0. [b 4 -c 4 -a 2 ( b 2 - c 2 ) + c 4 - a 4 - 6 2 (c 2 - a 2 ) + a 4 - b 4 - c 2 ( a 2 - fr 2 )] Ex. 3. In a triangle prove that {a + b + c) ^ tan j + tan ^ = 2c cot ^. The left-hand member (s - c) (s - a) J , by Art. 167, a) ' V s (s - b) - [>/£♦n/S] 2 Js {s -c) .c J(s-a)(s- b) , since 2 s = a + b + c, — 2c cot ^ . This identity may also be proved by substituting for the sides. We have, by Art. 163, a + & + c_sin^-|-sin B + sin G c ~ sin G .ABC! A B 4 cos — cos — cos — 2 cos — cos — J , as in Art. 127, = — 0 . G G 2 sin — cos — a 2 . G Sln 2 184 TRIGONOMETRY. Also 2 cot ^ u 0 CAB 2 cos - cos — cos - £ u A B tan — + tan — i u . C r . A B A . B~\ sm 2 L Sln 2 COS 2 +COS 2 Sm 2j „ ABC 2 COS ^ cos ^ cos ^ & . 0 . .d. 4* J5 sm — sin 2~ o ^ 5 2 cos - cos — 2 2 . G sm 2 We have therefore (2 4* & + c 2 cot ^ . 4 . R ’ tan - + tan — so that ( A B\ C tan — + tan — J = 2c cot - (Art. 69.) 4. If the sides of a triangle he in Arithmetical Progression , prove that so also are the cotangents of half the angles. We have given that a + c = 2b . and we have to prove that .A ,G . B cot-^ + cot- = 2 cot — z z z ( 1 ), ( 2 ). Now (2) is true if / S (s - a) j s($-c) „ j S{s- b) v (s-&Hs-c) v (s-a)(s-6) V (s — (s - c) (s - a) ’ or, by multiplying both sides by 4 (s - a) (s- h) ( s-c ) if i.e. if (s-a) + (s-c) = 2(s-fc), 2 s - (a + c) = 2s - 2b, i.e. if a+c = 2b, which is relation (1). Hence if relation (1) be true, so also is relation (2). SIDES AND ANGLES OF A TRIANGLE. 185 EXAMPLES. XXVII. In any triangle ABC, prove that , . B-G b-c A L sm- 2 - = —cos-. £ 2. ft (cos R + cos C) = 2 (6 + c) sin 2 — . 3. a (cos <7-cosR) = 2 (b-c) cos 2 — . a -\-b A 4- B A — B 4. -; = tan —- cot — 0 — . a - b 2 2 5. (b + c- a) ^cot ^ + cot ^ j = 2a cot ^ . 6. a‘ 2 + b‘ 2 + c 2 = 2 (be cos A +ca cos B + ab cos C). 7. (a 2 - b 2 + c 2 ) tan B = (a 2 + b 2 - c 2 ) tan <7. 8. c 2 = (ft-6) 2 cos 2 ^+(ft + 6) 2 sin 2 ^. 9. a sin (B - C) + 5sin (C - A) + csin (A -B) = 0. a sin (B-C) b sin (C - A) c sin (A - B) 10 . 13 . 14 . b 2 - c 2 c 2 - a 2 ft 2 -6 2 „ . A . B-C _ . B . C-A . C . A-B A 11. a sin - sin —^—1-6 sin- sin — - -f- c sm - sin =0. 12. ft 2 (cos 2 R - cos 2 C)+ 6 2 (cos 2 C-cos 2 A) + c 2 (cos 2 A - cos 2 B) = 0. c 2 - ft^ b 2 -c 2 . n , c 2 - a 2 . ^ a 2 -b 2 . a 2 sin 2A + 6 2 sin 2B + sin 2(7 = 0. ,A . B , c COt — -f* COt jr + cot 7 j- (a + b + c) 2 2 2 2 a 2 + b 2 + c 2 cot A + cot B + cot C * 15 . ft 3 cos (B - C) + 6 3 cos ((7 - A) +c 3 cos (A -B) = 3 ft&c. 16 . In a triangle whose sides are 3, 4, and ^/38 feet respectively, prove that the largest angle is greater than 120°. 186 TRIGONOMETRY. [Exs. XXVII.] 17. The sides of a right-angled triangle are 21 and 28 feet; find the length of the perpendicular drawn to the hypotlienuse from the right angle. 18. If in any triangle the angles be to one another as 1:2:3, prove that the corresponding sides are as 1 : s f 3 : 2. 19. In any triangle if A 5 A B 20 tan 2=6’ tan 2 = 37’ Q find tan —, and prove that in this triangle a+c = 2b. u 20. I n an isosceles right-angled triangle a straight line is drawn from the middle point of one of the equal sides to the opposite angle. Shew that it divides the angle into parts whose cotangents are 2 and 3. 21. The perpendicular AD to the base of a triangle ABC divides it into segments such that BD, CD and AD are in the ratio of 2, 3 and 6; prove that the vertical angle of the triangle is 45°. 22. A ring, ten inches in diameter, is suspended from a point one foot above its centre by 6 equal strings attached to its circumference at equal intervals. Find the cosine of the angle between consecutive strings. 23. If b 2 and c 2 be in a.p., prove that cot A, cot B and cot C are in a. p. also. A B 24. If a, b and c be in a. p., prove that cos A cot — , cos B cot — W Ji Q and cos C cot — are in a. p. A B C 25. If b and c are in h.p. prove that sin 2 , sin 2 — and sin 2 — are also in h.p. 26. The sides of a triangle are in a.p. and the greatest and least angles are 6 and 0; prove that 4(1- cos 6) (1 - cos 0) = cos 6 + cos 0. 27. The sides of a triangle are in a.p. and the greatest angle exceeds the least by 90°; prove that the sides are proportional to +1, and V?-l. [Exs. XXVII.] SIDES AND ANGLES OF A TRIANGLE. 187 28. If (7=60°, then prove that 1 1 3 a + c b + c a+b+c' 29. I n any triangle ABC if D be any point of the base BC , such that BD : DC :: m : n, prove that ( m + n) cot ADC — n cot B - mcot C, and (m + n) 2 AD' 2 = (m + n) (mb 2 + nc 2 ) -mna 2 . 30. If in a triangle the bisector of the side c be perpendicular to the side 6, prove that 2 tan A +tan C = 0. 31. In any triangle prove that, if d be any angle, then b cos d = c cos (A - 6) + a cos (C + 6). 32. If p and q be the perpendiculars from the angular points A and B on any line passing through the vertex C of the triangle ABC, then prove that a 2 p 2 + b 2 q 2 — 2abpq cos C = a 2 b 2 sin 2 C. 33. In the triangle ABC , lines OA, OB , and OC are drawn so that the angles OAB, OBC, and OCA are each equal to w; prove that cot w = cot A + cot B + cot C, cosec 2 oj = cosec 2 A + cosec 2 B + cosec 2 C. and CHAPTER XIII. SOLUTION OF TRIANGLES. 174. In any triangle the 3 sides and the 3 angles are often called the elements of the triangle. When any 3 elements of the triangle are given, provided they be not the 3 angles, the triangle is in general completely known, i.e. its other angles and sides can be calculated. When the 3 angles are given, only the ratios of the lengths of the sides can be found, so that the triangle is given in shape only and not in size. When 3 elements of a triangle are given the process of calculating its other 3 elements is called the Solution of the Triangle. We shall first discuss the solution of right-angled triangles, i.e. triangles which have one angle given equal to a right angle. The next four articles refer to such triangles, and C denotes the right angle. 175. Case I. Given the hypotlienuse and one side , to solve the triangle. RIGHT-ANGLED TRIANGLE. 189 Let b be the given side and c the given hypothenuse. The angle B is given by the relation • d & sm B = -. c L sin 5=10 4- log b — log c. Since b and c are known we thus have L sin B and therefore B. The angle A (= 90 c — B) is then known. The side a is obtained from either of the relations cos B = -, tan B = - , or a = J(c — b) (c + b). 0 CL 176. Case II. Given the two sides a and b, to solve the triangle . Here B is given by tan B = - , a so that L tan 5 = 10 + log b — log a. A b C Hence L tan B, and therefore B } is known. The angle A (= 90' — B) is then known. The hypothenuse c is given by the relation c = Va 2 +6 2 . This relation is not however very suitable for loga¬ rithmic calculation, and c is best given by § . ^ b b sm B = - , i.e. c = ——^. c sm B log c = log b — log sin B = 10 + log b — L sin B. Hence c is obtained. 190 TRIGONOMETRY. 177. Case III. Given an angle B and one of the sides a, to solve the triangle. Here A (= 90° — B) is known. The side b is found from the rela¬ tion - = tan B, B a and c from the relation a D - = cos B. c 178. Case IV. Given an angle B c, to solve the triangle. Here A is known and a and b are obtained from the relations - = cos B, and - = sin B. c c EXAMPLES. XXVIII. 1. In a right-angled triangle ABC, where C is the right angle, if a = 50 and B = 75°, find the sides, (tan 75° = 2 + ^3.) 2. Solve the triangle of which two sides are equal to 10 and 50 feet and of which the included angle is 90°; given that log 20 = 1-30103, and L tan 26° 33' = 9*6986847, diff. for l' = 3160. 3. The length of the perpendicular from one angle of a triangle upon the base is 3 inches and the lengths of the sides containing this angle are 4 and 5 inches. Find the angles, having given log 2 = -30103, log 3 = -4771213, L sin 36° 52' = 9-7781186, diff. for l'=1684, L sin 48° 35' = 9-8750142, diff. for l'=1115. 4. Find the acute angles of a right-angled triangle whose hypothenuse is four times as long as the perpendicular drawn to it from the opposite angle. and the hypothenuse A b C SOLUTION OF TRIANGLES. 191 179. We now proceed to the case of the triangle which is not given to be right angled. The different cases to be considered are; Case I. The three sides given ; Case II. Two sides and the included angle given; Case III. Two sides and the angle opposite one of them given; Case IV. One side and two angles given; Case V. The three angles given. 180. Case I. The three sides a , b, and c given. Since the sides are known, the semi-perimeter s is known and hence also the quantities s — a, s — b, and s — c. A B C The half-angles , —, and - are then found from the formulae tan — = v (s-b)(s-c) /(s-c)(s-a) s(8-a) ’ 2 V s(s-b) and , 0 tan —= (s — a)(s — b) s(s - c) Only two of the angles need be found, the third being known since the sum of the three angles is always 180°. The angles may also be found by using the formulae for the sine or cosine of the semi-angles. (Arts. 165 and 166.) The above formulae are all suited for logarithmic computation. 192 TRIGONOMETRY. The angle A may also be obtained from the formula 7)2 I /->2 _ a /2 cos -4 =--. (Art. 164.) 26c This formula is not, in general, suitable for logarithmic calculation. It may be conveniently used however when the sides a, 6, and c are small numbers. Ex. The sides of a triangle are 32, 40, and 66 feet; find the angle opposite the greater side , having given that log 207 = 2*3159703, log 1073 = 3-0305997, L cot 66° 18' = 9*6424341, tabulated difference for l'=3431. Here a = 32, 5 = 40 and c = 66, , 32 + 40 + 66 so that s = --— Hence cot = 69, s - a = 37, s - 5 = 29 and s - c = 3. r s(s-c) _ v (s -a (s -a) (s - 5) 69x3 37x29 207 1073 * L cot ^ = 10+ I [log 207 - log 1073] z z = 10 + 1 *15798515 -1-51529985 = 9-6426853. C . L cot — is therefore greater than L cot 66° 18', z so that is less than 66° 18'. z c Let then — = 66° 18'-#". z The difference in the logarithm corresponding to difference of x" in the angle therefore = 9-6426853 - 9-6424341 = -0002512 Also the difference for 60"=-0003431. TT x -0002512 Hence so that x = 60 -0003431 ’ 2512 ao w, x 60 = nearly 44. 3431 C .-. = 66° 18' - 44" = 66° 17' 16", and hence C = 132° 34' 32". z THE THREE SIDES GIVEN. 193 EXAMPLES. XXIX. 1. If the sides of a triangle be 56, 65, and 33 feet, find the greatest angle. 2. The sides of a triangle are 7, 4^3, and ^13 yards respectively. Find the number of degrees in its smallest angle. 3. The sides of a triangle are x 2 + x + 1, 2x + 1 and x 2 - 1; prove that the greatest angle is 120°. 4. The sides of a triangle are a , 5, and >Ja 2 + ab + b 2 feet; find the greatest angle. 5. If a = 2, b = ^6 and c = */3 - 1, solve the triangle. 6. If a = 2, 5 = ^/6 and c = /V /3 + l, solve the triangle. 7. If a = 9, 5 =10 and c = 11, find B, given log 2 = *30103, L tan 29° 29' = 9*7523472, and L tan 29° 30' = 9-7526420. 8. The sides of a triangle are 130, 123 and 77 feet. Find the greatest angle, having given log 2= -30103, L tan 38° 39'=9*9029376, and L tan 38° 40' = 9*9031966. 9. Find the greatest angle of a triangle whose sides are 242,188, and 270 feet, having given log 2 = -30103, log 3 = -4771213, log 7 = *8450780, L tan 38° 20' = 9*8980104, and L tan 38° 19' = 9 *8977507. 10. The sides of a triangle are 2, 3, and 4; find the greatest angle, having given log 2 = -30103, log 3 = -4771213, L tan 52° 14'= 10*1108395, and L tan 52° 15' = 10*1111004. Making use of the tables, find all the angles when 11. a = 25, 5 = 26 and c = 27. 12. a = 17, 5 = 20 and c = 27. 13. a = 2000, 5 = 1050 and c=1150. L. T. 13 194 TRIGONOMETRY. 181. Case II. Given two sides b and c and the included angle A. Taking b to be the greater of the two given sides, These two relations give us B-C 2 and B + C 2 ’ and therefore, by addition and subtraction, B and C. The third side a is then known from the relation a _ b sin A sin B ’ , . t 7 sin A which gives a = b ——^ , sin B and thus determines a. The side a may also be found from the formula a 2 = b 2 + c 2 — 2 be cos A. This is not adapted to logarithmic calculation but is sometimes useful, especially when the sides a and b are small numbers. 182. Ex. 1. If b = b, the right-hand member of (1) is greater than unity, and hence there is no corresponding value for C. If c sin B = b, the right-hand member of (1) is equal to unity and the corresponding value of G is 90°. If c sin B < b y there are two values of G having c sin B b as its sine, one value lying between 0° and 90° and the other between 90° and 180°. Both of these values are not however always admissible. For if b > Cy then B > G. The obtuse-angled value of G is now not admissible ; for, in this case, G cannot be obtuse unless B be obtuse also, and it is manifestly impossible to have two obtuse angles in a triangle. If b < c and B be an acute angle, both values of C are admissible. Hence there are two values found for A and hence the relation (2) gives two values for a. In this case there are therefore two triangles satisfying the given conditions. Since, for some values of b, c and B, there is a doubt or ambiguity in the determination of the triangle, this case is called the Ambiguous Case of the solution of triangles. 186. The Ambiguous Case may also be discussed in a geometrical manner. Suppose we were given the elements b , c and B and that we proceeded to construct, or attempted to construct, the triangle. 