'•i-i'i .PLANE TRIGONOMETRY AND APPLICATIONS BY E. J. WILCZYNSKI, Ph.D. THE UNIVERSITY OF CHICAGO EDITED BY H. E. SLAUGHT, Ph.D. THE UNIVEK8ITT OF CHICAGO ALLYN AND BACON ISoston Neto gorfe . Chicago ,/5^ COPYRIGHT, 1914, BY E. J. WILCZYNSKI PDb Nortooolr ^reaa J. 8. Gushing Co. — Berwick & Smith Co. Norwood, Mass., U.S.A. r- PREFACE The characteristic features of this book may be summarized as follows : 1. The method of presentation is thoroughly heuristic. This en- ables the student to get a firm grasp of the subject by teaching him to recognize the fundamental ideas which underlie and unify the separate steps of the mathematical argument, instead of con- fusing him by a disconnected dogmatic statement of isolated facts. 2. The book is divided into ttoo parts. The first part is devoted to the theoretical and numerical solution of triangles. The second part treats of the functions of the general angle, their addition theorems and other properties, together with applica- tions to simple harmonic curves, simple harmonic and wave motion, and harmonic analysis. Part One is also published sepa- rately and is well adapted for use in secondary schools. The complete book is intended for the freshman course in colleges. 3. The discussion of the solution of triangles, in Part One, is not interrupted by any digressions about coordinate systems, addition theorems, and the like. It has been thought desirable to post- pone to the second part the consideration of all of these matters, which are indeed important but unnecessary for the solution of triangles. 4. The definitions for the functions of an obtuse angle have been made to grow organically out of the needs of the problem of solv- ing triangles, in a way which seems both simple and natural, and which at the same time illustrates an important principle of mathematical procedure. 5. The ivhole theory of triangles has been unified by giving a central position to the area problem. As a consequence, almost all of the necessary equations present themselves spontaneously and in a connected fashion. The law of tangents is the only one which causes any difficulty in this respect. But the law of tan- gents also has been made to submit to a heuristic treatment, by 331458 iv PREFACE introducing the notion of the formrratio of a triangle, and combin- ing this notion with a direct geometric proof of the formulae for sin J. -f sin 5 and sin A — sin B. 6. Tlie numerical aspect of the work has been discussed very fidly. Directions for computation are given in great detail ; most of the common sources of error are pointed out ; and methods for detect- ing and correcting them are indicated. After a thorough discus- sion of the significance of the number of decimal places needed in a computation, the student is urged to train and use his judgment on this matter. He is given an opportunity to do this by supply- ing him with complete five- and three-place tables and a partial set of four-place tables. 7. The slide rule is explained with considerable detail and its use recommended. A number of other labor-saving devices are discussed. 8. The examples have been selected with great care. Examples without real significance have been avoided, and the numbers have been chosen so as not to lead to five-place calculations when such a show of accuracy would be absurd. Special efforts have been made to word the examples in such a way as to avoid ambiguity. 9. The applications cover a ivider field than usual, and include problems in heights and distances, surveying, navigation, engi- neering, astronomy, and physics. But the examples involving such applications are not, as in most texts, introduced at random and without previous explanation. Every notion which is re- quired for. the solution of any example in the book is fully explained on the spot or in some earlier portion of the text. 10. TJie use of a few new terms, such as the standard position of an angle, odd and even cardinal angles, has helped to simplify materially the statement of a number of important results. 11. The addition formulas, are presented in two different ways. The first, more elementary method, is made to yield the general result by the help of mathematical induction. The second method, based on the notions of directed lines, line-segments, and angles, appears here in a very simple and elegant form. 12. The articles on harmonic and laave motion tend to show the student that Trigonometry has other applications besides the solution of triangles. r PREFACE V 13. A considerable amouyit ofli istorical matter has been introduced^ not in the form of detached historical notes, but organically con- nected with the topic under discussion. Most of this matter was gathered from Braunmuhl's Vorlesungen ilber die Geschichte der Trigonometrie. Professors Cajori and Karpinski. have kindly answered some questions of a historical nature about which we were in doubt. 14. The type and the manner of spacing used in the tables are the results of a number of experiments, the object being to pro- duce a set of tables which should be as pleasant to the eye as possible. The tables are bound separately for various reasons. In order to make them easily legible, a certain size of page was necessary, and it was thought undesirable to use so large a page for the text itself. In the second place, it is a great advantage for the student if he can have his text and his tables open before him at the same time. In the third place, it is often desirable, in examinations, to allow students to use their tables without their books. Pinally, a separation of the tables and text makes it easy to use this text with other tables, or these tables with other texts, thus providing a maximum of elasticity in organizing a course. Many of the older texts on Trigonometry have been consulted during the preparation of this book, and the attempt has been made to learn from all of them. The works of Serret, Lijbsen, WiEGAND, Crockett, Moritz, Hall and Frink have been espe- cially helpful. A few purely numerical examples have actually been taken from these and other texts without change, so as to reduce somewhat the task of computing the answers, and at the same time to make the answers more trustworthy. Most of the examples, however, are new ; many of them are new in kind. In conclusion, the author and editor wish to acknowledge their indebtedness to their colleagues at the University of Chicago for various helpful suggestions and criticisms. E. J. WILCZYNSKI. H. E. SLAUGHT, Editor. CONTENTS PART ONE. SOLUTION OF TRIANGLES CHAPTER I The Object of Trigonometry. PAGB 1. Direct measurement of lines 1 2. Direct measurement of angles 3 3. The impossibility of finding all distances by direct measure- ment 6 4. The graphic method 8 5. The desirability of an arithmetical method for solving triangles 9 CHAPTER II The Trigonometric Functions of Acute Angles. 6. The necessity of introducing new ideas 11 7 . Definition of the trigonometric functions of an acute angle . 13 8. The practical need of tables giving the values of the trigo- nometric functions 16 9. Relations between the six trigonometric functions of an acute angle 18 10. Relations between the functions of complementary angles . 20 11. The values of the functions of 0^, '60°, 45°, 60°, 90° . . . 22 CHAPTER III Solution of Right Triangles by Natural Functions. 12. Arrangement and use of the table of natural functions . . 25 13. Solution of right triangles by means of the table of natural functions ... 29 vii viii CONTENTS CHAPTER IV PAOB Discussion of Some Devices for reducing the Labor involved in Numer- ical Computations. 14. The number of decimal places 34 15. The accuracy of a sum, difference, product, or quotient of two numbers obtained by measurement 36 16. Labor-saving devices 37 17. Definition of logarithms 38 18. The properties of logarithms 43 CHAPTER V Calculation by Logarithms. 19. Common logarithms 47 20. Properties of the mantissa 48 21. Determination of the characteristic . . . . . . 49 22. Arrangement and use of the table of logarithms ... 51 23. Cologarithms . . 54 24. Extraction of roots by means of logarithms .... 65 25. Logarithmic calculations which involve negative numbers . 56 26. The logarithms of the trigonometric functions .... 57 27. The accuracy of five-place tables 59 28. The trigonometric functions of angles near 0° or 90° . . 60 29. The logarithmic or Gunter scale 61 30. The shde rule 62 CHAPTER VI Application of Logarithms to the Solution of Right Triangles. 31. The general method 67 32. The preliminary graphic solution 67 33. The gross errors 68 34. Selection of formulae and checks 68 35. The framework or skeleton form 69 36. The computation 70 37. Revision of the computation when the check is unsatisfactory 70 38. Applications to simple problems of surveying, navigation, and geography 72 39. Right triangles of unfavorable dimensions . . . . 80 CONTENTS ix CHAPTER VII PAax Theory of Oblique Triangles. 40. The area of an oblique triangle in terms of two of its sides and the included angle 82 41. The law of sines , 84 42. The law of cosines. A generalization of the theorem of Pythagoras 85 43. Properties of the functions of an obtuse angle .... 88 44. Other formulae for the area of an oblique triangle ... 90 45. The radius and center of the inscribed circle .... 93 46. The half-angle formulae 95 47. The circumscribed circle .95 48. The form ratios of a triangle 97 49. The formulae for the sum and difference of two sines . . 98 50. The law of tangents 100 51. A second proof of the law of tangents and MoUweide's equations 102 CHAPTER VIII Solution of Oblique Triangles. 52. The fundamental problem 107 53. Case I. . Given one side and two angles 108 54. Case II. Given two sides and the included angle . . . 109 56. Casein. Given two sides and the angle opposite to one of them 111 56. Case IV. Given the three sides of the triangle . . .116 57. Problems in heights and distances, plane surveying, and plane sailing 118 58. Displacements, velocities, and forces 126 59. Reflection and refraction of light 128 PART TWO. PROPERTIES OF THE TRIGONOMETRIC FUNCTIONS CHAPTER IX The General Angle and its Trigonometric Functions. 60. The notion of the general angle 133 61. Initial and terminal side. Standard position of an angle . 135 62. The notion of the trigonometric functions of a general angle . 136 63. Rectangular coordinates 137 CONTENTS PAGB 64. Definition of the trigonometric functions of a general angle . 140 65. Discussion of the exceptional cases 141 66. The four quadrants 145 67. General character of the trigonometric functions. Their periodicity 146 68. Relations between the trigonometric functions of a general angle 148 69. Trigonometric identities which involve functions of a single angle 149 CHAPTER X Graphic Representations of the Trigonometric Functions. 70. Line representation of the trigonometric functions . . . 151 71. Graphs of functions, a number of whose numerical values are given 155 72. Graphs of simple algebraic functions . . . . .158 73. Graphs of the trigonometric functions 159 74. The natural unit of circular measurement. Definition of a radian 163 75. Relations between the functions of two symmetrical angles . 167 76. Relations between functions of two angles whose sum or differ- ence is a right angle 169 77. The quadrantal formulae 171 78. Properties of the sine and cosine curves 175 CHAPTER XI Relations between the Functions of More than One Angle. 79. The addition theorems for sine and cosine . . . . 180 80. The addition theorems for tangent and cotangent . . . 185 81. The subtraction formulae 186 82. Formulae for converting products of trigonometric functions into sums, and vice versa 187 83. Functions of double angles - 189 84. Functions of half angles 190 85 a. The Umit ^^^ and related limits 198 d 85 h. The auxiliary quantities S and T 198 CONTENTS XI CHAPTER XII PAOB Directed Lines and Directed Line-segments. 86. Plan of another proof for the addition formnlse . . . 201 87. Directed lines and segments 201 88. Angles between directed lines . . . " . . . . 204 89. Projections 204 90. Projection of a broken line 207 91. Direction cosines of a line 209 92. Formula for the cosine of the angle between two lines whose direction cosines are given 210 93. New proof for the addition and subtraction formulae . . 212 94. The generalized law of sines 212 CHAPTER XIII The Inverse Trigonometric Functions and Trigonometric Equations. 95. The problem of inverting the trigonometric functions . .214 96. Determination of all of the angles which correspond to a given value of one of the functions 214 97. The inverse trigonometric functions . . ... . 217 98. Trigonometric equations 222 99. The equation a sin ^ + 6 cos ^ = c 223 CHAPTER XIV w Applications to the Theory of Wave Motion. 100. Simple harmonic motion 226 101. The period and phase constant 228 102. Some illustrations of simple harmonic motion . . . 229 103. Simple harmonic curves 231 104. Amplitude 232 105. Wavelength 232 106. Phase constant 233 107. Wave motion 235 108. General harmonic motion . . . . . . . 239 109. General harmonic curves 241 110. Harmonic analysis or trigonometric interpolation . . . 243 111. Theorems leading to the general solution of the problem of trigonometric interpolation 249 112. General solution of the problem of trigonometric interpolation 257 PLANE TRIGONOMETRY AND APPLICATIONS PART ONE SOLUTION OF TRIANGLES CHAPTER I THE OBJECT OF TRIGONOMETRY 1. Direct measurement of lines. One of the most com- mon operations of practical geometry is that of measuring the distance between two points. In its simplest form this consists merely in the repeated application of some unit of measurement to the required distance. The units of measurement most frequently used for this purpose are a foot rule, a yardstick, a surveyor's chain, tape lines of definite length, etc. Fractional parts of the unit are usually read from a graduated sca/e, engraved or stamped on the standard used. A familiar illustration of this is the scale of inches on an ordinary foot rule. In spite of the apparent simplicity of this process, it is a matter of great practical difficulty to carry out such meas- urements with a high degree of precision. The sources of error are numerous and, in part, unavoidable. No instru- ment made by man is absolutely accurate. Thus, if we use a yardstick, it will not be absolutely straight, and it may be a trifle too long or too short. It will be very difficult to make sure that we are lajring off the second yard of our dis- tance exactly where the first yard ends. Consecutive posi- tions of the yardstick will form angles with each other, which are not exactly equal to 180°. In fact, it is almost impossible to run a straight line of considerable length, with 1 2 .-. \ ; , ; . . THE OBJECT OF TRIGONOMETRY any degree of accuracy, without the help of more complicated instruments, such as the transit described in Art. 2. The graduated scales, used for measuring fractional parts of the unit of length, are also affected by various sources of inaccuracy, and it will be difficult for the observer to esti- mate accurately a fractional part of the smallest visible divi- sion on the scale. Enough has been said to indicate just a few of the many difficulties encountered in the, apparently so simple, opera- tion of measuring the length of a line, and to emphasize the fact that we must always regard the result of such a meas- urement as an approximation^ even if the most refined instru- ments known to Science have been used. The difference between rough and fine measurements is one of degree only. The more refined the method, the smaller will be the " probable error " and the closer the approach to the truth. But we can never be sure that a quantity has been measured with absolute precision. EXERCISE I 1. What are some of the sources of inaccuracy in measuring the length of a table ? 2. If you wish to measure the distance diagonally across a table, by means of a foot rule, what additional sources of inaccuracy will appear ? Would a stretched cord be of some use in this connection ? 3. How would you measure the distance diagonally across a room from one of the floor corners to the opposite corner of the ceiling ? Do you know of any other method by which this distance might be obtained, except that of direct measurement ? 4. How would you join two given points by a straight line (say for the purpose of constructing a fence), over a level piece of ground, if the distance is too great to enable you to stretch a cord ? Do you know of any property of the line of sight which might help in the solution of such a problem ? 5. If you attempt to measure two different distances, of which one is about ten times as great as the other, using the same foot rule and the same method of measurement in both cases, which of these two dis- tances will probably be obtained with greater accuracy ? Why ? 6. What difficulties arise, and how would you attempt to meet them, if you were asked to measure the horizontal distance between two points on an uneven piece of ground ? DIRECT MEASUREMENT OF ANGLES 7. Suppose you have measured a distance (say the edge of a table) with great care and have found it to be equal to 4 feet and 9| inches. Is this the exact length of the table, or may it be a small fraction of an inch greater or less ? 8. If you were to repeat the measurement with still greater care, making use of a more perfect standard of length, is it likely that you would find exactly the same result as before ? 9. In any such measurement can you ever attain absolute accuracy ? If not, why not ? 10. Is there any way of knowing whether a measurement is absolutely accurate ? Is there any way of knowing whether it is accurate to within a certain limit of accuracy, say one tenth of an inch ? 11. If a distance has been measured by a process which may be relied upon to give a result accurate only to one one-hundredth of an inch, is it desirable, proper or honest, to give the result expressed in decimal parts of an inch to more than two decimal places ? Why not ? In per- forming calculations based upon such measurements, how many decimal places should we ordinarily carry? 12. In measuring distances by means of metal rods, when great ac- curacy is required, changes of temperature must be taken into account. Why? 13. With what units of length are you familiar ? 14. Gather from an encyclopedia what you can concerning the " standard yard " kept at Washington. 15. What is the metric system? What are the relations to each other of the units called millimeter, centimeter, decimeter, and meter? What is the length of a meter in inches? 16. Can you see any reason why the metric system should be prefer- able to the English system of weights and measures ? 2. Direct measurement of angles. The operation of measur- ing angles is scarcely less important than that of measuring distances. A pro- tractor is an instru- ment used for this purpose. In its simplest form, a pro- tractor consists of an arc of a circle graduated to de- ees (Fig. 1). A Fig. i THE OBJECT OF TRIGONOMETRY mere inspection of the instrument will enable the student to see how angles may be measured and constructed by means of it. The unit usually employed in measuring angles is the ninetieth part of a right angle, and is called a degree. A degree is divided into sixty equal parts, each of which is called a minute^ and each minute is subdivided into sixty seconds. Thus 60 seconds (60'^) = one minute (l^? 60 minutes (60') = one degree (1°), 90 degrees (90°) = one right angle. Very frequently, the angles smaller than one degree are expressed as decimal parts of a degree instead of in minutes and seconds. For the purpose of measuring angles in the field, surveyors make use of an in- strument called a transit or theodolite. The essential parts of this instrument are (cf. Fig. 2): 1. a horizontal graduated circle ; 2. a movable cir- cular plate adjusted so as to be capable of rotation around the center of the hori- zontal graduated cir- cle ; 3. an index at- tached to the movable v3lk pla-te in such a way FiQ, 2 as to enable the ob- DIRECT MEASUREMENT OF ANGLES 6 server to read off the amount of its rotation with reference to the fixed horizontal circle ; 4. two standards attached to the movable plate and carry- ing a horizontal axis to which is attached a telescope and also, in a complete instrument, a vertical graduated circle, used for measuring angles whose sides lie in a vertical plane. The transit is usually supported on a tripod. If we wish to measure the angle between two horizontal lines, the tripod is placed over the vertex of the angle and the telescope is pointed toward some point on one of the sides of the angle. The index will then point to a definite division on the horizontal circle. The operation of ascertaining the division of the circle toward which the index is pointing, is known as " reading the circle." After reading the circle and noting the result, the telescope is directed toward a point on the other side of the angle. The difference between the two readings of the horizontal cii'cle will give the magnitude of the angle. In a similar way the vertical circle, which is attached to the axis of the telescope, makes possible the measurement of angles whose sides lie in a vertical plane. Both circles are usually graduated to whole degrees. The index, in most instruments, is not a simple pointer, but a so-called vernier, an in- genious device which enables the observer to determine the reading of the circle to within a small fraction of a degree, even if the circle is graduated only to whole degrees. In the most accurate instruments, the vernier is replaced by a reading microscope. It is obviously very important that the horizontal circle be exactly horizontal, and that its center be exactly over the vertex of the angle which is to be measured. To help in making these adjustments, the surveyor uses a spirit level and a plumb line. Most transits are also supplied with a compass, which enables the observer to determine the absolute directions of the lines which he is surveying. EXERCISE II 1. Use a protractor to draw angles of 10°, 20'', 31°, 47° 30', 67°, 78°, 86°. 2. Draw five angles at random and measure them as accurately as possible with your protractor. 3. Draw a triangle at random, measure its angles and find their sum. What should this sum be ? If you have obtained a different result for the sum, what are the reasons ? 6 THE OBJECT OF TRIGONOMETRY 4. Construct, out of cardboard, an instrument embodying the prin- ciple of the transit, substituting for the telescope some other method of taking a sight. 5. Why is it important that the horizontal circle of a transit should be truly horizontal? How does the spirit level enable us to make it so? Study the article on the spirit level in some encyclopedia or in a treatise on surveying. 6. Study the articles on vernier, micrometer, reading microscope, compass, in an encyclopedia or in some appropriate treatise, and write an abstract of the same. 7. Describe the sources of inaccuracy which you think may arise in the measurement of an angle by means of a protractor or theodolite. 8. Deduce rules for converting minutes and seconds into decimal parts of a degree and vice versa. , 9. Apply this rule to the angles , \ ' 30°. 20', 10° 45', 8° 40' 20", 3°'8'V'. 10. Express the following angles in degrees, minutes, and seconds : 23°.14, 18°.^5, 46^235. 3. The impossibility of finding all distances by direct measurement. We have discussed briefly the direct methods for measuring distances and angles. It is clear from our discussion that it is very difficult to measure great distances in that way. But there are many cases in which it is al- together impossible to apply such direct methods. How, for example, should we proceed to find the distance through a mountain or across an extensive valley ? How shall we find the distance from New York to London, or from the earth to the moon, by direct measurement ? It is clear that, if we wish to answer such questions at all, we shall have to devise some method different from that of direct measurement. The attempt to solve, by indirect methods, problems whose direct solution is inconvenient or impossible usually leads to great advances in Science. The most important theorems of elementary geometry were probably first discovered by the Ancients in their attempts to devise con- venient and practical methods for the measurements which they found necessary for the purposes of their everyday life. It is generally believed, for instance, that the Egyptians became expert geometricians and sur- veyors at an early period of their history, because it was so important DIFFICULTY OF DIRECT MEASUREMENT 7 for them to be able to reestablish the boundary lines of their lands after the effacement of their marks by the annual inundations of the Nile. The earliest Greek philosophers and mathematicians were pupils of the Egyptians, and some of their first achievements were connected with problems of the particular kind which we are now discussing. Thus it is reported that Thalp:8 of Miletus (about 600 B.C.) measured the height of a pyramid by measuring the length of its shadow, at the instant when the shadow of a vertical stick by its side was exactly as long as the stick itself. The height of the pyramid would then be equal to the length of its shadow at that moment. This method has the inconvenient feature of compelling the observer to wait (many hours perhaps) for the right moment. According to a report by Plutarch, Thales also devised a second method which avoided this inconvenience. This method involves the use of the simplest properties of similar triangles and may be illustrated by means of the following example : We place a stick 5 feet high into the ground near a building whose height we wish to find. At any convenient moment we measure the 125 feet Fig. 3 length of the shadows of the stick and of the building. Suppose we find in this way that the shadow of the building is 125 feet long at the moment when the shadow of the stick is 10 feet long. If h denotes the height of the building, we shall have (Fig. 3) whence A: 125 h 5:10, 62.5 feet. According to Eudemus (about 300 B.C.), one of the earliest writers on the history of mathematics, Thales also invented a method for measuring the distance from the shore to a ship at sea. Although Eudemus does not describe Thales's method in detail, he says enough to lead us to conclude that it was essentially as follows : Let 5L be a tower, say a lighthouse on the shore, and 8 THE OBJECT OF TRIGONOMETRY let S be the ship at sea. Measure the angle BLS and the height of the tower. Construct a similar triangle B'L'S' on the drawing board and measure B'S' and B'L'. Then BS can be found from the proportion BS'.BL = B'S'.B'L'. 4. The graphic method. The examples discussed in Art. 3 show how we may solve many problems of practical geom- etry by indirect measurements. We did not measure directly the quantity which we were seeking, but some other related quantities, and ultimately, by means of these relations, we determined the desired quantity itself. But these same examples may serve to illustrate another point. The solutions which we gave are examples of the graphic method, that is, of a process which makes use of drawing instruments, of accurate geometric constructions and measurements, for the purpose of obtaining the values of the unknown quantities. Such graphical methods are often extremely valuable and have been developed in recent times, in connection with other parts of mathematics, so as to give rise to very important results. The graphic solution of any problem about triangles will finally re- duce to an application of certain theorems of geometry which state that a triangle is determined and may be constructed when certain ones of its six parts (sides and angles) are given. It is clear, then, that these theorems are particularly important for the graphic method. Some of the following questions have been chosen for the purpose of aiding the student to refresh his memory in regard to these matters. EXERCISE III In the following examples and throughout the book we shall usually denote the angles of a triangle hj A,B, C and the sides opposite to these angles by a, b, c, respectively. The student should be provided with a protractor, a pair of compasses, and a ruler divided decimally, say into centimeters and millimeters. All constructions and measurements should be made as carefully as possible. 1. Given a = 3.72, b = 4.91, c = 2.56. Find the angles. v^ 2. Given a = 4.27, B = 35°, C = 69°. Find the remaining parts of the triangle. 3. Given b = 5.63, c = 6.71, A = 27°. Find the remaining parts of the triangle. AN ARITHMETICAL METHOD 9 4. Given a = 4.23, b = 5.16, A = 55°. Find the remaining parts of the triangle. 5. When a, b, c are given at random, can we always find a triangle of which o, &, c are the sides, or is there some restriction on the possible values of a, &, c ? 6. If a, b, A are given, there may be one or two solutions, or no solution. Discuss these cases. 7. Is a triangle determined when we know the magnitudes of its three angles? Why? When the three angles of a triangle are given, have we obtained essentially more information than if only two of them are given ? Why ? Is the third angle of a triangle independent of the other two? 8. What do you mean by similar triangles? 9. Under what conditions are two triangles similar ? 10. Use the shadow method of Thales (Art. 3) to measure the height of some building in your neighborhood. 11. By means of a transit, or else by means of the home-made in- strument of cardboard suggested in Ex. 4 of Exercise II, find the dis- tance of some object situated on the opposite side of the street from your home, without crossing the street. Afterward check your result by direct measurement. 12. How may a person on board ship find his distance from a build- ing on the shore if he knows its height? ^j 5. The desirability of an arithmetical method for solving triangles. We have seen that it is a rather simple matter to solve a triangle by the graphic method. But we can hardly feel altogether satisfied with the graphic solution, for it will clearly not permit us to reach any great degree of accuracy. To be sure, by making our drawings on a very large scale, we might lay off distances with considerable precision, but we should still encounter the difficulty of accurately plotting angles. Clearly it would require extraordinary skill and exceedingly fine instruments to enable us to draw an angle so accurately that its error should not exceed one minute of arc. Many other circumstances combine to make a graphical solution unsatisfactory if a high degree of precision is required. The main value of the graphic method lies in furnishing a solution whose approximate correctness is apt to he apparent to 10 THE OBJECT OF TRIGONOMETRY the eye^ and which may therefore^ in almost all cases^ serve as a check on the more complete solution obtained in some other way. But the lack of accuracy is only one of the defects of the graphic method, although perhaps the most important one from a practical point of view. The other defect which we wish to emphasize is more of a theoretical nature. The parts of a triangle (its sides and angles) are usually given as numbers (so many feet, so many degrees). It seems natural, therefore, to suppose that there must exist some arithmetical process for the solution of triangles. Trigonometry enables us to find the unknown parts of a triangle by arithmetical processes. This statement must not be regarded as a complete defi- nition of trigonometry. We shall see later that the solution of triangles by arithmetical methods constitutes only a part, although an important part, of trigonometry. EXERCISE IV 1. What are the principal sources of inaccuracy in solving a triangle by the graphic method ? 2. We wish to construct an isosceles triangle, given the angle at the vertex and the altitude. Suppose that the error made in constructing the angle is 5'. Will the effect of this error on the base of the triangle, as obtained from the construction, be different for different altitudes? Will it be greater or less for the higher triangles ? 3. If you wish to find a point as the intersection of two straight lines, are you likely to obtain a more accurate result if the two lines are at right angles, or if they are nearly parallel? 4. A side and two adjacent angles of a triangle are given, and its other parts are to be found graphically. Is such a graphic solution likely to be accurate if both of the angles are very small ? If the sum of the given angles is very close to 180°? 