202 TRIGONOMETRY. We first measure an angle ABD equal to the given angle B . We then measure along BA a distance BA equal to the given distance c, and thus determine the angular point A. We have now to find a third point G y which must lie on BD and must also be such that its distance from A shall be equal to b. To obtain it, we describe with centre A a circle whose radius is b. The point or points, if any, in which this circle meets BD will determine the position of C. Draw AD perpendicular to BD, so that AD = AB sin B = c sin B. One of the following events will happen. The circle may never reach BD (Fig. 1) or it may AMBIGUOUS CASE. 203 touch BD (Fig. 2), or it may meet BB in two points C 1 and C 2 (Figs. 3 and 4). In the case of Fig. 1, it is clear that there is no triangle satisfying the given condition. Here b < AD, i.e. < c sin B. In the case of Fig. 2, there is one triangle ABB which is right-angled at B. Here b = AB = c sin B. In the case of Fig. 3, there are two triangles ABG l and ABC ' 2 . Here b lies in magnitude between AB and c, i.e. b is > c sin B and < c. In the case of Fig. 4, there is only one triangle ABC 1 satisfying the given conditions [the triangle ABC 2 is inadmissible; for its angle at B is not equal to B but is equal to 180 — B]. Here b is greater than both c sin B and c. To sum up : Given the elements b, c, and B of a triangle, (a) If b be < csinB, there is no triangle. (/3) If b = c sin B y there is one triangle right-angled. (7) If 6 be > c sin B and < c and B be acute, there are two triangles satisfying the given conditions. (8) If b be > c, there is only one triangle. Clearly if b = c, the points B and C 2 in Fig. 3 coincide and there is only one triangle. If B be obtuse, there is no triangle except when b >c. 187. The ambiguous case may also be considered algebraically as follows. 204 TRIGONOMETRY. From the figure of Art. 184 we have b 2 = c 2 -f a 2 — 2 ca cos B . .\ a 2 — 2ac cos B -f c 2 cos 2 B = b 2 — c 2 4- c 2 cos 2 J5 = b 2 — c 2 sin 2 5. .*. a — c cos £ = ± V 6 2 — c 2 sin 2 i?, i.e. a = c cos jB + V 6 2 — c 2 sin 2 5.( 1 ). Now ( 1 ) is an equation to determine the value of a when b, c and B are given. (a) If b < c sin B , the quantity V 6 2 — c 1 sin 2 B is imaginary and ( 1 ) gives no real value for a. (/?) If 6 = c sin jB. there is only one value, c cos B, for a; there is thus only one triangle which is right- angled. ( 7 ) If b > c sin B , there are two values for a. But, since a must be positive, the value obtained by taking the lower sign affixed to the radical is inadmissible unless c cos B — V 6' 2 — c 2 sin 2 B is positive, i.e. unless Vi 2 — c 2 sin 2 B < c cos B, i.e. unless b 2 - c 2 sin 2 B < c 2 cos 2 B , i.e . unless 6 2 < c 2 . 1 There are therefore two triangles only when b is > c sin B and at the same time < c. 188. Given b= 16, c = 25, and B = 33° 15', prove that the triangle is ambiguous and find the other angles, having given log 2 = -30103, L sin 33° 15' = 9-7390129, L sin 58° 56'= 9-9327616, Lsin 58° 57'= 9-9328376. and AMBIGUOUS CASE. 205 We have . c . 25 . 100 . 10- . Q1 „, sin (7 = sm B = —sm B = 7 -. sin B=—r sin 33° lo . 5 16 64 2 6 Hence L sin C=2 + L sin 33° 15' - 6 log 2 = 9*9328329. C therefore lies between 58° 56' and 58° 57', so that <7 = 58° 56' + x". For a difference of x" in the angle the difference in the log = 9*9328329 - 9*9327616= *0000713. For a difference of 60" in the angle the difference = 9*9328376 - 9*9327616 = *0000760. Hence x 60 x = *0000713 713 *0000760 760 * 6x713 76 = 56 nearly, so that L sin C=L sin 58° 56' 56". .*. 0 = 58° 56' 56" or 180° - 58° 56'56". Hence (Fig. 3, Art. 186) we have C x = 58° 56' 56", and C 2 = 121° 3' 4". .*. L BA G 1 = 180° - 33° 15' - 58° 56' 56"= 87° 48' 4", and Z BA C 2 = 180° - 33° 15' - 121° 3' 4" = 25° 41' 56". EXAMPLES. XXXI. 3 1. If a = 5, 5 = 7, and sin A = - , is there any ambiguity? 2. If = 30°, c=150, and 6 = 150^/3, prove that of the two triangles which satisfy the data one will be isosceles and the other right- angled. Find the greater value of the third side. Would the solution have been ambiguous had 5 = 30°, c = 150, and 6 = 75? 7. In the ambiguous case given a, b, and A, prove that the difference between the two values of c is 2 Jo? - 6- sin- A. 8. If a = 5, 6 = 4, and A = 4.5°, find the other angles, having given log 2 = -30103, L sin 33° 29'= 9-7520507, and L sin 33°30' = 9-7530993. 9. If a = 9, 6 = 12, and A = 30°, find c, having given log2=-30103, log 3 = -47712, log 171 = 2-23301, log 368 = 2-56635, L sin 11° 48'39" = 9-31108, L sin 41° 48'39" = 9-82391, and L sin 108° 11' 21" = 9-977774. 10. Point out whether or no the solutions of the following triangles are ambiguous. Find the smaller value of the third side in the ambiguous case and the other angles in both cases. (1) ^4 = 30°, c = 250 feet, and a — 125 feet; (2) A = 30°, c = 250 feet, and a=200 feet. Given log 2 = *30103, log 6*03893 = *7809601, L sin 38° 41' = 9*7958800, and L sin 8° 41' = 9*1789001. 11 . Given a = 250, b = 240, and A = 72° 4' 48", find the angles B and C, and state whether they can have more than one value, given log 2*5 = *3979400, log 2*4 = *3802112, L sin 72° 4' = 9 *9783702, L sin 72° 5' = 9*9784111, and L sin 65° 54' = 9*9606739. 12. Two straight roads intersect at an angle of 30°; from the point of junction two pedestrians A and B start at the same time, A walking [Exs. XXXI.] SOLUTION OF TRIANGLES. 207 along one road at the rate of 5 miles per hour and B walking uniformly along the other road. At the end of 3 hours they are 9 miles apart. Shew that there are two rates at which B may walk to fulfil this condition and find them. For the following 3 examples , a hook of tables will he required. 13 . Two sides of a triangle are 1015 feet and 732 feet and the angle opposite the latter side is 40°; find the angle opposite the former and prove that more than one value is admissible. 14 . Two sides of a triangle being 5374*5 and 1586*6 feet, and the angle opposite the latter being 15° 11', calculate the other angles of the triangle or triangles. 15 . Given A = 10°, a = 2308*7, and 5 = 7903*2, find the smaller value of c. 189 . Case IV. Given one side and tivo angles, viz. a, B, and C. Since the three angles of a triangle to two right angles, the third angle is given also. The sides b and c are now obtained from the relations b c __ a sin B sin C sin A ’ are together equal giving 7 sin B 1 sin C b = a —. —j , and c = a ——. sin A sin A 190 . Case V. The three angles A, B and C given. Here the ratios only of the sides can be determined by the formulae a _ b c sin A sin B sin C ’ Their absolute magnitudes cannot be found. 208 TRIGONOMETRY. EXAMPLES. XXXII. 17 1 If cos A-—- and cos C = — 7 , find the ratio of a : b : c. 22 14 2. The angles of a triangle are as 1 : 2 : 7 ; prove that the ratio of the greatest side to the least side is aJo + 1 : *Jo -1. 3. If A— 45°, B — 75°, and C = 60°, prove that a + c J2 = 2b. 4. Two angles of a triangle are 41° 13' 22" and 71° 19' 5" and the side opposite the first angle is 55; find the side opposite the latter angle, given log 55 = 1-7403627, log 79063 = 4-8979775, L sin 41° 13' 22" = 9*8188779, and L sin 71° 19' 5" = 9*9764927. 5. From each of two ships, one mile apart, the angle is observed which is subtended by another ship and a beacon on shore; these angles are found to be 52° 25' 15" and 75° 9' 30" respectively. Given L sin 75° 9' 30" = 9*9852635, L sin 52° 25' 15" = 9*8990055, log 1*2197= *0862530 and log 1*2198= *0862886, find the distance of the beacon from each of the ships. 6. The base angles of a triangle are 22^° and 112J°; prove that the base is equal to twice the height. For the following 5 questions a book of tables is required. 7. The base of a triangle being seven feet and the base angles 129° 23' and 38° 36', find the length of its shorter side. 8. If the angles of a triangle be as 5 : 10 : 21, and the side opposite the smaller angle be 3 feet, find the other sides. 9. The angles of a triangle being 150°, 18° 20', and 11° 40', and the longest side being 1000 feet, find the length of the shortest side. 10. To get the distance of a point A from a point B, a line BC and the angles ABC and BCA are measured, and are found to be 287 yards and 55° 32'10" and 51° 8'20" respectively. Find the distance AB. 11 . To find the distance from A to P a distance, AB, of 1000 yards is measured in a convenient direction. At A the angle PAB is found to be 41° 18'and at B the angle PBA is found to be 114° 38'. What is the required distance to the nearest yard ? CHAPTER XIV. HEIGHTS AND DISTANCES. 191 . In the present chapter we shall consider some questions of the kind which occur in land-surveying. Simple questions of this kind have already been considered in Chapter III. 192 . To find the height of an inaccessible tower by means of observations made at distant points. Suppose PQ to be the tower and that the ground passing through the foot Q of the tower is horizontal. At a point A on this ground measure the angle of elevation a of the top of the tower. Measure off a distance AB(=a) A from A directly toward the foot of the tower, and at B measure the angle of elevation (3. To find the unknown height x of the tower, we have to connect it with the measured length a. This is best done as follows: L. T. 14 210 TRIGONOMETRY. From the triangle PBQ we have x BP = sin/3.(1), and from the triangle PAB we have PB sin PA B sin a a sin BP A sin (/3 — a) . since Z BP A = Z QBP — Z QMP = /3 — a. From (1) and (2), by multiplication, we have x sin a sin /3 a sin (/3 — a) ’ ( 2 ), x = a sin a sin /3 sin(/3— a) ’ The height x is therefore given in a form suitable for logarithmic calculation. Numerical Example. If a =100 feet, a = 30°, and /3 = 60°, then sin 30° sin 60° JS A £ = 100-.————- = 100 x -^— = 86*6 feet. sin 30 2 193. It is often not convenient to measure AB directly towards Q. Measure therefore AB in any other suitable direction on the hori¬ zontal ground and at A measure the angle of elevation a of P, and also the angle PAB (= /3). At B measure the angle PBA (= y)- 8 In the triangle PAB we have then Z APB = 180 ° - Z PAB - Z PBA = 180 ° - (/? + y). HEIGHTS AND DISTANCES. 211 Hence AP sin PBA _ sin 7 a sin BP A sin (/3 + 7 ) ’ From the triangle PAQ , we have x = AP sin a = a sin a sin 7 sin(/3 + 7 )’ Hence x is found by an expression suitable for logarithmic calculation. 194. To find the distance between two inaccessible points by means of observations made at two points the distance between which is known , all four points being supposed to be in one plane. Let P and Q be two points whose distance apart, PQ, is required. Let A and B be the two known points whose distance apart, AB , is given to be equal to a. At A measure the angles PA B and QAB , and let them be a and /3 A respectively. At B measure the angle PBA and QBA, and let them be 7 and 8 respectively. Then in the triangle PAB we have one side a and the two adjacent angles a and 7 given, so that, as in Art. 163, we have AP given by the relation AP sin 7 __ sin 7 a sin APB sin (a + 7 ) In the triangle QAB we have, similarly, AQ _ sin 8 a sin (/3 + 8 ). 14—2 212 TRIGONOMETRY. In the triangle APQ we have now determined the sides AP and AQ ; also the included angle PAQ (= a — ft) is known. We can therefore find the side PQ by the method of Art. 181. If the four points A, B, P , and Q be not in the same plane, we must, in addition, measure the angle PAQ ; for in this case PAQ is not equal to a — /3. In other respects the solution will be the same as above. 195. Bearings and Points of the Compass. The Bearing of a given point B as seen from a given point 0 is the direction in which B is seen from 0. Thus if i 5 ■ the direction of OB bisect the angle between East and North, the bearing of B is said to be North-East. If a line is said to bear 20° West of North we mean that it is inclined to the North direction at an angle of 20°, this angle being measured from the North towards the West. HEIGHTS AND DISTANCES. 213 To facilitate the statement of the bearing of a point the circumference of the mariners compass-card is divided into 32 equal portions, as in the above figure, and the sub¬ divisions marked as indicated. Consider only the quadrant between East and North. The middle point of the arc between N. and E. is marked North-East (N.E.). The bisectors of the arcs between N.E. and N. and E. are respectively called North-North-East and East-North- East (N.N.E. and E.N.E.). The other four subdivisions, reckoning from N., are called North by East, N.E. by North, N.E. by East, and East by North. Similarly the other three quadrants are subdivided. It is clear that the arc between two subdivisions of the card subtends an angle of , i.e. 11 J°, at the centre 0 . EXAMPLES. XXXIII. 1. A flagstaff stands on the middle of a square tower. A man on the ground opposite the middle of one face and distant from it 100 feet just sees the flag; receding another 100 feet the tangents of elevation of the top of the tower and the top of the flagstaff are found to be ^ 5 and - . Find the dimensions of the tower and the height of the flagstaff, the ground being horizontal. 2. A man, walking on a level plane towards a tower, observes that at a certain point the angular height of the tower is 10° and after going 50 yards nearer the tower the elevation is found to be 15°. Having given L sin 15° = 9*4129962, L cos 5° = 9*9983442, log 25 *783 = 1-4113334 and log 25*784 = 1*4113503, find, to 4 places of decimals, the height of the tower in yards. 214 TRIGONOMETRY. [Exs. XXXIII.] 3. BE is a tower standing on a horizontal plane and ABCD is a straight line in the plane. The height of the tower subtends an angle 6 at A, 26 at B, and 3 6 at C . If AB and BC be respectively 50 and 20 feet, find the height of the tower and the distance CD. 4. A tower, 50 feet high, stands on the top of a mound ; from a point on the ground the angles of elevation of the top and bottom of the tower are found to be 75° and 45° respectively; find the height of the mound. 5. A vertical pole (more than 100 feet high) consists of two parts, the lower being ^rd of the whole. From a point in a horizontal plane o through the foot of the pole and 40 feet from it, the upper part subtends an angle whose tangent is i. Find the height of the pole. 6. A tower subtends an angle a at a point on the same level as the foot of the tower and at a second point, h feet above the first, the depression of the foot of the tower is /3. Find the height of the tower. 7. A person in a balloon, which has ascended vertically from flat land at the sea level, observes the angle of depression of a ship at anchor to be 30°; after descending vertically for 600 feet he finds the angle of depression to be 15°; find the horizontal distance of the ship from the point of ascent. 8. PQ is a tower standing on a horizontal plane, Q being its foot; A and B are two points on the plane such that the z QAB is 90°, and AB is 40 feet. It is found that 3 1 cot PAQ = — and cot PBQ=^. -LU ^ Find the height of the tower. 9. A column is E.S.E. of an observer and at noon the end of the shadow is North-East of him. The shadow is 80 feet long and the elevation of the column at the observer’s station is 45°. Find the height of the column. 10. A tower is observed from two stations A and B. It is found to be due north of A and north-west of B. B is due east of A and distant from it 100 feet. The elevation of the tower as seen from A is the complement of the elevation as seen from B. Find the height of the tower. [Exs. XXXIII.] HEIGHTS AND DISTANCES. 215 11 . The elevation of a steeple at a place due south of it is 45° and at another place due west of it the elevation is 15°. If the distance between the two places be a , prove that the height of the steeple is a(sJS-l) 2 4/3 * 12 . A person stands in the diagonal produced of the square base of a church tower, at a distance 2 a from it, and observes the angles of elevation of each of the two outer corners of the top of the tower to be 30°, whilst that of the nearest corner is 45°. Prove that the breadth of the tower is a (V10-V2). 13 . a person standing at a point A due south of a tower built on a horizontal plane observes the altitude of the tower to be 60°. He then walks to B due west of A and observes the altitude to be 45°, and again at C in AB produced he observes it to be 30°. Prove that B is midway between A and C. 14 . At each end of a horizontal base of length 2 a it is found that the angular height of a certain peak is 6 and that at the middle point it is 0. Prove that the vertical height of the peak is a sin 6 sin 0 J sin (0 + 6) sin (0 - 6) 15 . A and B are two stations 1000 feet apart; P and Q are two stations in the same plane as AB and on the same side of it; the angles PAB, PBA , QAB , and QBA are respectively 75°, 30°, 45°, and 90°; find how far P is from Q and how far each is from A and B. For the folloiving 4 examples a book of tables will be wanted. 16 . At a point on a horizontal plane the elevation of the summit of a mountain is found to be 22° 15' and at another point on the plane a mile further away in a direct line its elevation is 10° 12'; find the height of the mountain. 17 . From the top of a hill the angles of depression of two successive milestones, on level ground and in the same vertical plane with the observer, are found to be 5° and 10° respectively. Find the height of the hill and the horizontal distance to the nearest milestone. 18 . A castle and a monument stand on the same horizontal plane. The height of the castle is 140 feet and the angles of depression of the top and bottom of the monument as seen from the top of the castle are 40° and 80° respectively. Find the height of the monument. 216 TRIGONOMETRY. [Exs. XXXIII.] 19. A flagstaff- PN stands on level ground. A base AB is measured at right angles to AN, the points A, B and N being in the same horizontal plane, and the angles PAN and PBN are found to be a and ft respectively. Prove that the height of the flagstaff is • sin a sin /3 ~ - — — -- * • sin (a - /3) sin (a + £) If AB = 100 feet, a = 70°, and /3=50°, calculate the height. 20. A man standing due south of a tower on a horizontal plane through its foot finds the elevation of the top of the tower to be 54 c 16'; he goes east 100 yards and finds the elevation to be then 50° 8'. Find the height of the tower. 21. A man in a balloon observes that the angle of depression of an object on the ground bearing due north is 33°; the balloon drifts 3 miles due west and the angle of depression is now found to be 21°. Find the height of the balloon. 22. From the extremities of a horizontal base-line AB, whose length is 1000 feet, the bearings of the foot C of a tower are observed and it is found that z CAB = 56° 23', iCBA = 47° 15', and that the elevation of the tower from A is 9° 25'; find the height of the tower. 196. Ex. A flagstaff is on the top of a tower vjhich stands on a horizontal plane. A person observes the angles , a and ft, subtended at a point on the horizontal plane by the flagstaff and the tower ; he then walks a known distance a toward the tower and finds that the flagstaff subtends the same angle as before; prove that the height of the tower and the length of the flagstaff are respectively a sin ft cos (a + ft) and a sm a cos (a -P 2ft) cos (a 4- 2ft) ’ Let P and Q be the top and foot of the tower, and let PR be the flagstaff. Let A and B be the points at which the measurements are taken, so that zPAQ = ft and Z PAR = Z PBR = a. Since the two latter angles are equal, a circle will go through the four points A, B, P, and R. • HEIGHTS AND DISTANCES. 217 / l ! /T \ 1 P 'fry' Q To get the height of the flagstaff we have to connect the unknown length PR with the known length AB. This may be done by connecting each with the length AR. To do this, we must first determine the angles of the triangles ARP and ARB. Since A, B, P, and R lie on a circle, we have z BRP = Z BAP = 0, and Z APB = Z ARB — 6 (say). A -- '' b Also Z APR = 90° + z PAQ = 90° + 0. Hence, since the angles of the triangle APR are together equal to two right angles, we have 180° = a + (90° + /3) + (0 + 0), so that 6 = 90° — (a 4 - 2/3).( 1 ). From the triangles APR and ABR we then have (Art. 163), sin a sin RPA sin RBA sin 6 [It will be found in Chap. XV. that each of these quantities is equal to the radius of the circle.] Hence the height of the flagstaff a sin ol a sin a , . . by (1). = PR = sin 0 cos (a + 2/3) Again and = cos BPQ — cos (a + 0), PB sin PAB sin/3 a sin APB sin 6 . Hence, from (2) and (3), by multiplication, PQ sin 0 cos (a + 6) sin 3 cos (a + 0) , T- = - Z = co s (a + 2/3) ’ by (1) - ( 2 ). (3). 218 TRIGONOMETRY. Also BQ = PQ tan BPQ = PQ tan (a 4- ft) sin ft sin (a 4- ft) = a and AQ=a+BQ=a cos (a 4- 2ft) 9 cos (a 4- 2ft) + sin ft sin (a 4- /3) — a cos (a -b 2ft) cos /3 cos (a + ft) cos (a + 2ft) If a, a, and ft be given numerically these results are all in a form suitable for logarithmic computation. 197. Ex. A man walks along a straight road and observes that the greatest angle subtended by two objects is a ; from the point where this greatest angle is subtended' lie walks a distance c along the road and finds that the two objects are now in a straight line which makes an angle ft with the road; prove that the distance between the objects is n a + ft a — ft c sin a. sin ft sec — A~~ sec 0 . r 2 2 Let P and Q be the two points and let PQ meet the road in B. B HEIGHTS AND DISTANCES. 219 If A be the point at which the greatest angle is subtended then A must be the point where a circle drawn through P and Q touches the road. [For, take any other point A' on AB and join it to P cutting the circle in B' and join A'Q and B'Q. Then Z PA'Q < Z PB'Q (Euc. I. 16), and therefore < Z.PAQ (Euc. III. 21).] Let the angle QAB be called 6. Then (Euc. III. 32) the angle APQ is d also. Hence 180° = sum of the angles of the triangle PA B = 0 4- (a + 0) + /3, so that 6 = 90 — -- + ^ . From the triangles PAQ and QAB we have PQ __ sin a , AQ _ sin f3 _ sin/5 A Q sin 6’ an c sin A QB sin (6 + a) ' Hence, by multiplication, we have PQ sin a sin {3 c sin 6 sin (6 + a) ’ sin ol sin /3 a + /3 a — f3 * cos —^— cos - ^— 2 2 PQ = c sin a sin j3 sec a + /3 a —/3 sec 2 2 220 TRIGONOMETRY. EXAMPLES. XXXIV. 1. A bridge has 5 equal spans, each of 100 feet measured from the centre of the piers, and a boat is moored in a line with one of the middle piers. The whole length of the bridge subtends a right angle as seen from the boat. Prove that the distance of the boat from the bridge is 100^/6 feet. 2. A ladder placed at an angle of 75° just reaches the sill of a window at a height of 27 feet above the ground on one side of a street. On turning the ladder over without moving its foot, it is found that when it rests against a wall on the other side of the street it is at an angle of 15° with the ground. Prove that the breadth of the street and the length of the ladder are respectively 27(3 + V3) and 27 (^6 - ^2) feet. 3. From a house on one side of a street observations are made of the angle subtended by the height of the opposite house; from the level of the street the angle subtended is the angle whose tangent is 3 ; from two windows one above the other the angle subtended is found to be the angle whose tangent is - 3; the height of the opposite house being 60 feet, find the height above the street of each of the two windows. 4. A rod of given length can turn in a vertical plane passing through the sun, one end being fixed on the ground; find the longest shadow it can cast on the ground. Calculate the altitude of the sun when the longest shadow it can cast is 3J times the length of the rod. 5. A ship A observes another ship B leaving a harbour, whose bearing is then N.W. After 10 minutes A, having sailed one mile N.E., sees B due west and the harbour then bears 60° West of North. After another 10 minutes B is observed to bear S.W. Find the distances between A and B at the first observation and also the direction and rate of B. 6. A ship sailing north sees two lighthouses, which are 6 miles apart, in a line due west; after an hour’s sailing one of them bears S.W. and the other S.S.W. Find the ship’s rate. [EXS. XXXIV.] HEIGHTS AND DISTANCES. 221 7. A ship sees a lighthouse N.W. of itself. After sailing for 12 miles in a direction 15° south of W. the lighthouse is seen due N. Find the distance of the lighthouse from the ship in each position. 8. A man, travelling west along a straight road, observes that when he is due south of a certain windmill the straight line drawn to a distant church makes an angle of 30° with the road. A mile further on the bearings of the windmill and tower are respectively N.E. and N.W. Find the distances of the tower from the windmill and from the nearest point of the road. 9. An observer on a headland sees a ship due north of him ; after a quarter of an hour he sees it due east and after another half-hour he sees it due south-east; find the direction that the ship’s course makes with the meridian and the time after the ship is first seen until it is nearest the observer, supposing that it sails uniformly in a straight line. 10. a man walking along a straight road which runs in a direction 30° east of north notes when he is due south of a certain house; when he has walked a mile further he observes that the house lies due west and that a windmill on the opposite side of the road is N.E. of him; three miles further on he finds that he is due north of the windmill; prove that the line joining the house and the windmill makes with the road the angle whose tangent is 48-25^/3 IT • 11. A, B, and C are three consecutive milestones on a straight road from each of which a distant spire is visible. The spire is observed to bear north-east at A, east at B, and 60° east of south at C. Prove that rj _|_ g /g the shortest distance of the spire from the road is —miles. lo 12. Two stations due south of a tower, which leans towards the north, are at distances a and b from its foot; if a and (3 be the elevations of the top of the tower from these stations, prove that its inclination to the vertical is 4 __ 1 b cot a - a cot /3 COt -;- • o - a 13. From a point A on a level plane the angle of elevation of a balloon is a, the balloon being south of A ; from a point B which is at a distance C south of A the balloon is seen northwards at an elevation of (3; find the distance of the balloon from A and its height above the ground. 222 TRIGONOMETRY. [Exs. XXXIV.] 14. A statue on the top of a pillar subtends the same angle a at distances of 9 and 11 yards from the pillar; if tan a = ^- , find the height of the pillar and of the statue. 