5. Formulate, in your own words, the distinction between an arith- metical and a graphical process. CHAPTER II THE TRIGONOMETRIC FUNCTIONS OF ACUTE ANGLES 6. The necessity of introducing new ideas. At the close of Chapter I we formulated the problem of devising arith- metical methods for the solution of triangles. But, if we think of the many shapes which a triangle may assume, this problem appears to be a most formidable one. We shall, therefore, for the present, confine our attention to the com- paratively simple case of a right triangle. We have good reason to suppose that, if we succeed in solving our problem for all right triangles, we shall be able to deal later with the general case also, since every triangle may be decomposed into two right triangles. Let us then consider a triangle ABC^ right- angled at (Fig. 5), and let us denote the sides opposite the angles A^ B, C by a, 5, ^ (2) a^ + b^=(^. "^ H^<^^-^ The first of these equations enables us to calculate one of the acute angles if the other one is given. The second provides a method for calculating any one of the three sides if the other two are given. But we have nothing as yet which will enable us to find the angle A if two of the sides (say a and 6) are given, although the triangle is clearly determined by these sides and might be constructed by geometry. The equations (1) 11 12 FUNCTIONS OF ACUTE ANGLES and (2), unaided by other relations, are obviously inadequate for this purpose, since (1) is a relation between the angles A and B alone, while (2) involves only the sides of the tri- angle. Now, clearly, a statement which is only concerned with the angles of a triangle cannot convey any positive information about t\\Q sides ^ and vice' versa. In order that we may he able to solve a right triangle by arithmetical processes^ there must^ therefore^ be added to equa- tions (1) arid (2) certain other relations^ in which the sides and angles shall not be separated, but in which they shall occur simultaneously. The discovery of such relations and their adequate formulation is the foundation upon which all of trigonometry must finally rest. A simple illustration will make clear the nature of these new relations. The numerical measure for the steepness of an inclined plane, say a mountain road, may be given in two ways. We may say that the road makes a certam angle A (say 5°) with the horizontal plane, or we may say that it rises a certain number of feet (say 87.5 feet) in a horizontal distance of 1000 feet. 87 5 The quotient, — '—, is technically known as the slope, grade, or gradient of the road. The gradient is clearly connected with the angle A in such a way that, if A should vary, to every value of the angle there corresponds a definite value of the gradient and vice versa. Thus, if AB (Fig. 6) represents the road, and if a and b are both expressed in feet, the gra- dient of AB is equal to -. The value of this quotient changes with the b angle A, so that for different angles we find different values for the gradient. The general question before us may be formulated as fol- lows. If the acute angle A of the right triangle ABO be- comes larger or smaller, what effect will such a change have upon the sides ? And, conversely, if the sides of a triangle change, what effect will this have on the angles ? Now, since similar triangles have their corresponding angles equal, it is clear that there are some changes in the DEFINITIONS OF FUNCTIONS 13 lengths of the sides which produce no change in the angles. In fact, if all of the sides of a triangle are magnified in the same ratio, the angles are not changed at all. In the two right triangles ABC and A'B'Q' (Fig. 7) we have a _a! h _h' « _ «^ ^ c c' e c' h h' if the angle at A' is equal to that at A, But if the angle A' is different from A, the ratios —,, -, ^, etc., will not be equal to -, -,% e G c c etc., respectively. For if they were, corresponding pairs of sides of the two triangles would have the same ratio, the triangles would be similar, and, contrary to our hypothesis, angle A! would have to be equal to angle A. Consequently, while the lengths of the indivicbual sides of a right triangle have nothing to do with the size of its angles, the ratios of these lengths are connected with the magnitude of the angles in a very intimate fashion. In fact, so close is this relation that the values of the ratios -, -, -, etc., may be G c determined (by construction) as soon as the acute angle A has been chosen, and conversely ; if one of these ratios is given, we can find (by construction) one, and only one, cor- responding acute angle A. 7. Definitions of the trigonometric functions of an acute angle. We have seen that the values of the ratios -, -, etc., c c of the sides of a right triangle are closely bound up with the magnitude of the angle A. If the angle changes, each of these ratios changes and vice versa. Now, one variable quantity is called a function of another^ if they are so related that any change in the latter produces a corresponding change in the former. 14 FUNCTIONS OF ACUTE ANGLES Consequently, each of the six ratios -,-,-,-,-,-, deter- c c h a a h mined by the sides of the right triangle ABQ^ is -a, function of the angle A^ because any change in A produces a corre- sponding change in each of those ratios. Each of these functions has received a name and a symbol. The reason for choosing these names will appear later (see Arts. 10 and 70), and cannot be discussed with profit at the present moment. We proceed to give the formal definitions of the six trigonometric functions of an acute angle A. Construct any right triangle (cf. the Fig.), one of whose acute angles is equal to the given angle A.* Of the two legs of this right triangle^ that one which passes through the vertex of the angle A is said to he adjacent to A. The other leg is said to he opposite to A, and the third side of the triangle is called its hypotenuse. The sine of A is the ratio of the opposite side to the hypotenuse. The cosine of A is the ratio of the adjacent side to the hypotenuse. The tangent of A is the ratio of the opposite side to the adjacent side. The cotangent of A is the ratio of the adjacent side to the opposite side. The secant of A is the ratio of the hypotenuse to the ad- jacent side. The cosecant of A is the ratio of the hypotenuse to the opposite side. * We have seen in Art. 6 that the size of this right triangle is absolutely o-f no consequence, since any two triangles of this kind are similar, so that the corresponding ratios for the two triangles will be equal. DEFINITIONS OF FUNCTIONS 15 In symbols we may write these definitions as follows : sin ^ = - , cos ^ = - , tan ^ = ^ ^ (1) : : I cseA = — , sec A = ^f cotA = —» These symbols are written abbreviations of the names of the functions. In speaking, the symbols are pronounced as though the name of the func- tion had been written out in full. Thus, tan A is pronounced tangent of A or tangent A ; esc A is pronounced cosecant of A or cosecant A, etc. In defining the trigonometric functions, we made use of the numerical measures of certain line-segments, namely, of the sides of a right triangle. Now, the numerical measure of a line-segment changes if the unit of measurement is changed. One might, therefore, expect the values of the trigonometric functions of an acute angle to change with every change of the unit of length. But this is not the case. Owing to the fact that only ratios of these line-segments appear in equations (1), the functions sin A^ cos A^ etc., are found to have the same value whether the line-segments a, 6, c are measured in feet, inches, or in terms of any other unit of length. Consider, for instance, a right triangle for which a = 3 feet, & = 4 feet, c = 5 feet. According to (1) we have sin ^ = f , cos A = f, etc. Let us now introduce the inch as unit of length instead of the foot. The numerical measures of the sides of the triangle will now be a = 36, b = 4:8, c = 60. According to (1) we shall now find sin A = |^, cos A = If, etc. But |§ = f , |f = f, etc., so that we obtain precisely the same values for sin A, cos A, etc., whether the sides of the triangle be expressed in feet or inches. If the trigonometric functions were concrete numbers, that is, if they were the numerical measures of some kind of con- crete quantity such as a length, an area, or a volume, their values would change every time that a change is made from one unit of length to another. We have just seen that this is not the case. Therefore, the trigonometric functions are pure or abstract numbers.* * This same fact is sometimes (somewhat inadequately) expressed by the statement that the trigonometric functions are ratios. 16 FUNCTIONS OF ACUTE ANGLES 8. The practical need of tables giving the values of the trigonometric functions. The trigonometric functions just defined will enable us to find the unknown parts of a right triangle when certain parts are given, provided only that we can devise a practical method for actually obtaining the numerical values of these functions for any given acute angle. Now, the values of the functions have been calcu- lated by mathematicians and the results have been collected in tabular form for convenient use. From the practical point of view, therefore, it only remains to explain the arrange- ment and the use of the tables. To the more scientifically inclined student, however, the question will immediately suggest itself as to how these use- ful tables were actually computed. We shall reserve the answer to this question for the second part of the book. The following examples, however, will show how a table of trigonometric functions may be prepared by the graphic method, provided that no very high degree of accuracy be required. EXERCISE V 1. Find the functions of 40° by the graphic method. Solution. With the help of a protractor construct the angle PAQ (Fig. 8) equal to 40°. Choose any point B on AQ and drop a perpendic- ular BC from 5 to ^P. Measure the distances BC, AC, and AB. Then will (1) sin 40°=^, cos40°=4i- AB AB .P Although the point B might be chosen anywhere on AQ, it will be especially convenient to make AB equal either to one unit, ten units, or one hundred units. For, as equations (1) show, we have to divide by AB, and if AB is equal to 1, 10, or 100, we avoid the long division which would otherwise be necessary. Let us, therefore, make AB = 10 centimeters. We should then find by measurement, ^^ a a a r^ t^t •^ ' C5 = 6.4 cm., ^C = 7.7 cm. According to (1), therefore, sin 40° =M ^ 0.64, cos 40° = "^ = 0.77. 10 10 VALUES OF FUNCTIONS 17 Further we find tan 40° = — = 0.83, cot 40° = ^ = 1.20, sec 40° = — = 1.30, CSC 40° =— = 1.56. 7.7 6.4 We should obtain a more accurate result for tan 40°, and more con- veniently, if we were to use another triangle, making this time ^C = 10 cm. Measurement would then give BC = 8.4 cm., and tan 40°=— =0.84. 10 2. Find the functions of the following angles by the graphic method : (a) 10°. (6) 15°. (c) 20°. {d) 70°. Construct carefully each of the following right triangles, measure the angles, and find the six functions of each acute angle. 3. a = 3, 6 = 4, c = 5. 4. a = 5, 6 = 12, c = 13. ^ 5. a = 8, & = 15, c = 17. /;, ,. 6. Construct and measure an acute angle whose sine is equal to \. 7. Construct and measure an acute angle whose tangent is equal to f. 8. Construct and measure an acute angle whose cosine is equal to |. 9. Can you think of two right triangles (Fig. 7) with different angles A and A', for which the sides a and a' are nevertheless equal? 10. Can you conceive of two right triangles (Fig. 7) with different angles A and A'y for which the ratios - and — are nevertheless equal? b b' 11. Why, then, do we speak of the ratio ~ as a function oi Al Why do we not introduce a or db as a function of yl ? 12. Assuming that we have access to a table of the values of the trigonometric functions, show how we might solve each of the following problems. To find the remaining parts of a right triangle when the following parts are given. I. a, &, C = 90°. Til. a, c, C = 90°. II. a. A, C = 90°. IV. c,A,C= 90°. 13. Show that neither the sine nor the cosine of an acute angle can ever be greater than unity. 14. Show that the tangent of an acute angle may have any positive value whatever. Similarly for the cotangent. 15. What restrictions, if any, are there on the values which the secant and cosecant of an acute angle may assume ? 18 FUNCTIONS OF ACUTE ANGLES 9. Relations between the six trigonometric functions of an acute angle. The preceding discussion suffices to indicate the importance of constructing a table of the values of the trigonometric functions. The task of computing these tables may be abbreviated very considerably by noting that the six functions are not independent of each other. In fact, we have (Fig. 9) (1) sin^ cos^ = tan J. CSC J.= sec J. = cot^ = so that we obtain at once the relations csc^ = sin J. sinvl ,1 sec J. = cos J. = sec^' cot A' or, (2) sm^csc^ = l, cos^sec^ = l, tan^cot^ = 1.* But, two numbers whose product is equal to unity are called reciprocals of each other. Therefore, equations (2) are equivalent to the following statement : The sine and cosecant^ the cosine arid secant, and finally/ the tangent and cotangent, of an acute angle are reciprocals. Clearly, knowledge of this fact reduces greatly the labor of computing tables of the functions. For, having found the values of the three functions, sine, cosine, and tangent, the values of the remaining three can be obtained from these by computing their reciprocals. But there are other relations besides' (2) which enable us to reduce still further the labor involved in constructing, a table of the trigonometric functions. We have tan^ = f. * Note that sin A esc A is written for sin 4 x esc -4 just as ab is written for ax 6. RELATIONS BETWEEN FUNCTIONS 19 If we divide both numerator and denominator of the fraction - by c, we find b/c Bat we have, by definition (cf. equations (1)) - = sin ^, - = cos A^ c c and therefore (3) tan^ = ^i^. « Of course, since cot A is the reciprocal of tan A^ we also have The relations (2), (3), (4) enable us to calculate the values of all six functions when the sine and cosine are known. But it actually suffices to know the sine. For, if we divide both members of the familiar equation (5) a2 + 62 = c2 by c2, we find But, by definition, we have a • A h A — = sin A^ - = cos A^ c c so that (5) becomes (6) sm2^ + cos2^= 1, where sin^ J. and go^^ A have been written, as is customary, for (sin J.)^ and (cos J.)^, respectively. It follows from relations (2), (3), (4), and (6) that, if we know the value of a single one of the six trigonometric functions of an acute angle, the values of the remaining five may be computed. The detailed proof of this statement will be left to the student in some of the examples given below. ■s 20 FUNCTION'S OF ACUTE ANGLES EXERCISE VI In each of the following twelve examples, the value of one function of the acute angle A is given. Find the values of the remaining functions. 1. sin^ = |. 4. cot ^ = 2. 7. sin ^ = x, 10. cot^=a:. 2. cos^=T^3. 5, sec^ = ^f. 8. cos^=a:. 11. sec^=a:. 3. tah^ = l. 6. cscA=^^. 9. tan^=a:. 12. csc^=af. 13. Construct an isosceles right triangle and make use of this figure for the purpose of computing the functions of 45°. ■y 1^ Divide an equilateral triangle into two right triangles by dropping a perpendicular from one of its vertices to the opposite side. Make use of this figure for the purpose of computing the functions of 30° and 60°. 15. Collect in a table the results of Exs. 13 and 14. 16. Prove the formula sec^ A = l-\- tan^ A. 17. Prove the formula csc^ A = l + cof^ A.' 18. Prove that the sine, tangent, and secant of an angle increase when the angle grows from 0° to 90°. 19. Prove that the cosine, cotangent, and cosecant of an angle decrease when the angle grows from 0° to 90°. 20. Prove that tan ^ < 1, cot ^ > 1 if J < 45°, and that tan A > 1, cot^45°. 21. Show that each of the six functions may be expressed either as a product or as a quotient of two of the others. 10. Relations between the functions of complementary angles. As a result of the relations discussed in the preced- ing article, the problem of computing the values of the six trigonometric functions for every angle between 0° and 90° has been reduced to that of computing these values for a single one of these functions. But we may reduce the prob- lem still further by observing the relations between the functions of the two acute angles of the same right triangle. We have (cf. Fig. 10) sm-4 = -, cos^ = -, c c cosA = -, sni^ = -, c c COMPLEMENTARY ANGLES 21 cot A = -, tan ^ = -, a a sec^ = --, cscij=-, h b cscA = -, sec^ = -, a a and therefore ^^ ^^ sin A = cos j5, tan ^ = cot B, sec ^ = esc B^ cosJ. = sin^, cot J. = tan ^, esc ^ = sec ^. Since the angles A and B are complementary, we may write these equations as follows : sin (9^° ~ ^) = cos A,^ cot (90° - ^) = tan A, (2) cos (90° -A) = sin Zr^ sec (90° - A) = esc A. tan (90° -A) = cot A, esc (90° - ^) = sec ^. An easy way to remember these formulae is as follows: Let the six functions be grouped into three pairs : sine and cosine, tangent and cotangent, secant and cosecant. Let us speak of either, function of one of these pairs as the cof unction of the other. Then, the six formulae (2) are all included in the following statement. Any trigonometric function of the complement of an angle A is equal to the cof unction of A. It is apparent that this theorem will enable us to find the trigonometric functions of any acute angle greater than 45°, if we know the functions of all angles less than 45°. Thus, for instance, tan 75° is equal to cot 15°, sin 82° is equal to cos 8°, etc. As a consequence of this fact it is possible to reduce the space occupied by the tables of the function « to exactly half of what would otherwise be necessary. The relation between the functions of complementary angles is also important in another respect. It is this relation which has given rise to the words cosine, cotangent, and cosecant. The cosine is the sine of the complement. At a time when Latin was still the universal language of the scientific world, the cosine was therefore called complementi sinus. 22 FUNCTIONS OF ACUTE ANGLES This was later (in the seventeenth century) contracted to cosinus. The words cotangent and cosecant originated in the same manner. EXERCISE VII 1. Express as functions of the complementary angles . sin 37°, cos 62°, tan 13°, cot 75, sec 12° 15', esc 55° 37'. 2. If the table of values of the functions is so arranged as to give only the functions of angles less than 45°, how may we obtain the values of sin 57°, cos 63° 15', tan 75^ 12', cot 67° 18' ? 3. What acute angle is that whose sine is equal to the sine of its complement? 4. Find an acute angle for which tan^ = cot (45° + A). Hint. Substitute for tan^l its equal cot (90° — A) and note that two acute angles with the same cotangent are equal to each other. 5. Find an acute angle for which sin 2 A = cos (45° — A). 6. Find an acute angle for which cot 3 A = tan 2 A. 7. Find an acute angle for which cos^ = sin 6 ^1. 8. Find an acute angle for which sec 2 A = esc 7 A. 11. The values of the functions of o°, 30°, 45°, 60°, 90°. While we have decided to postpone the general question of the arithmetical calculation of the trigonometric functions, we have already performed this calculation for a few special angles, viz. : 30°, 45°, 60° (cf. Exs. 13, 14 of Exercise VI). The figures there suggested, and the results are as follows : (1) ^--^ (2) sin 45°= cos 45° = —^ tan 45°= cot 45° = 1, sec45°=csc45° = V2. sin 30° = cos 60° = 2-^ = i, c 2 cos 30°= sin 60° = J V5, tan 30°= cot 60° = -i^ = -L = 1 V3, cot 30 sec 30°= CSC 60° 1 \A =42°.39 (2) 6 = 4.271 (7) B = 47°.61 (8) a2 ^ 15.20 (8) sin A = 0.6742 (4) b^ = 18.24 (9) cos A = 0.7386 (5) a^ + b^ = 33.44 (10) | q^^^^ c2 = 33.44 (11) J Remark. The quantities a^ b^, c^ required for the check were obtained from the table of squares. When no such table is available, it is usually desirable to write the check equation in the form a2 = c2-62 = (c- b)(c-\-b), since this form of the equation reduces by one the number of multiplica- tions required. BY NATURAL FUNCTIONS 33 In this example, the check computation would then yield a2= 15.202, c-b = 1.512, c + b= 10.054, c^-b^= 15.202 Check. EXERCISE X In each of the following right triangles, right angled at C, two parts are given. Compute the remaining parts and check. Also check by means of a graphic solution to provide against gross errors. ^"1. a = 27, A = 25°.l. / 5. c= 604.5, A = 47°.53. i/2. a = 34.5, e = 52.8. 6. a = 8.695, b = 7.321. 3. a = 2.781, b = 3.056. 7. b = 62.78, c = 81.39. 4. & = 87.95, ^ = 55°.36. 8. jB=29°.58, c = 2354. 9. A gravel roof slopes three fourths of an inch per horizontal foot. What angle does it make with a horizontal plane ? 10. The pitch of a gable roof is the quotient obtained by dividing the height of the ridge-pole above the garret floor by the width of the garret. What is the pitch of a gable roof covering a garret 38 feet wide, if the ridge-pole is 15 feet above the garret floor, and what angle does the roof make with a horizontal plane ? 11. At a time when the sun was 55° above the horizon, the shadow of a certain building was found to be 112 feet long. How high is the building ? 12. The side of a regular decagon is 3.471 feet. Find the radii of the inscribed and circumscribed circles. 13. The side of a regular polygon of n sides is equal to a. Find formulae for the radii of the inscribed and circumscribed circles. CHAPTER IV DISCUSSION OF SOME DEVICES FOR REDUCING THE LABOR INVOLVED IN NUMERICAL COMPUTATIONS 14. The number of decimal places. As we attempted to point out in Chapter I, every number obtained as a result of measurement is really an approximation. If we measure the distance between two dots on our drawing board, by means of a carefully constructed scale which reads to the fiftieth part of an inch, we may still estimate half of ^his smallest scale unit with the naked eye. Let us assume that the divisions of the scale are reliable, that the ruler is very nearly straight, that the dots are very small, and that we are using the greatest of care in our measurement. We may then concede that the result of such a measurement (say 5.34 inches) is accurate to the nearest y^ of an inch. This means that no number, with two figures to the right of the decimal point, is as close to the true value as 5.34. It means that 5.33 is certainly too small and that 5.35 is certainly too large. It does not mean that the true value is exactly 5.34 inches, but that the true value lies between 5.335 and 5.345 inches. When we record the result of such a measurement, the number of decimal places which we write (two in this example) is an indication of the degree of precision which we claim for the result. In this connection, let us note emphatically that a zero, when obtained as the last digit of the measure of a quantity, should never be suppressed. Suppose, for instance, that in the above example we had obtained 5.30 inches as the result of our measurement. This means that we are certain that the true value of the distance lies somewhere between 5.295 and 5.305 inches. 34 ACCURACY OF RESULTS 35 If we were to record this result as 5.3 inches, suppressing the final zero, we should be giving the erroneous impression that we had measured the distance only to the nearest tenth of an inch and that it might have any value between 5.25 and 5.35 inches. Thus, the number of decimal places, which we use in record- ing the result of a measurement^ is an indication of the degree of precision which we attribute to this result. This being so, we are guilty of negligence whenever we express such a result by a decimal with fewer places than we are able to guarantee. For we are thus throwing away knowledge which we have actually had in our possession. But it would be dishonest to express our result with more decimal places than we can guarantee. For we should then be tending to mislead others into thinking our measurements more accurate than they really are. We may estimate the degree of precision of a number, as we have just done, on an absolute scale. But clearly it will usually be more reasonable to adopt as a measure of pre- cision the ratio of the ^' probable error " of the measurement to the total magnitude of the quantity measured. If we do this, an error of one foot in a thousand is to be regarded as of no greater importance than an error of ywq'Q ^^ ^^ inch in an inch. In either case we may say that it is an error of -^^ of one per cent. Whenever we properly record the result of a measurement by a number consisting of four digits, no matter where the decimal point may be placed, this means, in the light of our preceding discussion, that the error of the last digit is guar- anteed to be less than half a unit of the last decimal place. Now, the smallest number expressed by four digits is 1000. Let us suppose that the exact value of our unknown quan- tity is w = 1000 units, but that as a result of our measure- ment we have found a; = 999.5 units. Then our error is 2 oVo' ^^ A" ^^ ^^^ P®^ cent, of the total magnitude measured. The largest number expressible by four digits is 9999. If 36 NUMERICAL COMPUTATIONS the exact value of x is 9999 and our error of measurement is half a unit, the error will be to the total magnitude measured as \ is to 9999, or as 1 is to 19998. It will be an error of about 2^0 ^^ ^^® P®^ ^®^^ ^^ ^^ whole. Thus, four digits are certainly sufficient to express the result of a measurement if its degree of accuracy does not exceed ^^ of one per cent. Now, most of the ordinary operations of surveying fall well within this limit, so that four decimal places are usually sufficient to express the results obtained by the surveyor. 15. The accuracy of a sum, difference, product, or quotient of two numbers obtained by measurement. ' It is clear that the sum or difference of two numbers can have no greater precision than the less accurate of the two numbers. Con- sequently, it is useless and misleading to retain more decimal places in one term of a sum or difference than in the others. In forming a product we are apt to do a great deal of use- less work if we fail to remember that the factors, and there- fore the product, are mere approximations. Suppose we have measured the sides a and 5 of a rectangular field to the nearest hundredth of a foot and have found a =35.67 ft., 5 = 86.72 ft. To find the area of the field we' form the Oft 79 product ah. The ordinary method (shown in „_' _ the margin) gives the result 3093.3024 square ' feet. But if we allow the result to stand in -n/^or. thls form, we shall exhibit either our ignorance O2032 A ' ^ ^ ^ ^ ' • 17 . „^ or our desire to create a false impression, l" or 9fini R *^^^ would seem to indicate a result precise to ^OQ^ ^r>9J. ^^ nearest .^^^ ^ ^^ of a square foot, whereas it is uncertain by more than a whole square foot as we shall now show. In fact, the equations a =35.67 ft., 5 = 86.72 ft. merely mean that a and h are between the limits 35.665 , respectively, the scales A and D being on the rule, and B and Q on the slide. Fia. 16 The scale A is composed of two logarithmic scales such as that of Fig. 14, so that its right-hand end might be labeled 100, since log 100 = 2. On most slide rules, however, the first principal division on scale A after 9 is not labeled 10, * Oughtred's instruments were described in publications of William Fos- ter, one of his pupils, in 1632 and 1633. 64 CALCULATION BY LOGARITHMS as in Fig. 14, but 1, the next one is not labeled 20, but 2, and so on to the last one, which is again labeled 1 instead of 100 or 10. Thus, the two halves of scale A are exact copies of one another. This is done for precisely the same reason that the mantissas only are printed in our tables of logarithms. The slide rule also makes use of the mantissas only. The characteristics, or what amounts to the same thing, the position of the decimal point in the result, must be obtained by inspection or by special rules. Scale B is on the upper edge of the slide, in direct con- tact with scale A on the rule, and is an exact copy of scale A. These two scales together may be used for multiplica- tion and division as explained above. Scale I) is on the lower part of the rule. It is a single logarithmic scale, from 1 to 10, of the same length as the combined two scales of A. The logarithm of any number is therefore represented, on scale i>, by a distance twice as great as that which represents the logarithm of the same number on scale A, It follows from this that the number which is found on scale A^ vertically above any number of scale i>, is the square of the latter. Any number on scale i>, on the other hand, is the square root of the number ver- tically above it on scale A. Scale C is on the lower edge of the slide, in direct contact with slide D on the rule. It is an exact copy of scale D. These two scales together may be used for multiplication and division, according to the same rules which hold f(5^r scales A and B, Besides these four scales, the slide rule is supplied with a runner (cf. Fig. 16), which is useful in performing com- pound operations, and also in comparing two scales (such as A and 2)), which are not in direct contact with each other. The runner was made a permanent feature of the slide rule by Mannheim in 1851.* * Amedee Mannheim (1831-1906), a distinguished geometer of recent times. The runner had however been used occasionally, long before Mannheim, by a number of English mathematicians. THE SLIDE RULE 65 It often happens, in manipulating the slide rule, that the result is to be sought opposite a number of the slide which falls outside of the scale on the rule. In such cases, we may shift the slide, bringing the right-hand index to the place which the left-hand index occupied previously, and read off the result as before. For, such a shift has no influence on the mantissa, since it merely amounts to dividing the result by 10. On the Mannheim rule, this shifting of the slide may be avoided by working with scales A and B rather than with and D. Scales Q and i), however, have the advan- tage of greater accuracy. If the slide be withdrawn entirely, it will be found to have three other scales on its reverse side, two of which are labeled S and T. These are scales of logarithmic sines and tangents, respectively, and may be used for calculating such products as c sin A^ c tan A, The middle scale on the reverse side is used for finding the value of the logarithm of a number, and is important if we wish to compute a power of a number with a complicated fractional exponent. For more complete information concerning the slide rule, we must refer to the manuals which are usually presented to the purchaser of such an instrument.* Cheap slide rules, especially constructed for the beginner, may now be obtained of all dealers under the name Student's or Col- lege Slide Rule. Engineers and computers use the slide rule so extensively that the student will find it advis- able to make himself familiar with the instrument by actual use. The Mannheim slide rule, which we have described, admits of three-figure accuracy. In some (exceptional) cases, results correct to four decimal places may be obtained by its use. The Thacher and Fuller slide rules, more compli- cated instruments, but constructed on essentially the same * See also Raymond's Plane Surveying. 66 CALCULATION BY LOGARITHMS principles, admit of far greater accuracy. The* Eichhorn Trigonometric Slide Rule was invented for the purpose of solving triangles, and is especially adapted for this work. But, of course, it has not the wide range of usefulness of the ordinary slide rule. CHAPTER VI APPLICATION OF LOGARITHMS TO THE SOLUTION OF RIGHT TRIANGLES 31. The general method. We have shown, in Chapter III, how to solve right triangles by means of the natural functions, and we have become acquainted with the theory and use of logarithms in Chapter V. To solve a right triangle by loga- rithms, it suffices to combine the results of these two chapters. We use the same formulae as in Chapter III, but perform the multiplications and divisions by means of logarithms, using the table of logarithmic sines, cosines, etc., in place of the table of natural functions. In order to illustrate the various practical questions which arise in such a calculation, we shall give a rather extended discussion of the following example : ^ Example. The legs of a right triangle were found to be a = 527.38 feet and b = 621.24 feet. Calculate the hypote- nuse and the acute angles A and B, 32. The 'rolution. i preliminary graphic We first make a drawing, approxi- mately to scale, making a = 5.3 centimeters, b = 6.2 centimeters. We find by neasurement c A = 8,1 centimeters, = 40°. 5, 5 = 49°.5. This figure and these measurenaents serve two purposes. In the first place, the figure helps us pick out the formulae which we shall need for the trigonometric solution of our triangle. In the second place, comparison of the approxi- mate values of the unknown quantities obtained graphically, 67 68 SOLUTION OF RIGHT TRIANGLES BY LOGARITHMS with the final results as obtained by calculation, constitutes a valuable check. If the results obtained by the two methods should differ by more than can be accounted for by the in- accuracies of the graphic solution, we must look for a mis- take in our calculation. 33. The gross errors. Mistakes, which are large enough to be detected by means of a graphic check, are known as gross errors and are usually due to one of the following causes : 1. The lise of a wrong formula. 2. A misplaced decimal point or, what amounts to the same thing, an erroneous characteristic. 3. The use of a number taken from a wrong column in the tables, resulting, for instance, in erroneously using the log cos of an angle in place of the log sin. 4. Addition of two logarithms when subtraction is re- quired or vice versa. 5. Purely arithmetical errors of addition and subtraction. The errors of the first four classes can be avoided by the exercise of a sufficient amount of care. The student should not attempt to gain speed in calculation until he has first learned to be accurate. The errors of the last class are quite unavoidable, but they will usually be detected almost as soon as made if the ordinary arithmetical checks for addition and subtraction be applied every time that one of these operations is used. 34. Selection of formulae and checks. After completing the approximate graphical solution of a triangle, we pick out the formulae which we wish to use in the computation. In our example, these are the following : (1) tan^ =^, c = -^— = — ^, 5 = 90°-^, 6 sin ^ cos A d or, in logarithmic form, (2) log tan ^ = log a - log J, 5 = 90° - J, log c = log a — log sin A = log h — log cos ^ . We have two formulas for c, and in most cases it makes little differ- ence which one we decide to use. If we were to use both, the agi'eement ¥ SKELETON FORM OF SOLUTION 69 of the two results would constitute a partial check on the accuracy of our work. It would not be a total check however ; a mistake in the loga- rithm of a or 6 could not be detected by means of it.* It is hardly worth while therefore to use both formulae for c. That one is to be pre- ferred which has the greater denominator, as the result obtained from it is likely to be the more accurate. For a complete check, we may make use of the equation a^+b^ = c2, which, however, we prefer to write in the form (3) a = V^^-^Tp = V(6--|-&) (c-b), which is more convenient for logarithmic computation. It should not be necessary to write the formulae in the logarithmic form (2). Form (1) is shorter and more directly connected with the geometry of the problem. Moreover, it contains all of the information that is necessary for the solution of the problem, for anybody who has studied logarithms. || 35. The framework or skeleton form. Having selected the formulae, we proceed to plan the details of the computa- tion by providing a definite, properly marked place for every number which will be needed in the course of the work. Moreover, we shall plan these details in such a way that those numbers which are to be combined by addition or subtraction will have their places in the same vertical column next to each other. In our particular example we may adopt the following framework : k Given { ;: (1) log& = (4) (2) log cos A = (7) loga = (3) logc = (4) -(7) = (8) log& = (4) c = (9) log tan A = (3) -(4) = (5) b= (2) (A = (6) c-b = (9) -(2) = (10) Results i B = (16) c + b = (9) + (2) = (11) . c = (9) log(c-b) = (12) log (c + b) = (13) l0g(c2-&2)=: (12) + (13) = (14) Check ]l«g^(^^-^^) = I' log a = 1(14) = (15) (3) * Since such a mistake could be interpreted as leading to the correct solution of a triangle different from the given one, namely, that one whose sides a' and b' have as logarithms the values, which were, by mistake, assigned to a and b. 70 SOLUXrON OF RIGHT TRIANGLES BY LOGARITHMS The numbers in parenthesis merely indicate the order in which this skeleton form may be filled in, and how some of the results are obtained. The student should use these numbers only to aid him in understanding the construction of the framework and the plan of the computation. They should not be used in writing out the actual calculation. 