15. A tower and a spire on the top of the tower subtend equal angles at a point whose distance from the foot of the tower is a; if h be the height of the tower, prove that the height of the spire is 7 a 2 + 7i 2 1 a^h 2 * 16. A flagstaff on the top of a tower is observed to subtend the same angle at two points on a horizontal plane, which lie on a line passing through the centre of the base of the tower and whose distance from one another is 2 a, and an angle (3 at a point halfway between them. Prove that the height of the flagstaff is 2 sin (3 cos a sin (a - ft)' 17. An observer in the first place stations himself at a distance a feet from a column standing upon a mound. He finds that the column subtends an angle, whose tangent is ^, at his eye which may be supposed to be on the horizontal plane through the base of the mound. On 2 moving - a feet nearer the column he finds that the angle subtended is O unchanged. Find the height of the mound and of the column. a sin a x / 18. A church tower stands on the bank of a river which is 150 feet wide and on the top of the tower is a spire 30 feet high. To an observer on the opposite bank of the river the spire subtends the same angle that a pole six feet high subtends when placed upright on the ground at the foot of the tower. Prove that the height of the tower is nearly 285 feet. 19. A person, wishing to ascertain the height of a tower, stations himself on a horizontal plane through its foot at a point at which the elevation of the top is 30°. On walking a distance a in a certain direction he finds that the elevation of the top is the same as before, and on then walking a distance - a at right angles to his former direction he finds the O elevation of the top to be 60°. Prove that the height of the tower is either \/l a or \/i a - [Exs. XXXIV.] HEIGHTS AND DISTANCES. 223 20. The angles of elevation of the top of a tower, standing on a horizontal plane, from two points distant a and b from the base and in the same straight line with it are complementary. Prove that the height of the tower is Jab feet, and, if d be the angle subtended at the top of the tower by the line joining the two points, then sin 0 = ^—^ . 21. A tower 150 feet high stands on the top of a cliff 80 feet high. At what point on the plane passing through the foot of the cliff must an observer place himself so that the tower and the cliff may subtend equal angles, the height of his eye being 5 feet ? 22. A statue on the top of a pillar, standing on level ground, is found to subtend the greatest angle a at the eye of an observer when his distance from the pillar is c feet; prove that the height of the statue is 2 c tan a feet, and find the height of the pillar. 23. A tower stood at the foot of an inclined plane whose inclination to the horizon was 9°. A line 100 feet in length was measured straight up the incline from the foot of the tower, and at the end of this line the tower subtended an angle of 54°. Find the height of the tower, having given log 2 = -30103, log 114-122 = 2 -0584726, and L sin 54° = 9*9079576. 24. A vertical tower stands on a declivity which is inclined at 15° to the horizon. From the foot of the tower a man ascends the declivity for 80 feet, and then finds that the tower subtends an angle of 30°. Prove that the height of the tower is 40 (^6 - x /2) feet. 25. The altitude of a certain rock is 47° and after walking towards it 1000 feet up a slope inclined at 30° to the horizon an observer finds its altitude to be 77°. Find the vertical height of the rock above the first point of observation, given that sin 47° = *73135. 26. A man observes that when he has walked c feet up an inclined plane the angular depression of an object in a horizontal plane through the foot of the slope is a, and that, when he has walked a further distance of c feet the depression is /3. Prove that the inclination of the slope to the horizon is the angle whose cotangent is (2 cot /3 - cot a). 224 TRIGONOMETRY. [Exs. XXXIV.] 27. A regular pyramid on a square base has an edge 150 feet long and the length of the side of its base is 200 feet. Find the inclination of its face to the base. 28. A pyramid has for base a square of side a; its vertex lies on a line through the middle point of the base and perpendicular to it, and at a distance h from it; prove that the angle a between the two lateral faces is given by the equation sin a — 2h J 2a-+ ilf- a 2 + 4/i 2 29. A flagstaff, 100 feet high, stands in the centre of an equilateral triangle which is horizontal. From the top of the flagstaff each side subtends an angle of 60° ; prove that the length of the side of the triangle is 50^/6 feet. 30. The extremity of the shadow of a flagstaff, which is 6 feet high and stands on the top of a pyramid on a square base, just reaches the side of the base and is distant 56 and 8 feet respectively from the extremities of that side. Find the sun’s altitude if the height of the pyramid be 34 feet. 31. The extremity of the shadow of a flagstaff, which is 6 feet high and stands on the top of a pyramid on a square base, just reaches the side of the base and is distant x feet and y feet respectively from the ends of that side; prove that the height of the pyramid is tan a - 6, where a is the elevation of the sun. 32. The angle of elevation of a cloud from a point h feet above a lake is a and the angle of depression of its reflexion in the lake is 8 ; prove that its height is h S | D + . sin (/3 - a) 33. The shadow of a tower is observed to be half the known height of the tower and sometime afterwards it is equal to the known height; how much will the sun have gone down in the interval, given and log 2 = -30103, L tan 63° 24' = 10*3009994, diff. for l' = 3159? [EXS. XXXIV.] HEIGHTS AND DISTANCES. 225 34. An isosceles triangle of wood is placed in a vertical plane, vertex upwards, and faces the sun. If 2 a be the base of the triangle, h its height, and 30° the altitude of the sun, prove that the tangent of the angle at the apex of the shadow is - 0 . oh z - a 1 35. A rectangular target faces due south, being vertical and standing on a horizontal plane. Compare the area of the target with that of its shadow on the ground when the sun is /3° from the south at an altitude of a 0 . 36. A spherical ball, of diameter 5, subtends an angle a at a man’s eye when the elevation of its centre is (3; prove that the height of the centre of the ball is ^ 5 sin cosec ^. u u 37. A man standing a plane observes a row of equal and equi¬ distant pillars, the 10th and 17th of which subtend the same angle that they would do if they were in the position of the first and were respectively ^ and ^ of their height. Prove that, neglecting the height of the man’s eye, the line of pillars is inclined to the line drawn to the first at an angle whose secant is nearly 2*6. For the following 4 examples a hook of tables will be wanted. 38. A and B are two points on the opposite bank of a river 1000 feet wide and between them is the mast of a ship PN ; the vertical elevation of P at A is 14° 20' and at B it is 8° 10'. What is the height of P above AP? 39. AP is a line 1000 yards long; B is due north of A and from B a distant point P bears 70° east of north; at A it bears 41° 22' east of north; find the distance from A to P. 40. A is a station exactly 10 miles west of B. The bearing of a particular rock from A is 74° 19' east of north and its bearing from B is 26° 51' west of north. How far is it north of the line AB ? 41. The summit of a spire is vertically over the middle point of a horizontal square enclosure whose side is of length a feet; the height of the spire is h feet above the level of the square. If the shadow of the spire just reach a corner of the square when the sun has an altitude 0, prove that h^2 = a tan 0. Calculate /?, having given a = 1000 feet and 0 = 25° 15'. L. T. 15 CHAPTER XV. PROPERTIES OF A TRIANGLE. 198. Area of a given triangle. Let ABC be any triangle and AD the perpen¬ dicular drawn from A upon the opposite side. Through A draw EAF parallel to BC and draw BE and CF per- the area of the triangle ABC = If rectangle BF = \BC . CF = \a . AD. But AD = AB sin B — c sin B. The area of the triangle ABC therefore —\ca sin J5. This area is denoted by A. Hence A = ^ca sin B = ^ab sin C = Jbc sin A ...(1). By Art. 169, we have sin A = ^ Vs (s — a) (s — b) (s — c), so that A = \bc sin A = Vs (s — a) (s — b) (s — c). . .(2). This latter quantity is often called S. AREA OF A TRIANGLE. 227 EXAMPLES. XXXV. Find the area of the triangle ABC when 1. a — 13, 5 = 14, and c = 15. 2. & li i—* 00 5 = 24, and c = 30. 3. a = 25, 5 = 52, and c = 63. 4. a = 125, 5 = 123, and c = 62. 5. a = 15, 5 = 36, and c = 39. 6. a = 287, 5 = 816, and c = 865 7. CO II e 5 = 84, and c = 91. 8. a = s/3, b = *J2, and c = — . 9. If -B = 45°, (7=60°, and a = 2( x /3 + l) inches, prove that the area of the triangle is 6 + 2^3 sq. inches. 10. The sides of a triangle are 119, 111, and 92 yards ; prove that its area is 10 sq. yards less than an acre. 11. The sides of a triangular field re 242, 1212 and 1450 yards; prove that the area of the field is 6 acres. 12 . A workman is told to make a triangular enclosure of sides 51, 41, and 21 yards respectively; having made the first side one yard too long, what length must he make the other two sides in order to enclose the prescribed area with the prescribed length of fencing ? 13. Find, correct to -0001 of an inch, the length of one of the equal sides of an isosceles triangle on a base of 14 inches having the same area as a triangle whose sides are 13-6, 15, and 15-4 inches. 14. Prove that the area of a triangle is \o? sin B sin C sin A If one angle of a triangle be 60°, the area 10*/3 square feet, and the perimeter 20 feet, find the lengths of the sides. 15. The sides of a triangle are in a.p. and its area is - ths of an o equal triangle of the same perimeter ; prove that its sides are in the ratio 3 : 5 : 7, and find the greatest angle of the triangle. 16. In a triangle the least angle is 45° and the tangents of the angles are in a.p. If its area be 3 square yards, prove that the lengths of the sides are 3^/5, 6-^/2, and 9 feet, and that the tangents of the other angles are respectively 2 and 3. 15—2 228 TRIGONOMETRY. [Exs. XXXV.] 17. The lengths of two sides of a triangle are one foot and y/2 feet respectively and the angle opposite the shorter side is 30°; prove that there are two triangles satisfying these conditions, find their angles, and shew that their areas are in the ratio V3 + 1: V3-1. 18. Find by the aid of the tables the area of the larger of the tw T o triangles given by the data A = 31° 15', a = 5ins. and 6 = 7 ins. 199. On the circles connected with a given triangle. The circle which passes through the angular points of a triangle ABC is called its circumscribing circle or, more briefly, its circumcircle. The centre of this circle is found by the construction of Euc. IV. 5. Its radius is always called R. The circle which can be inscribed within the triangle so as to touch each of the sides is called its inscribed circle or, more briefly, its incircle. The centre of this circle is found by the construction of Euc. iv. 4. Its radius will be denoted by r. The circle which touches the side BC and the two sides AB and AC produced is called the escribed circle opposite the angle A. Its radius will be denoted by r x . Similarly r 2 denotes the radius of the circle which touches the side CA and the two sides BC and BA produced. Also r 3 denotes the radius of the circle touch¬ ing AB and the two sides CA and CB produced. 200. To find the magnitude of R } the radius of the circumcircle of any triangle ABC. Bisect the two sides BC and CA in D and E respec¬ tively, and draw DO and EO perpendicular to BC and CA. RADIUS OF THE CIRCUMCIRCLE. 229 By Euc. IV. 5, 0 is the centre of the circumcircle. Join OB and 0(7. The point 0 may either lie within the triangle as in Fig. I., or without it as in Fig. II., or upon one of the sides as in Fig. III. Taking the first figure, the two triangles BOD and GOD are equal in all respects, so that Z BOD = Z COD , .*. Z BOD = J Z BOG = z BAG (Euc. hi. 20), = A. Also BD = B 0 sin B 0D. a .*. g = Rsin A. If A be obtuse, as in Fig. II., we have Z BOD = iZ BOG = Z BLG = 180° - A (Euc. hi. 22), so that, as before, sin BOD = sin A , and R = a 2 sin A’ If A be a right angle, as in Fig. III., we have R=OA = OC=^ a 2 sin A , since in this case sin A = 1. 230 TRIGONOMETRY. The relation found above is therefore true for all triangles. Hence, in all three cases, we have 2 sin A 2 sin B 2 sin C (Art. 163). 201. In Art. 169 we have shewn that sin A = j- Js (s — a) (s — b ) (s — c) = be 2S be’ where S is the area of the triangle. Substituting this value of sin .4 in (1), we have abc giving the radius of the circumcircle in terms of the sides. 202. To find the value of r, the radius of the ineircle of the triangle ABC. Bisect the two angles B and C by the two lines BI and Cl meeting in /. By Euc. ill. 4 ,1 is the centre of the incircle. Join I A, and draw ID , IE and IF perpendicular to the three sides. Then ID=IE=IF=r. We have area of A IBC area of A IGA area of A I A B \ID . BG = ^r.a, \IE . GA = Jr. 6, \IF. AB = Jr.c- and RADIUS OF THE INCIRCLE. 231 Hence, by addition, we have . a + \r. b + Jr. c = sum of the areas of the triangles IBC, IGA , and IAB = area of the A ABC. i.e. so that r a -f- b -f- c 2 r ,s = 8. 203. Since the angles IBD and IDB are respectively equal to the angles IBF and IFB, the two triangles IDB and IFB are equal in all respects. Hence BD = BF, so that 2 BD = BD + BF. So also AE = AF, so that 2AE= AE + AF, and CE = CD, so that 20E = CE + CD. Hence, by addition, we have 2 BD + 2AE+2CE=(BD +CD)+ ( CE+ AE) + (AF+FB), i.e. 2 BD + 2AC = BC+CA + AB. 2BD -}- 2b = a + b -l - c — 2s. Hence so and ID BD BD = s — b = BF ; CE = 5 — c = CD, AF=s — a = AE. B = tan IBD = tan 2 ’ B r = ID = BD tan — = (s — b) tan & B 2 • Now 232 TRIGONOMETRY. So r = IE = CE tan ICE = (s — c) tan ^, and also r = IF = FA tan IAF = (s — a) tan 2 ’ Hence r = (s — a) tan ^ = (s — b) tan ? = (s — c) tan 204. we have A third value for r may be found as follows: a = BD + DC = ID cot IBD + ID cot ICD B G = r cot y + r cot — _ r B Ci COS -g cos — + “ sin B 2 sin C 2 J 5 . a a sm -s- sin — = r ’ . G B C . B' sm — cos -a + cos sm 77 2 2 2 2 ■B C\ r sm | + -5 ) = t sm 90°-^ A = r cos — r = a . B . G sm 7 , sm 77 2 2 A cos ^ Cor. we have J J Since a = 2 B sin A = 4i2 sin — cos , AT> . A . B . C r = 4 K sm — sm — sm — . 2 2 2 205. To find the value of r lf the radius of the escribed circle opposite the angle A of the triangle ABC. RADII OF THE ESCRIBED CIRCLES. 233 Produce AB and AC to L and M . Bisect the angles CBL and BCM by the lines BI 1 and CI 1 and let these lines meet in / 1# Draw I 1 D l , I 1 E 1) and I-JP -i perpendicular to the three sides respectively. The two triangles IJ)iB and IJ?JS are equal in all respects, so that I 1 F 1 = /iA* Similarly I 1 E 1 = / X A« The three perpendiculars /jA, /iA and I 1 F 1 being equal, the point I 1 is the centre of the required circle. Now the area ABIfi is equal to the sum of the triangles ABC and I X BC\ it is also equal to the sum of the triangles IjBA and I^CA. Hence A ABC + AI 1 BC= AI.CA+AI.AB. 8 + i^A • BC= i/.A. CA +. AB , i.e. S + . a = . b + . c. .*. S = r-t b + c — a — /v» ”6 + c 4- a ' 1 2 — ' 1 2 a = (5 — a). r, = s - a Similarly it can be shewn that S , S r 2 = -- , and r 3 = s — b s — c 234 TRIGONOMETRY. 206. Since AE l and AF X are tangents, we have, as in Art. 203, AE X — AF 1# i Similarly BF X = BD X , and GE X = GD X . 2AE X = AE X + AF X = AB + BF\ + AG + GE 1 = AB + BB 1 + AG+CD 1 = AB + BG + GA =2s. AE X = s = AF X . Also BD X = BF 1 = AF X — AB = ,s — c, and GD x = CE x = AE x -AG = s-b. I X E X = A E x tan I X AE X , . A i.e. rj = s tan —. / 207. A third value may be obtained for r x in terms of a and the angles B and G. For, since I X G bisects the angle BCE 1} we have Z I X GD X = 1(180° - G) = 90° - J. So Z I X BD\ = 90° - ^. a = BG= BD X + D x G = I X D X cot I X BD X + 1J-K cot I X GD X (tan ^ + tan \ 2 RADII OF THE ESCRIBED CIRCLES. 235 B C / . B C B . C a cos cos 77 — r, sin 77 cos -7 + cos -7 sin — 2 2 V 2 2 2 2 = r x sin C\ . / „ ^4\ A + ~2 )= r i sin ( 90 - j) = r i cos j ■ B G 2 r, = a B C COS 77 COS 77 2 2 A cos ^ Cor. Since a = 2 R sin A = 4R sin 77 cos 4 2 2 we have , D . A B G r, = 4 it sm 77 cos 77 cos 77 . 2 2 2 EXAMPLES. XXXVI. 1. In a triangle whose sides are 18, 24, and 30 inches respectively, prove that the circumradius, the inradius, and the radii of the three escribed circles are respectively 15, 6, 12, 18, and 36 inches. 2. The sides of a triangle are 13, 14, and 15 feet; prove that (1) R = Si ft., (2) r = 4 ft., (3) r-^lOJ ft., (4) r 2 = 12 ft., and (5) r 3 =14 ft. 5 3. In a triangle ABC if a = 13, 5 = 4, and cos C= - , find J-O R, r , r l9 r 2 , and r 3 . 4. In the ambiguous case of the solution of triangles prove that the circumcircles of the two triangles are equal. Prove that 5. r 1 + r 2 + r 3 -r=4E. 6. r 1 r 2 + r 2 r 3 + r 3 r 1 = s 2 . ABC 7. cot 2 -cot 2 -cot 2 -. 8. rr 1 r 2 r 3 = S 2 . 9. - + — + — = 0. 10 S = 2R 2 sin A sin B sin C. r i r 2 r 3 r 236 TRIGONOMETRY. [Exs. XXXVI.] 11 , 4 R sin A sin B sin C = a cos A + b cos B + c cos C. 0 ATi A B C 10 rrj . J 12. S = 4JJrcos g-cos^cos g . 13. 7f= tan 2 ’ 2 3 14. (s - a) = r 2 (s - 6) = r 3 (s - c) = rs = S. 15. « (n*i + r 2 r 3 ) = 6 (rr 2 + r 3 r x ) = c (rr 3 + r 2 r 2 ). 16 . a 2 -f Z > 2 + c 2 S 2 17 . rrjCot g-=jSL 18. (ri-r)(r 2 -r) (r 3 - r) = 4JRr 2 . 19. (rj + r 2 ) tan ■— = (r 3 - ?•) cot ^ = c. __ 1 1 1 1 __ r-, r« Vo 1 1 20. ~T + 1-1-— oV) ♦ 21. 7-1-1—r —- od * a& fcc ca 2Br be ca ah r 2 R 22 . r 2 + ?*i 2 + r 2 2 + r 3 2 = 16R 2 - a 2 - b 2 - c 2 . 208. Orthocentre and pedal triangle of any triangle. Let ABC be any triangle and let AK, BL , and Cilf be the perpendiculars from J., 5, and C upon the opposite sides of the tri¬ angle. It can be easily shewn, as in most editions of Euclid, that these three perpendiculars meet in a common point P. This point P B is called the orthocentre of the triangle. The triangle KLM, which is formed by joining the feet of these perpendiculars, is called the pedal triangle of ABC. 209. Distances of the orthocentre from the angular points of the triangle. PEDAL TRIANGLE. 237 We have PK = KB tan PBK = KB tan (90° — G) Q = AB cos B cot G = —— 7 . cos B cos G sm U = 2 R cos B cos G (Art. 200). Again AP = AK - PK = c sin B - PK = 25 sin (7 sin B — 25 cos B cos G = — 22? cos (5 -f G) = 22? cos A (Art. 72). So 5P = 22? cos B, and GP = 22? cos G. The distances of the orthocentre from the angular points are therefore 22? cos A, 22? cos B and 22? cos G; its distances from the sides are 22? cos B cos (7, 22? cos G cos and 22? cos A cos 2?. 210. To find the sides and angles of the pedal triangle . Since the angles PKG and PLG are right angles, the points P, P, (7, and K lie on a circle. .-. z PKL = Z P(7Z (Euc. hi. 21) = 90° - A. Similarly P, K , P, and if lie on a circle, and therefore z PKM= Z.PBM = 90 °-A. Hence Z MKL = 180° — 2 A = the supplement of 2 A. So Z KLM — 180° — 25, ZLMK= 180° -2(7. and 238 TRIGONOMETRY. Again, from the triangle ALM , we have LM _ AL _A5cosA sin A sin AML cos PML c cos A c cos A cos PAL sin (7 .\ LM— ——~ sin A cos A sm 0 = a cos A (Art. 163). So MK — b cos B, and KL = c cos 5. The sides of the pedal triangle are therefore a cos A, b cos B, and c cos C ; also its angles are the supplements of twice the angles of the triangle. 211. Let I be the centre of the incircle and I ly I 2 and / 3 the centres of the escribed circles which are opposite to A, B and G j respectively. As in Arts. 202 and 205 IC bisects the angle AGB, and I\G bisects the angle BGM. /. z ICI 1 = z IGB -f z IjCB = i aACB + ^zMCB = ±[zACB + zMGB ] = \ . 180° = a right angle. Similarly Z ICI 2 is a right angle. Hence IiCI 2 is a straight line to which IC is perpen¬ dicular. So JoA/ 3 is a straight line to which I A is perpen¬ dicular, and I%BI 1 is a straight line to which IB is perpen¬ dicular. CENTROID AND MEDIANS. 239 Also, since I A and I 2 A both bisect the angle BAC, the three points A , 7, and I 1 are in a straight line. Similarly BII 2 and CII 3 are straight lines. Hence IJlJ 3 is a triangle which is such that A, B, and G are the feet of the perpendiculars drawn from its vertices upon the opposite sides and such that 7 is the intersection of these perpendiculars, i.e. ABC is its pedal triangle and 7 is its orthocentre. 212. Centroid and Medians of any Triangle. If ABC be any triangle, and D, E , and F respectively the middle points of BC, CA , and AB, the lines AD, BE, and CF are called the medians of the triangle. It is shewn in any edition of Euclid that the medians meet in a common point G, such that AG = %AD, BG — \BE, and CG = \CF\ This point G is called the centroid of the triangle. 213. Length of the medians. We have, if AD = x, b 2 = AC 2 = AD 2 + DC 2 - 2 AD . DC cos ADC a* = x 2 + —r — ax cos A DC, 4 a A c 2 = AB 2 = x 2 + — — ax cos ADB 4 cr = x 2 H — T - + ax cos ADC. 4 and 240 TRIGONOMETRY. Hence, by addition, we have b 2 + c 2 = 2x 2 + . AD = x = J Jib 2 4- 2c 2 — a 2 . Hence also AD = | V& 2 + c 2 + 26c cos A (Art. 164). So also BE= \ J2c 2 + 2a 2 - 6 2 , and CD = 1 V2u 2 + 26 2 - c 2 . 214. Angles that the median AD makes with the sides. If the Z BAD = /3, and Z GAD = 7, we have sin 7 _ DC _ a sin G AD 2# * a sin G a sin C sm 7 = —-= -- _— . 2a? V26 2 + 2c 2 - a 2 Similarly sin /3 = a sin 5 V2& 2 + 2c 2 — a 2 Again, if the Z ADC be 0, we have sin 6 __ A (7 _ b sin (7 AD # ‘ . „ b sin G 2 b sin G sm 0 =-- = , -- . a J2b 2 + 2c 2 - a 2 The angles that AD makes with the sides are therefore found. 215. The centroid lies on the line joining the circum- centre to the orthocentre. CIRCUMCENTRE AND ORTHOCENTRE. 241 Let 0 and P be the circumcentre and orthocentre respectively. Draw OD and PK perpendicular to BC. A Let AD and OP meet in G. The triangles OGD and PGA are clearly equiangular. Also, by Art. 200, OD = R cos A and, by Art. 209, AP = 2R cos A. Hence, by Euc. vi. 4, AG = AP_ GD ~ 0D~ ' The point G is therefore the centroid of the triangle. Also, by the same proposition, OG = OD_l GP “ AP ~ 2* The centroid therefore lies on the line joining the circumcentre to the orthocentre and divides it in the ratio 1 : 2 . It may be shewn by geometry that the centre of the nine-point circle (which passes through the feet of the perpendiculars, the middle points of the sides, and the middle points of the lines joining the angular points to the orthocentre) lies on OP and bisects it. The circumcentre, the centroid, the centre of the nine-point circle, and the orthocentre therefore all lie on a straight line. 216. Distance between the circumcentre and the ortho- centre. L. T, 16 242 TRIGONOMETRY. If OF be perpendicular to AP, we have zOAP= 90° - z.A0F= 90° -0. Also Z PAP = 90°-0. zOAP = A-zOAP-zPAP = A - 2 (90° —C) = A + 2(7-180 = A + 20 - (A + P + G) = 0 - B. Also OA = P, and, by Art. 209, PA = 2R cos A. .-. OP 2 = OA 2 + PA 2 — 20A . PA cos OAP = JB 2 + 4P 2 cos 2 A — 4P 2 cos A cos (O — P) = P 2 + 4P 2 cos A [cos A — cos ((7 — 5)] = R 2 — 4P 2 cos A [cos (P 4- O) + cos ((7 — P)] (Art. 72), = P 2 — 8 R 2 cos A cos 5 cos (7. OP = R Vl — 8 cos A cos P cos (7. *217. To find the distance between the circumcentre and the incentre. Let 0 be the circumcentre and OF perpendicular to AB. perpendicular to AC. Then, as in the last article, Let I be the incentre and IE Z OAF =90° -C. zOAI = zIAF-zOAF A+P+O C-B 2 2 CIRCUMCENTRE AND INCENTRE. 243 Also AI = IE . A . san — sin ^ 2 2 —^ = 4 R sin ~ sin ^ (Art. 204. Cor.). .'. OP = OA 2 + AP — 20A . AI cos OAI = R 2 + 16 A 2 sm 2 — sin 2 —— 8 R 2 sin -- sin cos —-— 2 2 2 2 2 OP .. a . 2 B. 2 C . = 1 + 16sin- -sin 2 - ft • B • C — 8 sm -r- sm — 2 2 B G . B . C' cos y cos y + sin sm — = 1 - = 1 - 0 . B . 0 / B C . B . C' 8 sm — sm — cos cos -r-sinr- sm — 2 2 V 2 2 2 2 0 . B . G B + G 8 sm — sm — cos - — 2 2 2 8 sin ~ sin ^ sin ^ (Art. 69).( 1 ). .-. 01 = + - Q . B . 0 . A 1 — 8 sm —- sm — sm - 7 T . 2 2 2 Also (1) may be written OP = R 2 — 2 ii x 4i2 sin ^ sin ^ sin ^ ^ -j Zl = i £ 2 — 2 i£r. (Art. 204 Cor.) In a similar manner it may be shewn that, if 7 X be the centre of the escribed circle opposite the angle A , we shall have 0 I l = R \/ 1 + 8 sin ^ cos — cos , and hence 04 2 = i2 2 + 2i?n. (Art. 207. Cor.) 16—2 244 TRIGONOMETRY. 218. Bisectors of the angles. If AD bisect the angle A and divide the base into portions x and y, we have, by Euc. vi. 3, x AB c y AC b ’ x _y _ x + y a 6 ( 1 ), c o b+c 6 + c giving x and y. Also, if 8 be the length of AD and 6 the angle it makes with BC , we have AABD + AACD = A ABC. 1 A 1 A 1 P\ • . A 7 C\ • 7 • A - co sm — + - be sin — = - be sm A , 2 2 2 2 2 i.e. 8 = be sin A 6 + c . A sm- 26c A T ~— cos ^ 6 + c 2 ( 2 ). Also d=180°-a- 2 A A+ 2B A + B ^ 2 * * * We thus have the length of the bisector and its inclination to BC. EXAMPLES. XXXVII. If J, I l9 I 2 , and J 3 be respectively the centres of the incircle and the three escribed circles of a triangle ABC, prove that i dr A 1, AI=r cosec —. A ABC 2. I A . IB . IC = abc tan — tan — tan —. A Jj u 3. AI 1 = r 1 cosec ^ . 4 . I leasee-. [EXS. XXXVII.] PROPERTIES OF TRIANGLES. 245 5. 7. 9. 10 . I 2 I 3 = a cosec —. 6. II X . IZ 2 . ir 3 =16R 2 r. Li I.,I 3 2 =4JR (r 2 + r 3 ). 8. / W 2 = —^. II 2 + L?1.2 = 112 + 2 = J/,2 + *^2. ABC Area of AJ 1 J 2 J 3 = 81^ 2 cos — cos — cos — = Li Z Li abc 2r ’ 11 . IIj « LI3 Ho • JgZj ZZ3 . sin A ~~ sin R ~ sin C If I, 0, and P be respectively the incentre, circumcentre, and ortho¬ centre, and G the centroid of the triangle ARC, prove that 12. I0 2 = R 2 (3 - 2 cos A - 2 cos R - 2 cos C). 13. IP 2 = 2r 2 - 4R 2 cos A cos R cos C. 14. OG 2 = E 2 -ha 2 + 6 2 + c 2 ). y 15. Area of AIOP=2R 2 sin sin ^ ^ sin ~~ ♦ Li Li Li m a e rvn 2 ™ 16 . Area of a IPG = -R 2 sin ——sin —-—sin—-—. O Li Li Li 17. Prove that the distance of the centre of the nine-point circle from the angle A is — Jl + 8 cos A sin R sin C. Li 18 . DEF is the pedal triangle of ARC; prove that (1) its area is 2S cos A cos R cos C, JR (2) the radius of its circumcircle is — , Li and (3) the radius of its incirele is 2R cos A cos B cos C. 19. 0i0 2 0 3 is the triangle formed by the centres of the escribed circles of the triangle ARC; prove that A R C (1) its sides are 4R cos — , 4R cos — , and 4R cos —, Li Li Li . TT A 7T B , 7T C (2) its angles are - - -, 3 _ 2 ’ and 2 _ 2 ’ and (3) its area is 2 Rs. 246 TRIGONOMETRY. [Exs. XXXVII.] 