36. The computation. We are now prepared to carry out the computation. This should be done on paper ruled into squares of such a size that each figure may conveniently occupy one square. We obtain the following results : Given a = 527.38 h = 621.24 log a = 2.72212 log 6 1:^2.79326 log tan ^ = 9.92886-10 [ ^ = 40° 19'.69 Results \ B=^r 40'.31 c = 814.92 logZ> = 2.79326 log cos ^ = 9.88215 - 10 logc = 2.91111 c= 814.92 b= 621.24 c-b= 193.68 c + b = 1436.16 log (c- 6) =2.28709 log (c + &) = 3.15720 log (c2 - 62) = 5.44429 logVc2- &2 = 2.722151 Check. loga = 2.72212] Remark. We observe that the check is not absolute. The agreement is as close, however, as we should expect. The inevitable inaccuracies, arising from the neglected higher decimal places, often manifest them- selves by discrepancies amounting to several units of the fifth decimal place. Consequently, we may declare the check to be satisfactory.* 37. Revision of the computation when the check is un- satisfactory. If, in the solution of such an example, the results fail to check satisfactorily, the magnitude of the dis- crepancy will help us to locate the error. If the discrepancy is very great, the error must be one of the gross kind which we have discussed in Art. 33. In case of a comparatively small discrepancy, our error is probably due to one of the following causes : 1. Purely arithmetical errors of addition and subtraction in the last few decimal places. * If a and 6 differ considerably, use as check a = v(c — 6) (c + b) or 6 = Vic — a){c-{-a), according as 6 < a or & > a. EXERCISES ON RIGHT TRIANGLES 71 2. Inexact interpolation, which would ordinarily affect only the last decimal place. 3. Addition of the correction obtained by interpolation when it should be subtracted, or vice versa. This last mistake may be avoided by carefully inspecting the table after the interpolation has been completed, so as to make sure that the quantity calculated actually lies between the two numbers of the table between which it should fall. EXERCISE XV In each of the following examples (1-10), two parts of a right triangle are given in the usual notation. Find the other parts : 1. c = 627, A = 23° 30'. 6. a = 13.690, b = 16.926. 2. c = 934, B = 76° 25'. 7. a = 67.291, c = 110.970. 3. a = 637, A = 4:° 35'. 8. 6 = 618.42, c = 1843.70. 4. & = 48.532, B = 36° 44'.00. 9. a = 965.24, A = 75° 15'.2. 5. a = 38.313, h = 19.522. 10. a = 7.3298, b = 6.1032. An isosceles triangle may be divided into two equal right triangles by dropping a perpendicular from the vertex to the base. Using the notations of Fig. 18, find the missing sides and angles of the following isosceles triangles. (Exs. 11-13.) 11. b = 2.1452, B = 121° 14'.60. 12. A = 52° 10'.2, a = 600.20. 13. h = 7.447, A = 76° 14'.00. 14. Prove that the area S oi a right triangle is S= ^bc sin A = ^ ac cos A . 15. Prove that the area S oi a right triangle is S = I c^ sin A cos A . 16-25. Find the area of each of the right triangles in Exs. 1-10. 26-30. Find formulae for the area of the isosceles triangle of Fig. 18, in terms of b and h ; a and b ; a and Ji ; a and B ; a and A. 31-33. Apply the results of Exs. 26-30 to find the areas of the isosceles triangles of Exs. 11-13. 34. Show that the perimeter jt? of a regular polygon of n sides in- scribed in a circle of radius R is jo = 2ni2sini^, n n 72 SOLUTION OF RIGHT TRIANGLES BY LOGARITHMS that the radius r of the inscribed circle is and that the area S of the polygon is 5=wi22sin— cos n n 35. Find the radius of the inscribed circle, the perimeter and the area of a regular pentagon, if the radius of the circumscribed circle is 12 feet. 36. Find the perimeter, the length of one side, the radii of the in- scribed and circumscribed circles of a regular octagon whose area is 24 square feet. 37. Since the polygon of Ex. 33 approaches the circle of radius R as 1 80*^ 1 SO'' limit when n grows beyond all bound, what limits do n sin cos n n and 2 n sin , respectively, approach ? n 38. Applications to simple problems of surveying, navigation, and geography. The connection between surveying and trigonometry is so obvious as to require no further explana- tion. Moreover, we have already discussed this relation in Chapter 1. Many of the following examples are concerned with simple problems of surveying, and most of the technical terms which occur in them are self-explanatory. Nevertheless, we shall give a brief discussion of these terms, so as to make the appli- cations seem more concrete and vivid. The student who wishes to know more about the subject should consult a treatise on surveying.* A pltmib line is a cord to one end of which is attached a weight. If such a plumb line be suspended, by fastening the other end of the cord to a fixed support, it will oscillate to and fro, and finally come to rest in its position of equilibrium, which is called the vertical line of the place of observation. Since the earth is approximately spherical in shape and the plumb line points toward the earth's center, the vertical lines of different places are not parallel. But the angle between ♦For instance, Raymond's Plane Surveying. PROBLEMS OF SURVEYING AND NAVIGATION 73 the vertical lines of two stations which are not very far apart (say ten miles), is so small that for most purposes these lines may be regarded as parallel. When a tract of land (to be surveyed) is comparatively small, it is therefore legitimate to neglect the effect of the earth's curvature, and the problem becomes one of plane survejring. The more difficult problems connected with a geodetic survey^ in which the earth's spherical form is taken into account, require knowledge of the methods of spherical trigonometry. We are concerned with plane surveying only, so that we shall regard the vertical lines of all places which occur in such a survey as parallel. A vertical plane is one which contains a vertical line. A horizontal line or plane is one which is perpendicular to ta vertical line. I An inclined line or plane is one which is neither vertical nor horizontal. An angle is said to be a horizontal or vertical angle, accord- ing as the plane of its sides is a horizontal or vertical plane. Both horizontal and vertical angles may be measured by means of the transit (Art. 2). Inclined angles may be measured by means of an instrument known as a sextant. Horizontal line D=rAngle of depression Fig. 19 Horizontal line JS7=AngIe of elevation Fig. 20 The angle which the line of sight from the observer to an object makes with a horizontal line, in the same vertical plane, is called the angle of elevation or the angle of depres- 74 SOLUTION OF RIGHT TRIANGLES BY LOGARITHMS Fig. 21 sion, according as the object is above or below the horizontal plane of the observer (cf. Figs. 19 and 20). The angle subtended by a line is that which is obtained by joining the extremities of the line to the eye of the observer.. The direction or bearing of any horizontal line is usually described by means of the angle which it makes with the north-south line or meridian, the latter being located approximately with the help of a surveyor's compass. Surveyors always meas- ure the bearing of a line as an acute angle from the north or south end of the meridian toward the east or west point, as the case may be. Thus, in Fig. 21, the bearing of OA is N 45° E, that of OB is S 10° W, that of 0(7 is N 30° W. Surveyors usually measure dis- tances by means of a Gunter's chain, which is 4 rods or 6Q feet long, and is divided into 100 links. For this reason, the operation of measuring the length of a line in the field is frequently called chaining. In order to measure the difference of level between two places, A and B (cf. Fig. 22), the observer at first makes his telescope point in a horizontal direction by means of a spirit level attached to the telescope. An assistant holds a graduated rod, i2, in a vertical position at A, and the ob- server at reads, by means of the telescope, the division on the graduated rod where it is struck by the horizontal line of sight. He repeats this operation with the rod at B. The difference between the two readings gives the difference of level between A and B. Of course, if the difference of level between A and B exceeds the length of the rod, intermediate stations must be introduced. This operation is known technically as leveling. Fig. 22 PROBLEMS OF SURVEYING AND NAVIGATION 75 The navigator does not always express bearings in the same language as the surveyor. He divides the circum- ^ = ^. ference into 32 equal parts, called points of the com- pass. Thus one point of the compass is an angle of llj degrees. The division points are named, as indi- FiG. 23. — The points of the Mariner's Compass cated in Fig. 23, with obvious reference to the four cardinal points of the compass, — north, south, east, and west. The navigator also makes use of the terms departure (to denote the east and west component of ,a course), and differ- ence in latitude (to denote the north and south component). These terms are illustrated in Fig. 24. EXERCISE XVI In solving the following problems, the student should exercise his judgment in regard to the number of decimal places to be used in the calculation. (See Chap. I, Arts. 1, 2; Chap. IV, Arts. 14, 15.) Many of these problems may be solved by means of three place tables (Tables IX, X, and XI of our collection), or by means of the slide rule. (See Chap. V, Arts. 29, 30.) In all of the examples the slide rule may be used as a check. 1. At a point 180.00 feet away from the base of a tower and in the same horizontal plane with it, the angle of elevation of the top was found to be 65° 40'.5. Find the height of the tower. 2. From the top of a cliff 120 feet above the level of a lake, the angle of depression of a boat was found to be 27° 40'. What is the air line distance from the top of the cliffc' to the boat? 76 SOLUTION OF RIGHT TRIANGLES BY LOGARITHMS 3. In order to measure the width of a river, a base line AC is meas- ured along one bank 215.6 feet long. By means of a transit, a point B is located on the opposite bank such that ACB is a right angle. The angle BAC is found to be 55° 16'. 2. What is the width BC of the river ? 4. From the top of a mountain 2653 feet above the floor of the valley, the angles of depression of two farmhouses in the level valley beneath, both of which were due east of the observer, were found to be 25° and 56°. What is the horizontal distance between the two houses ? 5. From the top of a hill, the angles of depression of two consecu- tive milestones on a straight level road, running due south from the ob- server, were found to be 22° 31' and 48° 15'. How high is the hill? Hint. Treat this problem as one involving two unknowns: 1st, the height of the hill; 2d, the horizontal distance from one of the milestones to the foot of the perpendicular dropped from the top of the hill to the horizontal plane of the road. 6. Three lighthouses A, B, C, are situated as in Fig. 25, the triangle ABC being right-angled at A. At the moment when a ship S is crossing the line AC, the angle A SB is found to be 75°. If the distance between the lighthouses A and B is 12 miles, what are the dis- tances from S to A stwd. B'i 7. A light on a certain steamer is known to be 35 Fig. 25 feet above the water. An observer on the shore, whose instrument is 5 feet above the water, finds the angle of elevation of this light to be 5°. What is the distance from the observer to the steamer? 8. What angle does a mountain slope make with a horizontal plane, if it rises 200 feet in a horizontal distance of one tenth of a mile ? 9. The cable of a captive balloon is 835 feet long. Assuming the cable to be straight, how high is the balloon when all of the cable is out if, owing to the wind, the cable makes an angle of 25° with a vertical line ? s^ g 10. A ship is sailing due west at the rate of 8.9 miles per hour. A lighthouse is observed due south at 10 p.m. The bearing of the same lighthouse at 11 : 55 p.m. was S. 34° E. Find the distance from the lighthouse to the ship at the time of the second observation. Hint. In Fig. 26, of this perpendicular may fall on the c A c D B . A c B D Fig. 35 Fig. 36 Fig. 37 line segment AB (Fig. 35), to the right of B (Fig. 36), or to the left of A (Fig. 37). In all of these cases we have, (1) 8=lch, if S denotes the area of the triangle. Now h can be expressed in terms of the given quantities, 5, c, and A^ as follows. In Figs. 35 and 36, in which A is an acute angle, we have -■= ^inA ov h = h sin A^ and therefore, by substituting this value of h in (1), (2) aS' = ^ he sin A. If A is an obtuse angle (Fig. 37), we have ^ = sin Z 7)^(7 = sin (180° - ^), or ^ = 5 sin (180° - A\ h and consequently, by substituting this value of ^ in (1),- (3) aS'= i^>csin(180°-^). 82 AREA OF AN OBLIQUE TRIANGLE 83 Thus^ the area of a triangle is equal to 1- h)i sin A or J he sin (180° - A) according as A is\an acute or an obtuse angle. While we have found a complete solution of our problem, the result is not quite as convenient as it might be, since we have two different formulae for aS' alccording as A is an acute or an obtuse angle. Is there any way in which we might dvoid the distinction between these two cases ? The expression i , . ^ ^ be sin A is meaningless, from our present point of view, if A is an obtuse angle. For we have, as yet, given no definition for the sine of an obtuse angle, the definitions of Art. 7 being applicable to acute angles only. Clearly, however, it will be desirable to attach a meaning to the symbol sin A, also in the case when A is an obtuse angle, now that we are dealing with oblique triangles, some of whose angles may be obtuse. We may define the sine of an obtuse angle in any way we choose, so long as it is not inconsistent with the definitions al- ready agreed upon, and we naturally choose our definitions and notations in such a way as to reduce to a minimum the number of formulae and theorems which must be remembered. Now we can make a single formula do the work of both (2) and (3), by adopting the following definition for the sine of an obtuse angle. T^e sine of an obtuse angle A is equal to the sine of the acute angle 180° — A, which is supplementary to A; or in symbols, (4) sin A = sin (180° - A), A being obtuse. As a consequence of this definition, equation (3), which gives the expression for the area of the triangle when A is obtuse, reduces to r< -, , . . JS=:f be sm A, so that formula (2) may be used whether A be acute or obtuse. The same formula is obviously true when A is a. 84 THEORY OF OBLIQUE TRIANGLES right angle ; for in that case sin ^ = 1, and the formula re- duces to S^lhc where c is the base and h the altitude. We therefore have a single formula S =^hcmiA for the area of a triangle in terms of two sides and the included angle^ whether the latter he acute, right, or obtuse. 41. The law of sines. Since the triangle has three angles and three pairs of including sides, we may write three differ- ent expressions for the area of the same triangle, viz. •- S=^ ho sin A = ^ ca sin B — ^ ah sin C. The equality of these three expressions is a very importajit fact, since it gives rise to the following relations between the sides and angles of any triangle : he sin A = ca sin ^ = ah sin C. We may write these relations in a somewhat simpler form, by dividing all three members of the continued equation by the product ahe. We find in this way sin A _smB _ sin O a h c ' or /'I \ CT _ ft _ c ^^) 8in^~siiiS sinC' whence (2) a sin A a sin A h sinB ; , : --^ h sin B c sin O c sin Q These formulae contain the so-called law of sines, which may be expressed in words as follows : any two sides of a triangle are to each other as the sines of the opposite angles. The first explicit statement and proof of the law of sines, known at the present day, is to be found in a treatise on trigonometry by the Persian, NasTr Addin, or Nasir Eddin (1201-1247 a.d.). Nasir Addln's treatise may also be regarded as the first in which trigonometry- was treated as a separate science, independent of its applications to astron- omy. THE LAW OF COSINES 85 EXERCISE XVIII 1. What becomes of the law of sines when one of the angles (say C) is a right angle ? 2. Prove the law of sines directly from Figs. 34, 35, 36, by comput- ing the value of h in each of the two right triangles into which ABC is divided by the altitude. 3. Show that the law of sines may be used to solve the following problem : Given two angles of a triangle and one of its sides ; to find the other sides and the remaining angle. 4. The formula, gin (180° - A) = sin A, holds when A is an obtuse angle, as a consequence of the definition adopted for the sine of an obtuse angle. Show that the same equation is also true if A is an acute or right angle. 42. The law of cosines. A generalization of the theo- rem of Pythagoras. We have learned to recognize the im- portance of the theorem of Pythagoras in the theory of right triangles, and the question naturally arises : what takes the place of this theorem in the case of an oblique triangle ? Most students will remember that the answer to this ques- tion is contained in the following two propositions of geom- etry : . Theorem 1. In any triangle the square of the side oppo- site an acute angle is equal to the sum of the squares of the other two sides diminished by twice the product of one of those sides and the projection of the other upon that side. Theorem 2. In any obtuse triangle, the square of the side opposite the obtuse angle is equal to the sum of the squares of the other two sides increased by twice the product of one of those sides and the projection of the other upon that side. The proof of Theorem 1 (repeated from Geometry) is as follows : Let A be an acute angle. The triangle ABO will have the form represented in Figs. 38 or 39, accord- 86 THEORY OF OBLIQUE TRIANGLES ing as the angle B is acute or obtuse. In either case we P^* BQ=a, QA^h, AB^c, and AD = m, Q so that m is the projection of h ^^x-^y i upon c, or upon c produced. ^^c A i-c\ The right triangle BOB gives ^ m Dc-mB A B Fig. 38 Fig. 39 ^2 _ ^2 _j_ ^jf' in both cases. Now BB = c — m in Fig. 38, and BB = m — c in Fig. 39. Therefore we find, in either case, a2 = ^2 + c2 _ 2 (?m + m^. The right triangle AQB gives in both cases. If this value of W' be substituted in the equation above, we find (1) «2 __ J2 _|. ^ _ 2 ^/7j (^A being an acute angle}^ which proves Theorem 1. To prove Theorem 2, we refer to Fig. 40. We have, in this case, a^ = A2 -f- ^& = A2 + (6? 4- w)2 or a2=A2 + c2 + 2cm + m2, and Z2 12 2 D mAv B h^ = h^—m^, Fig. 40 whence (2) a^ = P -{- c^ -\- 2 cm {A being an obtuse angle), which proves Theorem 2. From either Fig. 38 or 39 we obtain, by observing the right triangle A OB, m = AB = b cos A, In Fig. 40 we have instead, ^: CAB = 180° -A, m = b cos CAB = b cos (180° - A). If we substitute these values in (1) and (2), we find (3) a2 = /)2 _|_ ^ _ 2 l)c cos A (if A is acute), (4) ^2 = 52 + ^ _^ 2 be cos (180° - A^ (if A is obtuse). THE LAWS OF COSINES 87 m Just as in Art. 41, we have found two different theorems 'and two different formulse for the two cases when A is an acute or an obtuse angle. Can we again find a single theorem and a single formula to do the work of both ? Equations (3) and (4) show that this may indeed be done, provided that we define the cosine of an obtuse angle to he a negative number^ numerically equal to the cosine of its supple^ ment (which is of course an acute angle). For, with this definition, we shall have (5) cos A-- cos (180° — A) {if A is an obtuse angle), so that (4), as a consequence of (5), assumes the same form as (3). But formula (3) holds also when ^ is a right angle, for in that case cos ^ = 0, and the formula reduces to the theorem of Pythagoras. Thus, one and the same formula (3) a^ = b^ + c^-2bc cos A will he applicable to all cases if it be understood^ in accordance with our definitions^ that cos A is positive^ zero^ or negative according as the angle A is acute^ rights or obtuse. Equation (3) is generally known as the law of cosines, and completely replaces Theorems 1 and 2 of this Article. The law of cosines obviously enables us to compute the third side of an oblique triangle when two sides and the included angle are given. But it also enables us to find the angles of a triangle when its three sides are given. For, we find from (3), by transposition, 2hccosA=b^+(^~a^ and therefore The two geometric theorems (Theorems 1 and 2 of this article) to which the law of cosines is equivalent, were well known to the Ancients ; and the problem of finding the angles of a triangle, when its three sides are given, was solved by Ptolemy (2d century a.d.) of Alexandria in his Almagest by means of these theorems. The explicit formulation of the law of cosines, however, seems to be due to the great French mathe- matician, Francois Viete (also known as Vieta), (1540-1603). 88 THEORY OF OBLIQUE TRIANGLES EXERCISE XIX Solve the following triangles, using the tables of squares and natural functions : 1. a = 2, b = 3, C = 30°. 3. c = 2.34, a = 4.31, B = 116°. 2. b = 3.5, c = 2.4, A = 52°. 4. a = 3, 6 = 6, c = S. 5. a = 1.0, h = 2.0, c = 1.5. ^ 6. The relation cos^ = — cos (180° — ^) is true for all obtuse an- gles^ as a consequence of the definition of the cosine of an obtuse angle. Prove that this formula is also true if A is any acute angle or a right angle. •^ 7. Show that the relation sin^ A + cos^ A =1 holds for obtuse as well as for acute angles. 8. If A is an acute angle, tan A = ?H^, cot^ = ^?^, sec^ = -^— , esc A = -'^ cos A sin^ cos J. sin J. Let us define tan A, cot A, sec A, esc A by means of these same equa- tions when A is an obtuse angle. Show that, as a consequence of these definitions, we have tan ^ = - tan (180° -A), cot ^ = - cot (180° -A), secA = - sec (180° - A), esc A = esc (180° - A), for any obtuse angle A. " 9. Show that the equations of Ex. 8 are .valid also if the angle A is acute. 10. The law of cosines gives three equations for any triangle. Equation (3) of Art. 42 is one of these. Write the other two. 11. Write the equations for cos^, cos B, and cos C in terms of the tfl^ sides of the triangle. 12. Show that in any triangle, a2 + &2 + c2 - 2 6c cos^ - 2 ca cosB - 2 ab cos C = 0. 43. Properties of the functions of an obtuse angle.* We have defined in Art. 40 the sine, in Art. 42, the cosine, and in Ex. 8, Exercise XIX, the remaining functions of an ob- tuse angle. As a consequence of these definitions we have the following system of equations : (-, *Some iQstructors may prefer to change the order of topics by passing to the discussion of the general angle, Art. 60 et seq., and returning later to Art. 44. There is no reason why this should not be done. FUNCTIONS OF AN OBTUSE ANGLE 89 §rsin (180° -A) = sin^, cos (180° -Jl) = -cos .4, ) tan (180° - ^) = - tan J., cot (180° - ^) = - cot^, .sec (180° -A)= - sec A, esc (180° - A) = esc A, which are valid whether the angle A be acute, right, or ob- tuse. (Cf. Exercise XVIII, Ex. 4, Exercise XIX, Exs. 6, 8, and 9.) But an obtuse angle B may be written either in the form B = 180° -^, or ^= 90° + A\ both A and A' being acute angles. Moreover, from ^=180°-^ = 90°+A' follows A' = 90° - ^, or ^ = 90° - A', that is, the angles A and A^ are complementary. Let us substitute J. = 90° - A' in (1). We find that the left mem- ber of the first equation of (1) becomes sin (180° -A) = sin [180° - (90° -A')]= sin (90° + A'). The right member of the same equation assumes the form sin A = sin (90° - ^0 = cos J.^ (Art. 10.) Consequently, this first equation of system (1) becomes sin (90° + J.0 =cos^^ ^ In the same way we may prove the remaining equations of the following system: (sin (90° + A'^ = cos J-', cos (90° + J.') = -sin J.', (2) J tan (90° +^') = - cot^^ cot (90° -{-A') = -ta.nA\ I sec (90° + ^')= -cscJ.^ csc(90° + J.O=secJ.'. The student should carry out the details of these substi- tutions, and note the close resemblance between these for- mulae and the formulae of Art. 10, for sin (90° — A), cos (90°-^), etc. This resemblance is so close as to make it easy to remember equations (2). Their right members differ from the right members of the corre- sponding equations for sin (90° — A), cos (90^^ -A}, etc., at most in sign. This remark suffices to help us remember that sin (90° + A) 90 THEORY OF OBLIQUE TRIANGLES is either equal to + cos A or to — cos^, that cos (90° -\- A) is eithei equal to + sin A or — sin ^, etc. In order to choose between these alternatives we may argue as follows. Let A be an acute angle. Then sin A and cos A are both positive. But since 90° + A will then be ob- tuse, sin (90° + ^) is positive and cos (90 + ^) is negative. Therefore, of the two alternatives sin (90° + A) = cos A, or sin (90° + A) = -cos A, we must discard the latter, since its left member is positive and its right member is negative. Similarly, of the two alternatives cos (90° + A) = sin A, or cos (90 + A) = - sin A, we must discard the former. For it involves the contradiction that the negative number cos (90° + A) should be equal to the positive number sin A. This argument fixes the two formulae sin (90° + A) = cos A, cos (90° + ^) = - sin ^ in our memory. The remaining formulae of system (2) may be remem- bered in similar fashion. The first two formulae of system (1), namely, sin (180° - A) = sin A, cos (180° - A) = - cos A, are easy to remember, since it was upon these equations that we based our definitions of the sine and cosine of an obtuse angle. The remain- ing formulae of system (1) follow directly from these two. EXERCISE XX 1. Express the following quantities as functions of acute angles; sin 100°, cos 115°, tan 162°, cot 99°, sec 120°, esc 175°. 2. By means of the table of natural functions, find sin 98°.5, cos 176°.3, tan 124°.7, cot 134°.6. 3. Explain how equations (1) and (2) of Art. 43 provide two differ- ent methods of finding the functions of obtuse angles from the table for acute angles. 44. Other formulae for the area of an oblique triangle. The methods of Art. 40 suffice to determine the area of a tri- angle if one side and the corresponding altitude ( = 3, C = 30°. 4. a = 3, A= 30°, B = 75°. 2. b = 3.5, c = 2.4, A = 52°. 5. a = 3, 6 = 6, c = 8. 3. c = 5, A = 30°, 5 = 75°. 6. a = 1.0, 6 = 2.0, c = 1.5. 45. The radius and center of the inscribed circle. The fundamental basis for all of the different formulse which we have obtained for the area of a triangle, so far, was the equation S=lch, from which all of the others were derived. But this formula is unsymmetrical, since it singles out one of the sides of the triangle and subjects it to a treatment different from that accorded to the other two. We may avoid this lack of symmetry by picking out a point M anywhere inside of the triangle ABO (see Fig. 41), and joining M to the three vertices. The area of the given triangle will then appear as the sum of the areas of the three triangles BMC, CMA, AMB, whose altitudes are respec- tively equal to MD, ME, and MF. Clearly, there is one position of the point iltf which is better adapted for this purpose than any other, namely the center of the inscribed circle. For the distances from to the three sides of the triangle are equal to each other, so that the three triangles BOO, 00 A, and AOB will have equal altitudes. Let (Fig. 42) be the center of the inscribed circle and let r be the radius of this circle. Then the area of the triangle is S=B00^- OOA+AOB =z ^ ar + ^ br -{- \ cr Fig. 42 = J (« + 6 + c) T, Fig. 41 94 THEORY OF OBLIQUE TRIANGLES or finally (1) S=sr, " where s denotes the half -perimeter of the triangle as in Art. 44. Since we also have S =.^ %{8-a){^-h){s-c) (Equation (9), Art. 44), we find from (1), substituting this value of S and dividing both riiembers of the resulting equation by «, (2) ^^^ {s-a){s-b)(8-c) ^ which enables us to compute the radius of the inscribed circle when the sides of the triangle are given. If we wish to know, not merely the radius of the inscribed circle, but also the position of its center, we must compute the lengths of the line segments AF^ BF, etc. in Fig. 42. Now we know from Geometry that (3) AF = AU, BB = BF, CE = CD, Moreover, the sum of these six line-segments is equal to the perimeter 2s of the triangle. Therefore the sum of three of these segments, one chosen from each of the three equations (3), will be equal to half of the perimeter. That is, AF-i- BB-^ OB = 8, But BB+ CB = a, and therefore (4) AF=AE=8-a, In the same way we find (5) (BB = BF=8-b, 1 CF= 0B = 8-c. The three equations, (4) and (5), enable us to locate the points D, F, F in which the inscribed circle touches the sides of the triangle and consequently determine completely the position of the center of the circle. RADIUS OF CIRCUMSCRIBED CIRCLE 95 46. The half -angle formulae. Since the center of the inscribed circle is the point of intersection of the three angle bisectors of the triangle (see Fig. 42), the angle FAO is equal to | A. In the right triangle FAO^ we have therefore But FO is the radius r of the inscribed circle, and AF we have just found to be equal to 8 — a. (Equation (4), Art. 45.) We find therefore and in the same way r . laii - iv = — 8-b (2) tan|l^ = -f-,tan|c = — -, the value of r being given by equation (2) of Art. 45. These three equations, usually known as the half-angle formulae, are very important in providing a second method for computing the angles of a triangle when its sides are given. They are far more convenient for this purpose than the law of cosines, if logarithms are to be used. In fact, the law of cosines is so cumbersome from the point of view of logarithmic calculation, that we shall seek to find a substi- tute for it also in the only other case in which we have pro- posed to make any use of it, namely in the solution of a triangle when two sides and the included angle are given. (See Art. 42, p. 96.) We should not neglect to note, however, that the law of cosines has in recent times again come into practical use, especially in engineering practice, the calculations being performed not with logarithms, but with the help of tables of squares and products or with a calculating machine. The Eichhorn Trigonometric slide rule, mentioned on page 66, is based entirely upon the law of cosines. 47. The circumscribed circle. The center of the circum- scribed circle is the point of intersection of the three per- pendicular bisectors of the sides of the triangle. Therefore, 96 THEORY OF OBLIQUE TRIANGLES if iVis the middle point of the side AB of the triangle (see Fig. 43), the line OiV^will be perpendicular to AB. Now angle AOB is measured by the arc AB^ and the angle AOB is measured by half this arc (being an inscribed angle). Therefore ZAON=ZAOB=^ O since each of these angles is equal Ib to half of z AOB. Consequently we find, making use of the right triangle A OiV", Fig. 43 sin 0= sin A0N=4^, AO If we denote AO, the radius of the circumscribed circle, by B, this becomes sm (7=2^ = -—— B 2B ,or2i2 = sin whence, making use of the law of sines (Equation (1), Art. 41), a b c (1) 2B = sin A sin B sin O'' thus completing the law of sines by providing a simple geo- metrical interpretation for the common value of the three equal ratios a h c sin^' sin^' sin O^ namely, the diameter of the circumscribed circle. Since the area of the triangle is S= ^hc sin A, and since we find, from (1), sm A = — — , 2B we obtain the following remarkable formula abc (2) for the area of the triangle. JS = 4.B THE FORM RATIOS OF A TRIANGLE 97 EXERCISE XXII 1. Compute the radius of the inscribed circJe and the lengths of the line-segments into which this circle divides the sides of the triangle whose sides are a = 3, 6 = 6, c = 8. 2. What relation will there be between the sides of a triangle if the inscribed circle bisects one of its sides ? If it touches one of the sides at a trisection point ? 3. Find formulae for the distances from the center of the inscribed circle to the three vertices of the triangle. 4. Making use of the results of Ex. 3, show that sin ^ ^ = V (s — &)(s - c)lhc and cos ^ ^ = v.s(a- — a)lhc. 5. Find a formula for the area of the inscribed circle, and for the ratio of this area to that of the triangle itself. 6. By means of the formulae of Ex. 5, show that "^^ ^r <^^ 7. Show that r = s tan | A tan ^ B tan ^ C. 8. A circle is said to be escribed to a triangle, or inscribed externally, if it touches one of the sides externally and the prolongations of the other two sides (Fig. 44). There are three such circles for every triangle. Let r^ be the m^^ radius of that one which touches the side a externally. Show that AM=AN=s and A B~ r, = s t an i ^ = -^ = ^:^ . Fig . 44 s — a s — a 9. If Ta, rj, ?'c, are the radii of the three escribed circles, and r de- notes the radius of the inscribed circle, show that r ra n r^ . 10. Show that r-a + r-fe + re - r = 4 jR, if R is the radius of the circumscribed circle. 48. The form ratios of a triangle. If two sides of a triangle are equal, the triangle is said to be isosceles ; if no two sides are equal, it is usually called a scalene triangle. But if the difference between two sides of a triangle is small, 98 THEORY OF OBLIQUE TRIANGLES as compared with the combined length of these two sides, the triangle will differ but little from the isosceles form. Con- sequently, the ratio of the difference between two sides of a triangle to their combined length may be taken as a nu- merical measure for the departure of the triangle from the isosceles form, or as its form ratio with respect to those two sides. There are three such form ratios for every triangle. If we assume that the notation be so chosen that a>h^c, these three form ratios are a — b a— c b — c a+ b a-{- c b -{- c Clearly, one of them will be equal to zero if the triangle is isosceles, and all three will vanish for an equilateral tri- angle. By means of the law of sines, each of these form ratios may be expressed in terms of the angles of the triangle. In fact, according to Art. 47, equation (1), we have a=2EsmA, b=2BsinB, c=2BiimO, so that we obtain the expression r^\ a — b _ sin ^ — sin ^ ^ a + b sin ^ -f sin jB for the form ratio with respect to the sides a and b. Simi- lar expressions hold, of course, for the other form ratios ; ^^« and l^. a + c b -\- c ' 49. The formulae for the sum and difference of two sines. The fraction which occurs in the right-hand member pf equation (1), Art. 48, is not adapted to logarithmic calcula- tion. But we shall show that both numerator and denom- inator of this fraction, namely, the expressions sin J.