20. DEF is the triangle formed by joining the points of contact of the incircle with the sides of the triangle ABC ; prove that A B C (1) its sides are 2 r cos — , 2 r cos —, and 2r cos —, Si Si u (2) its angles are ~ | and | and 1 ( 3 ) its area is , i.e. ^ ^ S. abcs 2 lx 21 . D , E , and F are the middle points of the sides of the triangle ABC ; prove that the centroid of the triangle DEF is the same as that of ABC and that its orthocentre is the circumcentre of ABC. In any triangle ABC , prove that 22. The perpendicular from A divides BC into portions which are proportional to the cotangent of the adjacent angles, and that it divides the angle A into portions whose cosines are inversely proportional to the adjacent sides. 23. The median through A divides it into angles whose cotangents are 2 cot A + cot C and 2 cot A + cot B , and makes with the base an angle whose cotangent is \ (cot C - cot B). Si 24. The distance between the middle point of BC and the foot of the J) 2 _ q2 perpendicular from A is —-— . SCL 25. 0 is the orthocentre of a triangle ABC; prove that the radii of the circles circumscribing the triangles BOC , CO A, AOB and ABC are all equal. 26. AD, BE and CF are the perpendiculars from the angular points of a triangle ABC upon the opposite sides; prove that the diameters of the circumcircles of the triangles AEF , BDF and CDE are respectively a cot A, b cot B , and c cot C, and that the perimeters of the triangles DEF and ABC are in the ratio r : R. 27. Prove that the product of the distances of the incentre from the angular points of a triangle is 4 Rr 2 . 28. The triangle DEF circumscribes the three escribed circles of the triangle ABC; prove that EF FD DE [Exs. XXXVII.] PROPERTIES OF TRIANGLES. 247 29. If a circle be drawn touching the inscribed and circumscribed circles of a triangle and the side BG externally, prove that its radius is 30. If a, b, c be the radii of three circles which touch one another externally and and r 2 be the radii of the two circles that can be drawn to touch these three, prove that 1 12 2 2 — 4-= —f 7 4— • r 2 a b c 31. If A 0 be the area of the triangle formed by joining the points of contact of the inscribed circle with the sides of the given triangle, whose area is A, and A x , A 2 , and A 3 the corresponding areas for the escribed circles, prove that A x 4* A 2 + A 3 — Aq — 2 A. 32. If the bisectors of the angles of a triangle ABC meet the opposite sides in A', B', and (?', prove that the ratio of the areas of the triangles A'B'C' and ABC is 2 sin A . B . C — sin - sin — : cos Ji Ji A-B B-C C-A 2 cos 2 C0S -T'* 33. Through the angular points of a triangle are drawn straight lines which make the same angle a with the opposite sides of the triangle; prove that the area of the triangle formed by them is to the area of the original triangle as 4 cos 2 a : 1. 34. Two circles, of radii a and 6, cut each other at an angle 9. Prove that the length of the common chord is 2 ab sin 9 Ja? 4- b 2 + 2 ab cos 9 35. Three equal circles touch one another; find the radius of the circle which touches all three. 36. Three circles whose radii are a, b and c touch one another and the tangents at their points of contact meet in a point; prove that the distance of this point from either of their points of contact is / abc \h \a + b + c) 248 TRIGONOMETRY. [Exs. XXXVII.] 37. In the sides BC , CA, AB are taken three points A\ B', C’ such that BA' : A’C=CB' : B'A = AC' : C'B = m : n\ prove that if AA', BB\ and CC' be joined they will form by their inter¬ sections a triangle whose area is to that of the triangle ABC as (m - n) 2 : m 2 + ran + n 2 . 38. The circle inscribed in the triangle ABC touches the sides BC , CA , and AB in the points A lf B lf and C 1 respectively; similarly the circle inscribed in the triangle A 1 B 1 C 1 touches the sides in A 2 , B 2 , C 2 respectively and so on; if A n B n C n be the nth triangle so formed, prove that its angles are and | + (-2 )-"(c-|). Hence prove that the triangle so formed is ultimately equilateral. 39. A 1 B 1 C l is the triangle formed by joining the feet of the perpen¬ diculars drawn from ABC upon the opposite sides; in like manner A (2 B 2 C 2 is the triangle obtained by joining the feet of the perpendiculars from A 19 B 1 , and C 1 on the opposite sides and so on. Find the values of the angles A n , B n , and C n in the nth of these triangles. CHAPTER XYI. ON QUADRILATERALS AND REGULAR POLYGONS. 219. To find the area of a quadrilateral which is inscribable in a circle . Let ABCD be the quadrilateral, the sides being a, b, c and d as marked in the figure. The area of the quadrilateral = area of A ABC-{-area of A ADC = Raisin B+\cd sin D (Art. 198.) = \ (ab + cd) sin B , since, by Euc. in. 22, ZB = 180°- ZD, and therefore sin B = sin D. .. We have to express sin B in terms of the sides. We have a 2 + b 2 — 2ab cos B = AC 2 = c 2 + d 2 — 2 cd cos D. But cos D = cos (180° — B) = — cos B . 250 TRIGONOMETRY. Hence a * 2 4 - b 2 — 2 ab cos B = c 2 + d ' 2 -f 2 cd cos B , so that Hence cos £ = a 2 4 - b 2 —c 2 — d 2 2 (ab 4- cd) sin 2 5 = 1 — cos 2 B — (a 2 4 - b 2 — c 2 — c ? 2 ) 2 {2 (ab 4 - cd)} 2 {2 (ab 4 - cd)} 2 — [a 2 4 * b 2 — c 2 — d 2 } 2 4 (ab 4- cd) 2 {2 (ab +cd)+(a 2 +b 2 —c 2 —d 2 )} [2(ab -\-cd)—(a 2 +b 2 —c 2 —d 2 )} 4 (ab + cd) 2 { (a 2 4-2a6 4- b 2 )—(c 2 — 2cd -f d 2 ) } {(c 2 4- 2 cd 4 - d: 2 ) —(a 2 4- b 2 —2 ab) j 4 (ab 4 - cd) 2 {(a 4- b) 2 — (c — c?) 2 } {(c 4- e £) 2 — (a — 6 ) 2 } 4 (a& 4- cd) 2 [((X + b + c — d) (a+b —c 4 "^)| [(c-\-d 4 " a — b)(c-\-d — a -(- b)} Let 4 (ab 4- cd) 2 Oj 4 - b 4 - c 4 - d = 2s, so that J27 to *J32. 7. Prove that the radius of the circle described about a regular pentagon is nearly -^ths of the side of the pentagon. 8 . If an equilateral triangle and a regular hexagon have the same perimeter, prove that their areas are as 2 : 3. 9. If a regular pentagon and a regular decagon have the same perimeter, prove that their areas are as 2:^5. 10 . Prove that the sum of the radii of the circles, which are respec¬ tively inscribed in and circumscribed about a regular polygon of n sides, is. 7r 2n ’ where a is a side of the polygon. 11 . Of two regular polygons of n sides, one circumscribes and the other is inscribed in a given circle. Prove that the three perimeters are in the ratio 7T 7T 7T . sec - : - cosec -: 1 , n n n 7T and that the areas of the polygons are in the ratio cos 2 - : 1 . 12 . Given that the area of a polygon of n sides circumscribed about a circle is to the area of the circumscribed polygon of 2 n sides as 3 : 2, find n. 13. Prove that the area of a regular polygon of 2 n sides inscribed in a circle is a mean proportional between the areas of the regular inscribed and circumscribed polygons of n sides. 14. The area of a regular polygon of n sides inscribed in a circle is to that of the same number of sides circumscribing the same circle as 3 is to 4. Find the value of n. 15. The interior angles of a polygon are in a. p. ; the least angle is 120° and the common difference is 5°; find the number of sides. XXXIX.] REGULAR POLYGONS. 259 16. There are two regular polygons the number of sides in one being double the number in the other, and an angle of one polygon is to an angle of the other as 9 to 8 ; find the number of sides of each polygon. 17. Show that there are eleven pairs of regular polygons such that the number of degrees in the angle of one is to the number in the angle of the other as 10 : 9. Find the number of sides in each. 18. The side of a base of a square pyramid is a feet and its vertex is at a height of li feet above the centre of the base ; if 6 and 0 be respec¬ tively the inclinations of any face to the base, and of any two faces to one another, prove that 19. A pyramid stands on a regular hexagon as base. The perpendi¬ cular from the vertex of the pyramid on the base passes through the centre of the hexagon and its length is equal to that of a side of the base. Find the tangent of the angle between the base and any face of the pyramid and also of half the angle between any tw T o side faces. 20 . A regular pyramid has for its base a polygon of n sides, each of length a, and the length of each slant side is l ; prove that the cosine of the angle between two adjacent lateral faces is 4 1 2 cos-b a 2 n 4 1 2 - a 1 17—2 CHAPTER XVII. TRIGONOMETRICAL RATIOS OF SMALL ANGLES. AREA OF A CIRCLE. DIP OF THE HORIZON. 227. If 6 be the number of radians in any angle , which is less than a right angle, then sin 6, 0 and tan 6 are in ascending order of magnitude. Let TOP be any angle which is less than a right angle. With centre 0 and any radius OP describe an arc PAP' meeting OT in A. Draw PN perpendicular to OA and produce it to meet the arc of the circle in P\ Draw the tangent FT at P to meet OA in T and join TP'. The triangles PON and P'ON are equal in all respects, so that PN = NP' and arc PA = arc AP\ Also the triangles TOP and TOP' are equal in all respects, so that TP = TP'. SIN 0 < 0 < TAN 0 . 261 The straight line PP' is less than the arc PAP', so that NP is < arc PA. We shall assume that the arc PAP' is less than the sum of PT and TP', so that arc PA < PT. Hence NP, the arc AP, and PT are in ascending order of magnitude. Therefore NP OP’ arc AP OP and PT OP are in ascending order of magnitude. But NP OP = sin A OP = sin 0, = num ber of radians in Z.AOP = 0 (Art. 21). and PT OP = tan POT = tan AOP = tan 6. Hence sin 0, 0, and tan 0 are in ascending order of magnitude, provided that 7T 228. Since sin 0 < 0 < tan 0, we have, by dividing each by the positive quantity sin 0, 1 < 1 cos 0 * Hence -—^ always lies between 1 and-. sm 0 J cos 0 This holds however small 0 may be. Now when 0 is very small cos 0 is very nearly unity, and the smaller 0 becomes, the more nearly does cos 0 become unity, and hence the more nearly does become unity. cos 0 262 TRIGONOMETRY. Hence when 0 is very small the quantity 0 sin 6 lies between 1 and a quantity which differs from unity by an indefinitely small quantity. In other words, when 0 is made indefinitely small the quantity , and therefore , is ultimately equal to unity, i.e . the smaller an angle becomes the more nearly is its sine equal to the number of radians in it. This is often shortly expressed thus; sin 0 = 6, when 0 is very small. So also tan 6 = 6, when 6 is very small. a Cor. Putting 6 = -, it follows that, when 6 is indefi¬ n nitely small, n is indefinitely great. . a sm - n . Hence-is unity, when n is indefinitely great. a n a So n sin - = a, when n is indefinitely great. 229. In the preceding article it must be particularly noticed that 6 is the number of radians in the angle considered. The value of sin a°, when a is small, may be found. For, since if = 180°, we have sin a 0 = sin by the result of the last article. it a 180’ RATIOS OF SMALL ANGLES. 263 230. From the tables it will be seen that the sine of an angle and its circular measure agree to 7 places of decimals so long as the angle is not greater than 18'. They agree to the oth place of decimals so long as the angle is less than about 2°. 231. ife be the number of radians in an angle , which is less than a 6 s - 6 2 right angle , then sin 6 is > 6 - and cos 6 is > 1 - — . By Art. 227 we have .60 6 sin - > ~ cos - Hence, since sin 6 = 2 sin - cos » , we have But since, by Art. 227, . 6 0 sm - < - therefore Again, cos 6 = 1-2 sin 2 ^ ; z we have therefore, since It will be found in a later chapter that 232. Ex. 1 . Find the values of sin 10' and cos 10'. Since 264 TRIGONOMETRY. we have sin 10 ' = sin 7T 7r 180x6 180x6 3-14159265... = -0029089 nearly. Also 180 x 6 cos 10 '= sj 1 - sin 2 10 ' = [1- -000008468.. .]* = l-i [-000008468.. .]> z approximately by the Binomial Theorem, = 1--000004234... = •9999958.... Ex. 2. Solve approximately the equation sin 9 = -52. Since sin 6 is very nearly equal to i 6 must be nearly equal to Cl 7T 7r . 77 cos x + cos 77 sin# b Let then 6 = - + x, where x is small, b .-. -52 = sin -f- x^j =sin^ 1 J3 . = - cos x + sm x. z z Since x is very small, we have cos#=l and sin x — x nearly. - — Hence 2 /3 C .-. # = -02 x radians = =1-32° nearly. sj 5 75 0 = 31° 19' nearly. EXAMPLES. XL. Taking k equal to 3*14159265, find to 5 places of decimals the value of 1. sin 7'. 4. cos 15'. 2 . sin 15". 5 . cosec 8 ". 3 . sin 1 '. 6 . sec 5'. k i«x> [Exs. XL.] EXAMPLES. 265 Solve approximately the equations 7 . sin 6 = *01. 9. 8 . sin 6 = ' 48. 10. cos 6 = *999. 11 . Find approximately the distance at which a halfpenny, which is an inch in diameter, must be placed so as to just hide the moon, the angular diameter of the moon, that is the angle its diameter subtends at the observer’s eye, being taken to be 30'. 12. A person walks in a straight line toward a very distant object and observes that at three points A, B , and C the angles of elevation of the top of the object are a, 2a, and 3a respectively ; prove that AB = SBC nearly. 13. If 6 be the number of radians in an angle which is less than a right angle, prove that cos 6 is < 1 - 6 1 6 1 2 + 16' I [ 14. Prove the theorem of Euler, viz. that 6 6 6 sin 6 = 6 . cos - . cos — . cos .ad. inf. tt? -i . Q 0 . 6 6 2 . 6 6 6 We have sin 6 = 2 sin - cos - = 2 2 sin — cos cos - . e e e e = sin 2 3 cos - 3 cos 22 cos ^ =. n „ . 0 66 6 6 = 2 n sm 2^ x cos g • cos 22 . cos 23.cos^. Make n indefinitely great so that, by Art. 228 Cor., 2 n siu-^- = 0. 2 n 6 6 6 ”1 Hence sin 6=6 . cos - . cos —, cos —.ad inf. 2 2- J 15. Prove that ^1 - tan 2 0 ^1 - tan 2 ^ ^1 - tan 2 ^ } ^.ad inf. = 6 . cot 6. 266 TRIGONOMETRY. 233. Area of a circle. By Art. 225 the area of a regular polygon of n sides, which is inscribed in a circle of radius R, is W r> 2 • 27r ~ A 2 sin — . z n Let now the number of sides of this polygon be inde¬ finitely increased, the polygon always remaining regular. It is clear that the perimeter of the polygon must more and more approximate to the circumference of the circle. Hence, when the number of sides of the polygon is infinitely great, the area of the circle must be the same as that of the polygon. -x T n . 27t n Now - R 2 sm — = - R 2 2 n 2 sm 27T \7T n sm 27r n 2tt n = ttR- n 2ir n = vR-.^, where 6-^. 6 n When n is made infinitely great the value of 6 becomes sin 0 infinitely small and then, by Art. 228, ^ — is unity. The area of the circle therefore = 7TR 2 = it times the square of its radius. 234. A rea of the sector of a circle . Let 0 be the centre of a circle, A B the bounding arc of the sector, and let Z A OB — a radians. By Euc. VI. 33, since sectors are to one another as the arcs on which they stand, we have area of sector A OB _ arc AB area of whole circle circumference Ret a 2ttR 27r AREA OF A CIRCLE. 267 .*. area of sector AOB = — x area of whole circle L7T = tt- x 7 tR 1 — — R 2 . oc. 2i t 2 EXAMPLES. XLI. 1. Find the area of a circle whose circumference is 74 feet. 2. The diameter of a circle is 10 feet; find the area of a sector whose arc is 22J°. 3. The area of a certain sector of a circle is 10 square feet; if the radius of the circle be 3 feet, find the angle of the sector. 4. The perimeter of a certain sector of a circle is 10 feet; if the radius of the circle be 3 feet, find the area of the sector. 5. A strip of paper two miles long and *003 of an inch thick is rolled up into a solid cylinder ; find approximately the radius of the circular ends of the cylinder. 6 . A strip of paper, one mile long, is rolled tightly up into a solid cylinder, the diameter of whose circular ends is 6 inches; find the thick¬ ness of the paper. 7. Given two concentric circles of radii r and 2r; two parallel tangents to the inner circle cut off an arc from the outer circle ; find its length. 8 . The circumference of a semicircle is divided into two arcs such that the chord of one is double that of the other. Prove that the sum of the areas of the two segments cut off by these chords is to the area of the semicircle as 27 is to 55. ['-?■] 9. If each of 3 circles, of radius a, touch the other two, prove that 4 the area included between them is nearly equal to — a 2 . 268 TRIGONOMETRY. [Exs. XLI.] 10 . Six equal circles, each of radius a, are placed so that each touches two others, their centres being all on the circumference of another circle ; prove that the area which they enclose is 2a 2 (3V3-7r). 11. From the vertex A of a triangle a straight line AD is drawn making an angle 6 with the base and meeting it at D. Prove that the area common to the circumscribing circles of the triangles ABD and ACD is j (b 2 y + c 2 /3 - 2 be sin A) cosec 2 $, where $ and y are the number of radians in the angles B and C respec¬ tively. 235. Dip of the Horizon. Let 0 be a point at a distance h above the earth’s surface. Draw tangents, such as OT and OT\ to the surface of the earth. The ends of all these tangents all clearly lie on a circle. This circle is called the Offing or Visible Horizon. The angle that each of these tangents OT makes with a horizontal plane POQ is called the Dip of the Horizon. Let r be the radius of the earth and let B be the other end of the diameter through A. We then have, by Euc. in. 36, OT*=OA.OB = h(2r + h ), so that OT =*Jli (2 r + h ). This gives an accurate value for OT. In all practical cases, however, h is very small com¬ pared with r. [r = 4000 miles nearly and h is never greater, and generally is very considerably less, than 5 miles.] DIP OF THE HORIZON. 269 Hence h 2 is very small compared with hr. As a close approximation we have then The dip OT = V 2hr. = Z TOQ = 90° -z COT = z OCT. Also + nnw \/2hr tan OC1 = =- l, then ---and therefore according to our defini- 1 - ab tion tan " 1 ^ + A i s a negative angle. Here y is therefore a negative angle and, since tan ( 7 r + 7 ) = tan 7 , the formula should be tan -1 a + tan -1 b = tt + tan -1 -^ . 1 - ab Ex. 6. Solve the equation Here we have where , x +1 , x - 1 _ , tare 1 -v + taie 1 - —tan~ 1 (- 7). x -1 x and Since i.e. a + p = y, a = tan~ ] and hence tan a = X - + , x - 1 x - 1 i3 = tan“ 1 ^-and hence tan B =-—- , x x 7 = tan _i (— 7 ) and hence tan 7 = - 7 . tan (a + /3) = tan 7 , tan a + tan 8 - - —— — 7 1 - tan a tan ft x+ 1 x - 1 - - -] - X-l x ^ X +1 X - 1 5 X-l X i. e. so that 2x 2 - x 4-1 1 - x = -7, x = 2. This value makes the left-hand side of the given equation positive, so that there is no value of x strictly satisfying the given equation. The value x = 2 is a solution of the equation tan x + 1 x-l + tan — = 7 r-r tan -1 ( - 7). x INVERSE CIRCULAR FUNCTIONS. 277 EXAMPLES. XLIII. Prove that . . 3 . , 8 . , 77 1. S in-^ + sm-i- = sm-i- 85* • , 5 . , 7 , / 253' 2. sm-i-+smi r cosi - 3 - ,4 . ,3 A .27 . ,4 ,12 .‘33 3 . cos -1 + tan -1 - = tan -1 —. 4 . cos - 1 - + COS - 1 t5= C0S 5 o 11 5 13 bo 1 + x 5 . cos -1 x = 2 sin -1 = 2 cos 1 n n ,3 , . 16 1 .7 6 - 2 cos V13 + ° 0t 63 + 2 C0S 25 = 7r ' 7. tan -1 i + tan -1 |i = sin -1 -4~ + cot -1 3 = 45°. ^ o V5 ox ,1 x , 1 , ,2 ,2 1 . 12 8 , tan -1 - + tan - 1 — = tan -1 -. 9 . tan " 1 - = ^ tan " 1 —. __ , .1 .2 1 3 10 . tan - 1 + tan - 1 - = ^ eos“i 5 11 . 2 tan -1 1 +tan -1 i + 2 tan -1 5 = ^. o 7 8 4 -10 x i 3 x , 3 8 7 r 12. tan - 1 - + tan - 1 - - tan " 1 — = - . 13. tan -1 ^ + tan -1 ~ + tan -1 \ + tan -1 ^ = 7 . o o 7 8 4 14. 3tan- 1 J + tan- 1 l = |-tan-i I L^. 15. 4tan- 1 |-tan- 1 L + t an- 1 4 = |. 16. tan_i iS =2 sin_l A • __ , . m , . ra - n tt 17. tan - 1 -tan -1 -= — . n m + n 4 TO X lx X , 2t ± .3 t-t 3 18. tan 1 1 + tan " 1 - 2 = tan " 1 ^ , 1 3t-fi 1 t<- 7 3 and = . + ta n- 1 l— 2 if t>^. 278 TRIGONOMETRY. [Exs. XLIII.] 19 . tan -'-Ji (a + b + c) be + tan + b + c) ca + tan + b + c) ab = 7 r. . ,ab + l , .be+ 1 , .ca + l A a-b b-c c-a 21 . tan -1 n + cot -1 (n +1) = tan -1 (n 2 + n + 1). 22. cos ^2 tan -1 ^ = sin ^4 tan -1 ^ . 23.2 tan-! [tan (45° - a) tan §] = cos'! ’ 24. tan” 1 x = 2 tan -1 [cosec tan” 1 x - tan cot” 1 x]. 25. 2 tan- 1 ftangtanfe-fTL tan” 1 sin Q C ° S . L 2 \4 2 J J sin (3 + cos a 26. Shew that , /a-x . . /x-b . . a - cos -1 \/ - r = sm” 1 a / --- = cot” 1 \ — V a-b \ a-b X x - -x b 1 . ,2 J (a - x) (x -b) = - sm” 1 - -- 2 a - b 0C \l 27 . If cos” 1 - + cos” 1 ^ = a, prove that a x 2 2a:?/ . _ -s- r°os a+ f„ = sin 2 a. a 2 ao 0 2 Solve no 4. _1 s/l + tf 2 - s/i-OJ 2 28 . tan” 1 ^7-=—■ ; = / 3 . Vi+^W 1 -* 2 ,-ifLtl- 29. tan” 1 2a: + tan -1 3a; = ^ . 30. tan 1 ——- + tan' 4 a:- 2 a; + 2 4 3 31. tan” 1 [x + 1) + cot” 1 [x - 1) = sin -1 ~z + cos -1 -. o o Q 32. tan -1 a: + tan _1 (x - l) = tan“ 1 — . OX 33 . 2 tan” 1 (cos x) = tan -1 (2 cosec x). [Exs. XLIII.] INVERSE CIRCULAR FUNCTIONS. 279 34. 36. 37. 38. 39. 41. 42. 43. 44. tan -1 x + 2 cot -1 x = - tt. O cot -1 x - cot -1 (x + 2) = 15°. 35. tan cos -1 # = sin cot -1 cos -l X 2 -1 + tan -l 2x 2 tt x 2 + l ' x 2 -l 3 * cot -1 x -f cot -1 (n 2 - x +1) = cot -1 (n - 1). 7 r sin -1 x + sin -1 2x = - . o An • -1 S . T 12 7r 40. sin 1 - + sin -1 — = - . x x 2 , a , ,6 , , C , . <2 7T tan -1 - + tan -1 - + tan -1 - + tan -1 - = - . x x x x 2 sec -1 -| - sec -1 y = sec -1 b - sec -1 a. a b cosec -1 x = cosec -1 a + cosec -1 b. 2 tan -1 x = cos -1 1 —a 2 1 + a? - cos -1 1 -b 2 l + b 2 • CHAPTER XIX. ON SOME SIMPLE TRIGONOMETRICAL SERIES. 241 . To find the sum of the sines of a series of angles, the angles being in arithmetical progression . Let the angles be a, a + / 3, a + 2/3 ,. {a -f (n — 1) /3}. Let S= sin a 4* sin (a + /3) + sin (a + 2/3) .. . + sin {a 4- (n — 1) /?}. By Art. 97 we have O . . ft 2 sm a sin — 2 cos I a — 2 cos ( a + ^ ), ft" cos ( a + - — cos a + 2 sin (a + /3)sin^ /3 / 3/3' 2 sin (a + 2/8) sin ^ = cos ( a + ) — cos [ a + 3/8' 5/3' ft 2 sin {a + (n—2) /8} sin^ = cos {a + (n —f) /3} — cos {a+ (?i and /3 2sin {<*+(«—l)/3}sin^ =cos{a+(n—§)/3} —cos{a+(n By adding together these n lines, we have -M- 0 • ft O 2 sm 5 • w = cos SIMPLE TRIGONOMETRICAL SERIES. 281 the other terms on the right-hand sides cancelling one another. Hence, by Art. 94, we have O Ol Y*> P Q- 2 sin^.$=2sin ja + sin^~, i. e. S = sin ■{a + (^ 2 ^) f sin ^ . /3 sin i Ex. By putting /3 = 2a, we have sin a + sin 3a + sin 5a + ... + sin (2 n - 1) a _sin {a + (n-l) a} sin na _ sin 2 ??a sin a sin a 242. To find the sum of the cosines of a series of angles , the angles being in arithmetical progression . Let the angles be a, a -f 0, a + 2/3, ... a -f (n — 1) /3. Let S= cosa + cos (a-f /3) 4- cos(a+2/3) + ... + cos {a + (n — l)/3}. > By Art. 97, we have o • 0 • ( 2 cos a sm ^ = sm (a + - I — 3/3' 2 cos (a + /3) sin ^ = sin ^a + ) — sm ( a + 2 cos (a + 2/3) sin ^ = sin fa + ^ ] — sm | a + & 2 )’ 3/3' 9 2 cos ■{ a+(ft—2) /3) sin £ 2 ! =sin{a + (ft—|)/3j—sin{a+(ft —1)/3}, 282 TRIGONOMETRY. and ,8 2cos{a + (w —1)/3} sin^ =sin[a + (^—i)/3}-sin{a4-(^—f)/3}. By adding together these n lines, we have 2 S x sin ^ = sin {a + (n — J) 8 } — sin ja “ > the other terms on the right-hand sides cancelling one another. Hence, by Art. 94, we have nci -fin f n— 1 2S x sm - = 2 cos < a H-~— o\ • nfi fir sm — , e. S = n - 1 0 ) . nfi cos^gh- ——py sin sm g 243 . Both the expressions for S in Arts. 241 and 242 7l8 . 7 x 8 vanish when sin-^- is zero, i.e. when is equal to any multiple of 7r, i.e. when where p is any integer, i.e. when Hence the sum of the sines (or cosines) of n angles, which are in arithmetical progression, vanishes when the common difference of the angles is any multiple r 27T 01 - . n Exs. / 2tt\ f 47r\ cos a + cos I a S-+ cos a H-+ V n ) \ n ) to n terms = 0, SIMPLE TRIGONOMETRICAL SERIES. 283 and 47r\ . ( Sir'. sin a + sin | a + — J +sin I a + — ) + ... ton terms = 0. 244 . ex. l. Find the sum of sin a - sin (a + /3) + sin (a + 2/3) - ... to n terms . We have, by Art. 73, sin (a + /3 + tt)= - sin (a + /3), sin (a + 2/3 + 27r) = sin (a + 2/3), sin (a + 3/3 + Sir) = - sin (a + 3/3), Hence the series = sin a + sin (a + /3 + tt) + sin {a+ 2 (/3 + 7r)} + sin {a + 3 (/3 + 7r) ]- +... n-1 s ) . n (/3 + 7r) sin ia (/3 + 7r)| sin . /3 + 7T Sm ~2~ , by Art. 241, sm -••• n Hence, if r be the radius of the circle, we have PA x = 2r sin PA 2 = 2r sin PA 3 = 2r sin POA 1 2 POA = 2r sin 2 POA - = 2 r sin - 3 = 2r sin e 2 ’ Hence the required sum 0 2 • p sm b „ r • 6 = 2r sin - + sin = 2r 0 7r \ . / 'd 2 7T j + sin ( — 2 n ,2 n n _ 1 7T 1 . n 7r + - 2 sin . nj 2 n sin 7 r 2 n = 2r cosec 7T . [~TT 6 7T~ 2n ■ sm [_2 + 2 “ 2 raJ = 2rcosee^cos (| - £) . (Art. 241) EXAMPLES. XLIV. Sum the series : 1. cos 6 + cos 30 + cos 56+ ... to n terms. 2. cos — + cos 2A + cos —+ ... to n terms. Li Li sin a + sin 2a + sin 3a + ... + sin na , n +1 3. -h -= tan a - cos a + cos 2a + ... +cos na 2 286 TRIGONOMETRY. [Exs. XLIV.] 4. 5. 6 . 7. 8 . 9. 10 . 11 . 12 . 13. 14. 15. 16. 17. 18. 19. 20 . 21 . deduce sin a + sin 3a + sm 5a + ... +sm (2 n- 1) a -----t-t -=tan na. cos a + cos da + cos oa + ... + cos (2n - 1 ) a 7 r 3 tt 57r cos --, + cos - -- + cos --r + ... to n terms. 2 n + l 2 »t + l 2 n +1 cos a - cos (a + /3) + cos (a+ 2/3) - ... to 2n terms. sin a - sin (a + /3) + sin (a + 2(3) + ... to n terms cosa-cos(a + /3) + cos(a + 2/3) + ... to n terms = tan ja + ^-g— (tt+ /3)j . 71 — 4 71 — 6 sin 0 + sin-- 0 + sin- - d-r ... to 7 1 terms. 71-2 71—2 cos x + sin 3x + cos 5x + sin lx + ... + sin (4 n - 1) x. sin a sin 2a + sin 2a sin 3a + sin 3a sin 4a + ... to 7 i terms. cos a sin 2a + sin 2a cos 3a + cos 3a sin 4a + sin 4a cos 5a + ... to 2 n terms. sin a sin 3a + sin 2a sin 4a + sin 3a sin 5a+... to n terms, cos a cos jS + cos 3a cos 2/3 + cos 5a cos 3(3... to n terms. sin 2 a + sin 2 2a + sin 2 3a + ... to ti terms. sin 2 6 + sin 2 (6 + a) + sin 2 (6 + 2 a) + ... to n terms. sin 3 a + sin 3 2 a + sin 3 3a + ... to ti terms. sin 4 a + sin 4 2a + sin 4 3a + ... to n terms. cos 4 a + cos 4 2 a + cos 4 3a + ... to n terms, cos 6 cos 2 6 cos 30 + cos 26 cos 36 cos 40 + ... to n terms, sin a sin (a + j 8 ) - sin (a + /3) sin (a + 2/3) + ... to 2 n terms. From the sum of the series sin a + sin 2a + sin 3a + ... to n terms, (by making a very small) the sum of the series 1 + 2 + 3 + ... +77. 22. From the result of the example of Art. 241 deduce the sum of 1 + 3 + 5... to n terms. 23. If prove that 2 (cos a + cos 2a + cos 4a + cos 5a) and 2 (cos 3a + cos 5a + cos 6a + cos 7a) are the roots of the equation x 2 + x - 4 = 0. [Exs. XLIV.] SIMPLE TRIGONOMETRICAL SERIES. 287 24. ABCD... is a regular polygon of n sides which is inscribed in a circle, whose centre is 0 and whose radius is r, and P is any point on the arc AB such that BOA is 6 . Prove that PA . PB + PA . PC + PA . PD + PB . PC +... 25. Two regular polygons, each of n sides, are circumscribed to and inscribed in a given circle. If an angular point of one of them be joined to each of the angular points of the other then the sum of the squares of the straight lines so drawn is to the sum of the areas of the polygons as o • 2?r 2 : sin —. n 26. ^i, A 2 ...A 2n+l are the angular points of a regular polygon in¬ scribed in a circle and 0 is any point on the circumference between A 1 and A 2n+1 ; prove that OA j -f- OA s + ... + OA 2n +\ — OA 2 + OA 4 + OA 2n . 27. If perpendiculars be drawn on the sides of a regular polygon of n sides from any point on the inscribed circle whose radius is a, prove that 2 V 2 2 3 CHAPTER XX. ELIMINATION. 