— sinB, and sin A -f sin 5, may be written in the form of products, thus making it easy to compute their values by logarithms. THE SUM AND DIFFERENCE OF TWO SINES 99 To discover the product form of these expressions, we must construct a figure in which the angles A and B^ the sines of these angles, and the sum and difference of their sines, shall appear. We construct first (Fig. 45) the angles Z XOF = ^ and Z XOQ= B. With their common vertex as center, and any convenient radius r, we draw a circle whose intersec- tions with the sides of the angles we denote by X, P, and Q, respectively. Let PM and QN be perpendicular, and QR parallel to OX. Then (1) r sin ^ = MP, and r sin j& = NQ. But MP and NQ are the two bases of the trapezoid MPQN. If S is the middle point of the chord PQ, and US be drawn perpendicular to OX, US is the median of this trapezoid, and we shall have, US = l(iMP-^NQ\ whence, according to (1), (2) r (sin^ + sin J5) = MP+WQ = 2 US. On the other hand. TS=iPB I (MP ^Q}. and hence, (3) r(sin^ - sin5) = MP - NQ = 2 TS. In order to express the right members of (2) and (3) in terms of r, A and B, we consider the right triangles OSU and SQ_T in which US and TS occur. It is apparent from the figure that USOQ=lQOP=iCA-B}, [Z UOS = 5 + I ( A- 5) = 1 (^ +5), (4) 100 THEORY OF OBLIQUE TRIANGLES and (5) ATSQ^Z. UOS = H^ + i5), since each of these angles is complementary to the angle USD. Hence we have from the right triangles OiS'ZJand SQT, US = OS&in UOS = OS sin ^{A + B\ ^ ^ TS=SQ cos TSQ = SQ cos -i (^ + ^), and from the triangle OQS, OS = OQ cos SOQ = r cos -| (^ - ^), ^ ^ ^§ = OQ^inSOQ = rsin 1 (^ - ^). Substituting (6) and (7) in (2) and (3) and dividing by /•, we find sin^. + sin^ = 2sm^ (^ + B) cos^ (^ -JB), (8) 1 1 ^ ^ sin^ - sin J5 = 2cos| (^ + B) mn\ {A - B), which are the desired formulae. The student should observe that our proof of equations (8), based on Fig. 45, remains valid even if A is an obtuse angle, as long as MP is greater than or equal to NQ ; that is, as long as ^ + ^ does not exceed 180°. We may therefore apply equations (8) whenever A and B are two angles of the same triangle, since the sum of two angles of a triangle can never exceed 180°. 50. The law of tangents. Let us now return to the expres- sion (1) of Art. 48 for the form ratio of the triangle with respBct to the sides a and 6, of which sides we shall assume a to be the greater. We had found a — b _ sin ^ — sin ^ a -\- h sin A 4- sin B If now we make use of equations (8) of Art 49, we find g - 5 ^ 2 cos jr C^ 4- ^) sin 1 (^ - -^) a + b 2 sin 1- (^ + B) cos ^ (A - B)' THE LAW OF TANaii:XTS v 101 whence - - I a + since we have, for any angle M, sin M . Ttyr cos M . -mm- — = tan M^ = cot M. cos M sin M But the tangent and cotangent of the same angle are re- ciprocals, so that we may write finally a-b tan I (^ - B) a very important formula, generally known as the law of tangents. We find in the same way : (2) a — c _ tan \ {A — O) a + (?"~tanl (J. 4- Oy 5-g^ tanK-g- O h + c tan^ (^ + oy The law of tangents seems to have been expressed in this form for the first time by Vieta, to whom we also owe the modern form of the law of cosines. It had however been stated, in a more complicated but equiva- lent form, about ten years earlier by the Dutch mathematician Thomas Fink or Finchius (1561-1656) in his Geometria rotundi. It was also Fink who first introduced the names tangent and secant for the functions which we now call by these names. Many previous authors had used the name umbra = shadow for the function which we now call tangent, on account of the relation of this function to the shadow cast by a vertical stick. (Compare Art, 3 and solve the problem there attributed to Thales by trigonometry.) Fink not only discovered the law of tangents, but pointed out its principal application; namely, to aid in solving a triangle when two sides and the included angle are given. The possibility of such an ap- plication will appear from the following Illustrative Example. Given a, b, and C. To find A, B, and C. Solution. The law of tangents (Equation (1)) gives (3) tan 1 (^ - B) = ^—\ tan ^(.4 -\- B), a + 102 TfiEOftY OF OBLIQUE TRIANGLES aiV equaiiioii iiv which the right member is completely known, since a, h, and C are given, and since (4) \{A +B) = K180° -C)= 90°-i a Thus we can compute tan l(A — B) by means of (3) and then find l(A — B) from the table. Hence, knowing \(^A — B) and \{A -\- B) we can find (5) A =\{A +B)+l{A -B), B = i(A +B)~l(A -B). The sine law now enables us to find c, since (6) sin C gives c = a sin C sin A sin A If we wish to solve the same problem by the law of cosines, we first compute c from the equation (7) c2 = a2 + &2 _ 2ab cos C and afterward determine the angles A and 5 by using the law of sines. The first method has the advantage over the second that formulae (3), (4), (5), (6) are in a convenient form for logarithmic computatipn, while equation (7) is not. 51. A second proof of the law of tangents and Mollweide's equations. The law of tangents may also be proved directly from a figure without making use of the for- mulae for sin A + sin B and sin A — sin B. Let BC = a and CA = h (Fig. 46) be Fig. 46 two sides of the triangle ABO^ and let a>h. Draw a circumference with C as center and radius equal to 6, and let I) and E be the points in which this circumference meets BC and BO pro- duced. Let F be the point of intersection of the circum- ference with AB. Then we have SECOND PROOF OF LAW OF TANGENTS 108 (1) BD^a-h, BE=a + h. Since Z. EC A is an exterior angle of the triangle ABO^ the opposite interior angles being A and jB, we have Z EGA = A + B, so that (2) Z EBA = liA + B) since the latter angle, being inscribed in the circumference, is measured by half the arc which it subtends. It remains to find an angle in Fig. 46, equal to \(^A — ^). In order to do this, let us draw CF. Since A OF is an isos- celes triangle, we have AAFQ=^iFA'C=A and, since Z AFC is an exterior angle of the triangle BFC^ A= /. FOB + B, whence Z FOB = A- B. Since Z DAF is an inscribed angle subtending the same arc, (3) Z BAB = Z BAF = i(^ - ^). Let us apply the law of sines to the two triangles ABB and ABE, We find, from the first triangle, a-h ^ ^inBAB ^ sinl(^-^) ^ sinl(J.-^) c sin ABB sin (1S0° - EBA} sin EBA or finally .^. a-5 ^ sin-K^--^) ^ ^ (? sinl(^ + ^)* Similarly the triangle ABE gives a-\-h ^ sin EAB c ~ sin AEB' But Z ^^5 = Z J5;^2) H- Z i)^J5 = 90° + J (^ - ^)' Z ^J5^^ = 90° - Z JS'i)A = 90° - 1 (^ + B}, owing to (2) and (3) and the further fact that Z EAB is a right angle, since it is inscribed in a semi-circumference. 104 THEORY OF OBLIQUE TRIANQLES Therefore a + h ^ sin [90° + K^ - -g)1 c sin [90° - K^ + ^)]' which may be written ^ ^ cosi(^+^y as a consequence of equations (2), Art. 43, and of Art. 10. If now we divide equations (4) and (5), member by mem- ber, we find a-b ^ sin^ (A- B) cos l(A-\- B) ^ tan i ( J. - jg) . a-\-b sinl(J. + 5) gos^Ia-B) tan i ( JL + 5) ' that is, the law of tangents. Incidentally we have found two new formulse, (4) and (5). They assume a somewhat more serviceable form by means of the relation A-\- B + (7= 180°, which gives and therefore sin J (A + ^) = cos J O, cos | (^A-\-B) = sin ^ 0^. If we introduce these values in equations (4) and (5), they become a_6 sin^C^-S) ^^i, cosliA-B) (6) cos^C ^ smlc These formulce^ known as the Mollweide equations, are particularly convenient for the purpose of checking the accuracy of the numerical solution of a triangle. For each of these equations contains all of the six parts of the triangle, so that an error in any one of these parts would be likely to make itself felt by a lack of agreement between the two members of one of these equations. THIRD PROOF OF LAW OF TANGENTS 105 Of course there are two other pairs of equations of the rm (6), t h — c h -\- c form (6), the left members of which are — - — , 7^ , and , respectively. a a It is not justifiable historically to call equations (6) MoUweide's equations. The formula for ^ is to be found in Newton's Arith- c metica Universalis. Both equations (6) are given in Simpson's Trigo- nometry, Plane and Spherical (1748), and also in F. W. von Oppel's Analysis Triangulorum (1746). All of these works antedate considerably the publication of these equations by Mollweide in 1808. EXERCISE XXIII 1. Show that the law of tangents may also be obtained from Fig. 46 by drawing through D a line parallel to AE and meeting the side AB in G, and then computing tan| (A — B) and tan^ {A + J5) from the right triangles ADG and A ED. 2. Still another proof of the law of tangents proceeds as follows. In Fig. 47, draw CD, the bisector of angle C, and drop perpendiculars, AL and BM, from A and B to CD. Let N be the point of intersection ot AB with CD. Then ZBCD = ZDCA =iC and (1) ZLAN=ZNBM =z90° -Z BNM. ^^^' *^ But Z BNM is an exterior angle of the triangle BCN, so that Z BNM =B-hiC. Moreover, 90'^ = l(A -{-B + C) and therefore, by substitution in (1), ZLAN = i(A + £ + C) - (5 + i C) = K^ - B). Now, from the right triangles ANL and BNM, we see that 4. ^,A Dx LN MN tani(^-5)= — =— , and therefore also LN + MN (2) tan i(A - B) LA -^MB 106 THEORY OF OBLIQUE TRIANGLES But LN + MN = CM - CL = a cos^C-b cos ^ Cz= (a - b)cos ^ C, LA + MB = 6 sin i C + a sin ^ C = (a + J) sin ^ C, which gives, on substitution in (2), tan i(A -B) = -^=-^cot ^ C, a -\-h and this is equivalent to the law of tangents, since C = 180° -(A + B). CHAPTER VIII SOLUTION or OBLIQUE TRIANGLES 52. The fundamental problem. Each of the laws found in Chapter VII contains four of the six parts of a triangle, and thus suggests the possibility of computing any one of these four parts when the other three parts which occur in that law are given. On the other hand, the relation (1) A-^B+ 0= 180°, which is true for every triangle, contains only three of its parts. The three angles of a triangle are therefore not inde- pendent of each other, as is the case with the three sides. The familiar fact, that a triangle is not determined by its three angles, is to be regarded as a consequence of this. For, on account of relation (1), the three angles of a triangle are not independent data. Now there exists no other equation which^ like (1), contains no more than three parts of the triangle and which is true for all triangles. For, if there were such an equation, for instance, between a, 5, and c, it would be impossible to find a triangle in which more than two of these quantities have arbitrarily assigned values since two of the three quantities would then determine the third. But this is contrary to well-known facts of geometry ; since a triangle may be constructed for which all three sides a, 5, and c have arbitrarily assigned values, provided only that a ■\-h> c. The values of a and h do not determine the value of c ; while the values of A and B do determine the value of C. k Our illustration shows incideutally that there may, and, in fact, do exist, besides the equation (1), certain inequalities between three parts of triangle which must be satisfied by all triangles. The inequality a + ft > cis an instance. 107 108 SOLUTION OF OBLIQUE TRIANGLES Any three parts of a triangle, unless all three of them are angles, may therefore be regarded as independent data. Consequently there arises the following fundamental problem : To find the remaining parts of a triangle when any three in- dependent parts are given. The discussion of this problem leads to a division into four, cases. Case I. One side and two angles are given. Case II. Two sides and the included angle are given. Case III. Two sides and the angle opposite to one of them are given. Case IV. All three sides are given. 53. Case I. Given one side and two angles. Let 5, we use the for- mula (1) (A + B) = 90= iO i. * Remember that sin 110° 52'. 9 = sin (90° + 20° 52'. 9) = cos 20° 52'. 9. t Remember the rule for finding a cologarithm. (Art. 23.) (Art. 43.) 110 SOLUTION OF OBLIQUE TRIANGLES to find ^ ( J. 4- ^), and the law of tangents, (2) tanlU'-^)=^^tanl(^+^)=^:=4cotl(7, to find \(^A—B). We then find A and B from (3) A = liA + B^ + ^{A-B\ B = ^A + B)~^iA-B\ and apply the law of sines to find c, giving (4) c=^^^. sin J. As checks we use the relations (5) ^ + 5+ (7= 180°, and one of the Mollweide equations, in the form .g. a-h _ sinl(J.- ^) ^ ^ c ~^ cosi O ' . The law of sines furnishes another relation which might be used as a check instead of (6). But (6) is more reliable. If a < 6, we interchange the letters a and 6, and also A and B in all of the above formulae, to avoid the appearance of negative angles. EXERCISE XXV Example 1. Given a = 469.71, h = 264.37, C = 96° 57'.6. Find A, B, and c and check the results. Solution. C = 96°57'.6 (1) j^ C = 48° 28\8 (6) Given a = 469.71 (2) logcot^ C = 9.94711 - 10 (8) [ b = 264.37 (3) log (a -Z/) = 2.31247 (10) a + 6^=734.08 (4) colog(a + 6) = 7.13425 - 10 (11) a-b = 205.34 (5) log tan ^(A -B)= 9.39383 - 10 (12) ^(A+B) = 41° 31'.2 (7) loga = 2.67183 (18) ^ (^ _ b) = 13° 54'.6 (13) log sin C = 9.99679 - 10 (19) A = 55° 25'.8 (15) cologsin^ = 0.08437 (20) B = 27° 36'.6 (16) logc = 2.75299 (21) C = 96° 57'.6 (1) c = 566.23 TWO SIDES AND AN OPPOSITE ANGLE 111 Checks. ,log (a -&)= 2.31247 logc 1=2.75299 log^^ = 9.55948 c (10) log sin \{A-B)^ 9.38091 - 10 (14) (21) logcos^C=: 9.82144 -10 (9) 10 (22) logfi^lMA^:?) 3Z 9.55947 - 10 (23) cos^ C A+B+C = 180° O'.O (17) The checks consist in the result (17) and the close agreement of (22) and (23). Solve and check the following triangles : 2. b = 472, c = 324, A = 78° 40'. 3. a = 748, b = 375, C = 63° 35'.5. 4. a = 42.38, b = 35.00, C = 43° 14' 40". 5. h = 0.941, c = 1.256, A = 35° 17' 28". 6. a = 12.3460, b = 5.7213, C = 65° 30' 10". 7. a = 25.384, c = 52.925, B = 28° 32' 20". 8. & = p. 14367, c = 0.11412, A = 42° 14'.6. 9. a = 138.65, b = 226.19, C = 59° 12' 54". 10. b = 1436.7, c = 1141.2, A = 42° 14' 35". 55. Case III. Given two sides and the angle opposite to one of them. G-eometric discussion. Let J., a>h^inA. (Fig. 50.) There are two solutions in this case. The triangles ABQ and AB'C both satisfy all of the requirements of the prob- lem. IV. a'^h. (Fig. 51.) If a> b^ there is one solu- tion only ; namely, ABO. The Fig. 51 triangle AB' does not con- tain the given angle A^ but its supplement. If a = b, ABC is isosceles, and AB' O reduces to a triangle of zero area. Thus we may say that there is one solution only ii a>b. The following additional cases may present themselves if A is an obtiise angle. ^- A B Fia. 52 V. ab, (Fig. 53.) There is one solution; namely, the triangle ABO. The triangle AB O does not contain the given angle ^, but its supplement. Trigonometric discussion. How do the equations of trigo- nometry bring into evidence these various cases ? The law of sines gives ,1V ' r> b sin A (1) sin B = . TWO SIDES AND AN OPPOSITE ANGLE 113 If A^ «, h are giveu, we may compute the right member of this equation. Suppose first that A is an acute angle. If the right member of (1) is greater than unity, the triangle is impossible, since the sine of an angle is never greater than unity. This is Case I of the geometric discus- sion. If = 1, 5 must be a right angle. (Case II.) a If the given values of a, 5, and the acute angle J., are such that (2) Jjin^^l^ a we can find not merely one angle to satisfy equation (1) but two such angles, which are supplementary to each other, one acute and one obtuse. For if K is the acute angle which satisfies equation (1), the obtuse angle 180° — K^ which has the same sine as K (See Art. 40), will also satisfy equa- tion (1). Thus we find in the first place, from (1), the two pos- sibilities (3) B = K (acute angle), B = 180° - K (obtuse angle). We must remember, however, that A-\- B must be less than 180°. If, as we have supposed, A is an acute angle, A-\- K is certainly less than 180°. But A + 180° — ^ is less than 180° if and only '\i A< K\ that is (see Fig. 50, where Z. K= /. ABQ^^ if and only if a a >h sin A^ in accordance with Case III of the geometric discussion. If A is acute, but if J. -f- 180° - Z> 180°; that is, if A>K and therefore a>b, the obtuse angle solution 180° — IC for B becomes inadmissible and the problem has only one solution, in agreement with Case IV of the geo- metric discussion. If J. is an obtuse angle, the obtuse angle solution 180° — K for B is never admissible, since a triangle can contain at 114 SOLUTION OF OBLIQUE TRIANGLES most one obtuse angle. The acute angle solution B = K is admissible if and only if A-\- K is less than 180°. This distinction gives rise to the two remaining Cases, V and VI, of the geometric discussion. In the trigonometric solution of a numerical problem of this kind, it is essential to remember ithe following facts : 1. If^on computing the sine of an angle^ we find its value to he greater than unity ^ the triangle is impossible. 2. If the sine of an angle is found to be a positive proper fraction^ there are two possibilities for the corresponding angle. One of these angles is acute and the other^ the supplementary angle, is obtuse. 3. 77ie sum of any two angles of a triangle must be less than 180°. The sine of B having been found from the law of sines, as indicated above, it will become apparent from the correspond- ing values of B whether the number of solutions is 0, 1, or 2. If there is one solution, we find O from A-{-B-\- (7=180° and c from the law of sines. We may check by one of Mollweide's equations or by the law of tangents. If both values of B are admissible, we use each of them in succession, so as to find the remaining parts of the two triangles which are solutions of the problem. EXERCISE XXVI Example 1. Given A = 15^^ 32'.7, a = 103.21, b = 152.37. Find the remaining parts of the triangle or triangles determined by these data. Solution. Formulce : sin B = ^.mA, c = 180° -(A-{-B),c = ^^}J^. a am A Check: 6_a = ^iiilii|^. COS l C A = 15° 32'.7 (1) fB = 23° 18'.4 B' = 156° 41.6 Given \ a = 103.21 (2) Results C = 141° 8'.9 C = 7° 45.7 b = 152.37 (3) i c = 241.58 c' = 52.01 TWO SIDES AND AN OPPOSITE ANGLE 115 Computation. log 6 = 2.18290 (4) log a = 2.01372 (5) log sin A =: 9.42813 (7) log sin C = 9.79748- 10 (15) cologa = 7.98628-10 («) colog sin A = 0.57187 (17)* logsin5 = 9.59731-10 (8) log c = 2.38307 (18) Bz= 23°18'.4 (9) c = 241.58 (20) B' = 156° 41'.6 . (10) ]oga = 2.01372 (5) B + A= 38° 5r.l (11) log sin C = 9.13050- 10 (16) B' + A= 172° 14'.3 (12) colog sin ^ = 0.57187 (17)* C = Ur 8'.9 (13) log c' = 1.71609 (19) C = 7°45'.7 (14) c' = 52.01 (21) Check. B -A = 7°45'.7 (22) B' -A = 141° 8'.9 (23) KB- -A) = 3° 52'.9 (24) KB>- .A) = 70° 34'.5 (25) log c = 2.38307 (18) log sin K^-^) = 8.830.56-10 (26) colog cos ^(7=0.47811 (28) (30) (32)t log (6- a) = 1.69174 &-a = 49.17 log c' = 1.71609 (19) log sin K^'- -4) = 9.97455-10 (27) colog cos ^C' = 0.001 00 (29) log(6- a) =1.69164 (31) 6-a = 49.16 (33)t 5-a = 49.16 (34)t Example 2. Given A = 15°32'.7, a = 10.321, b = 152.37. Find the remaining parts of the triangle or triangles determined by these data. Solution. Given A = 15° 32\7 log b = 2.18290 - a = 10.321 log sin A = 9.42813-10 [ b = 152.37 colog q = 8.98628-10 log sin B = 0.59731 Since log sin 5 has the characteristic zero, sin B is greater than unity. Therefore, the triangle is impossible. Example 3. Given A = 15° 32^7, a = 167.38, b = 152.37. Find the remaining parts of the triangle or triangles determined by these data. Solution. ^4 = 15° 32'.7 Given a = 167.38 b = 152.37 \B Results C 14° 7'.2 150° 20'.1 c = 309.11 * Obtained from (7) . t Obtained from the logarithm above. ^. Obtained by subtraction from (2) and (3). 116 SOLUTION OF OBLIQUE TRIANGLES Computation. log h = 2.18290 loga = 2.22371 log sin A = 9.42813-10 log sin C = 9.69454-10 colog a = 7.77629-10 colog sin A = 0.57187 log sin B = 9.38732-10 log c = 2.49012 B= 14° 7'.2 c = 309.11 B'=165°52'.S B-h A= 29° 39'.9 B' +A =181°25'.5* C-150°20M Check. log c = 2.49012 A -B= 1°25'.5 log sin l(A -B) = 8.09516-10 1{A-B)= 0°42'.8 colog cos 1 C = 0.59180 1 C = 75° lO'.l log (a -/0= 1-17708 a-b = 15.01 < 2 - 6=15.03 Find out whether the triangles corresponding to the following data are possible, how many solutions there are, and what are the values of the missing parts. 4. a = 98, b = 100, A = 120°. 5. a = 767, b = 242, A = 36° 53' 2''. 6. a = 3541, b = 4017, A = 61° 27'. 7. a = 67.53, b = 56.82, A = 77° 14' 19". 8. a = 9.4672, c = 14.433, A = IV 14'.3. 9. a = 413.28, b = 378.19, B = 50° 16' 25". 10. a = 345.46, b = 531.75, A = 26° 47' 32". 56. Case IV. Given the three sides of the triangle. We use the formulae tsinlA= , t'diihB = -, tan J (7= , 8 — a 8 — 8 — c and the check A-hB-\- 0=180''. A ♦Therefore B' is inadmissible. There is only one solution, as/might have been foreseen, since a > 6. / GIVEN THE THREE SIDES 117 EXERCISE XXVII Example 1. Given a = 34.278, b = 25.691, c = 30.175. Find the angles A, B, C. Solution. [a = 34.278 Given & = 25.691 I c = 30.175 2^=90.144 .^ = 45.072 s-a = 10.794 s-& = 19.381 s-c = 14.897 2. s = 90.144* logr=0.91987 log (.9- a) = 1.03318 logr: log (5 -6): \A= 75°13'.0 Results B= 46°26'.6 [ C= 58° 20^2 Check. ^+^ + C=179°59^8 colog .9 = 8.34609 -10 log(s-a) = 1.03318 log (s- 6) = 1.28737 ]og(g-c) = 1.17310 log r2= 1.83974 1 :0.91987 logr = 0.91987t 1.28737 log (5 -c) = 1.17310 log tan M = 9.88669 - 10 log tan ^ 5 = 9.63250 - 10 log tan ^ C = 9.74677 - 10 ^ = 37° 36'.5 ^B = 2^° 13'.3 ^C=29° lOM Remark. In this problem some time may be saved by writing log r on the lower margin of a slip of paper and placing it above log (s — a) to find log tan ^ A, above log (s — h) to find log tan J B, and above log {s — c) to find log tan ^ C. A similar device is often useful in similar cases. Most computers also save time by omitting the — 10 attached to logarithms with negative characteristics. This omission can never give rise to serious misunderstanding. Find the angles of the" triangles whose sides have the following values : 2. a = 79.3, b = 94.2, c = 66.9. 3. a = 0.785, b = 0.850, c = 0.633. 4. a = 312, b = 423, c = 342. 5. a = 25.17, b = 34.06, c = 22.17. 6. a = 93146, b = 176530, c = 95768. 7. a = 12.653, b = 17.213, c = 23.106. 7/ /3y , If oniy one'of the three angles is to be calculated, it may be more con- venienji to make use of the formulae for sin ^ ^ or cos ^ ^, which may be found ^rom the indications given in Example 4, Exercise XXII. * Obtained by adding s, s — a, s — b, s — c, as a check on the additions and subtractions required to find these quantities. t Obtained by adding the four logarithms above log r^, since ^.._ (s-a)(s-b)is-c) _ t Obtained by taking one half of log r^. 118 SOLUTION OF OBLIQUE TRIANGLES 57. Problems in heights and distances, plane surveying, and plane sailing. Some of the following problems are direct appli- cations of the methods which have just been explained. In others, it will be necessary to consider several triangles in succession or simultaneously. It will always be advisable to draw a figure, approximately to scale, to denote the known as well as the unknown sides and angles of the figure by properly chosen letters, and to write down the formulae to be used in their general form, leaving the substitution of the numerical values to the last. Much blundering and much unnecessary work may be avoided by adopting this plan. The student should use his judgment in regard to the number of decimal places used in the numerical part of the work. Use three-place tables whenever possible. Many of the problems may be solved, wholly or in part, by the slide- rule. EXERCISE XXVIII 1. In order to find the height of a tower, the angles of elevation of its top are measured from two stations, A and B, in the same horizontal line with its base, and on the same side of the tower. If the angles of elevation of the tower from A and B are 32° and 65° respectively, and if the distance AB is 500 feet, find the height of the tower. 2. Find the distances from the two stations of Ex. 1 to the foot of the tower. 3. If the angles of elevation of the tower from A and B, in Ex. 1, are L and M respectively, and if the distance AB is equal to d feet, show that the height of the tower is \ , _ d sin L sin M ~sin(M-Z)* 4. An obstacle (a house) was found to interfere with the running of a straight line from A in the direction AB. (See Fig. 54.) An angle ABE was turned at B, equal to 123°, and the distance BE was measured equal to 150 feet. The angle BEC was made equal to 63°. How long must the distance EC be, and what angle must be turned at C, in order that CD may be the prolongation of Fig. 54 AB? APPLICATIONS INVOLVING OBLIQUE TRIANGLES 119 5. How may angles at B and E be chosen, in Ex. 4, so as to avoid all computation ? 6. Two straight railroad tracks intersect at an angle of 75°. What will be the distance, at the end of 20 minutes, between two trains which start from the crossing at the same instant, their speeds being 30 and 40 miles per hour respectively ? 7. Each of two battleships passing each other fired a salute. A person on shore observed that the interval between the flash and the report of the gun was 4 seconds for one ship and 6 seconds for the other. The angle, at his eye, subtended by the two ships, just as the salute was fired, was 55°. The velocity of sound is about 1140 feet per second. Find the distance between the ships. 8. Two lighthouses are 2.789 miles apart, and a certain rock is known to be 4.325 miles from one of them. The angle subtended by the two lighthouses at the rock is 16° 13'. How far is the rock from the other lighthouse? How many solutions are there to this problem? Can we make a choice between these solutions if we know which of the two light- houses is nearer to the rock ? 9. Find the radius of the largest cylindrical gas tank which can be constructed on a triangular lot whose sides measure 73, 82, 91 feet respec- tively, and locate the center of its circular base. 10. An observer measures the angle of elevation of a cloud due south of him at the moment when the sun also is due south (at apparent noon). ^ The angle of elevation of the sun was 65°, that of the cloud 75". If the V shadow of the cloud falls 550 feet north of the observer, how high is the cloud ? 11. At 9 P.M. two lights, known to be 8 miles apart, are observed to be due east from a certain vessel. At 10 p.m. one of these lights bears N.E. and the other N.N.E. If the course of the ship was due south, what was its rate ? 12. A tower is situated on top of a conical hill whose sides make an angle of 15° with the horizontal plane. At a distance of 120 feet from the foot of the tower (the distance being measured along the slope) the tower subtends an angle of 20°. Find the height of the tower. 13. If, in Ex. 12, the side of the hill makes an angle / with the hori- zontal plane, and if the angle subtended by the tower, at a distance of d feet from its foot, is A, show that the height of the tower is c? sin ^ A = cos {A + /) 120 SOLUTION OF OBLIQUE TRIANGLES 14. A tower 54 feet high, situated on top of a conical hill, subtends an angle of 15° 30' at a point 120 feet from the foot of the tower (the distance being measured along the slope). What angle does the side of the hill make with the horizontal plane ? 15. To find the slope of a railroad embankment, one end of a pole 12 feet long was placed on the level ground 6 feet from the foot of the em- bankment, and the other end was found to fall at a point 7.5 feet up its face. What angle does the embankment make with a horizontal plane ? Remark. A transit cannot conveniently be used to measure an angle formed by two walls, the angle formed by an embankment or buttress with a horizontal plane, etc. In such cases, as in this example, it is more convenient to measure distances and deter- mine the angles by calculation. 16. Two capes, A and B (Fig. 55), were ob- served from a ship at sea; one of them bore N.N.E. and the other N.W. It was found from the chart that the second cape bore W. by N. from the first and was 25.3 miles distant from it. What was the distance of the ship from each of the two capes ? 17. A battleship leaves port ^, on a due easterly course, at the rate of 16 miles per hour. A dispatch boat starts from B at the same moment. The port B bears S.S.W. of port A and is 25 miles distant from it. If the dispatch boat has a rate of 22 miles per hour, what should be the direction of its course so that it may meet the battleship, if neither ship alters its rate or course ? At what time will they meet ? Hint. In Fig. 56 we have AC = Ut, BC = 22 t, if t denotes the time (in hours) which passes between the time of sailing and the moment of meeting, and if C repre- sents the place of meeting. 18. The angle of elevation of the top of a tower, at a point in the same horizontal plane with its base, is equal to ^. At a point h feet directly above the first the angle of depression of the foot of the tower was found to be equal to B. Prove that the height of the tower is equal to h tan A cot B. 19. A valley has the cross section shown in Fig. 57, the angles L and M and the distance AB = 1 having been obtained by a survey. It is planned to connect the points A and JB by a Fig. 56 APPLICATIONS INVOLVING OBLIQUE TRIANGLES 121 bridge, supported by a pier at C. How high must this pier be made? 20. Show that the area of a quadrilateral is \dd' sin A^ii d and d' are the lengths of its diagonals and if A is one of the angles which the diagonals make with each other. 21. Two sides of a parallelogram are 3.41 and 2.60 feet long and the shorter diagonal is 1.58 feet long. Find the length of the other diagonal. 22. The sides of a field ABCD 2^re. AB = b1 feet, 5C = 43 feet, CD =45 feet, DA = 47 feet ; and the distance from ^ to C is 50 feet. Find the area of the field. 23. Two streets intersect at an angle of 75°. The corner lot has frontages of 150 feet and 115 feet on the two streets, and the remaining two boundary lines of the lot are perpendicular to the two streets. What is their length, and what is the area of the lot ? 24. In order to measure the distance between two pumping stations, A and B, in Lake Michigan, a base line CD = 11.1 chains was measured along the shore. (See Fig. 58.) The follow- ^ ^ B ing angles were measured : ^Ci)= Ci = 132°29', _ ACB = 82° 20', ^^P ^:^^^^'^H CDA =Di = 45° 59', CZ>5 = A = 124°48. .^'9 ■•' -1^.-7 Fig. 58 ' Compute the distance AB. (Fig. 58 is not drawn to scale.) 25. Devise a plan for finding the distance between any two inacces- sible points, A and B, in the same horizontal plane if two points, C and D, can be found in the same plane, from both of which A and B are visible. 26. Devise a plan for finding the dis- tance between two inaccessible points, A and B, if both are visible from only one ac- cessible point C. Hint. In Fig. 59, select a point D from which A and C are visible, and a point E from which B and C are visible. Measure CD, CE, and the angles D, Ci, C2, C3, E. 27. To compute the distance between two accessible points, A and B, if no point can be found from which both A and B can be seen. (For instance, if A and B are points on opposite sides of an inaccessible mountain.) Take a point C from which A may be seen and a point D from which B is visible. If C is visible from D, measure the angles ACD and CDB and the distances AC, CD, DB. Show how to calcu- late tlie distance AB from these data. 122 SOLUTION OF OBLIQUE TRIANGLES Fig. 60 28. If the points A and B of Ex. 27 are inaccessible, the distances A C and BD cannot be found by direct measurement. In such a case (see Fig. 60), select points C, D, E, F so that A, C, E shall be visible from D, and D, F, B from E. Measure the angles C, Di, Z>25 El, E2, F, and the distances CD, DE, and EF. Show how to find the distance AB from these measurements. 29. A tower is situated on top of a conical hill as in Ex. 12. Two points A and A' are chosen on the side of the hill at distances c? = 43 feet and d' = 98 feet respectively from the top, the points' being the lower one and the distances d and d' being measured along the slope. The angles subtended by the tower at A and A' were A = 42°, A' = 23° re- spectively. Find the height of the tower and the angle of inclination of the side of the hill. 30. Two observers, A and B, are 3 miles apart, A being due west of B and in the same horizontal plane. Both observers meas- ure, at the same instant, the bearing and angle of elevation of a balloon. A finds that the balloon bears N. 47° E. and that its angle of elevation is 23°. B finds that the balloon bears N. SS"" W. and that its angle of elevation is 19°.5. Find the height of the balloon above the horizontal plane of A and B. 31. To find the horizontal distance AD and the vertical distance DC from A to an inaccessible point C (Fig. 62) when it is not convenient to measure a base line in the same vertical plane with C. Measure a horizontal base line AB, k feet long, in any direction through A. Let D be the foot of the perpendicular from C to the horizontal plane, MN, of AB. Measure the hori- zontal angles BAD = A and ABD = B, and the vertical an- gles DAC = A' and DBC = B'. k sin A Fig. 61 Fig. 62 Show that AD = h = ksin B sin(^ + 5)' A: sin B tan A' ^^ sin (^+5) k sin A tan B' sin {A + B) sin {A + B) APPLICATIONS INVOLVING OBLIQUE TRIANGLES 123 The two formulse for h should give the same result in any numerical problem of this kind. Lack of agreement indicates inaccuracy in one of the observed angles or in the computation. 32. Two points, A and B, in the same horizontal plane and separated by a ridge, are to be connected by a straight level tunnel. In order to find the distance between thern, the surveyors measured the inclined angle, BCA, subtended by A*B from the top C of a neighboring hill, whose height, CD = h, above the plane of A and B is known. They also measured the angles of depression of A and B from C. Devise a method for computing the length of the tunnel AB. 33. Prove, by means of the trigonometric formulse, that the angles of a triangle can be found when the ratios of the three sides of the triangle are given, even if the absolute values of the three sides are not known. Prove the same fact by geometry. 34. Let ^, i?, C be three points of a horizontal line, whose mutual distances, AB = b, AC = c, BC = b -\- Cy are known. (See Fig. 63.) To find the horizontal distances, p, q, r, of an inaccessible point, E, from these three points, and the height h, it suffices to measure the three angles of elevation. A, B, C, of E from A, B, C respectively. For the figure gives (1) p = hcotA, q = hcotB, r = h cot C, and also q^=:b'^ + p^-2bp cos BAD, r^ = c'^ + p'^ + 2cp cos BA D ; whence Substitute the values of p, q, r from (1) and solve the resulting equa- tion for h^. We find (2) A2 = ^^(^ + ^) ^^ (cot^ C - cot^ A)b + (cof^B-cot^ A)c After computing h from (2), p, q, and r may be found from (1). 