245 . It sometimes happens that we have two equa¬ tions each containing one unknown quantity. In this case there must clearly be a relation between the constants of the equations in order that the same value of the unknown quantity may satisfy both. For example, suppose we knew that an unknown quantity x satisfied both of the equations ax + b = 0 and cx 2 + dx -f e = 0. From the first equation we have b and this satisfies the second if c + 6 — 0 , i.e. if b 2 c - abd -f a 2 e = 0. This latter equation is the result of eliminating x between the above two equations, and is often called their eliminant. ELIMINATION. 289 246 . Again, suppose we knew that an angle 0 satisfied both of the equations sin 3 0 = b, and cos 3 0 = c, so that sin 0 = b 3 , and cos 8 = cK Now we always have, for all values of 8 , sin 2 8 + cos 2 0 =1, . . 2 2 so that m this case b 3 + c 3 = 1. This is the result of eliminating 0. 247 . Between any two equations involving one unknown quantity we can, in theory, always eliminate that quantity. In practice a considerable amount of artifice and ingenuity is often required in seemingly simple cases. So between any three equations involving two un¬ known quantities we can theoretically eliminate both of the unknown quantities. 248 . Some examples of elimination are appended. Ex. 1 . Eliminate 0 from the equations a cos 9 + b sin 6 = c , and b cos 0+ c sin 0 = a. Solving for cos 9 and sin 9 by cross multiplication, or otherwise, we have cos 0 _ sin 0 _ 1 c 2 - ab a 2 -be ac-b 2 ' i.e. 1 = cos 2 0 + sin 2 0 = (c 2 - ab) 2 4- {a 2 - be) 2 (ac - ft 2 ) 2 9 (a 2 -bc) 2 + (c 2 - ab) 2 = (b 2 - ac) 2 . L. T. 19 290 TRIGONOMETRY. Ex. 2. Eliminate 9 between ax by cos 9 sin 9 — a 2 — b 2 and ax sin 9 by cos 9 „ + ~ =0, cos 2 6 ' sin 2 6 From (2) we have ax sin 3 9= - by cos 3 9. sin 9 _ cos 9 _ J sin 2 9 + cos 2 9 (ax)* J (1). ( 2 ). (by)* + (ax)* (Hall and Knight’s Higher Algebra , Art. 12) 1 (by)% + (ax)% Hence i V (62/) 3 + (ax)3 sin 9 (biy)i and so that (1) becomes 1 _ ^{by) ® + (ax)% ^ e ~ (^)i d 2 - 6 2 = V(6^)1 + ( ax )i [~ax . —I— - by -[-hi L (ax)* y (byi*’ J i.e. = (by)* + (ax)z((ax)z + (by )$} = {(ax)* + (by)i}%, (( ax + (by)% — (a 2 - b 2 )^. The student who shall afterwards become acquainted with Analytical Geometry will find that the above is the solution of an important problem concerning normals to an ellipse. Ex. 3. Eliminate 9 from the equations x n y . . - cos 9 - 7- sin 9 = cos 29 a b ( 1 ). and - sin 9 + \ cos 9 — 2 sin 29 a b ( 2 ). ELIMINATION. 291 Multiplying (1) by cos 0, (2) by sin 0, and adding, we have - = cos 0 cos 20 + 2 sin 6 sin 26 a = cos 6 + sin 6 sin 20 = cos 0 + 2 sin 2 0cos 0.(3). Multiplying (2) by cos 0, (1) by sin 0, and subtracting, we have % = 2 sin 20 cos 0 - cos 20 sin 0 6 = sin 20 cos 0 + sin 0 = sin 0 + 2 sin 0 cos 2 0.(4). Adding (3) and (4), we have - + \ = (sin 0 + cos 0) [1 + 2 sin 0 cos 0] a b v = (sin 0 + cos 0) [sin 2 0 + cos 2 0 + 2 sin 0 cos 0] = (sin 0 + cos 0) 3 , ( \ l. ^ + ^ J . (5). Subtracting (4) from (3), we have ^ - |= (cos 0 - sin 0) (1 - 2 sin 0 cos 0) = (cos 0 - sin 0) 3 , so that cos 0 - sin 0 = ^ .(6). Squaring and adding (5) and (6), we have EXAMPLES. XLV. Eliminate 0 from the equations 1. a cos 0 + 6 sin 0 = c, and 6 cos 0 - a sin 0 = d. 2. # = a cos (0 - a), and y = b cos (0 - /3). 3. a cos 20 = 6 sin 0, and c sin 20 = d cos 0. 4. cl sin a - 6 cos a = 26 sin 0, and a sin 2a -6 cos 20 = a. c • a a sin 2 0 cos 2 0 1 5 . x sin 6 -y cos 0= x /* 2 + */ 2 > and —— + = ~=—; 2 . a- b 2 x 2 -\-y 2 292 TKIGON OMETRY. [Exs. x:lv.] 6 . x cos 9 + a y sin 9 b and x sin 9 - y cos 9= Jd 1 sin 2 9 + 6 2 cos 2 9. 7. sin 9 - cos 9 =p, and cosec 9 - sin 9 = q. 8. If m = cosec 9 - sin 9, and ?i = sec 9 - cos 9, s 2 2. __ J2 prove that m 7 + n‘ 6 — (mri) y . 9. Prove that the result of eliminating 9 from the equations x cos (9 + a) + y sin (9 + a) = a sin 29, and y cos (9 + a) - x sin (9 + a) — 2 a cos 29, 2 2 2 is (x cos a + y sin a)^ + (x sin a - y cos a)* = (2 a)^. Eliminate 9 and 0 from the equations 10. a cos 2 9 + b sin 2 0 = c, b cos 2 0 + a sin 2 (p = d, and a tan 0 = 6 tan 0. 11. cos# + cos0 = a, cot0 + cot0 = b, and cosec 9 + cosec

y 2 . 23. — and +. 24. — and -. n_a 2 -b 2 V ~a 2 TF - ’ — and —. (1) 2^7r+y and 2mr + ; (2) 2mr + — and 2mr+ ^ ; 4: 4 4 4 (3) 2mr - j and 2mr + ^ ; (4) 2mr+ j and 2w7rH- 29. Vlll TRIGONOMETRY. 30. (1) ( 2 ) (3) 2nir - v and 2mr + % ; 4 4 2mr + ^ and 2mr + ~ : 4 4 2mr + — and 2 nir + ^. 12 . 13. 1. 3. 4. 5. 6 . 8 . 10 . 11 . 12 . 14. 16. XIX. (Pages 129, 130.) The sine of the angle is equal to 2 sin 18 7T -tr 5 + 10 ' XXI. (Pages 142, 143.) mr 7T 4 ° r J( 2n7r± 3 ± ^ or 2 ti 7 t. 2 . 1 7r 1 7T 2 n - ) — or ( 2 ti ± — ) —. 2/ 4 3/ 3 2ti =t 2717T I)^or ^ + (-l)*|. 1 3 727T or ( ti + -r 7 r or 2n - 1 7T. 3 or ( 2 n *l)r n 3 ° r ( n * 3 ) 7r ’ 7. 71 =b^ ) 7T Or 2?17T ± 9. 2mr ; (^- + ^ t r. : 7r 3 * W7T+ (- l) ,l ^i n7r + (-l) M ^; W7T-(-l)’ 37r vW 10 10 ‘ 1 7 r 2 ) 8 ’ * 2 ) 2 ' mir ; ' 71-1 2nr rmr — (— 1 ) m 7T 6 13. 2rmr; m + n m — n mir 2rir^). 15. 1 m n hmr tT±1 * 2 ttt ± 7717T 1 7T\ ;-ri -(2 rair*~). 71-1 71 \ 2/ LOl ^ ANSWERS. IX 17. 18. 20 . 22 . 23. 24. 26. 28. 30. 32. 34. 37. 38. 2n7r ~^; 1 ( 2wtt - £ ) . 19. mr + (-\) n 1- 2 4 7T 21. 2mr + — ±A. 4 / i \ m 7T + (- !) 4 - 3 • «7T+g + (- 1)»£. - 21°48' + W7r + (- l)* 1 [68° 12']. 2 w7r + 78°58'; 2 twt+ 27°18'. [N.B. cos 25° 50' = -9.] n-K + 45°; mr + 26°34'. 25. 2«tt; 2mr + ^. 7r 2ri7r ; 2iTi7r + — . ' V 7T 7T 27. 2mr + —; 2w7r + ^ . 2i7l7T + 77 ; 2t17T — — . o 2 sin # = 717-1 8 29. mr. 31. cos 0 = J 17-3 7T 7T 7fc7T — • 2717T ± - . O Jj 33. 2ri7r =fc : 2mr ± ^ 3 4 7T n + 4j2- 7T 35. ri7r = 1 = —. 4 36. W7T + 7 r 4* 0 = mr ir . „ a 1 — or mr =t —; also 6 — mr ± —, where cos a -- -. 2 3 2 3 1\ 7T w+ 3h- 7r 39. W7T + 7T . XXIII. (Pages 157, 158.) 1 . f-90309; 3-4771213; 2-0334239; T’4650389. 2. -1553361 ; 2-1241781 ; -5388340; 3-0759623. 3. 2; 2; 0; 4; 2 ; 0; 3. 4. -312936. 5. 1-32057; 5-88453; -461791. 6 - (1)21; (2)13; (3)30; (4) the 7th; (5) the 21st; ( 6 ) the 32nd. TRIGONOMETRY. 4 log 3 log 2 + 2 log 3 log 7+4 log 3 — log 2 ' 4 log 7 — 3 log 3 — 2 log 2 " 7 log 3 + 4 log 2 3 I og 3 + log 2 — 2 log 7 ' •22221. 9. 8-6414. 10. 9-6192. 1-6389. 12. 4-7161. 13. -41432. XXIV. (Pages 167—169.) 4-5527375; 1-5527394. [N.B. log 35706 = 4-5527412.] 4-7689529; 3-7689502. 478-475; -004784777. 4. 2-583674; -0258362. (1) 4-7204815; (2) 2-7220462 ; (3) 2-7240079 ; (4) 5273-63 ; (5) -05296726 ; (6) 5-26064. •6870417. 7. 43° 23'45". •8455104; -8454509. 9. 32° 16'35"; 32°16'21". 4-1203060; 4-1218748. 4-3993263; 4-3976823. 12. 13°8'47". 9-9147334. 14. 34°44'27". 9-5254497 ; 71°27'43". 16. 10-0229414. 18° 27'17". 18. 36° 52'12". XXV. (Pages 171, 172.) 13°27'31". 2. 22°1'28". 1-0997340 ; 65°24'12-5". 9-6198509; 22°36'28". 10° 15'34". 6. 44° 55'55". (1) 9-7279043; (2) 9-9270857; (4) 10-0757907 ; (5) 10-2001337 (6) 10-0725027 ; (7) 9-7245162. (1) 57° 30'24"; (2) 57°31'58"; (4) 57° 6' 39". •53736037. (3) 10-1958917 ; (3) 32° 29'15"; ANSWERS. XI 10 . ( 1 ) cos (x — y) sec x sec y ; ( 2 ) cos (x + y) sec x sec y ; (3) cos (x — y) cosec x sec y; (4) cos (x -f y) cosec x sec y ; (5) tan' 2 #; ( 6 ) tan x tan y. 1 . 2 . 3. 4. 6 . 17. 1 . 2 . 3. 4. XXVI. (Pages 179, 180.) 11 . 9 5’ 2’ and 7* 8 4 5 an( l _ • _ _ 741 ’ 5 ’ 5 ^/ 41 ’ 41 ’ 25 34 IT 5 , g, and 1 . 40 24 , 496 , and 1025 12 12 9 5 and 00 . 7 . 287 - and 81fJ . 41 c 4 56 _ 12 5 ' 5’ 65 and 13' 7. 60°, 45°, and 75' XXVII. (Pages 185—187.) 16|ft. 19. 5 22 . 313 338 XXVIII. (Page 190.) 186*60... and 193*18. 26°33'54" ; 63°26'6"; 10^/5 ft. 48°35'25" 36°52' 12" and 94°32'23". 75° and 15°. XXIX. (Page 193.) 1. 90°. 2 . 30°. 4. 120' D • 5. 45°, 120° and 15°. O 40 CD 60°, and 75°. 7. 58° 59'33''. 8 . 77° 19' 11". 9. 76°39'9" 10 . 104° 28'39''. 11 . 56° 15' 4", 59°51'11" and 63° 53'45". 12 . 38°56'33", 47°41'7" and 93° 22' 20". ■■ 13. 130°42'20'5'', 23° 27'8-5'' , and 25° 50'31". Xll TRIGONOMETRY. XXX. (Pages 197—200.) 1 . 63° 13'2"; 43°58'28". 2. 117°38'45"; 27°38'45". 3. 8 V '7; 79° 6'24"; 40° 53'36". 4. 87° 27'25-5"; 32° 32'34-5". 5. 40° 53'26"; 19° 6'24"; J7 : 2. 6. 71°44'30" : 48° 15'30". 7. 78°17'41"; 48°36'19". 8 . 108° 12'25-5"; 49°27'34-5". 9. .1 = 45°; 5 = 75°; e = J6. 10. v /6 ; 15°; 105°. 11. - 8965. 14. 40 yds.; 120°; 30°. 15. 7-589467 ; 108°26'6"; 18°26'6"; 53°7'48". 16. 226-87; 73°34'50"; 39°45'10". 17. 2-529823. 18. A = 83°7'39" ; £ = 42°16'21"; c= 199-099. 19. B = 110°48'15"; C = 26° 56'15" ; a = 93-5192. 20. 73° 1'51" and 48° 41'9". 21. 88°30'1" and 33°30'59". XXXI. (Pages 205—207.) 1. There is no triangle. 2. B l = 30°, Ci = 105°, and b, = ; #, = 60°, C 2 = 75°, and b. 2 = J§. 3. B x = 30°, C l = 120°, and 5, = 100; B 2 = 90°, C 2 = 60°, and b 2 = 200 . 5. 4V3=t2V5. 6. 100^/3; the triangle is right-angled. 8 . 33°29'30" and 101°30'30". 9. 17-1 or 3-68. 10. (1) The triangle is right-angled and 71= 60". (2) 2?! = 8°41' and C 2 = 141° 19'; B 2 = 111° 19' and 38°41'. 11. 65° 54' and 42°1'12". 12. 5-988... and 2-6718... miles per hour. 13. 63°2'12" or 116°57'48". 14. 62°31'23" and 102° 17'37", or 117°28'37" and 47°20'23". 15. 59266-1. ANSWERS. Xlll XXXII. (Page 208.) 1. 7:9:11. 4. 79*063. 5. 1 mile; 1*219714... miles. 7. 20 97615... ft. 8. 6*85673... and 5*4378468... miles. 9. 404*4352 ft. 10. 233*2883 yards. 11. 2229*02 yards. XXXIII. (Pages 213—216.) 1. 100 ft. high and 50 ft. broad; 25 feet. 2. 25*783414 yds. 3. 33*07... ft.; 17^ ft. 4. 18*3... ft. 5. 120 ft. 6. li tan a cot /3. 7. 1939*2... ft. 8. 100 ft. 9. 61*224... ft. 10. 100^/2 ft. 15. PQ = BP = BQ = 1000 ft.; AP = 500 (^6 - J2) ft; AQ = 1000 J2 ft. 16. *32119 miles. 17. *1736482 miles; *9848078 miles. 18. 119*2862 ft. 19. 132*266 ft. 20. 141*682 yds. 21. 1*42771 miles. 22. 125*3167 ft. XXXIV. (Pages 220—225.) 3. 20 ft; 40 ft. 2 4. I cosec y, where y is the sun’s altitude; sin y — -. 5. 3*732... miles; 12*342... miles per hour at an angle, whose tangent is J 3 + 1, S. of E. 6. 10*2426... miles per hour. 7. 16*3923... miles ; 17*394... miles. 8. 2*39 miles; 1*366 miles. 9. 13. 14. 2 9 It makes an anode whose tangent is — ; — hour. ° ® 3 52 c sin /3 cosec (a + /3 ); c sin a sin /3 cosec (a + /?). 9 yds. ; 2 yds. 17 3 ’ 3 ' xiv TRIGONOMETRY. 21. 7573 ft. 22. c (1 — sin a) sec i 23. 114-122 ft. 25. 1069-745645 ft. 27. The angle whose tangent is - . 30. 45°. A 33. 18° 24'6". 35 . tan a sec /3 : 1. 38. 91*896 ft. 39 . 1960*95 yds. 40. 2*45832 miles. 41. 333*4932 ft. XXXV. (Pages 227, 228.) 1. 84. 2. 216. 3. 630. 4. 3720. 5. 270. 6. 117096. 7. 1470. 8. 1 ’183_ 12. 35 yds. and 26 yds. nearly. 13. 14-941... inch. 14. 5, 7, and 8 ft. 15. 120°. 17. 45° and 105°; 135° and 15°. 18. 17• 1064... sq. ins. XXXVI. (Pages 235, 236.) 3. 81, l\, 8, 2, and 24 respectively. XXXVII. (Pages 244—248.) 35. 2-1547... or -1547 times the radius of each circle. 39. ^. = J + (-l)V2".^-j) . XXXVIII. (Pages 253, 254.) 1. (1) 3 ^105 ft.; (2) 10^/7 ft. 3. l-fand2ift. XXXIX. (Pages 257—259.) 1. 77*98 ins. 2. 3*215. 3. (1) 1*720... sq. ft.; (2) 2*598... sq, ft.; (3) 4*8284... sq. ft.; (4) 7*694... sq. ft.; (5) 11*196... sq. ft. 4. 1*8866... sq. ft. 5. 3*3136... sq. ft. 6. 2 + v /2:4; JYTjl : 2. 12. 3. 14. 6. 15. 9 or 16. 16. 20 and 10. ANSWERS. XV 17. and 5, 12 and 8, 18 and 10, 22 and 11, 27 and 12, 42 and 14, 54 and 15, 72 and 16, 102 and 17, 162 and 18, 342 and 19 sides respectively. 19. V 6 - 1. 4. 7. 10 . 1. 3. 5. 7. 1. 3. 5. 1. 2 . 5. 6 . 9. 10 . •00204. •99999. 34'23". 2° 26'15". XL. (Pages 264, 265.) 2. -00007. 3. 5. 25783-100... 6. 8. 28° 41'7". 9. 11. 114-59... inches. •00029. 1 - 0000011 , 39'34". XLI. (Pages 267, 268.) 435*77 sq. ft. 2. 4*9087... sq. ft. 127° 19'26". 4. 6 sq. ft. 11*0004 inches. 6. *00044625 inch. 2 3 ?rr * XLII. (Pages 269, 270.) ri0'22". 2. 17*14 miles. *61 miles; 1°48' nearly. About 61800 metres = about 384 miles. 6. 3960 miles. XLIV. (Pages 285—287.) i sin 2 n6 cosec 6. A ?>n — 1 . . ?>n . 3 cos —-— A sin — A cosec - A. 4 4 4 1 . 2717T 2lV - sin-- 2 n+1 . cosec n + 1 * sin a 4- U-- -|)^] si nw/3 £ sec 7^ • 9 8. — sin nO ^2' sin 2 nx (cos 2 nx + sin 2 nx) (cos x + sin x) cosec 2x. j [(n +1) sin 2a — sin (2 n + 2) a] cosec a. XVI TRIGONOMETRY. 11 . 12 . 13. 14. 15. 16. 17. 18. 19. 1. 2 . 3. 5. 7. 10 . 11 . 12 . 13. i sin (n + 2) a . sin na cosec a. n 1 , . - cos 'la — - cos (w+ o) a sin cosec a. cos (2?ia - a) cos (ft + 1) (3 - cos (2na + a) cos n(3 + cos a (1 - cos (3) 2 (cos /3 - cos 2a) i [(2^ + 1) sin a — sin (2 n + 1) a] cosec a. ^ cos [2 6 4- (n - 1) a] sin na cosec a. 1 1 3 . n + 1 . na a 1 . n +1 . 3 na . 3a - sm — — a sin — cosec - - 7 sin 3 —— a . sin — — sin —. 42 224 2 22 ^ [3 n - 4 cos (n + 1) a sinna cosec a + cos (2 n + 2) a sin 2fta cosec 2a]. | [3ft + 4 cos (ft + 1) a sin na cosec a + cos (2ft + 2) a sin 2;m cosec 2a]. n 1 . w# 4 Sm 2 l/o ^ + 3 ^ ?& + 7 , COS — 9 — C7 + cos —-— 0 + cos —— 0 cosec 30 1 . 3 nO 3n + 9 + sm — cos —-— 0 cosec ^ . 4 2 2 2 — - sin (2a + 2 nfS) sin 2 n/3 sec /5. l XLV. (Pages 291, 292.) a 2 + 6 2 = c 2 + d i . _ cl (2c 2 — d 2 ) — bote. 4. a sin a + 6 cos a = ^/26 (a + 6). K+£=i. 6. ^ + ?* = a + 6. cr 6 2 a b (p 2 + 1) 2 + 2q (p 2 + 1) (p + q) = 4 (p + g) 2 . a 2 (a — b){a — c) — b 2 (6 — c) (6 — a?). 86c = a {46 2 + (6 2 — c 2 ) 2 }. y J(cl + b + c)(— a + b + c)=as (c 2 - a 2 — 6 2 ) J(a—b + c)(a 4 6 - c). b 2 [as (6 2 — a 2 ) + a (a 2 + 6 2 )] 2 = 4c 4 [6 2 as 2 + a 2 ?/ 2 ]. CAMBRIDGE: PRINTED BY C. J. CLAY, M.A. AND SONS, AT THE UNIVERSITY PRESS. tO | UNIVERSITY OF ILLINOIS 516.24L848P r nm unr plane trigonometry c M URBANA 30112017254175