35. Given the mutual distances of three points A, B, C,to find the distances AD, BD, CD from a fourth point D in the plane of ABC to each of the given three points, when the angles are given which the lines AB, BC, CA subtend at D. This problem, usually called Pothenofs problem, may be solved as fol- lows : 'Ua^^S) T; 124 SOLUTION OF OBLIQUE TRIANGLES In Fig. 64, let BC = a, CA =h, AB = c be the given mutual distances of A, B, and C. Of course the angles of the triangle ABC may then also be regarded as known. Let Z CDB = L, Z. CD A = M, A BDA = N be the angles subtended by the sides of the triangle from D. These angles' we regard as known by meas- urement and, of course, N = M -\- L. The problem of finding the distances ADy BD, CD may evidently be regarded as solved, if we can find the two angles Z CAD = P and Z CBD = Q. We shall show how to compute these angles. Applying the law of sines to the triangles A CD and BCD, we find b sin P a sin Q (1) whence CD = sinM sinP sinL ' asmM Ct A- sin Q & sin Z Hence, by the theory of proportion, sin P — sin Q _ a sinM b sin L sin P + sin Q a sin M -{- bainL which gives (Equations (8), Art. 49), 1 - (2) ' tanKP-Q) -, tanl(P+Q) 1^ b sinL a sin M b sin L a sin M On the other hand we have P+Q+C + iV^= 360°, and therefore (3) i(P + Q) = 180° - KC + N). Since the angles C and N are known, (3) gives the value of ^(P + Q). If this value be substituted in (2), we may obtain from (2) the value of j(P _ Q). From l(P + Q) and |(P - Q), we find P and Q themselves by addition and subtraction. The distance CD may then be computed by (1) in two ways, providing a check for the correctness of the work. Remark 1. The problem obviously becomes indeterminate, if the point D should happen to be on the circumference of the circle deter- mined by the three points A, B, C. For, as the point D moves along this circumference, the angles L and M do not change, so that the posi- tion of the point D on the circumference is not determined by the value APPLICATIONS INVOLVING OBLIQUE TRIANGLES 125 of .these angles. It is evident, then, that if the point 7) is close to the circumference of the circle circumscribed about the triangle ABC, its position cannot be determined by this method with any considerable degree of accuracy- Remark 2. The name Pothenot's problem cannot be justified his- torically. A complete solution of this problem was given by Snellius in his Doctrince triangulorum canonicce libri quatuor, which appeared in 1627, almost seventy years before Pothenot presented his solution of the same problem to the Paris Academy of Science. 36. Show that the problem of Pothenot may also be solved by draw- ing a circle through the three points A, B, D, and by making use of the triangle ABE, where E is the second point of intersection of CD with this circle. 37. In surveying a harbor, a submerged rock was located, for chart- ing purposes, by sighting three known objects A, B, C on land from a boat immediately above the rock. The known distances were BC = a = 312 feet, CA =b = 520 feet, and the angle C was known to be 65^ 27'. The angles obtained by observation were L = CDB = 23° 25' and M = CD A = 32° 52'. Find and check the distance from the rock to C and the angle which this line makes with the side AC ot the known triangle. In Exs. 38 to 45, we use the notations of Chapter VII; A, B, C for the angles, a, b, c for the sides, 2 s for the perimeter, r and R for the radii of the inscribed and circumscribed circle respectively, and S for the area of the triangle. Show how to find all of the sides and angles of the triangles determined by the following data. 38. a -^ b, c, A — B are given. 42. r, A, B are given. 39. a - b, c, A — B are given. 43. S, A, B are given. 40. R, a,b are given. 44. aS^, a, b are giVen. 41. R, A, B are given. 45. s, R, a are given. 46. A furnace maker receives an order for a number of furnaces, some 40 inches and some 42 inches in diameter. These furnaces are to be fitted on the outside with an iron casting whose inside length, measured along the arc, is 26 inches. In order to avoid the necessity of making two dif- ferent castings, the manufacturer considers the possibility of making a single casting, to fit exactly a furnace 41 inches in diameter but, of course, not fitting exactly either of the sizes ordered. His experience tells him that such a casting will serve the purpose, if no point of its inner surface is more than a quarter of an inch from the outer surface of the furnace after being placed in position. Will it be necessary to make separate castings for the two different sizes ? 126 SOLUTION OF OBLIQUE TRIANGLES 58. Displacements, velocities, and forces. If a body is transported from one place in the plane to another, and we wish to describe its change of position or displacement^ it is clearly not sufficient to state how far the body has been moved. We must also include in our descrip- tion a statement concerning the di- rection of the displacement. N w- s Fig. 65 Thus, the displacement from to P (Fig. 65) may be described completely by stating its magnitude (the length of the line-segment OP), and its direction (either the bearing NOP of the line OP or the angle EOP or some other angle which fixes the direction of OP). If the displacement is in a horizontal plane and the direc- tions from to iV, S^ E^ TTin Fig. Qb represent north, south, east, and west respectively, the projections OP^ and OP" of OP on OE and ON are called the easterly and northerly components of the displacement. If the displacement is not horizontal, we define, in a similar manner, its horizontal and vertical components. Let us suppose that a point M is displaced from to P. (See Fig. 66.) The displacement may be represented in magnitude and direction by the directed line-segment OP, (The arrowhead indicates that the displace- ^ ment is from toward P, and not from P toward 0.) Let OQ represent a second displacement. If the point M^ originally at 0, be made to undergo both of these dis- placements in succession (in either order), it will ultimately arrive at i2, where R is the fourth vertex of the parallelogram determined by OP and OQ. For this reason, the displacement OR is said to be the resultant of the displacements OP and OQ. We may think of the two displacements OP and 0§ as taking place simultaneously. An instance of this sort is Fig. 66 DISPLACEMENTS, VELOCITIES, AND FORCES 127 furnished by a passenger shifting his position on board of a moving ship. His total displacement, in space, is the result- ant of that which is due to the motion of the ship and of that caused by his own muscular efforts. The velocity of a train, of a bullet, or of any uniformly moving object, is measured by the distance which it describes in a unit of time, that is, by a displacement. Therefore a velocity, like a displacement, has direction as well as magni- tude. The case of a passenger's stroll on board a moving ship suffices to illustrate the phrase : resultant of two velocities. Since the velocity of a uniformly rdoving body is its dis- placement in a unit of time, the resultant of two velocities is found by the same method as a resultant of two displace- ments, i.e. by the parallelogram construction illustrated in Fig. 66. It is one of the fundamental facts of mechanics (proved by countless experiments) that two forces acting upon the same material point combine into a single resultant force accord- ing to this same parallelogram law. This fact is generally known as the law of the parallelogram of forces. Owing to the fact that displacements, velocities, and forces are directed quantities which combine according to the parallelogram law, these three classes of things have many properties in common. There are many other instances of quantities of this same character and, on account of their importance, they have received a special name. Directed quantities^ which combine in accordance with the parallelogram law, are called vectors.* By a proper choice of the units, every vector may be represented by a directed line-segment or, what amounts to the same thing, by a displacement. Thus, the line-segments of Fig. 66 may be interpreted as forces, if the directions of these line-segments coincide with the directions of the forces, and if each segment is made to contain as many length units as there are force units in the corresponding force. * From the Latin vector, meaning one who carries or conveys. 128 SOLUTION OF OBLIQUE TRIANGLES EXERCISE XXIX 1. A steamer is moving N.N.E. with a velocity of 16 miles pei hour. Find the northerly and easterly components of its velocity. 2. A horizontal force of 10 lb. and a vertical force of 24 lb. are acting simultaneously on a point. Find the magnitude and direction of the resultant. - 3. A schooner is sailing due west at the rate of 6 miles per hour. A sailor is crossing the deck, from south to north, at the rate of 3 miles per hour. What is the magnitude and direction of his velocity in space ? 4. A force of 250 lb. is acting on a body in a direction which makes an angle of 17° with the horizontal plane. How much of this force tends to lift the body, and what part of it tends to move the body in a hor- izontal plane ? 5. Two forces, of magnitudes 350 lb. and 510 lb., respectively, act upon the same point, in directions which make an angle of 35° with each other. Find the magnitude of the resultant, and the angles which it makes with each of the component forces. 6. A force of 216 lb. is resolved into two components, which make angles of 27° and 32° respectively, with the direction of the original force. Find the magnitude of each component. 7. A man wishes to reach a point on the opposite side of the river, 250 yards upstream. The velocity of the current is 2.5 miles per hour and the width of the river is 300 yards. If the man's rate of rowing (in still water) is 4 miles per hour, in what direction must he point the head of the boat in order that his course may be a straight line ? 59. Reflection and refraction of light. The path of light, in a homogeneous medium like air, is rectilinear. But if a ray of light meets the polished surface of a sheet of metal or glass, its direction is changed in accordance with the law that the angle of incidence is equal to the angle of reflection. This law is illustrated in Fig. 67, where the ray A strikes the reflecting surface of the mirror at and is Reflected in the direc- tion OB. The line OiV, perpendicular to the reflecting surface at 0, is called its nor- mal. The angle NOA, or i, is called the angle of incidence^ and the angle NOB, or r, is the angle of reflection. According to the law of reflection of light (veri- REFLECTION AND REFRACTION OF LIGHT 129 fied by thousands of experiments) these two angles are equal. When a ray of light AO, after passing through air, meets the bounding surface of a second transparent medium like glass (see Fig. 68), only a part of the light is reflected. Another portion of the light enters the second medium and continues on its way in a path OB^ which makes a certain angle with the direction AO oi the original ray. If NN' is the perpendicular or nor- mal to the bounding surface at 0, the angle NOA^ or ^, is called the angle of incidence and N'OB^ or r, is the angle of refraction. As a result of numerous experiments, it has been found that the quotient Fig. 68 sin sm r will have the same value, for a given kind of glass, for all different values of ^. In other words, as the angle of inci- dence changes the angle of refraction also changes, but in such a way as to leave the quotient (1) sin ^ sinr = n unchanged. This quotient n is called the index of refraction of the glass with respect to air. Its value is different for different kinds of glass. For ordinary crown glass n is about equal to 1.5. If the ray of light again emerges into air, fJs indicated in Fig. 68, after having passed through a sheet of glass with exactly parallel sides, careful measurements show that the ray 5(7 is parallel to the original ray AO. In other words, the direction of a ray of light is not changed by passing through a sheet of plate glass whose two faces are exactly parallel to each other. Therefore : the index of refraction of 130 SOLUTION OF OBLIQUE TRIANGLES air with respect to glass is the reciprocal of the index of refrac- tion of glass with respect to air. The law of the refraction of light was first discovered by Snellius in 1618. The simple formula (1) was given by Descartes in 1637. EXERCISE XXX 1. A light is placed on the line perpendicular to a plane circular mirror at its middle point. The distance from the light to the mirror is 15.76 inches, and the mirror is 8.32 inches in diameter. Consider two rays which strike the mirror in the extremities of one of its diameters. What angle will they make with each other after reflection ? 2. In an experiment, a source of light is to be placed at ^, a mirror at B, and a photographic plate at C. The distances are: AB = 5.367 meters, BC = 6.329 meters, CA = 7.361 meters. What angle must the mirror at B make with the line AB so that the light, reflected at B, may pass through C ? 3. Two billiard balls, A and B, have been placed at distances a and b inches respectively from the same cushion. The line joining them makes an angle L with the cushion. Let K be the angle at which the first ball must strike the cushion, so as to hit the second after rebound- ing. Show that taniir = ^-±^tani. b-a Remark. Billiard balls not endowed with a lateral rotation (without " English ") rebound in accordance with the law of reflected light. 4. Find the index of refraction from the following observations. (WuUner) Angles of incidence, i . . . 40'^ 60° 80° Angles of reffaction, r . . . 24° 24' 33° 38' 38° 57' The three values obtained for n will disagree slightly owing to inac- curacies in the measurements. 5. A ray of light strikes a plate of crown glass at an angle of inci- dence of 37°. Find the angle between the reflected and the refracted ray, if the index of refraction is 1.559. REFLECTION AND REFRACTION OF LIGHT 131 Fig. 69 6. A ray of light ABCD passes through a glass prism whose cross section (see Fig. 69) is an equilateral triangle. If the index of refraction is 1.559, what must be the angle of incidence in order that the path of the light in the prism may be parallel to one of its faces? What angle will the ray CD make with its original direc- tion AB, after emerging from the prism? ,^_^ 7. A ray of light, ABC, etc., enters a glass prism, whose cross section is an isosceles right triangle and whose index of refraction is 1.5, at the point B (Fig. 70), at right angles to the face of the prism. At C no part of the ray can be refracted (Why ?), and all of it is reflected in the direction CE. Such Fig. 70 a prism is called a total reflecting prism. THE GREEK ALPHABET a, A Alpha V, N N.u A B Beta I 5 Xi 7. r Gamma 0, O micron 5, A Delta TT, n Pi €, E Epsilon P^ p Rho r, z Zeta 0-, 9 ,2 Sigma '?» H Eta T, T Tau e, n ,@ Theta y, T Upsilon t, I Iota ^ c|> Phi /c, K Kappa x^ X Chi X, A T /ambda ^, ^ Psi /x, M Mu G), a Omega 182 PART TWO PROPERTIES OF THE TRIGONOMETRIC FUNCTIONS CHAPTER IX THE GENERAL ANGLE AND ITS TRIGONOMETRIC FUNCTIONS 60. The notion of the general angle. In elementary geome- try we usually think of an angle as ready-made. We there think of two lines as given and understand by the angle be- tween them a measure of their difference of direction. But many reasons urge us to oppose to this static idea what might be called the dynamic concept of angle, which presents an angle, not as the ready-made difference of direction between two fixed lines, but as something which is generated by the rotation of a straight line around a fixed point as pivot. Thus, for instance, we shall say that the minute hand of a clock describes, or generates, an angle of 90° in fifteen minutes, an angle of 360° in one hour, an angle of 1890° in five hours and a quarter. Although the minute hand points to the same place on the face of the clock after any number of complete revolutions, we are not likely to make the error of ignoring these complete revolutions.* If we did, we should be ignoring the distinction "between 1 o'clock, 2 o'clock, 3 o'clock, etc. If we were to say that an angle of 360° is the same as one of 0°, or that an angle of 450° is equal to one of 90°, we should be committing the same error. We see that, while an angle in the sense of elementary geometry can never be greater than 180°, our new concept of angle permits us to speak of angles of any magnitude. * It is the purpose of the hour hand to record the number of complete revolutions. 133 134 FUNCTIONS OF THE GENERAL ANGLE Our notions will be enriched in still another way, if we adopt the dynamic instead of the static concept of angle. The line, whose rotation generates the angle, may revolve in either of two opposite directions, clockwise or counter- clockwise; and we must distinguish between these two kinds of rotation, just as we distinguish between two motions in opposite directions on a straight line. This distinction may be made by ascribilig to every angle, not merely a magnitude^ but also a sign depending upon the direction of the rotation by which the angle is generated. It 18 customary/ to speak of counterclockwise rotations as positive, and of clockwise rotations as negative. The reason for this convention * will appear later (Art. 63). EXERCISE XXXI / Using a protractor, combine the following angles graphically and check the results arithmetically. 1. 30° + 60°, 50° - 30°, 30° - 60°, 50° + (- 30°), 30° + (- 60°). 2. 25° + 15° - 35°, 13.5°- (- 25°)+ 150°. 3. 225° + 345° - 185°, 30° + (3 x 15°). 4. What angle does the minute hand of a clock describe in 3 hours and 25 minutes? in 5 hours and 13 minutes? 5. Suppose that the dial of a clock is transparent so that it may be read from both sides. Each of two persons, stationed on opposite sides of the dial, observes the motion of the minute hand for fifteen minutes. Upon comparing notes, they find that they do not agree in regard to the angle described by the minute hand during this period of time. In what respect do they differ? 6. What is the magnitude of the angle described by a spoke of a carriage wheel, 3 feet in diameter, when the carriage travels a distance of 500 feet? Note. Think of the wheel as if it were turning on the axle while the carriage is standing still. * The word convention is here used in a special sense, meaning an arbitrary agreement. INITIAL AND TERMINAL SIDE 135 7. The earth describes an approximately circular orbit about the sun as center in 365 days. What angle will the line joining the sun to the earth (the earth's radius vector) describe in 415 days? 8. Two wheels, A and B, are joined by a belt as in Fig. 71. The diameter of A is twice that of B, and A is moving in counterclockwise direction. What angle will a spoke of B describe while A rotates through an angle of 300°? 9. If the two wheels of Ex. 8 are joined by a crossed belt, what angle will a spoke of B describe when A rotates through an angle of 300'^ ? Fig. 71 10. If n wheels are connected by gears, what kind of a number must n be in order that the first and last wheel may rotate in the same direction? in opposite directions? 61. Initial and terminal side. Standard position of an angle. Our new concept of an angle, as a measure for the /(7 amount of rotation of a line, leads us y^ ^/ to distinguish between the initial and o, ^ p "^ terminal sides of an angle. Thus, in Fig. 72, we have three angles whose senses of rotation are indicated by \q^ curved arrows. The angles A 0\B and COzD Pjq y2 *r® positive, for the rotation is counter- clockwise. Angle EO3F is negative. The initial sides of these angles are A Oi, CO2, EO3, and their terminal sides are BO^, DO2, and FO3 respectively. If an angle is thought of as generated hy the rotation of a straight line, the initial and final positions of this line are called the initial and terminal sides of the angle respectively. If we wish to compare two parallel directed line-segments in regard to magnitude and sign, we usually think of one of them as being moved, until its initial point coincides with the initial point of the other. In the same way, in order to com- pare two angles we usually place them so that their vertices and initial sides shall coincide. It is customary, for purposes of comparison, to place all '^, 136 FUNCTIONS OF THE GENERAL ANGLE angles in such a position that their initial sides are on a horizontal line and pointed toward the right. An angle placed in this way shall be said to be in its standard position. Thus, Fig. 73 represents the three angles '^F of Fig. 72 in their standard positions ; the ^^(^■'^^ angles A OB, AOD, and AOF oi Fig. 73 being equal to the three angles AO^B, CO^D, and EO^F of Fig. 72 respectively. EXERCISE XXXII Place the following angles in their standard positions : 1. 15°, 225°, 415^ 768°. , 2. - 25°, - 275°, - 615°, - 365°. 3. If two angles differ by an integral multiple of 360° and both angles are placed in standard position, how will the terminal sides of the two angles be situated with respect to each other ? 4. If two angles, placed in standard position, have the same terminal side, what is the relation between them? 5. If the sum of two angles is an integral multiple of 360°, how will the terminal sides of the two angles be situated with respect to each other, both angles being placed in standard position ? -^ J'<^ o 62. The notion of the trigonometric functions of a general angle. Having formulated the notion of a general angle^ it be- comes necessary to revise our definitions of the trigonometric functions, since our original definitions are applicable to acute angles only. To be sure, we have already made some prog- ress in this direction by defining the functions of an obtuse angle. (See Arts. 40 and 42.) But those definitions were provisional, and it will be advisable to reopen the whole ques- tion, so as to gain a larger and more adequate point of view. Our new concept of an angle, as a measure for the amount of rotation of a line, practically forces upon us the following considerations which automatically suggest the new defi- nitions of the trigonometric functions. RECTANGULAR COORDINATES 137 Let us draw a positive acute angle 6 (see the Greek alpha- bet on page 132) in its standard position (Fig. 74 a). Let us choose a point P anywhere on its terminal side and from P drop a perpendicular PM to the initial side. Then, in Accordance with the definitions of the functions of an acute angle, we have MP n OM (1) sin e OP cos e OP Fig. 746 Let US now think of the angle as growing. Nothing remarkable happens until 6 reaches 90°. At that moment, and as the motion continues, our original definitions cease to be applicable, because the right triangle POM, of which ^ is an interior angle, ceases to exist. But we may think of PMQ (in Fig. 74 a) as a plumb line attached to a point P on the moving terminal side of the angle. If there is no obstacle at 0, this line, re- maining always vertical, will pass from the right to the left oi 2^ 6 grows from an acute into an obtuse angle (see Fig. 74 5), and the line-segment OM will change its direc- tion. To indicate this change of direction of OM, we rep- resent OM by a positive number in Fig. 74 a and by a negative number in Fig. 74 J. If we count the distance OP (which does not change) as positive, in all positions of the moving line, and if we retain equations (1) as definitions for sin 6 and cos 6, we see that cos 6 becomes negative when becomes obtuse. The line-segment PM does not change its direction until 6 grows beyond 180°. Therefore the sine of an obtuse angle, like that of an acute angle, is positive. The sine of an angle, however, becomes negative when the angle lies between 180° and 360°. i 63. Rectangular coordinates. All of these things may be stated more briefly by the introduction of rectangular coordi- 138 FUNCTIONS OF THE GENERAL ANGLE nates^ a notion of utmost importance, not merely in trigonom- etry, but in other branches of mathematics. Let us draw two lines, unbounded in length and perpen- dicular to each other. We shall usually think of one of them as horizontal and call it the a5-axis, and call the other, which is vertical, the ^/-axis. The point 0, in which the two axes intersect, is called the origin of coordinates. We adopt a unit of length, and denote the distances from any point P to these two axes by x and y re- spectively. In Fig. 75 we have NP = Oiltf = X, MP = 0N=^ y, —^ where the notation is chosen in such a way that x is measured on or par- allel to the 2;-axis, and y on or parallel to the y-axis. We call x the abscissa and y the ordinate of the point P. Both numbers together are called the coordinates of P. If we take into account only the magnitudes and not the directions of the lines OM^ OiV", etc., that is, if x and y are regarded as numbers without sign, there will be four points which have the same coordinates. For instance, the points P, P', Pi', P'", in Fig. 76, would all correspond to a; = 3, y = 2. M Fia. 75 + V p > + 2 P + 1 . 3 — 2 -.1 +1 +2 ♦3 1 pj 2 F'" In order to avoid this inconvenience, we intro- duce the convention that the abscissas of all points to the right of the y-axis shall be positive, and of those to the left negative ; that the ordinates of all points above the rr-axis shall be positive, and of those below negative. Fig. 76 RECTANGULAR COORDINATES 139 The coordinates of the four points in Fig. 76 are now different from each other. The coordinates of P are a; = + 3, 3/ = + 2, The coordinates of P' are a; = — 3, y = -\- 2, The coordinates of P" are a: = — 3, y = — 2, The coordinates of P'" are a: = + 3, y =- 2. The positive directions of the x- and y-axes^ which have now been defined, will hereafter he indicated by a plus sign (as in Fig. 76). The distance from the origin to any point P is usually denoted by r and is called the radius vector of that point. We shall always regard the radius vector as positive^ and clearly we shall always have (see Fig. 77), r = 4- Va;2 -I- y"^. Let us think of OP (Fig. 77) as rotating around as a center. If OP originally coincides with the positive a;-axis, it will require a counterclockwise rota- tion of 90°, or a clockwise rotation of 270°, to bring it into coincidence with the positive 2/-axis. We naturally think of the numerically smaller angle first, and define the positive sense of rotation to be that one which enables us to turn the positive a;-axis into the position of the positive ?/-axis by means of a rota- Fig. 77 tion of only 90'^, But this implies that the positive direction of rotation is counterclockwise. (Cf. Art. 60.) EXERCISE XXXIII Plot the points whose coordinates are given in Exs. 1 to 5. Find, by measurement, to the nearest degree for the angles and to the nearest tenth of a unit for the distances, the radius vector of each point and the posi- tive angle which it makes with the positive a:-axis. Find the same results by calculation, making use of three-place tables. 1. a: = + 3, 2/ = + 4. 3. a; = - 2.4, y = -\- 5.5. 2. ar = -t- 12, y = - 5. 4. a: = - 2.6, y = - 2.1. 5. x = + 1.27,2/ = -2.18. + £0 140 FUNCTIONS OF THE GENERAL ANGLE In the following examples, r denotes the radius vector of a point P, and the positive angle which this radius vector makes with the posi- tive direction of the a:-axis. Plot the points. Find, by measurement to the nearest tenth of a unit, the abscissas and ordinates of these points. Find the same results by calculation with three-place tables. 6. r = 2,e = 30°. 8. r = S, = 210°. 7. r = o,e = 135°. 9. r .= 4, 6 = 285°. 10. r = 2.56, ^ = 310° 20'. 64. Definition of the trigonometric functions of a general angle. We are now in a position to give the definitions of the functions of a general angle in a compact manner. Place the angle 6 in its standard position ; that is, with its vertex on the origin and its initial side on the positive a;-axis of a system of rectangular coordinates. (See +37 Fig. 78, where ^ is a positive acute angle.) Piek out a 'point P^ differ- ent from the origin^ anywhere on the terminal side of the angle. Then we adopt the folloiving definitions : rr^, . £ ^ n ordinate of P - a 1/ The sine of angle a = — r— , sm = - . radius vector oiP ^ mi ' £ ^ a abscissa of P a « The cosine ot angle a = — -— , cos = - . radius vector ot P ^ rr^i , ^ £ 1/1 ordinate of P ^ a 1/ The tangent of angle & = -- — : r— -, tan = ^» abscissa oi P -^ The cotangent of angle 6 = — ^^— ^^ — — , cot = — • ordinate of P ^ rr«u . £ 1/1 radius vector of P ^^^ a r The secant ot angle u = — : —— — , sec = - • abscissa ot P **' rr.1 ^ £ 1/1 radius vector of P ^^^ a r The cosecant of angle 6 — r-zr — , esc = — • ordinate of P *' EXCEPTIONAL CASES 141 As Fig. 78 shows, these definitions reduce to the familiar definitions of Art. 7 if ^ is an acute angle. In the case of an obtuse angle, they give results which agree with the defi- nitions of Arts. 40, 42, and 43. But in our present defini- tions, 6 is not restricted either in magnitude or sign ; it may be a positive or negative angle of any magnitude. The quantities x and y may be positive, zero, or negative, but r is always positive. EXERCISE XXXIV Construct carefully the following angles and, by measurement, find approximate values of the six trigonometric functions, correct to two significant places, paying particular attention to their signs. 1. 25°. 2. 320°. 3. 110°. 4. - 130°. 5. +.725°. 6. - 10°. 7. In Art. 11, the exact values of the functions of 30°, 45°, and 60° were expressed by means of radicals. In a similar way find the values of the functions of the following angles : 120°, 135°, 150°, 210°, 225," 240°, 300°, 315°, 330°. 8. What are the signs of the trigonometric functions of the follow- ing angles : 150°, 320°, 1000°, - 625°. Find, by construction and measurement to the nearest degree, the values of the angles for which 9. >^^' V sin^ = -t, cos^ = + |. >- .:(& 10. V T/ tan ^ = 1, cosB = 1_ V2' 11. Show that the values of the trigonometric functions of a general angle, as given by the definitions of Art. 64, will not be changed if, instead of the point P, any other point«»^'~6n the terminal side of the angle be chosen. /^ /u . / \ :l 65. Discussion of the exceptional cases. Each of the trigonometric functions is defined in Art. 64 formally^ as a quotient of two numbers. This formal definition will have a real significance whenever the two numbers actually have a quotient. Now we know from Algebra that two numbers, B (the dividend) and d (the divisor), always have a unique 142 FUNCTIONS OF THE GENERAL ANGLE quotient q if the divisor is different from zero. That is, there exists a number q such that or what amounts to the same thing, such that (2) D = dq, whenever d is different from zero. Now, let us discuss the case where d (the divisor) is equal to zero, while D (the dividend) is not. In this case there exists no number q which satisfies equation (2). For this equation now becomes (3) i> = 0.^, and its right . member is equal to zero no matter what number we substitute for ^, while its left member is, b}^ hypothesis, different from zero. Consequently it involves a contradiction to assume that a number has been obtained by dividing another by zero, and the operation of dividing by zero is therefore excluded from Algebra. The formal definitions of Art. 64 involve divisions by x^ y, and r, and therefore lose their significance in any case in which one of these divisors is equal to zero. Now r= OP is the radius vector of a point P which may be chosen any- where on the terminal side of the angle except at 0. (Com- pare the wording of the definitions in Art. 64.) Therefore r is never equal to zero. From this fact and the equation (4) a;2+?/2 = r2, we conclude further that x and y cannot both be equal to zero at the same time. It will now be clear that the formal definitions of Art. 64 fail to provide the symbols ... tan 90°, sec 90°, tan 270°, sec 270°, ^ ^ cot 0°, CSC 0°, cot 180°, CSC 180°, with any actual meaning. But they do define each of the six trigonometric functions of all positive angles less than 360° with the eight exceptions just mentioned. EXCEPTIONAL CASES 143 If the angle is greater than 360°, or if it is negative, other exceptional cases appear. But their relation to the eight exceptional cases (5) is so simple that we may leave it to the student to complete this discussion. Although the tangent of 90° is not defined, our definitions are clearly applicable to the tangent of an angle 6 which differs from 90° by the slightest conceivable amount. Let us see how the tangent of an angle behaves when 6 ap- proaches 90° as a limit. We have, in Fig. 79, +05 Fig. 79 where P may be any point different from on the terminal side of the angle. For our present purpose it will be convenient to select the point P in the following manner. Draw a line RS parallel to the rc-axis at any convenient distance from OM^ and let P be the intersection of the terminal side of the angle 6 with this fixed line. Then, as 6 approaches 90°, MP = y will remain constant, while OM=x approaches the limit zero. The quotient ^ = tan will therefore become larger X and larger. Since 0M= x may be made as small as we please by taking the angle 6 = MOP close enough to 90°, while the ordinate MP always remains the same, we see that the quotient can be made as large as we please. In other words, the angle 6 can be made to differ so little from 90° that its tangent will become larger than any number whatsoever. This is what is meant by the statement that tan becomes infinite when 6 approaches 90° as a limit. We sometimes express this same statement by writing (6) tan 90° =00, a symbolic equation which should be interpreted as a short- hand account of the situation which has just been described. It is not a definition of tan 90°. For oo is not a number, and 144 FUNCTIONS OF THE GENERAL ANGLE the symbolic equation (6) is not at all concerned with what happens to tan 6 when 6 is equal to 90°. It merely tells us, in symbolic form, what happens when 6 approaches 90° as a limit ; namely, that tan 6 then increases without hound. In the preceding discussion we considered a variable angle MOP which approached 90° as a limit. In order to see what happened to its tangent, we chose the point P on the terminal side of the angle in such a way that its distance y from the a;-axis remained constant. We may obtain the same result in a slightly different way, which may, to some students, appear more conclusive. Let the angle MOP approach 90° as before. Then (Fig. 80), UnMOP = ^ = ^. OM X Let us, this time, choose the point P on the terminal side of the angle in such a ■+£c way that its distance x from the y-axis remains unchanged while d approaches 90° as a limit. It is evident from the figure that MP = y will then increase without bound. Therefore, we obtain again the result that tan d = y/x becomes infinite when 6 approaches 90° as a limit. So far we have tacitly assumed that is an acute angle increasing toward 90° as a limit. What happens when 6 starts as an obtuse angle to decrease toward 90° as a limit? Since the tangent of any angle between 90° and 180° is negative, an argument precisely similar to that just carried out shows that the numerical value of tan 6 again grows beyond all bounds when 6 approaches 90°, remaining, how- ever, always negative. We see therefore that the following statements are both true : 1. When 6 is acute and increases toward 90° as a limit, tan 6, remaining always positive^ grows numerically beyond bound. 2. When 6 is obtuse and decreases toward 90° as a limit, tan ^, remaining always negative^ grows numerically beyond bound. THE FOUR QUADRANTS 145 These two statements are frequently summed up in the symbolic formula (7) tan 90° = ± 00 . A precisely similar discussion will show that tan 6 again becomes infinite when 6 approaches 270°, that sec 6 becomes infinite when approaches 90° or 270°, and that cot 6 and CSC 6 become infinite when 6 approaches either 0° or 180°. The functions sin and cos 6 are always finite. 66. The four quadrants. The x- and ^-axes divide the plane into four portions called quadrants. The quadrant bounded by the positive x- and «/-axes is usually called the first quadrant. If we start from the first quadrant and describe a path around the origin in the counterclockwise direction, we traverse in order the 1st, 2d, 3d, and 4th quadrants. An angle is said to be in the first, second, third, or fourth quadrant according to the quadrant in which its terminal side falls when the angle is in its standard position, that is, with its initial side upon the positive a;-axis. The cardinal angles 0°, 90°, 180°, 270°, etc., may be regarded as belonging to either one of the two quadrants upon whose boundaries they lie. The following table gives the signs of the trigonometric functions of an angle in the various quadrants : I II III IV Sine + + - - Cosine .... + - - + Tangent . . . + - + - Cotangent . . . + - + - Secant .... + - - + Cosecant . . . + + - - 146 FUNCTIONS OF THE GENERAL ANGLE EXERCISE XXXV 1. Prove each of the following symbolic statements and explain its significance in words. cotO°=±oo, tan 90° = ±00, cot 180° = ± oo , tan 270° = ± cx> , csc0°=±oo, sec 90° =±00, cscl80° = ±oo, sec 270° = ± oo . 2. Show that the numerical value of the sine or cosine of an angle can never exceed unity. 3. Show that I) / d may have any value whatever if T> and d are both equal to zero, and hence that the symbol ^ is wholly indeterminate. Why can no one of the trigonometric ratios ever have this form ? 4. Determine the quadrants of the following angles and the signs of their trigonometric functions. 325°, 710°, 1045°, 609°, 412°, - 52°. 5. In what quadrant is an angle if its sine and cosine are both posi- tive? If its sine is positive and its tangent negative ? If its secant and tangent are both positive ? 6. If we know that the sine and cosine of an angle have the same sign, what can we say about the quadrant of the angle ? 7. Is there an angle whose tangent is positive and whose cotangent is negative ? 8. If we are told that the tangent and cotangent of an angle are both positive, does this enable us to determine the quadrant of the angle ? 67. General character of the trigonometric functions. Their periodicity. We are now in a position to und^iiP^nd liow the functions change with Mie angle. For the purposes of this discussion it will be convenient to think of the radius vector r as constant. This ' means that the point P, which ac- cording to our definition must be selected on the terminal side of the angle, describes the circumference of Fig. 81 a circle as 6 changes from 0° to 360°. (See Fig. 81.) It is easy to verify the following statements by reference to the figure : PERIODICITY OF FUNCTIONS 147 As increases from 0° to 90°, sin 6 increases from to 1. As 6 increases from 90° to 180°, sin 6 decreases from 1 to 0. As 6 increases from 180° to 270°, sin 6 decreases from to —1. As 6 increases from 270° to 360°, sin 6 increases from — 1 to 0. It is also evident that the function sin 6 repeats its values in exactly the same order if P moves around the circumference a second, third, ••• nth. time. The same thing is true of the other trigonometric functions, a very important fact which may be expressed as follows : Each of the six trigonometric functions is periodic and its period is equal to 360°. That is, each of these functions repeats its values at intervals of 360°, so that sin (O + n- 360°) = sin 6, cos (^ + n • 360°) = cos 0, etc., where n is any positive or negative integer or zero. The behavior of each of the six functions in the neighbor- hood of each of the four cardinal angles may be recapitulated for convenience of reference in the following table : Sine COSINB Tangent Cotangent Secant Cosecant 0° + 1 TOO + 1 Too ' 90° + 1 ±00 ±00 + 1 180° -1 Too -1 ± 00 270° - 1 ±20 Too - 1 The student should use this table to describe in words the variation of each of the six functions as 6 changes from 0° to 360°. EXERCISE XXXVI Discuss the variation of the following functions as varies from 0° to J60°. 1. cos^. 6. sin 2^. 11. sin (- ^). 2. tan e. 7. 2 sin 6. 12. cos {-6). 3. cot^. 8. sin 3^. 13. sin^^. 4. sec^. 9. sin 4^. 14. sin ^ ^. 5. csc^. 10. tan 4^. 15. sin (^ + 25°). 148 FUNCTIONS OF THE GENERAL ANGLE 68. Relations between the trigonometric functions of a general angle. The rel-ations which we found in Art. 9, between the functions of an acute angle, still hold without alteration for an angle of any magnitude. In fact the equa- tion between the abscissa, ordinate, and radius vector of a point P, that is, , ^ _|_ ^2 _ ^.2^ is true, no matter in what quadrant the point P may be situated. This is due to the fact that only the squares of x, y, and r occur in this relation. If wx divide both members of the above equation by r^, we find (^^(y}\^i But, by the definitions of Art. 64, we have, in all quadrants, - = cos 6^, ^ = sin (9, r r so that the preceding equation becomes (1) sin2e + cos2e=l. Since we have, by definition. sin (9 = ^, r CSC (9 = -, y COS ^ = -, r sec 6' = -, X tan l9 = ^. cot (9 = -, X y we find at once (2) sin 6 CSC 6 = 1, cos sec = 1, tan cot = 1. We have also , ^ y _ y/r _ sin 6 tan u — — — — — — -x-, X x/r cos V which, combined with (2), gives the further relations (3) tan = -, cot = -. ^ ^ COS0 sin0 If we divide both members of (1) by cos^ 9 and make use of (2) and (3), we find (4) 1+ tan2 =.sec2 0. I TRIGONOMETRIC IDENTITIES 149 In a similar fashion, if we divide both members of (1) by sin^ ^, we see that (5) 1 + cot2 e = csc2 e. When the angle 6 is acute, all of its functions are positive. Consequently if oneTof its functions is given, all of the others may be found without ambiguity by means of the above rela- tions. (Cf. Art. 9 and Exercise VI, Exs. 7-12.) But if we do not know in what quadrant an angle lies and are given the value of merely one of its functions, the angle itself and its other functions are not determined uniquely. If we are told, for instance, that sin 9 = ^, equation (1) only tells us that cos2 ^ = 1 - (J)2 = 3 . so that cos B = ±l V3, where either sign may be taken. In fact there are two angles between O'' and 360° whose sines are equal to J; namely, 30° and 150°. We may distinguish between them by stating whether the cosine is positive or negative. EXERCISE XXXVII Find the other functions of the angle as determined by each of the following conditions : 1. sin ^ = — ^ and ^ is in the third quadrant. 2. sin ^ = — I and B is in the fourth quadrant. 3. tan B = + 2 and B is in the third quadrant. 4. cot ^ = — 3 and sin B is positive. 5. sec ^ = + 2 and tan B is negative. /' 6. Find the values of the other functions if sin B = a. Are all pos- sible values of a admissible ? State a reason for your answer. 7. If tan B= m, find the values of the other functions. 8. If sec B = k, find the values of the other functions. Are all values of k admissible in this problem ? State a reason for your answer. 69. Trigonometric identities which involve functions of a single angle. By means of the relations of the preceding article, an expression which involves the trigonometric func- tions of y , angle may be written in a great many different forms. /^ ,1"^ often important to be able to recognize that two trigonometric expressions, although different in form, 150 FUNCTIONS OF THE GENERAL ANGLE are really identical. This may frequently be done by inspec- tion. In more eomplicated cases it is advisable to express each of the two quantities, whose identity we wish to estab- lish, in terms of some one of the six functions (the sine, for example). It will then become evident as a mere matter of algebra whether or not the two quantities are really identical. EXERCISE XXXVIII 1. Show that sec ^ — tan 6 sin 6 — cos 0, for all values of 6 for which tan 6 and sec are defined. (See p. 142.) Solution. We bave, for all values of 6, for which tan 6 and sec 6 are defined, a^^A fo«^a;«/l 1 sin ^^. /J 1 — sin2^ cos^^ ^ sec u — tan a sm a = ;: ;: sin u = — = ~ = cos u, cos ^ cosp cos^ cos which proves the truth of the original assertion. 2. Prove that tan ^ + cot ^4 = sec A esc A is an identity.* Solution. Denote the quantity on the left member by L and that on the right member by R. Then Z = tan^+cot^=5ia^ + ^-2iii = «.iB!A±^2^= 1 , COS A sm A sin A cos A sin A cos A R = sec A CSC A = cos A sin A sm A cos A Therefore L = R\ that is, tan A + cot A = sec ^ esc A. q. e. d. Prove that the following statements are identities : 3. cos ^ tan ^ = sin 0. 7. cos^^ - sin^ A =2 cos^ A - I. 4. sin <^ cot <^ = cos . 8- (cse^ - 1) sin^ = cos^ $. J 5. sin2 4- sin2 $ tan2 $ = tan2 $. i^ 9. 1 + tan2 Q = ?— — . 1 — sin2 d 6. cos2^-sin2^=l-2sin2^. cot (9 - sin Q tanjg^ i + sin ^ cos B. CSC d — sec ^ 14. J1-ZL!HL^ = sec ^ - tan 6, if Q is an acute angle. ^ 1 + sin d *In other words, show that the left member is equal to the rig'^'^^member for all values of A for which the functions tan ^, cot ^, sec A, c^^^ -^ have been defined ; that is, for all values of A except A = 0°, 90°, 180°, 36( '^ etc. CHAPTER X GRAPHIC REPRESENTATIONS OF THE TRIGONO- METRIC FUNCTIONS 70. Line representation of the trigonometric functions. The trigonometric functions were defined as ration or abstract numbers, not as lines. (See Art. 7.) This does not, how- ever, preclude the possibility of representing them as lines. An abstract or concrete quantity of any hind may be repre- sented as a line-segment, by choosing arbitrarily a certain line- segment to represent a unit of the same kind. Thus we may represent as lines the populations of the various states of the Union, taking a line-segment one inch long to represent a popula- tion of 1,000,000. The populations of New York and^IUinois will then be represented by line-segments 9.11 and 5.64 inches in length respec- tively. Thus, although a population obviously is not a line-segment, it may be represented by a line-segment. In the same way we may repre- sent the values of the trigonometric functions by lines, although they are not lines, but abstract numbers. The following is a convenient method for obtaining a representation of the values of the trigonometric functions as line-segments. We construct a circle with the origin of coordinates as center. An angle whose vertex is at the center of this circle will subtend an arc whose numerical measure, in degrees, minutes, and seconds, is equal to that of the angle. We may therefore speak indifferently either about the functions of the angle or of the functions of the arc. The point in which the initial side of the angle meets the circle is called the origin of the arc. The point in which the terminal side of the angle meets the circle is called the terminus, or the end of the arc. 152 GRAPHIC REPRESENTATIONS OF FUNCTIONS If an angle is placed in its standard position, the origin of the subtended arc will be at A, the point in which the positive a;-axis meets the circle. We shall call this point the primary origin of arcs. The point B^ in which the positive ?/-axis intersects the circle is called the secondary origin of arcs. Let us choose any convenient unit of length, say an inch, and let us agree to measure all distances in terras of this unit. We then construct the circle, the so-called unit circle^ whose center is at the origin of coordinates and whose radius is equal to the unit of length. In Fig. 82, \Qi AOQ = 6 he, any angle in its standard posi- j^y tion, and AP the arc which it sub- Is tends on the unit circle. Then (1) sm 0^y_:=K^y^MP, r 1 since the distance r = OP is equal to the unit of length, so that r = 1. Now X = OM and y — MP are the coordinates of P, the terminus of the arc AP. Conse- quently we may express our result as follows : If any angle 6 is placed in its standard position^ the value of its sine is equal., in magnitude and sign., to the ordinate of the terminus of the arc ivhich the angle subtends on the unit circle. The value of its cosine is equal to the abscissa of the terminus of this arc. In order to find a line representation for tan 6 and cot ^, we draw tangents to the unit circle at A and B and denote by T and T' the points in which the terminal side of the angle intersects these two tangents. (See Fig. 83.) Then we find tan. = ^=f=^2', cot^ = tan^Or' BT' ^ BT' OB 1 = BT', LINE REPRESENTATION OF FUNCTIONS 153 since the radius of the circle If is an obtuse angle, OP will have to be prolonged backward in order to intersect the tangent at A. Moreover the point of intersection T will then be below A. " ,j,^Q Now the tangents BT and --i^-^ I ^-^^+it AT are parallel to the x- and i/-axes respectively. Let us agree to give signs to the line-segments meas- ured on these two tangents as though they were ab- scissas or ordinates of a point. That is, let AT \tQ positive or negative according as T is above or below A^ and let BT be positive or negative according as T^ is to the right or left of B. This convention is indicated in Fig. 83 by the two -f- signs at the ends of the two tan- gents. The student may now verify that the equations Fig. 83 (2) tan ^= at; col e = BT\ which we have obtained from Fig. 83 in the case of an acute angle, will give correct results in magnitude and sign^ no matter in what quadrant the angle may happen to fall. We may formulate our results as follows : If any angle 6 is placed in its standard position^ the value of its tangent is equal, in magnitude and sign, to the ordinate of the point in which the terminal side of the angle, prolonged backward if necessary, intersects the tangent to the unit circle at the primary origin of arcs. The cotayigent of the angle is equal, in magnitude and sign, to the abscissa of the point in which the terminal side of the angle, prolonged backward if necessary, intersects the tangent to the unit circle at the secondary origin of arcs. 154 GRAPHIC REPRESENTATIONS OF FUNCTIONS (3) Referring once more to Fig. 83, we have a OT OT rim OT' or = 0T\ csGd = sGcBOr= ^_ OjB 1 These line representations for the secant and cosecant will hold, in magnitude and sign, not merely for acute angles, but for angles in any quadrant, if we agree to make the following conventions in regard to sign. OT shall be positive if T is on the same side of as P, i.e. if ^ is on the terminal side of the angle 6. OT shall be negative if Tis on the terminal side of the angle 6 prolonged backward. OT' shall be positive or negative according as T' falls on the terminal side of the angle 6 or on the terminal side prolonged back- ward. We leave it to the student as an exercise to verify these statements in detail and to formulate the contents of equations (3) in words. Figures 84 to 87 illustrate the line representation of the GRAPHS OF FUNCTIONS 155 trigonometric functions for an angle in each one of the four quadrants. In each of these figures MP = sin e, AT= tan <9, 0T= sec 0, 0M= cos e, BT' = cot 6, 0T' = esc 0. These line representations of tau 6 and sec 6 suffice to explain why the names tangent and secant were chosen for these functions. The word sine is not capable of such a simple explanation and has a long and complicated history. The Greeks did not use the six functions which we have introduced. In place of the sine of an angle they made use of the chord PQ, subtended by the angle POQ, on a circle of known radius. (See Fig. 88.) If the circle has a unit radius, Fig. 88 shows that this chord PQ is equal to twice QR, or Fig. 88 PQ = 2 sin ^ POQ. Thus the chord, used by the Greeks, is essentially twice the sine of half the angle. Aryabhata, a famous Hindoo mathematician (born 476 a.d.), was apparently the first to introduce the sine of the angle in place of the chord, and he, quite naturally, spoke of it as the half-chord or jyd-ardhd, where J yd is the Sanskrit for chord or bowstring and ardhd for one half. For the sake of brevity the adjective ardhd was soon omitted and the sine was called simply jyd. The Arabs, who far more than any other people cultivated the sciences during the Middle Ages, took over this word from the Hindoos, but changed its spelling to jiba, so as to make the spelling accord with the pronunciation in the sense of their own language. But in written Arabic the consonants only are represented by definite characters, the vowels being merely indicated by dots which are fre- quently omitted altogether. As a consequence of this practice, the Hin- doo word jlba was soon corrupted into jalb, a genuine Arabic word meaning bosom, heart, or pocket according to the context. In the twelfth century, when the Arabic texts were translated into Latin, the word Jaib was translated literally by the Latin word sinus meaning bosom. Thus, a foreign word was first converted by the Arabs into a word of their own language having a similar sound but an entirely different meaning, and later.this Arabic word was translated literally into Latin. Of course the derivation of the English word sine from sinus is obvious. 71. Graphs of functions, a number of whose nixmerical ralues are given. There is a second way of representing 156 GRAPHIC REPRESENTATIONS OF FUNCTIONS graphically the values of the trigonometric functions, which is even more important than that which has just been dis- cussed. For it gives us, in a still more vivid fashion, a picture of all the most essential properties of these functions ; and it has the further advantage of being applicable, not merely to the trigonometric functions, but to all of the other functions which naturally arise in pure and applied mathe- matics. In order tor lead the student to appreciate fully the power of this new method, we shall first illustrate it by a number of examples taken from fields other than trig- onometry. 85 = Date y = Population 1790 3,929,214 1800 5,308,483 1810 7,239,881 1820 9,638,453 1830 12,860,692 1840 17,063,353 1850 23,191,876 1860 31,443,321 1870 38,558,371 1880 50,155,783 1890 62,947,714 1900 75,994,575 1910 91,972,266 1790 1800 1810 1820 1830 1840 1850 I860 1870 1880 1890 1900 1910 Fig. 89 The population of the United States is determined every ten years by a national census. The table in the margin gives the results of these censuses. It is customary to represent the contents of this table graph- ically by laying off the dates horizontally (i.e. as abscissas), and erect- ing for each of these dates a vertical line (ordinate), which shall give by its length in terms of an appropriately chosen unit the population at that time. Figure 89 gives such a graphic representation of the facts contained in this population table and presents these facts in a more easily intelligible form than the table itself. Moreover if we join the endpoints of the ordinates by a smooth curve (the population curve), GRAPHS OF FUNCTIONS 157 we may draw some fairly reliable conclusions as to the state of the population in the years 1805, 1815, etc., in which no census was taken. If we plot the population curves of two or more countries upon the same sheet, a great many interesting matters may be brought out by comparison. Clearly we may adopt such a graphic method, whenever we have a table giving a relation between two variables ; that is, a table which shows that to certain numerical values of a first quantity x there correspond certain numerical values of a second quantity y. EXERCISE XXXIX 1. The following table gives the population of the cities of New York, Chicago, and Philadelphia for the years named : New York . Chicago . . Philadelphia 1850 515,547 28,269 340,045 805,651 109,206 585,529 1870 942,292 298,977 674,022 1,206,299 503,298 847,170 1890 1,515,301 1,099,850 1,046,964 1900 3,437,202 1,698,572 1,293,697 4,766,883 2,185,283 1,549,008 Make a graph illustrating this information,* and from the graph find the probable population of each of these cities in 1908. 2. Make a population table and a population curve for the city and state in which you live. 3. Let the student provide himself with a railroad time-table, giving the names of the various stations, their distances from the starting point, and the times at which a certain train leaves these stations. Draw a distance-time diagram for one or several trains, plotting the times as abscissas and the distances as ordinates. 4. On April 3, 1912, the following temperatures were observed in Chicago : 11 A.M. 12 M. 1 P.M. 2 P.M. 3 P.M. 4 P.M. 5 P.M. 6 P.M. 3 a.m. 32° 4 a.m. 32° 5 A.M. 32° 6 A.M. 31° 7 A.M. 32° 8 A.M. 33° 9 A.M. 34° 10 A.M. 36° 38° 7 P.M. 34° 39° 8 P.M. 33° 39° 9 P.M. 34° 38° 10 P.M. 34° 37° 11 P.M. 33° 36° 12 P.M. 33° 35° 1 A.M. 32° 35° 2 A.M. 31° Represent graphically. * In all such work, involving plotting of curves, it is advisable to use cross- section paper. 158 GRAPHIC REPRESENTATIONS OF FUNCTIONS 72. Graphs of simple algebraic functions. The relation between the variables x aiid ?/, instead of being given by a table as in the previous examples, may be given by an equation. We may then use the equation for the purpose of constructing a table, and then draw a graph as before. Example 1. Find the graph oi y = 2x — b. Solution. If we substitute a: = in the 5 ; for a; = 1, we find y - _ 2 - 1 + 1 4-2 + 3 + 4 + 5 -9 -7 — 5 -3 -1 + 1 + 3 + 5 y given equation, we find — 3 ; etc. We construct in this way the table printed in the margin. If we plot the points x=-2, ?/=-9; x--\, y = -1\ etc., obtained in this way, we find the points marked in Fig. 90 with a little cross. All of these points are found to be on a straight line. This observation makes it seem likely that all of the points whose coordinates satisfy the equation y =^2x — b, not merely those which we happened to compute, are on this same straight line. It is not difficult to prove that this is so, but the proof will not be given here. +x * / , 4-10 1 T + 9 J \ 1-8 / \ ■♦■7 / \ 1-6 / \ +5 / \ ■♦•4 Y \ +3 / \ ■•■2 / . . . \: ^7 Fig. 90 • 2-1 *■! +2 t3+4 Fig. 91 + a' Example 2. Find the graph oi y = x^. Solution. As before, we construct the table in the margin by comput- ing the values of y which correspond to the values a: = — 3, — 2, — 1, 0, -2 -1 + 1 + 2 + 3 GRAPHS OF TRIGONOMETRIC FUNCTIONS 159 y + 1, + 2, + 3. We then plot the points obtained in this — way, the points marked with a little cross in Fig. 91, and ^ unite them by a smooth curve. 4 1 The principle involved in these examples, that to every equation between two variables x and y there ^ corresponds a curve and vice versa, is at the founda- tion of Analytic G-eometry, which is one of the most important developments of modern mathematics. The great merit of having introduced this idea into mathematics is due to Descartes (1596-1650) and Fermat (1601-1665). EXERCISE XL Find the graphs of the following equations : 1. y = x. 8. y = 2 r2 - 3 a; + 1. 2- y = -x. 9. y = x\ 3. y = -2x + \, ^Q ^^i^3_i. ^. y = 2x\ 5. y = -x\ ' ^~^' 6. y = x"^ — 5. 1 12. y=±-l. 7. y = x^ — 5 X. X 73. Graphs of the trigonometric functions. We now pro- ceed to apply this method to the relation y = sin X. We choose an arbitrary line-segment on the a;-axis to repre- sent one degree and another arbitrary line-segment on the ?/-axis to represent the unit value of the sine, that is, the abstract number 1. If we are using millimeter paper, or a metric scale, it will be convenient to make a distance of one millimeter on the aj-axis stand for one degree, and to measure the ordinates in terms of a unit 10 centimeters or 100 mil- limeters long. As this is a rather large scale it will probably be necessary to paste several sheets together in order to be able to construct the whole curve. 160 GRAPHIC REPRESENTATIONS OF FUNCTIONS From the table of natural sines we obtain the table in the margin, in which the values of the angle x as well as the corresponding values of sin x are expressed in milli- metres in accordance with the adopted scale, which makes 1 mm. on the a:-axis stand for 1°, and 100 mm. on the y- axis stand for the unit value of the sine. This table enables us to plot ten points of our curve represent- ing ten values of the function sin x in the first quad- rant. (See Fig. 92, which is a reduced copy of such a curve.) The definition of the sine of a general angle (Art. 64) and the line representation of the sine in the unit circle (Art. 70), both show very clearly that two angles like 80° and 100°, or 70° and 110°, which X y = smx 10 17 20 34 30 50 40 64 50 76 60 87 70 94 80 98 90 100 Fig. 92. — The Sine Curve differ by the same amount from 90° but in opposite direc- tions have the same sine. Consequently that portion AB of our curve, which represents the values of sin x for angles in the second quadrant, will be a symmetric counterpart of the first portion OA^ which corresponds to angles in the first quadrant. (See Fig. 92.) The unit circle also makes it evident that the sines of two angles which differ by 180° are numerically equal .but opposite in sign. Consequently that portion BCD of our curve, which corresponds to angles in the third and fourth quadrants and all of whose ordinates are negative, may be obtained easily from the known part OAB. The parts OAB GRAPHS OF TRIGONOMETRIC FUNCTIONS 161 and BCD of the curve are in fact congruent, but are situated on opposite sides of the a;-axis. We have already noted that the sine function repeats its values at intervals of 360°. It is a periodic function with a period of 360° (Art. 67). This manifests itself in the graph + 1/ + ar Fio. 93. — The Cosine Curve by the fact that the picture in each of the intervals from 360° to 720°, etc., from - 360° to 0°, etc., is an exact copy of that piece of the curve which lies between 0° and 360°. The curve obtained in this way from the sine function is called the sine curve. The cosine curve, which is the graph of the function y — cos 2;, is of the same general character as the sine curve. Its form is given in Fig. 93, and may be obtained by applying to the cosine function an argument exactly similar to that which has just been carried out for the sine. We may apply the same method to the function y = tan x. However, the graph obtained in this way differs very essen- tially from the sine and cosine curves. In fact we know that when x approaches 90° from below, that is, if X assumes a succession of values like 89°, 89°.9, 89°.99, 89°.999, etc., 162 GRAPHIC REPRESENTATIONS OF FUNCTIONS the tangent of a;, remaining always positive, will grow numeri- cally beyond all bound. If on the other hand x approaches 90° from above, through a sequence of values like 91°, 90°.l, 90°.01, 90°.001, etc., -I the tangent of ic, remaining always negative, again grows numerically beyond all bound. (Cf. Art. Qb.^ We see, therefore, that the difference between the values of tan (90° + K) and tan (90° - 7i) grows larger and larger as the angles 90° + h and 90° - h themselves come closer and closer together. We express this by saying that the function tan x is discon- tinuous for X = 90°. The corresponding property of the graph is an interruption or break in the otherwise continuous curve. There are no such breaks in the sine or cosine curves. We can think of a material point (say the point of a lead pencil) as actually describing a sine curve without interrup- tion. If we were to attempt to do the same thing for the tangent curve, we should have to interrupt the path of the point at a; = 90°, at a; = 270°, etc. We meet here the important distinction between continuous and discontinuous functions, the precise formulation of which must be left to a later point in the student's career. The tangent is, of course, a periodic function and repeats its values at intervals of 360°. But we may now observe that, unlike the sine or cosine, it repeats its values after the shorter interval of 180°. To recapitulate: the tangent is a periodic function of period 180°, and is discontinuous for X = 90° and for all values of x which differ from 90° b^ integral multiples of 180°. Figure 94 shows the form of the tangent curve. EXERCISE XLI 1. Plot the curves y = 2 sin x, y = dsinx, y = 4 sin x. 2. Plot the curves y = 8m2 x, y = sin 3 ar, y = sin 4 x. RADIAN MEASURE 163 3. How are the curves of Exs. 1 and 2 related to the curve 2^ = sin X ? 4. Show that the curve y = cot x is discontinuous for x = 0°, 180°, 360°, etc., and has the form indicated in Fig. 95. +y / Fig. 94. —The Tangent Curve Fig. 95. — The Cotangent Curve 5. Show that the curves ?/ = sec x and y = esc a; have the forms indi- cated in Figs. 96 and 97. 0° 3C0° Fig. 96. — The Secant Curve 270° 3C 90" IS Fig. 97. —The Cosecant Curve L 74. The natural unit of circular measurement. Definition of a radian. In constructing the graphs of the trigonometric functions, the student may have observed that the units of measurement on the x- and z/-axes were both chosen arbi- trarily, and might have been selected in infinitely many different ways, thus altering materially the appearance of the resulting curve. This mutual independence of the two scales, on the two coordinate axes, is a natural consequence of the fact that 164 GKAPHIC REPRESENTATIONS OF FUNCTIONS the two quantities x and y were regarded as different in kind. One of them was regarded as an angle measured in degrees, and the other as an abstract number. Having chosen a certain horizontal line-segment as representative of the unit of angles (1°), it was still admissible to choose arbitrarily another (vertical) line-segment to represent the unit of abstract numbers. Whenever x and y represent two quantities of different kind, the x- and y-scales are, in the nature of things, independent of each other. Even if x and y are quantities of the same kind, it is often more convenient to choose the lengths of the units different on the two scales. If this were not done, the resulting curve might fail utterly to serve the purposes for which it was intended. Thus, when we draw a profile map of an extensive country (showing the elevations of various points in a certain vertical cross section), the vertical scale must be chosen much larger than the horizontal scale. Otherwise the differences of elevation, as depicted on the map, would become so small as to be unnoticeable. Nevertheless it will usually be desirable to choose the hori- zontal and vertical units equal to each other whenever x and y may be regarded as quantities of the same kind, provided that the resulting curve does not thereby lose its usefulness as it would in the example just quoted. Now the considerations of Art. 70 show that, from a cer- tain point of view, the quantities x and y which occur in such an equation as ., _ -^ ^ » y — Sill X may be regarded as quantities of the same kind. In fact, we observed in Art. 70 that we might think of the number x as the measure of the arc AP (Fig. 82) in- stead of as the measure of the corresponding angle AG P. We saw further that certain line-segments could be con- structed whose lengths, in terms of the radius of the circle as unit, were equal to the values of sin x^ cos x^ etc. If then we measure the length of the arc x in terms of the radius of the circle as ninit, instead of in degrees, we shall have the I RADIAN MEASURE 166 arc and its trigonometric functions expressed in terms of the same unit. This unit of arc measure^ an arc of a circle whose length is equal to the radius of the circle^ is called a radian. One advantage gained by measuring arcs in radians is this : the arc and its trigonometric functions will then be expressed in terms of the same unit. If r is the radius of a circle, the length of its circumfer- ence is equal to 2 7rr. Therefore a circumference may be said to contain 2 tt radians. Since it also contains 360 de- grees, we have (1) 2 TT radians = 360°, whence 1 ,. 360° 180° 1 radian = — — = . 2 TT TT Since tt = 3.14159265, we find, to seven decimal places, (2) ^ 1 radian = 57°.2957795. On the other hand we find from (1) • ' (3) l° = j|^ radians, or (4) 1° = 0.0174533 radian. These equations make it easy to find the number of de- grees in an angle or arc when its measure is given in radians, or vice versa. Clearly it follows, from the definition of a radian, that the length of the arc which an angle of one radian intercepts on the circumference of a circle of radius r is itself equal to r. Then, an angle of half a radian at the center will intercept an arc on the circumference whose length is equal to ^r. In general : an angle of 6 radians at the center of a circle of radius r intercepts an arc s upon the circumference whose length is (5) ' s = re. 166 GRAPHIC representatio:ns of functions In the simplicity of this formula lies a second great ad- vantage of measuring angles in radians rather than in de- grees. EXERCISE XLII Convert into radians the following angles: ~L. 90°. /^. 30°. 5. +693° 20'. ^2. 270°. 4. +25° 15'. 6. - 1030° 0'. Convert into degrees the following angles which are given in radians, state their quadrants and the signs of their trigonometric functions : >^. -. (^9. ^. 11. 0.7691. 2 16 l«. E. 10. 3.14159. 12. 5.3214. 4 1^ 13. Prove that the area of a sector of a circle of radius r is equal to \ r^O if 0, the angle at the center, is measured in radians. 14. Prove that a segment of a circle of radius r, whose arc is equal to 6 radians, has the area ^r2(^-sin^). 15. Compute the area of a circular segment of radius 11 feet, if its arc is equal to 52°. 16. How will the formulae of Exs. 13 and 14 be modified, if the angle 6 is expressed in degrees ? 17. A cord is stretched around two wheels, a large one of radius r and a smaller one of radius r' feet, the distance between the centers of the wheels being d feet. If the cord is not crossed and if ^ is the angle of inclination, expressed in radians, of the free part of the cord to the line of centers of the wheels, show that sin = '-=^, / .; 2 p cos ^ + /'I + ^) r + (l - ^^'J, where I is the entire length of the cord. 18. If the cord in Ex. 17 is crossed, show that its length I may be found by means of the formulae 6=1+^,1 d 2[rfcos^ + (|+^](r + r')]- 19. Find the length of a belt which is to be stretched around two wheels 3 and 2 feet in diameter respectively, if the distance between the centers of the two wheels is 5 feet : (a) if the belt is crossed, (b) if it is not crossed. FUNCTIONS OF SYMMETRICAL ANGLES 167 20. Draw the graphs of the trigonometric functions sin a:, cos x, tan x, cotar, if x is expressed in radians, using the same unit of length for x distances and y distances. 75. Relations between the functions of two symmetrical angles. Let us consider two angles, like 90° — ^ and 90° -f^, which differ from one of the cardinal angles by the same amount but in opposite directions. If we place two such angles in the standard position, their terminal sides will be symmetrically situated with respect to one of the two coordinate axes, so that this axis will bisect the angle be- tween them. As a consequence of this fact the trigo- nometric functions of the two angles are related to each other in a very simple fashion. In order to obtain these relations we shall consider each of the four cardinal angles separately, making use of the line representation of the functions given in Art. 70. Let us begin with the cardinal angle 0° or radians. Figure 98 represents the unit circle and the two angles ^ = ZAOPand - 6 = Z. AOP\ each in its standard position. The arcs AP ' and AP' subtended by Fio. gg these angles on the unit circle are symmetrical with respect to A. Therefore the ordinates of P and P' (the termini of these arcs) are numerically equal and opposite in sign, while their abscissas are equal in magnitude and sign. But the ordinate and abscissa of the terminus of the arc AP are respectively equal to the sine and cosine of 6, The ordinate and abscissa of P' are respectively equal to the sine and cosine of — 6. (See Art. 70.) mk Consequently we have ^Kl) sin ( — ^) = — sin ^, cos ( — ^) = cos 6, 168 GRAPHIC REPRESENTATIONS OF FUNCTIONS Consider next the case of two angles 90° — 6 and 90° + B symmetric with respect to the cardinal angle 90°. In this case (see Fig. 99) P and P' have the same ordinate, while their ab- scissas are numerically equal but opposite in sign. Consequently sin (90° - ^) = sin (90° + (9), 4 y A ^\ I ^ r (2 a) cos(90°-^) = -cos(90° + 6>), Fig. 99 or, if the angles are measured in radians, (2 5) sing - ^)= sin(| + 6^, cos(| - 6^= - cos(| + ^). By a precisely similar argument, the details of which we leave to the student, we find (8 a) or, (3 6) sin (180° -6)=- sin (180° + (9), cos (180° - (9) = cos (180° + (9) ; sin (tt — ^) = — sin (tt + 0), cos(7r — ^) == cos (tt + ^), according as the angle is measured in degrees or radians, and also (4 a) or, (4 5) sin (270° -0)= sin (270° + <9), cos (270° -6)=-^ cos(270° + (9) ; .„(^-.)..i.f^.«). Of course we shall also have sin (360° - ^) = - sin (360° + (9), cos (360° - <9) = cos (360° -f Oy I RELATIONS AMONG THE FUNCTIONS 169 But these equations are really repetitions of (1) if we remember that, on account of the periodicity of the sine and cosine, sin (360° + ^) = sin <9, sin (360° - 0) = sin ( - (9), cos (360° + 6)= cos 6, cos (360° - (9) = cos (- (9). The relations, which correspond to (1), (2), (3), (4) for the remaining trigonometric functions, may easily be obtained by expressing the tangent, cotangent, secant, and cosecant in terms of the sine and cosine. (See Art. 68.) Thus, for instance, tan(- 0)^ ^^"^- ^) = - '^'f =- tan 0. ^ ^ cos(-^) cos^ The student should actually work out the sixteen equations obtainable in this way and combine them in tabular form with the eight equations (1) to (4). 'The student should also observe that, although we have constructed the figures for the case when ^ is a positive acute angle, our proof of formulae (1), (2), (3), (4) remains valid word for word, if ^ is a positive or negative angle of any magnitude. A good way to convince one's self of this fact is to think of the angle as variable and to follow out mentally the changes which would take place in such a figure as Fig. 98 or Fig. 99 when the angle 6 increases or decreases. The fact that equations (1) to (4) are universally valid will thus be rendered intuitive. 76. Relations between functions of two angles whose sum or difference is a right angle. If is an acute angle, we know, from Art. 10, that " (1) sin (90° -6)= cos 0, cos (90° - d)= sin 0. If we combine these equations with (2 a) of Art. 75, we find further (2) sin (90° + 6}= cos (9, cos (90° + 0} = - sin 6. We wish to show that the four equations (1) and (2) are true, not merely when 6 is an acute angle, but when ^ is a 170 GRAPHIC REPRESENTATIONS OF FUNCTIONS positive or negative angle of any magnitude. This may be done by the method of mathematical induction. We begin by proving the following theorem. If equations (1) and (2) are true for a certain angle 0, they are also true for the angle d' = 90° + 0. Proof. By hypothesis, equations (1) and (2) are true for the angle 6. Therefore we have sin 6' = sin (90° + ) = - sin d, and cos (90° - (90 = cos [90° - (90° + ^)] = cos (- l9) = cos 6, (Art. 75, equations (1)), which proves that (3) sin (90° -^0= cos (9^ cos (90° - 6") = sin (9', since both members of the first equation are equal to — sin 6^ and both members of the second are equal to cos 6. Since equations (2 a) of Art. 75 are true for all angles, we now find sin (90° + 6'^ = sin (90° -6')= cos 6', ^ ^ 'cos (90° + (90 = - cos (90° -0')=- sin (9^ Since equations (3) and (4) are the same as (1) and (2), with 6' in place of ^, we have actually proved our theorem; namely, if equations (1) and (2) are true for the angle ^, they are also true for the angle 6' = 90° + 0. We know that equations (1) and (2) are true for all posi- tive acute angles. As a consequence of the theorem just proved, they are successively seen to be true for all positive angles in the second, third, or fourth quadrant, and conse- quently for all positive angles whatever. But they are also true for all negative angles. For let be a negative angle. Let n • 360°, where w is a positive in- teger, be the lowest integral multiple of 360° which makes 0' = e + n' 360° THE QUADRANTAL FORMULAE 171 a positive angle. Then, on account of the periodic character of the sine and cosine, we shall have (5) sin e' = sin (l9 + w • 360°)= sin 6, cos d' = cosle 4- n ' 360°) = cos (9, and similarly sin(90°-6>') = sin(90°-^), cos(90°-^') = cos(90°-^), ^^ sin(90° + l9') = sin(90°+(9), cos(90° + l9') = cos(90° + 6'). Since 0' is a positive angle, we have sin (90° - ^0 = cos l9^ cos (90° - ^') = sin 0', sin (90° + (90 = cos (9', cos (90° + (9') = - sin 6^. If in these equations we substitute the values (5) and (6), we find sin (90° - <9) = cos (9, cos (90° - <9) = sin ^, sin (90° + ^) = cos ^, cos (90° + ^) = - sin ^. Therefore e^'wa^zo /IS (1) and (2) are true for positive and nega- tive angles of any magnitude. The formulae for tan (90° + ^), sec (90° + (9), etc., may be found by expressing these functions of 90° + ^ in terms of sin (90° + 6) and cos (90° + 6) and making use of (2) ; for instance, we find tan (90° + ^) = ^^^CQQ° + f> = -^^ = - cotk ^ ^ cos (90° + (9) -sin (9 77. The quadrantal formulae. If we unite equations (1) of Art. 76 with the corresponding formulae for the remaining four functions, we obtain the following system of equations: sin (90° - <9) = cos 0, cos (90° - (9) = sin d, (1) tan (90° - <9j) = cot d, cot (90° - ^) = tan (9, sec (90° -6)= CSC d, esc (90° - 6>) = sec 6. In the same way we find, from equations (2) of Art. 76, the system: [ sin (90° + (9} = cos 0, cos (90° + ^) = - sin 6, (2) tan (90° -h 6*) = - cot 6, cot (90° + ^) = - tan 6, \ sec (90° + <9) = - CSC d, esc (90° + ^) = + sec 6. I 172 GRAPHIC REPRESENTATIONS OF FUNCTIONS Since these equations are true for angles of any magnitude and not merely for acute angles, we conclude from (1) and (2) that sin (180° -0) = sin (90° + 90° -0)= cos (90° - ^) = sin 6, cos(180°-^) = cos(90° + 90°-6>)=-sin(90°-^) = - cos (9, etc., giving rise to the further system of equations : r sin (180° - ^) = sin (9, cos (180° - (9) = - cos 6, (3) tan (180° - (9) = - tan 0, cot (180° - (9) = - cot (9, i sec (180° -6) = - sec (9, esc (180° -6) = esc 0. In similar fashion we find sin (180° + ^) = sin(90° + 90^ + (9) = cos (90° + (9) = - sin (9, cos (180° + (9) = cos (90° + 90° + 0) = - sin (90° + (9) = —cos 6, etc., so that we obtain (4) f sin (180° + (9) = - sin (9, cos (180° + ^) = - cos (9, tan (180° + <9) = tan 6, cot (180° + 6)= cot 0, sec (180° + 0)=- sec (9, esc (180° + 0) = - esc ^. Again we have sin (270° -0)= sin (90° + 180° - 6) = cos (180° - (9) = — cos 6, etc., ' whence ' sin (270° - ^) = - cos (9, cos (270° - (9) = - sin (9, (5) tan (270° -0)= cot (9, cot (270° - <9) = tan ^, . sec (270°- 0) = - CSC ^, esc (270° -6) = - sec (9, and similarly f sin (270° + J- P" -/ C' V' 4' jjf' °/ ^ A Px B M"\C -A- l^TT -^ X ~ 2 y M EM" 2 3fr 7^ ' F' p> +x Fig. 100. — The Sine Curve. Natural Scale represent one radian on the a^-axis as that which represents the abstract number 1 on the i/-axis. Figure 100 represents the sine curve y = sin a: constructed in accordance with this choice of units. Let us consider two points of this curve which are at the same distance from the ^-axis but on opposite sides, such as P and P^ The ordinates of these points are obviously numerically equal but opposite in sign. Denote the abscissa of P by 2 ; then, that of P' will be — 2, and we find that (1), sin ( — 2) = — sin z. If we draw a line parallel to the y-axis through the point A for which a; = — , this line clearly divides the curve into two symmetrical portions. Consequently two points, such as P and P'^ at the same distance from this line but on opposite sides of it, will have equal ordinates. That is, sin Oif = sin OM^K But if we denote MA by z, we have OM^'^-z, OM" so that (2). sin(|-2) = sin(| + ^). TT + z, THE SINE AND COSINE CURVES 177 Let the curve be divided into two portions by a line parallel to Oy through the point B whose abscissa is equal to tt. Points of the curve at equal distances from this line but on opposite sides of it, such as P^' and P"\ have ordinates numerically equal but opposite in sign. Therefore (3), sin(7r— 2) = — sin(7r -f 2). In similar fashion we find (4). sin {-^ - 2 j = sin (^ + ^j, (5), sin(2 7r— 0) = — sin(2 7r + 2). The points M and M" were equidistant from A. Conse- quently the distances OM and M'^B are equal. Since the ordinates MP and M"P" are equal, we shall have sin OM" = sin OM. If we put 0M= 2, we shall have M" B = z^ and therefore OM'' = TT - 2, so that the preceding equation becomes (6), sin Qir — z) = sin z. By similar considerations in connection with the cosine curve, we find a system of relations which correspond com- pletely to the above equations (1), to (6),. They are (l)c cos ( — 2) = cos 2, (2)c cos(|-^) = -cos(| + 2), (3)c COS (tt — 2) = cos (tt + ^), (4\ cosf-^-2J = - cosf-^H-2J, (5)c cos(2 7r — 2) = cos(2 7rH- 2), (6)c cos (tt — 2) = — cos 2. The truth of all these equations which are thus suggested by the curves has been established, in a slightly different notation, in Arts. 75-77. 178 GRAPHIC REPRESENTATIONS OF FUNCTIONS We can hardly fail to notice the striking similarity between the sine and cosine curves. In order to put into evidence the relations between them, we construct Fig. 101 which contains them both. + + 07 Fig. 101 This figure suggests that, if we displace the cosine curve TT toward the right through a distance of — units, it will coin- cide with the sine curve. If this is true, a point P of the cosine curve whose coordinates are (a) 0M= 2, il!fP = C0S2J, and which by our displacement will be brought into coin- cidence with a point P^ whose coordinates are iV) OM' = z + ^, M'P' = MP, should, in its new position, be a point on the sine curve. Therefore the coordinates, OW and M^P\ of this point P^ should satisfy the relation y — sin x which is satisfied by the coordinates of all points of the sine curve. This gives (O MP^ = sin [z + 1^, and, on combination with (a) and (5), (7), sin ^2 + I") = M^P^ = MP= cos z. THE SINE AND COSINE CURVES 179 Thus, equation (7), must be true if the geometric relation between the sine and cosine curves suggested by Fig. 101 is actually based on fact. But this equation coincides with formula (2) of Art. 76, except for the notation, so that its validity is no longer open to question. Consequently, the sine and cosine curve differ only in posi- tion and may he brought into coincidence by a displacement of — units parallel to the x-axis. EXERCISE XLIV 1. What geometric property of Fig. 101 corresponds to the relations sin f - — 2 ] = cos z, cos (^ — 2=) =sin2? 2. How are the relations , -, ^ sin f — + 2 j = - cos 2, cos ( -^ + 2; J = sin 2 to be obtained from Fig. 101 ? 3. Plot the tangent and cotangent curves and discuss these graphs in a fashion analogous (so far as possible) to the discussion of Art. 78. CHAPTER XI RELATIONS BETWEEN THE FUNCTIONS OP MORE THAN ONE ANGLE 79. The addition theorems for sine and cosine. In Art. 77 we expressed sin ( ^ + ^ J, cos ( ^ + ^ )i etc., in terms of sin 6 and cos 0, The angle was an angle of any magnitude, but the angle added to it was always an integral multiple of — radians or 90°. The question now arises whether it is pos- sible to find similar formulse for sin (a + /3) and cos (a H- ^), where both a and /3 are angles of any magnitude. Let us assume, to begin with, that a and ^ are both positive acute angles whose sum a + yS is also acute. We place the acute angle a in its standard position xOA. (See Fig. 102.) We then place the angle ^ with its initial side upon OA (the terminal side of the angle a), so as to make Z A OB equal to yS. Then ZxOB= « + A and moreover this angle is in its stand- ard position. Therefore if we take any point P, different from 0, on its terminal side OB and drop a perpendicular PM from P to the a;-axis, we shall have (1) ' sin (« + yS) = ^, cos (« + ^) =M; Let us drop perpendiculars FQ, §iV, and QE from P to OA, from Q to the a;-axis, and from Q to MP. Then we have MP ^x FiQ. 102 \ (2) sin (« + yS) = OP NQ-\-RP ^ NQ RP OP OP OP' 180 ADDITION THEOREMS FOR SINE AND COSINE 181 NO Now — ^ is a ratio of sides of two different right triangles^ namely, ONQ and OPQ. But these triangles have the side OQ m. common, and this common side may be used to trans- form — ^ into a product of two ratios, each of which contains two sides of the same right triangle and is therefore a trigo- nometric function of the acute angles of this triangle. In fact we find .o. NQ NQ OQ . Q In the same way we find for the second term of (2) RP^RP FQ OF PQ ' OP' But PO HP —^ = sin /9, and — — = cos BPQ = cos a, since the sides of the angle BPQ are respectively perpen- dicular to those of a. Consequently we find (4) -^ = cos a sin fi. ) If (3) and (4) be substituted in (2), we obtain the impor- tant formula (5) sin (a + yS) = sin a cos yS + cos a sin yS. Referring once more to Fig. 102, we have (6) cos(.4-^)=^=^^--^g = ^-^. ^ ^ ^ ^^ OP OP OP OP If we again transform each of these ratios into a product of two others, we find ON ON OQ r, RQ RQ PQ . . r, 182 FUNCTIONS OF MORE THAN ONE ANGLE so that (7) cos (a 4- ;Q) = cos a cos yS — sin a sin yS. It is easy to see that equations (5) and (7) remain valid if a and yS are acute angles, even if their sum is greater than a right angle. In that case we shall have the situation represented in Fig. 103. If we make precisely the same constructions as before, the proof of the formula for sin (a + yS) will remain applicable word for word. But since a + /3 is now in the sec- ond quadrant, its cosine is negative ; that is, , ^r,^ MO COS (a + /5) = - — , where by MO we mean merely the positive number express- ing the length of the line-segment MO^ so that —MO is a negative number. Now the figure shows that MO = MN- 0N= RQ- ON, so that MO^ ON-RQ ^ON RQ OP OP' cos (a -I- y3) = OP OP From this point on, the proof proceeds exactly as in the pre- vious case, beginning from equation (6). Thus, we have proved that equations (5) and (7) are cer- tainly true if a and y8 are positive acute angles, even if their sum is greater than 90°. We may now show that these formulae are true for two angles in any quadrant. In o«^er to do tir ,. we first prove the following theorem. If the formulcB (8) sin (a -f- y3) = sin a cos yS 4- cos a sin ^, cos (a -f- yS) = cos a cos /8 — sin a sin y9, are true for two angles a and yS, they remain true if either of these angles he increased hy 90°. ADDITION" THEOREMS FOR SINE AND COSINE 183 Proof. Let us assume that equations (8) are true for a certain pair of angles a and /3. Put a' = 90°+ a, so that . «' + ^ =(90° + «)+ ^. We shall then have (Art. 77, equations (2)), sin {a! + /3) = sin (90° + a + yS) = cos (a + y8) (9) = cos a cos /3 — sin a sin yS, cos (ce^ + /3) = cos (90° + a + /8) = - sin (a + yQ) = — sin a cos yS — cos a sin )S, and also sin a! = sin (90° + a) = cos a, cos a! = cos (90° + a) = — sin a, whence sin a = — cos a', cos a= sin a'. If we substitute these values in (9), we find sin (a' -{- y8) = sin a' cos yS + cos a' sin yS, cos (a' -}- y8) = cos a' cos yS — sin a' sin yS. But these formulae are of the same form as equations (8), with a' = a + 90° in place of a. The same process would show that equations (8) would still be satisfied if we replaced y(3 by y8^ = 90° + y8. Consequently our theorem is proved. We know already that equations (8) are true if a and yS are any two positive acute angles. On account of the theorem just proved they will still be true if a is in the second quadrant and y8 in the first, and hence also if both a and y8 are in the second quadrant, and hence if one of these angles is in the third quadrant, etc. Therefore equations (8) are true fo^ dl positive angles. But they are also true if either or both angles are negative. For instance, let a be a negative angle while yS is positive. By adding to a a sufficient number n of complete positive revolutions, we shall obtain I a' = a + 71. 360°, 184 FUNCTIONS OF MORE THAN ONE ANGLE a positive angle. But then sin a' = sin a, cos a' = cos a, sin (a' + y8) = sin (a + yS), cos (a' -h y8) = cos (a + yS), so that sin (a + y8) = sin (a' + /3) = sin a' cos /8 + cos a' sin /3 = sin a cos yS -j- cos a sin /3, and similarly for cos (a + z^)- If yS is negative, we may proceed in the same manner. Consequently we obtain the following important theorem: If a and ^ are two positive or negative angles of any magni- tude^ the sine and cosine of their sum are always given hy the formulcB sin (a + P) = sin a cos p + cos a sin p, (8) COS (a + p) = cos a cos p — sin a sin p. The method of mathematical induction employed for proving the general validity of these formulae is of fundamental importance in all parts of mathematics. We have already made use of it in Art. 76. Equations (8) are usually known as the addition theorems for sine and cosine. EXERCISE XLV 1. Cornpute the sine and cosine of 75*^. Solution. We have 75° = 30° + 45°, sin 30° = ^ cos 30° = ^ V3, sin 45'=' = cos 45° = Therefore sin 75° = sin 30° cos 45° + cos 30° sin 45° = ? cos 75° = cos 30° cos 45' - sin 30° sin 45° = ? Compute the sines and cosines of the following angles : 2. 120°. 4. 210°. 6. 195°. 3. 150°. 5. 105°. ~7. 165°. ^8. Find formulae for sin (45° + 6) and cos (45° + 6). \J 9. Find formulae for sin (30° + &) and cos (30° -\- &). THEOREMS FOR TANGENT AND COTANGENT 185 x^O. Find formulae for sin (60° + 0) and cos (60° + 0). 11. Show that those of the quadrantal formulae of Art. 77, which in- volve the sine and cosine, are special cases of the addition theorems for the sine and cosine. Prove that the following equations are true for all values of the angles which appear in them. That is, prove that they are identities. ^12. sin (a + (3) cos /3 + cos (a + /3) sin ^ = sin {a + 2 (3). ^13. cos (a + y8) cos ^ — sin (a + /3) sin yS = cos (a + 2 /?). f 14. Show how sin (a + /8 + y) and cos (« + /3 + y) may be expressed in terms of the functions of a, (3, and y. V Simplify the following expressions : 15. sin (1 + w) ^ cos (1 - n) 6 + cos (1 + n) ^ sin (1 - n) 0. ' 16. cos (1 -f n) ^ cos (1 -n) ^ - sin (1 -{- n) 6 sin (1 - n) 6. 80. The addition theorems for tangent and cotangent. The addition formulae for the tangent and cotangent may be obtained as consequences of those for the sine and cosine. We have tan C 4- 3^=: sii^ (« + )^) _ sin a cos /3 + cos ct sin /3 ^ cos (a + /3) cos a cos /5 — sin a sin /3 The fraction in the right member of this equation may be expressed in terms of tan a and tan yQ by dividing both numerator and denominator by cos a cos yS. We obtain in this way sin a cos ff cos a sin /S ^ ON COS a cos S cos a cos 8 tan (a + yS) = ^ 3 ^, -J _ sin a sin yo cos a cos yS or, finally, (1) tan(a + B)= tang + tanp^ ^ '^^ 1-tana tan p By a similar process we find r9\ ^^4- r I /D\ cot a cot yS — 1 (2) cot(a + ^) = ^ . cot a + cot yO 186 FUNCTIONS OF MORE THAN ONE ANGLE EXERCISE XLVI Find the tangents and cotangents of the following angles : '^ 1. 75°. 3. 210°. 2. 105^ 4. 150°. Prove that the fdllowing equations are identities: 1 + tan ^ 5. tan (45° + 6) 6. cot (45° + 6) 1 - tan d cot ^ - 1 cot ^ + 1 7. Find formulae for tan (30° + 0) and cot (30° + 6). 8. Find formulse for tan (60° + 0) and cot (60° + 0). 81. The subtraction formulae. The formulse for sin (a — /S), cos (a — ^), etc., may be obtained from those for sin (a + 13)^ cos (a + yS), etc. For, since the addition formulse are known to be true for all positive and negative values of a and jS, we may write ^^ _ , ,m;S -^^-^ ^>^^i'i^ sin (a - /3) = sin [a + (- yS)] = sin a cos (— yS) + cos a sin (— yQ), cos (a — fi) = cos [a + ( — ^)] = cos « cos (— /3) — sin a sin ( — yS). But, on account of equations (9), Art. 77, we have sin ( — /3) = — sin A cos ( — yS) = cos ^, so that we find sin (a — p) = sin a cos p — cos a sin p, (1) cos (a — p) = cos a cos p -f sin a sin p. In the same way we find ^ox ^ ^ Q\ tana — tan p (2^ tan ( a — B) = ^t ^ ^ V r/ i+tanatanp and xox 4. ^ o\ cot a cot yS + 1 (3) cot(a— p)= ^-- . ^^ ^ . "^^ cot y8- cot a Of course equations (2) and (3) may also be obtained from (1) by division. PRODUCTS AND SUMS OF FUNCTIONS 187 EXERCISE XLVII 1. Let the student devise a geometric proof for equations (1), in the case when a and y8 are positive acute angles, a being the larger of the two. Note. If Fig. 102 be described in words, but if the angle there denoted by fi be instead regarded and constructed as a negative angle, the same description will apply to both figures and the same method of transforming the ratios which was used in Art. 79 will be effective in the present example. 82. Formulae for converting products of trigonometric fimc- tions into siuns, and vice versa. Before the invention of logarithms the calculations of products and quotients was a very laborious process. In those days, then, it was considered a great simplification if a formula whose numerical evaluation required multiplication could be transformed into another requiring only addition or subtraction.* The addition and subtraction formulfe will enable us to accomplish this for a product of two sines, of two cosines, or of a sine and cosine. In fact, we have the two equations sin (a -j- )S) = sin a cos ^ -h cos a sin ^, "" sin (a — yS) = sin a cos y8 — cos a sin yS, from which we derive, by addition and subtraction, ^1 X sin (a + /3) -|r sin (a — /3) = 2 sin a cos y8, sin (a + /3) — sin (a — /S) = 2 cos a sin yS. In the same way, from cos (a 4- /3) = 60S a cos yS — sin a sin /8, cos (a — /3) = cos a cos y8 + sin a sin /3, we find >^2\ cos (a + /3) + cos (a — /S) = 2 cos a cos yS, cos (a + /3) — cos (a — yS) = — 2 sin a sin yS. From these last equations we find, by trUnsposition and division by 2, ^gv cos a cos y8 = 1 [cos (a — /3) + cos (a + y8)], sin a sin /8 = J [cos (a — y8) — cos (a + /3)], * This was the so-called prosthaphxretic method. 188 FUNCTIONS OF MORE THAN ONE ANGL^ whereas, from the first equation of (1), we find (4) sin a cos yS = ^ [sin (« — yS) + sin (a + y8)] . The result obtained from the second equation of (1) is the same as (4) except for an interchange of the letters a and /3. liquations (3) and (4) are important in that they enable U8 to transform a product of two sines^ two cosines^ or of a sine and a cosine into a sum or difference. For many purposes this is very important even nowadays,* although not for the purposes of numerical calculation. In fact, at the present time, for numerical work we prefer formulae which involve multiplication to those involving addition, because the former process is more easily performed by logarithms. The above formulae, slightly modified, may also be used for the purpose of converting sums and differences of sines and cosines into- products. Let us put a + yS = ^, a-ff=B, sothat a = i(^ + ^), /3 = l(^-5). Then, equations (1) and (2) yield the following four for- mulie : 'sin ^ + sin ^ = 2 sin i- (A + B) cos J (A - J?), sin ^ - sin ^ = 2 sin ^{A — B} cos i(A-\- j5), cos ^ + cos ^ = 2 cos -|- (A — B) cos -J (A + B), cosA-GosB = -2 sin^^A - B) sin -|- (J. + B). We have seen the first two of these equations before, hav- ing derived them directly from a figure (in Art. 49) for the purpose of proving the law of tangents. Our proof, at that time, did not permit us to affirm that the formulae were true for all angles A and B. That such is the case, however, has now been made evident, since the present proof was obtained without placing any restrictions on the values of the angles A and B. * For instance, in harmonic analysis (see Arts. Ill and 112) and in the integral calculus. (5) FUNCTIONS OF DOUBLE ANGLES 189 EXERCISE XLVIII Prove the following equations: 1. sin 3 ^ + sin ^ = 2 sin 2 6 cos 6. 2. sin (^+x] + sml--x\ = 2 sin - cos a: = V2 cos x. « sin 6 a -f sin 4 a , e 3. ^ = tan 5 a. cos 6 a + cos 4 a . sin a — sin /3 2 sin a + sin yS ~ ^^^ a + ^' 2 ^ 5. cos a — cos 3 a = 2 sin a sin 2 a. 6. sin 3 a + cos a = sin 3 ct + sin (90° - a) = ? 7. By generalizing the process observed in Ex. 6, derive a formula for sin a + cos ft. /^. Derive a formula for sin a — cos ft. Reduce the following products to sums or differences : 9. sin 4 a cos 2 a. 12. cos 2 ^ cos 8 6. 10. sin 6 ^ sin 4 9. 13. sin 5 a cos 3 a. 11. cos 2/3 sin 4^. -i-14. sin^^cos^. 15. Making use of the formulae (5) of Art. 82, show how to derive the law of tangents from the law of sines. 83. Functions of double angles. If we put yS= a in equa-N tions (8) of Art. 79, we find (1) sin 2 a = 2 sin a cos a, and (2) cos 2 a = cos^ a — sin*'* a. ' On account of .the relation sin^ a + cos^ a = 1, the latter equation may also be written in either of the fol- lowing two forms : (3) cos 2 a = 1 — 2 sin^ «, or (4) cos2a= 2cos2a— 1. 190 FUNCTIONS OF MORE THAN ONE ANGLE If we put fi = ain equations (1) and (2) of Art. 80, we find 2 tan a (5) tan 2 a = (6) cot 2 a = 1 — tan^ a cot^ a — 1 2 cot a which equations may, of course^also be derived from (1) and (2) by division. 84. Functions of half angles. In Art. 83 we regard the functions of a as known, and we learn how to compute the functions of 2 a. We shall now invert the problem by regarding as known the functions of 2 a, the problem being to calculate the functions of a, the half angle. To put the character of the problem more clearly into evidence, we shall which merely amounts to thinking of any angle ^ as a double angle; namely, as double its half. With this change of notation, equations (3) and (4) of Art. 83 become (1) cos 6> = 1 - 2 sin2 |, cos (9 = 2 cos2 ^ - 1. e If we solve the first of these equations for 2 sin^ q ^^^ ^^® second for 2 cos^ -, we find (2) 2sin2^ = l-cos(9, 2 cos2^= 1 + cos (9, whence ro\ z Q . ^ /i - cos e e , ^/i +cose (3) sin — =±\/ , cos-=±\/^ ^^ 2^2 2^2 The ambiguous signs on the right members are determined 9 by the quadrant of the angle -• If ^ is a positive angle not 6 greater than 180°, - is in the first quadrant and the + sign must be chosen in both of the equations (3). FUNCTIONS OF HALF ANGLES 191 From (3) we find, by division, C4) tan 1 = ± Ji^^^, cot ^ = ± V^^t^^, ^ 2 Vl + cosG 2 ^1-cose the appropriate sign being again determined by the quadrant of the angle J 0, According to (1), Art. 83, we have 6 6 (5) * 2 sin - cos - = sin 6. Let us divide each member of the first equation of (2) by the corresponding member of (5). We find ^^.x , 1 — cos (P) tan- = — r-^ — 2 sm d By a similar process we obtain, from the second equation of (2), (7) cot| = l±^. 2 sm ^ These last two formulae might also have been derived from (4). But the proof we have given is preferable since it avoids the necessity of discussing the ambiguous sign, a dis- cussion which would be necessary if we had followed the other method. The equations for sin -, cos -, etc., are of very great impor- tance, because theymay he used for the purpose of computing a table of trigonometric functions. In fact, we have already shown how to calculate the values of the functions of 0°, 30°, 45°, 60°, and 90° (Art. 11). By means of the addition and subtraction formulae (Arts. 79, 81) we are therefore in a posi- tion to find also the functions of 15° and 75°. We can now calculate the values of the functions of one half of 15° or 7°.5, of one half of 7°. 5, or 3°.75, etc. By continuing this process of bisection and combining the results by means of the addition theorem, we may obviously compute the values of the functions for a set of angles between 0° and 90° as ,%^ 192 FUNCTIONS OF MORE THAN ONE ANGLE close together as we please. By interpolation we may then find the functions of 1°, 2°, 3°, etc. The method, of which we have just given an outline, is essentially the same as that employed by Ptolemy (second century a.d.).* Ptolemy, however, also made use of the inscribed pentagon (cf. Exercise XLIX, Ex. 12), and his table was a table of chords, not of sines. (See Art. 70.) His table gives the values of the chord for each half degree of arc with a degree of accuracy somewhat greater than that which would correspond to a modern five-place table. The earlier tables of Hipparchus and Menelaus are not extant. The Hindus followed the method which we have outlined even more closely. In fact, the table given by Aryabhata (born 476 a.d.) gives the values of the sine at intervals of 3° 45'. As we have seen, this is precisely the interval which would arise as a result of continued bisec- tion of 30°. Essentially the same method was used in subsequent improvements and enlargements of these tables, especially by Rheticus (1514-1574) and PiTiscus (1561-1613). Other far more powerful methods have since been developed, based essentially on the notions of the calculus and the theory of infinite series. EXERCISE XLIX 1. From the functions of 30° find those of 60°. 2. From the functions of 60° find those of 120°. 3. From the functions of 90° find those of 45°. • 4. From the functions of 30° find those of 15°. 5. From the functions of 15° find those of 7°.5. i 6. Find formulae for sin 3 a, cos 3 a, tan 3 a. "^HiNT. Put 3 a = 2 a + oT^ 7. Find formulae for sin 4 a, cos 4 a, tan 4 a. 8. Find formulae for sin 5 a, cos 5 a, tan 5 a. 9. Prove formula (1), Art. 83, by means of a figure. I ^10. Given tan $ = ^, being in the first quadrant. Find the func- tions of 2 ^ and ^ 0. * Ptolemy's great work on Astronomy, usually known as the Almagest, re- mained in undisputed authority until the time of Copernicus. The so-called Ptole- maic system of astronomy, as opposed to the more modern Coperuicau system was named after him for this reason. I CERTAIN LIMIT RELATIONS 193 '' 11. If is in the third quadrant and sin = — —, find the functions Df 2 e. ^5 12. In a circle of radius 1, inscribe a regular pentagon. Show that, by means of this construction the trigonometric functions of 72"^ and 18° may be computed. In particular, show that 8inl8° = |:(V5-l). 13. Making use of the results of Ex. 12, compute the functions of 12°. 14. Making use of the results of Ex. 13, compute the functions of 6°. Assuming the truth of the law of cosines, and setting s = l{a + 1) -^ c), prove the following formulae for the functions of the half angles of a triangle. 15. sinM=A/ ^'^~^^^'~"^ ^ be 16. cosi^=V'^^^^^ ^ he 17. taDM=\ K-'-''>^-''-"> - ^ s(s - a) 18. Prove formulae (6) and (7) of Art. 84 by means of equations (4). 85 a. The limit ^^- and related limits. Although the method sketched in Art. 84 for calculating a table of the values of the trigonometric functions is adequate, it involves far more labor than is actually necessary. The following theorem, whose truth is almost self-evident, is of great importance in this connection as it enables us to calculate the sine of a very small angle with a minimum of effort. If an angle or arc is expressed in radians^ the quotient — ^— u approaches the limit 1 when the angle itself approaches zero as a limit. In symbols I limit ?i5L?=:l. 194 FUNCTIONS OF MORE THAN ONE ANGLE Fig. 104 But In order to prove this theorem, let us draw an acute angle A OF = ^ as in Fig. 104, and symmetrically the angle AOQ also equal to 6. With the vertex as center and any convenient radius r, draw the circular arc PAQ and join FQ, intersecting OA in M. The tan- gents FT and QT, at F and Q, will meet in a point T of OA prolonged. Obviously we shall have (1) FQ < arc FAQ < FT+ TQ. FQ = 2 FM= 2 r sin 0, arc FAQ = 2aTGAF = 2 rd (Art. 74, equation (5)), FT+TQ=2FT^2ri2.nd, where the truth of the second equation depends essentially upon our assumption that 6 is expressed in radians. If these values be substituted in (1), we find 2 r sin (9 < 2 r ^ < 2 r tan ^, or, after division by the positive quantity 2 r, (2) sin ^ < (9 < tan 6, If we divide all three members of this inequality by the positive number sin ^, we find e 1 (3) 1< sin 6 cos We know that cos 6 approaches the limit 1 when 6 ap- proaches zero. Since the value of — — -^ according to (3), sin , which latter quantity itself ap- lies between 1 and cos 6 proaches 1, (4) e sin 6 must also have 1 as its limit. That is, limit ^ 0*0 sin = 1, CERTAIN LIMIT RELATIONS 195 From (3) we have further ^ ^ sin ^ ^ /, 1 > —5— > cos 6, u so that, by a similar argument, we find (^^ limit ?HL?_i If we divide all members pf (2) by the positive quantity tan ^, we find cos e < — ^ < 1, tan a so that (Q^ limit ^ = limit ^^^ ^ = 1 ^ ^ «=^o tan (9 «^o Since sin ( — ^) _ — sin 6 _ sin ^ :^^e~~ -e ~~r' tan ( — ^) _ — tan 6 _ tan 6 equations (4), (5), and (6) will still remain true if 6 ap- proaches zero through negative instead of positive values. These formulae have many important applications. For the present, we shall mention only the one indicated at the beginning of this article. The content of equation (5) may be formulated as follows. If we write (T) !i^=l-S, the angle 6 (expressed in radians) may be chosen so small that h i the difference between 1 and — ^— j will become less than any previously assigned quantity. Since we find from (7) ^ ^ ^^ we see that we can make the angle 6 (expressed in radians) so small that the difference between 6 and sin 6 becomes less than any previously assigned small fraction of 0. 196 FUNCTIONS OF MORE THAN ONE ANGLE Suppose, for instance, that we wish to compute the sine of a small angle to 5 decimal places, that is, with an error which shall be less than 5 units of the sixth decimal place, or .000005. We now know that the angle may be chosen so small that its sine may be equated to the radian measure of the angle itself with an error of less than 5 units of the sixth decimal place. In other words, the equation (8) sin 6=0 (in radians) will be true up to five decimal places for all angles which are sufficiently small. Of course, our method does not inform us just how small 6 must be in order that equation (8) may be true up to five decimal places. It would take us too far afield to investigate this question, a complete answer to which is beyond the scope of this book. The student may convince himself, however, by actual comparison with the tables, that equation (8) is true to five decimal places for all angles less than 2°. In all numerical work, then, involving such small angles, no error noticeable in five-place calculations is introduced by putting sin 6=6 (in radians). Since we have cos ^ = 1 - 2 sin2- (Art. 84, equation (1)), we may put (9) cos^=l-2(|J=l-l^, a formula which will certainly be true up to the fifth decimal place for all angles less than 2°. In fact equation (9) holds to five decimal places even for angles much larger than this and may serve for the purpose of computing the cosines of such angles. Since 6^ is small as compared with 6, if 6 itself is small, we shall even be justified in equating cos ^ to 1 for very small angles. Our tables show that no error is intro- duced in five-place calculations by putting cos ^ = 1, if ^ is less than 0° 16'. CERTAIN LIMIT RELATIONS 197 Since (8) is true to five decimal places if ^ < 2°, we see that we shall have sin 26 = 26 with the same degree of approximation if ^ < 1°. More generally the formula (10) sin n6 = n6 (6 in radians) is correct to five decimal places if n6 is less than 2°. The results deduced in this article make it very easy to compute the functions of very small angles. By combining these results with the methods of Art. 83 an extensive table of the trigonometric functions may be constructed with com- parative ease. EXERCISE L Compute the values of the following functions of small angles to five decimal places by the method of Art. 85 a and compare with the values obtained from the table : 1. sin 12'. 3. sin 1°. 5. tan 1°. 2. tan 15'. 4. cos 1°. 6. cot 1°. 7. What will be the angle subtended by a lamp-post 10 feet high at a distance of one mile? 8. In order to find the distance from the earth to the moon, the following plan may be adopted. Two astronomers stationed at A and B respectively (Fig. 105) observe at the same instant the angular distance of the moon's center M from their respective zeniths (their overhead points), Z and Z'. This gives the angles a = Z^M and /3 = Z'BM. For the sake of simplicity as- Fig. 105 sume that both stations A and B are on the equator, that the moon is in the plane of the equator, and let E be the center of the earth. Then /. A EB = X is equal to the differ- ence between the longitudes of the two stations and may be regarded as known. We may now compute ^EAM= 180° -a, /. EBM = 180° - /3, ZAEB = X. 198 FUNCTIONS OF MORE THAN ONE ANGLE Now the sum of the four angles of the quadrilateral AEBM is four right angles ; that is, ^M+ 360° - a - y8 + A. = 360=, whence From the value of M obtained in this way, it is easy to compute the angle subtended by the earth's radius at the center of the moon. This angle is called the moon's parallax. Find the moon's distance from the earth if the moon's parallax is 57' and if the earth's radius is 4000 miles. 9. The apparent diameter of the moon as seen from the earth is about 31'. Making use of the result of Ex. 8, what is the moon's diameter in miles? 10. The sun's parallax is about 8".8. Assuming 4000 miles as the length of the earth's radius, find the distance from the earth to the sun. 85 b. The aixxiliary quantities S and T. We have seen in Art. 28 that the ordinary tables of sines and tangents be- come inconvenient for very small angles. To avoid this in- convenience, we constructed an additional table (Table III), giving the values of the sines and tangents of such small angles directly for every second of arc. But we may accom- plish the same purpose in another way, by means of an auxiliary table occupying far less space than the additional table just mentioned. This second method is based on the fact that the quotients sin J tan 6 change very slowly if ^ is a small angle. We have just seen that each of these quotients has unity as its limit when 6 approaches zero, provided that the angle is measured in radians. Let us instead express 6 in minutes of arc. Let 6' denote the number of minutes and B^^^ the number of radians 'contained in the angle 6. Then, accord- ing to Art. 85 a, (1) limit 5yLi = limit i5]2i=l. AUXILIARY QUANTITIES 5 AND T 199 Since (Art. 74) 1° = -^ radians = 0.0174533 radian, 180 and therefore y ^ 0.0002909 radian, the angle 6^ which contains 6' minutes, will contain l9(^) =0.000290919' radian. " Consequently we find sin ^ . n 0^^ n nAAonr^n sin e = sin ^ -^ TT-^r^::^— — = 0.0002909 6' ' 0.0002909 * ^(^ ' and therefore, on account of (1), (2) limit !HL^ = limit i^ = 0.0002909. In other words, if 6' is an angle expressed in minutes, the common limit toward which and — — — tend, when 6' 6 6' approaches zero, is a number whose first seven decimal places are given by 0.0002909. Let us write ^Q\ sin 6 . tan 6 (3) « = -^, « = -^- These quantities change their values very slowly for small , values of 0. In fact we have just seen that, for angles which are small enough, we shall have log 8 = \ogt = log 0.0002909 = 6.46373 - 10. For ^ = 2° = 120' we have ^ sin 2° ^ tan 2° ^~ 120 ' 120 ' which gives, if we look up the logarithms from the tables, log sin 2° = 8.54282 - 10 log tan 2° = 8.54308 - 10 log 120 = 2.07918 log 120 = 2.07918 log 8 = 6.46364 - 10 log t = 6.46390 - 10 Therefore, while changes from 0° to 2°, log s changes only by 9 units and log t by 17 units of the fifth decimal place. 200 FUNCTIONS OF MORE THAN ONE ANGLE Table IV enables us to find the values of iS=\ogs and T= log t for every angle between 0° and 2°. To find the logarithm of the sine or tangent of such an angle we have the formulae (immediate consequences of (3)) .^. log sin (9 = log 5>' + log 8 = lojf ^' + aS', ^ ^ log tan = log 6' + log ^ = log 6' + T, If the log sin or log tan of a small angle is given, to find the angle, we may do this by means of one of the equations .rN log 6^ = log sin 6 — S, ^ ^ log 6' = log tan - T, obtained from (4) by transposition. Of course the quantities aS' and T are available, not only for sines and tangents of small angles, but also for cosines and cotangents of angles close to 90°. EXERCISE LI 1. Find the sine and tangent of 1° 13'.21 by using the auxiliaries S and T. Solution. Since 6=1° 13'.21, we have 0' = 73'.21. log 0' = 1.86457 log 6' = 1.86457 S = 6.46369 - 10 7^^6.46379 -10 log sin d = 8.32826 - 10 log tan ^ = 8.32836 - 10 2. Given log sin 9 = 8.24798 - 10. Find 6. Solution. We find from Table IV cori:esponding to log sin ^=8.24798 - 10, 5 = 6.46370. Formula (5) leads to the calculation log sin = 8.24798 - 10 S = 6.46370 - 10 log 6' = 1.78428 .'.6' = 60'.85 = 1° 0'.85. Find the values of the logarithms of the following functions by means of the auxiliaries S and T: I 3. sinl°21'.63. 5. cos 89° 13'.21. | 4. tan0°32'.61. 6. cot 88° 21'.75. Find the angles determined by the following functions by means of S and T: 7. log sin $ = 7.76345 - 10. 9. log cos = 8.42371 - 10. 8. log tan = 8.50731 - 10. 10. log cot = 8.53729 - 10. ■r^ CHAPTER XII DIRECTED LINES AND DIRECTED LINE-SEGMENTS * 86. Plan of another proof for the addition formulae. When we proved the addition theorem in Art. 79, we found it necessary to divide the proof into a number of cases accord- ing as the angles were in the first, second, third, or fourth quadrants. To be sure, by making use of the method of mathematical induction we found it a fairly simple matter to make an exhaustive discussion covering all cases. Neverthe- less we feel that it must be possible to devise a method enabling us to prove this theorem at one stroke for angles of any magnitude. The key to the solution of this problem is found to be a careful formulation of the notions of a directed line and a directed line- segment. Since these notions are of very great importance, not only in this connection, but in many other parts of pure and applied mathematics, we shall find it worth our while to speak of them, even if they are not absolutely indispensable for the proof of the addition theorem. 87. Directed lines and segments. A straight line is infi- nite in extent and is determined by any two distinct points upon it. We may, however, think of one and the same straight line as having either of two opposite directions, in which case we speak of it as a directed line. Since we can never draw more than a finite portion of a line, we may indicate the direction of a directed line by placing a -h sign near one end of that portion which actually appears in the figure. In Fig. 106 we have thus indicated the direction of * This chapter may be omitted in a first course if the time is insufficient. 201 202 DIRECTED LINES AND LINE-SEGMENTS the directed line I which is to be thought of as pointing toward the upper right-hand corner of the page. This method of indicating the direction of a di- rected line has ah'eady been used in this book to indicate the positive direction of the x-axis and y-axis of a system of rectangular coordinates. (See Art. 63.) These are directed lines. A line -segment is a finite portion of a line and may be described by naming its end points, such as AB in Fig. 106. But again, we may think of it as a directed line-segment, thus distinguishing between AB and BA. When a directed line-segment lies upon a directed line, its direction or sense may be the same as that of the line or else opposite to it. If a directed line-segment on Z is 5 units long and if its djrectioil is the same as that of Z, we may represent it by the number +5. A line-segment of the same length and opposite direction will be represented by — 5. In general, a directed line-segment, which lies on a directed line, shall be counted positive or negative according as it has the same or the opposite direction as the directed line. For such line-segments, we always have BA = —AB, or AB + BA = 0. We are now ready to prove the following theorem. If A, B, C, are any three points on a directed line, then (1) AB + BC=AC, where AB, BC, and AC are directed line-segments. Proof. 1. Let J. (7 be positive and let B be between A and O. Then AB and BC are also positive and the truth of the theorem, in this case, is obvious. (See Fig. 107.) 2. Let AC hQ positive, but let C be be- tween A and B. Then AB is positive, but BC is negative and equal to — CB. Thus AB + BC=AB-CB = AC (See Fig. 108.) 3. Let AC still be positive, but let A be be- tween B and C Then (Fig. 109), AB -f BC= -BA^- BC= BC-BA = AC. fiq. io9 Fig. 107 BX Fig. 108 C^* DIRECTED LINES AND SEGMENTS 203 Ji AOis negative, there are again three cases according as the order of the three points is CBA^ BOA, or CAB. But the above relation (1) will be found to be true in all cases. These last three cases may, of course, also be reduced to the former three by reversing the direction of the line I. The following is a simple corollary of the above theorem, if A, B, C, i>, are any four points of a directed line^ we have the relation (2) AB + BC+CD:=AD between the directed line-segments AB^ BO^ CD^ and AD* In fact, by the theorem just proved, we have AB + BQ^AC, AC-hCI) = AD, so that we find by addition AB + BC+ AC -\- CD = AC -\- AD, which reduces to (2) if we subtract A (7 from both members. It may now be proved by induction that, in general, if A, B, (7, ••• M, I^ are any finite number of points on a directed line, then (3) AB-hBC+CD-{- ... -\-MJS'=AN. Equations (1), (2), (3) may also be proved by algebra. On the directed line I, let us introduce a point as origin or zero point of a scale, whose positive readings are on that side of which corresponds to the positive direction of the line I. This is precisely what we did when we established scales upon the a:-axis and ?/-axis of a coordinate system (Art. 63). Denote by Ij^ the reading of the scale which corresponds to the point A, by l^ that which corresponds to B. The differ- ence Iff — Ij^ will give the length of the line-segment AB, affected with a plus or minus sign according as AB is a positive or negative line-segment in the sense of our pre- vious definition. 204 DIRECTED LINES AND LINE-SEGMENTS If then we have any three points A, B, Q on the directed line, we shall have AB^l^-h. BQ^la-lB. AO=la-h, and therefore AB + BC= Ib-Ia + Ic-Ib=Ic-Ia=-AO, which is the same as (1). In the same way we may also prove equations (2) and (3). 88. Angles between directed lines. Let I and m be two directed lines, and let us denote the angle between their positive directions by (Z, m) or (w, I) according as we think of Z or m as the initial side of the angle. To be perfectly specific, we understand by (Z, m) the angle, less than 360°, through which it would be necessary to rotate the directed line I in the counter- clockwise direction, in order to make its positive direction coincide with the posi- tive direction of the directed line m. In Fig. Ill Fig. Ill, the angle (I, m) is marked d. Similarly (m, V) is the angle through which m would have to be turned in the positive (counter- clockwise) direction in order to make the positive direction of m coincide with that of I. In Fig. Ill (m, V) is marked 6' . We see that we shall always have (1) 0, m) + (w, Z)=360°, so that (m, 0=360°- (I, m) and therefore (see Art. 77, equations (7)), (2) sin (m, V)= — sin (?, w), cos (w, Z) = cos (Z, m). 89. Projections. The projection of a point P on a line Z is the foot of the perpendicular dropped from the point to the line. The projection of a line-segment AB on a line Z (see Fig. 112) is the line-seg- ment A B' of Z bounded by the projections of A and B. If AB is a directed line-segment, so is A'B'. -' i^^ A I ' 1 "lO. 11 3> 2 PROJECTIONS 205 We wish to solve the following problem. G-iven a directed line-segment AB on a directed line p ; to find the magnitude and sign of its projection upon any second directed line I. The line-segment AB may have the same sense as the directed line p or else the opposite sense ; it will be repre- sented by a positive number in the first case and by a nega- tive number in the second (Art. 87). If we denote by I AB I the positive number which represents merely the length (regardless of direction) of the line-segment AB, we shall have (1) AB = \AB |, read AB = length AB, or (2) AB = — I AB |, read AB = minus length AB, according as the direction of AB agrees with that of the directed line p or not. Let us consider first the case (1) in which AB is positive. (See Figs. 113 and 114.) Let OM be the projection of AB on L Choose as origin and I as the a;-axis of a system of ^^ Fig. 113 Fig. 114 O^ coordinates, so that the positive a;-axis coincides with the positive direction of the line I. The line m through 0, per- pendicular to I and with its positive direction as indicated in the figures, will then be the «/-axis. Through draw a directed line p' parallel to p and let P be its intersection with the line MB through B perpendicular to I* The line- * The word parallel in this theory means, not only that the lines are parallel in the ordinary sense, but that their positive directions are the same. The word anti-parallel is sometimes used for two parallel directed lines whose positive' directions are opposite. 206 DIRECTED LINES AND LINE-SEGMENTS segments AB and OP have the same length and direction, and the same directed line-segment OM on I as projection. But, by the definition of the cosine of a general angle (Art. 64), OM cos (I, p) = cos (I, p') = — , since OMis the abscissa and OP the radius vector of a point P on the terminal side of the angle (Z, p'^, this angle being in its standard position and OP = AB being positive in the case under consideration. Consequently we have 0M= OP cos (Z, p) = AB cos (Z, p). Since OM is the projection of AB on Z, we may write this as follows : (3) 0M= proji AB = AB cos (Z, p) = AB cos (jd, Z) ;* for, according to equation (2), Art. 88, the angles (Z, p) and (^, Z) have the same cosine. Thus the projection upon the directed line Z, of the positive line-segment AB of the directed line p^ is equal to AB mul- tiplied by the cosine of the angle between Z and p. The projection is a directed line-segment on Z, positive if (Z, p} is p in the first or fourth, negative if (Z, jt?) is in the second or third quadrant. We have proved equation (3) when AB is positive. Let us con- sider now the case when AB is a negative line-segment of p, (See Fig. 115.) The projection oiAB, as well as that of OP, is OM. We may think of OP, which is a negative line-segment on p\ as a positive line-segment of a directed line p'^, coincident with p' but having the opposite direction. But this latter directed line makes an angle with Z just 180° greater than (Z, p') ; that is. +m Fig. 115 (4) + 71 . 360°) = sin (180° - ^) = s, where n is any positive or negative integer or zero. We see therefore that all angles, such as (1) 71 . 360° + (9 = 2 71 . 180° + <9, or (2) n . 360° + 180° - ^ = (2 7i + 1) • 180° - 6, have the same sine as the angle 6. This may be expressed as follows. All of those angles which can be obtained by add- ing a given angle to any even multiple of 180° or else by subtracting the given angle from any odd multiple of 180°, have the same sine. That these are the only angles which have the same sine follows easily from the fact that two distinct angles in the same quadrant cannot have the same sine. If the given value of 8 is negative, nothing essential is changed in the above argument, except that will then be in the third or fourth quadrant instead of being an acute angle. It will still be true that all of the angles given by formulae (1) and (2), and no others, have the same sine as 6. 216 INVERSE TKIGONOMETRIC FUNCTIONS Now we may include all of the angles (1) and (2) in the single expression (3) m . 180° ± e, where m may be any integer, even or odd, and where the + or — sign is to be used according as m is even or odd. We may avoid this awkward distinction by writing in place of (3) (4) ^.180° + C-1)"^6>, since (— 1)"* is equal to + 1 or — 1 according as m is even or odd. Thus, if 6 is one angle whose sine is equal to a given number «, the most general angle which has the same sine is ^.180°+ (-1)"* (9, where m is any positive or negative integer or zero, A brief way of indicating this fact is given by the follow- ing equations: (5) sin [m • 180° + (_ 1)^ 6^] = sin 6 or sin [mir + ( — 1)"" ^] = sin 6, where we use the first or the second form according as 6 is expressed in degrees or in radians. Since the cosecant of an angle is the reciprocal of its sine, we have also (6) CSC [m . 180° -h ( - 1)"^ 0^ = esc 6 or CSC [WTT + ( — 1)"* ^] = CSC 0. We find, by precisely similar considerations, cos (2 m . 180° ±6}= cos (9, cos (2 wtt ± (9) = cos 0, (^^ sec (2 m - 180° ±6)= sec (9, ^^ sec (2 mir ±6)= sec 0, and tan (m • 180° + 6) = tan 6, ^ tan (mir + 6) = tan B, ^^) cot (m . 180° + ^) = cot 6, ^^ cot (rmr + 6>) = cot 6, according as the angles are expressed in degrees or in radians. INVERSE TRIGONOMETRIC FUNCTIONS 217 EXERCISE Llll Without making use of the trigonometric tables, find all of the angles which satisfy the following conditions: 1. sin^=f 3. tan^=l. 5. cot ^ = VS. 7. tan ^ = oo. /2. cos^ = -|. 4. sec^ = V2. 6.'^csc^=l. 8. ^sin ^ = 0. Making use of the tables, find all of the angles which satisfy the fol- lowing equations : 9. sin ^ = - .4721. 11. tan 6 = 1.7269. 10. cos ^ = + .3216. 12. sec ^ = 2.7213. 97. The inverse trigonometric functions. In the equation (1) X = sin y, we now propose to regard y, the angle or arc, as a function of 27, the sine. We may express this new way of looking at the relation (1), by saying that y is an angle whose sine is equal to x or (2) y is an arc whose sine is equal to 05, a statement which is usually written in the contracted form (3) y — arc sin oc. It should be noted that (1) and (3) are merely different ways of expressing the same relation between x and y. They differ only in one respect. In (V) y is regarded as given and x is to be found, while in (8) x is regarded as given and y is to be found. The relation between the func- tions (1) and (3), that of being inverses of each other, is of the same kind as in the more familiar case of the function X = y'^^ which has as its inverse y = ± ^x. In the same way we define the equation y = arc cos a? to mean that y is an arc whose cosine is equal to x. There- fore this equation is equivalent to X = cos y. 218 INVERSE TRIGONOMETRIC FUNCTIONS Similarly, if a; = tan y^ we write y = arc tan x, and in the same way we define the symbols arc cot sc, arc sec oc, arc esc oc. Let us return to equations (1) and (3). We know that the sine of an angle is never numerically greater than 1. Consequently we can find no angle whose sine is equal to a; if a; is numerically greater than 1. We may express, this as follows: 77ie function arc sin x is defined only for those values of x which are not numerically greater than 1 . If X is numerically less than unity, there exists not merely one angle whose sine is equal to a?, but the number of such angles is unlimited. (See Art. 96.) If x is positive, one of the corresponding angles is a positive acute angle. If x is negative, the smallest corresponding 'positive angle is in the third quadrant. But in this case there is a negative acute angle whose sine is equal to the given negative value of x. Let us use the symbol Arc sin nc, distinguished from arc sin x hy the use of the capital letter A, to indicate the numerically smallest angle or arc whose sine is equal to x. The function Arc sin x, like arc sin x, is defined only for the values of x between — 1 and + 1 ; that is, for those values of X for which ^ ^ . -l0; a:>0,y<0; a;<0,2/<0. 98. Trigonometric equations. It often happens that angles are to be determined by means of equations which they must satisfy. Such equations usually contain the trigonometric functions of the unknown angle and are then known as trigonometric equations. Let us confine our attention to the case where one unknown angle is to be found as a solution of one equation. Such an equation may have one of the following forms. I. It contains 6 algebraically, but does not contain the trigonometric functions of 0. Example. e^-~ = 0. II. It contains the trigonometric functions of the angle d in algebraic combinations, but does not contain the angle itself explicitly. Examples, sin^ ^ — cos ^ = 0, 2 tan ^ + cot ^ = — 3. III. It contains the angle and its trigonometric func- tions simultaneously. Example. ^ — - sin ^ = ^ . equ ^se Clearly the equations of the first type are merely algebraic equations, and their discussion properly belongs to a trea- se on algebra, t may be shown that the equations of the second type can also be reduced to algebraic equations, although it is often easier to effect their solution without so reducing them. The equations of the third type will not be considered in this book. The solution of such equations is a difficult matter and can be accomplished in a satisfactory manner only in a few cases. It should be mentioned, however, that TRIGONOMETRIC EQUATIONS 223 the method of graphs usually provides an approximate solu- tion for such equations. We shall now discuss a few simple examples of equations of the second type, in such a way as to illustrate the fact that their solution may be reduced to that of algebraic equations. EXERCISE LV 1. Solve the equation 2 cos^ x — 5 + 7 sin x = 0. Solution. We have cos^ x = 1 — sin^ x. Therefore the given equation becomes - 3 - 2 sin2 x + 7 sin a; = 0. Consequently, if we put sin x = s, we obtain the quadratic equation iors, 2 s2 _ 7 s + 3 = 0. The solution of this equation gives s = 3 or s = |. The first solution must be discarded, since s = sin x cannot be numerically greater than unity. The second solution tells us that sin X = ^, so that X = 30° and x = 150° are the only positive angles smaller than 360° which satisfy the equation. In examples 2 to 5, find all positive angles less than 360° which satisfy the given equations. This may be done without using any tables. 2. cos 2 ^ + cos ^ = - 1. 4. tan 2 ^ = - 2 sin 0. 3. cot 2 ^ + tan ^ = - f V3. 5. sec2 $ + csc^ = 4:. In solving the following equations, make use of the tables : 6. sin a: tan a; ::= — ^^^. 7. cos a; cot a: = — f . 8. Solve the equation cos 3 a: = sin 2 x. 99. The equation a sin 8 + 6 cos 9 = c. The equations of the form (1) a sin -\-b cos 6 = e may be solved by the method of the preceding article. But, unless the numbers a, 6, c are especially simple, it is far more convenient to proceed as follows : We introduce two auxiliary quantities I and X, such that .ry. I sin L = a, I cos L = b, 224 INVERSE TRIGONOMETRIC FUNCTIONS where, moreover, I is assumed to be positive. That it is always possible to find a positive number I and an angle L which satisfy equations (2) be- comes obvious if we think of a and h as the rectangular coordi- nates of a point P. (See Fig. 126.) Equations (2) show that I is the radius vector OP of P and that L is the angle which OP makes with the positive direction of the ic-axis. Our figure also shows us that +a? Fig. 126 (3) tanX = ^, l = +^a^ + h\ Of course these same equations also follow directly from (2) without the use of geometry. If now we substitute the values (2) for a and h in (1), we ^^^ I (sin i sin ^ + cos L cos 6) = c, or by Art. 79, equation (8), (4) Zcos(l9-X)=(?. Therefore, in order to solve (1) we may first determine I and L from (2) and then find 6 from (4). EXERCISE LVI 1. Solve the equation 2.1346 sin 6 - 3.0526 cos = 0.9875. Solution. a = + 2.1346 h = - 3.0526 c = + 0.9875 log a = log (I siu L) = 0.32932 (4) log b = log (I cos L) = 0.48467 n (5) log tan Z = 0.81465 n (7) = (4) -(5) L = 145° 2M5 (8) log sin i = 9.75820 (9) log cos Z = 9.91355 n (10) log/ = 0..57112 (11) = (4) -(9) = (5) -(10) logc = 9.99454 i^) log cos (d-L)= 9.42342 (1) (2) (3) = ( ) ) = (12) = (6) -(11) TRIGONOMETRIC EQUATIONS 225 e^- L= 74° 37'.61 (13) ^2 -i = 285° 22'. 39 (14) L = 145° 2M5 (8) ^1 = 219° 39'.76 (15) = (13) + (8) $2= 70°24'.54 (16) = (14) + (8) Remarks. The numbers in parentheses indicate the order in which the results are written down and how some of them are obtained. The — 10 has not been written down in the case of negative characteristics. The n which follows the logarithms of b, tan L and cos L indicates that the corresponding numbers are negative. (See Art. 25.) L is chosen in the second quadrant because, I being positive, sin L is positive and cos L is negative. The values 6^ — L and 6^ — L, both obtained from (12), are the two values, less than 360°, which — L may have so as to corre- spond to the given value (12) of log co8(^ — Z), Check. By substitution in the original equation, log a =0.32932 log & = 0.48467 n log sin ^1 = 9. 80500 n log cos 6^ = 9.88639 n log (a sin 6^) = 0713432 n log {h cos ^i) = 0.37106 a sin ^1 = - 1.3624 h cos e^ = + 2.3500 a sin e^ + h cos $]_ = + 0.9876 c = 4- 0.9875 This calculation checks 6i and therefore also L. The relation between ^1, 62-, and L is so simple as to make unnecessary a separate check for 9^. 2. Solve the equation - 3.2471 sin 6 + 5.7469 cos ^ = - 6.3271. 3. Solve the equation 2.1725 sin 6 + 3.2749 cos $ = 5.7216. CHAPTER XIV APPLICATIONS TO THE THEORY OF WAVE MOTION 100. Simple harmonic motion. When we began to study trigonometry, it was for a very practical purpose. We wished to find an arithmetical method for solving triangles. We accomplished this purpose by means of the trigonometric functions and by using tables of the numerical values of these functions. Later we generalized the notion of the trigono- metric functions more than was strictly necessary for the simple problem of solving triangles, and we found it to be an interesting task to investigate these trigonometric functions and their various properties for their own sake. We shall now find that these properties, aside from their theoretical interest, have in their turn most important applications. The fundamental reason for the great importance of the trigonometric functions lies in their periodicity. (Cf. .Art. 67.) Many natural phenomena are periodic in character, and whenever the attempt is made to represent such a phe- nomenon by a mathematical expression, the trigonometric functions are found to be indispensable. The simplest periodically recurring motions are connected with uniformly \a~'^^ rotating bodies. Let the point (Fig. 127) describe a circular path o\ radius a around the point as center| and let us assume that it is moving with a constant angular velocity of radians per second. Let us assume further that P starts its motion at the time ^ = from th( point (7, which is so located that Z ^ 0(7 is equal to a radians. 226 SIMPLE HARMONIC MOTION 227 If P is the position of the point at the time <, that is, t seconds after the motion has begun, we shall have (1) e = a}t+a, where 6 denotes the angle A OP expressed in radians. The point will describe its circular path in counterclockwise or clockwise fashion according as © is positive or negative. While the point P is moving in its circular path, its pro- jection M upon the a;-axis will oscillate to and fro between the points A and A\ reaching its greatest speed when at 0, gradually slowing down until it reaches A\ when it reverses the direction of its motion, returns with gradually increasing speed to 0, after reaching which point it slows down again until it reaches A and again reverses its motion. To find an analytic expression for the motion of the point M^ we observe that OM^x, OP=a, A0P = 6, 0M= OP cos AOP, whence, making use of (1), (2) a:=a cos (a)^ + a). In the same way we see that iV, the projection of P upon the ^-axis, moves in accordance with the equation (3) ^ = a sin (ft)^+ a). Equations (2) and (3) are so closely related that it will suffice to study one of them. In fact, if we put in (3) a = ^ + a^ it becomes (cf. Art. 77, equations (2)), y = a sin la)t-\-a^-{- — j = a cos (^cot + a'), which is of the same form as (2). We shall therefore con- fine our attention to equation (3). Since any diameter of the circle may be chosen as y-axis, we may express our result as follows. If a point describes a circular path with uniform velocity/, its projection upon any fixed diameter of the circle moves in accordance with an equa- tion of the form (3). Such a motion is called a simple har- monic motion. The quantity a which measures the maximum 228 THEORY OF WAVE MOTION distance of the point iV from its mean position is called the amplitude. 101. The period and phase constant. When the angle B has increased from its initial value a by 2 tt radians, the point P will have described a complete circumference and the motion of the point N will have passed through all of its phases. The time T^ which is required to accomplish this, is called the period of the simple harmonic motion. The period is determined by the condition that the angle (oT de- scribed by the point P in the time T must be equal to 2 tt. Therefore we find (1) G)^=2 7r, T=^. If we wish to put the period into evidence in the equation of a simple harmonic motion, we observe that (1) gives 27r T so that we may write, in place of (3), Art. 100, (2) ^ = «sin^-^ + aJ. This equation represents a simple harmonic motion of period T and of amplitude a. The quantity a is called the phase constant. The phase constant is an angle and enables us to calculate the distance from to the position occupied by the moving point iVat the time ^ = when the motion began. Thus, if a = 0, the point N starts from as its initial position and begins to move upward ; if « = -, the initial position of N is at B, etc. Let us think of two different points oscillating up and down along the line BB\ each in accordance with an equation of the form (2), the amplitudes and periods of the two mo- tions being the same while the phase constants are different. Then these two points will move in exactly the same way, only that one will always be ahead of the other. The sec- ond point will appear to be imitating the motion of the first, ILLUSTRATIONS OF HARMONIC MOTION 229 lagging behind it in a perfectly definite fashion. This is what is meant by saying that the phases of the two motions are different, the phase of the motion (2) at the time t being equal to ^ ^ T + a. The use of the word phase in this connection is not merely accidental. The appearance of the moon at a given instant (its phase) depends upon the place in its orbit around the earth which it happens to occupy at that time. By analogy we speak of a periodic phenomenon as passing through all of its phases in the course of a period, and of course two otherwise identical periodic phenomena may have their corresponding phases occur at different times. It is this difference which manifests itself in the different values of the phase constant. We have already observed that the substitution « = — +«' converts (3), Art. 100, into an equation of form (2), Art. 100. We may now express this fact as follows: The two equations Vi = a sm -^, ^2 = « cos -^= a sin (^^r+ 2J represent two simple harmonic motions of amplitude a and period T^ which differ only in phase^ the phase difference being equal to — radians or 90°. The time interval which elapses between corresponding phases of these two motions is \ T^ that is, a quarter period. 102. Some illustrations of simple harmonic motion. The notion of simple harmonic motion is of fundamental impor- tance in many problems of applied mathematics. The mo- tion of a simple pendulum, the vibrations of a tuning fork, and many of the motions of elastic bodies may be described conveniently in terms of simple harmonic motion. The vertical motion of a particle of a water wave is approximately of the same type, and the whole theory of sound and light is based on the idea of harmonic motion. 230 THEORY OF WAVE MOTION EXERCISE LVII In Exs. 1-5 the unit of time is one second and the nnit of length one inch. Describe completely the simple harmonic motion given by each of these equations; that is, determine their amplitudes, periods, and phase constants, assuming equation (2) of Art. 101 as the standard form for the equation of such a motion. 1. 2/ = 2.3745 sin ?-^. ^ 5 2. y = 3.7216 sin (4 < + ^y 3. y = 1.4712 sin (2.7215 1 + 1.7291). 4. y= 11.7261 cos (7 0- 5. 6. A certain pendulum has a period of oscillation of 5 seconds. Its motion is started by displacing the bob from its position of equilibrium three inches toward the right and then releasing it. Write the equation of its motion. What will be the corresponding equation for a point halfway between the lower end of the pendulum and its point of support ? Hint Since the amplitude of the oscillation is small as compared with the length of the pendulum, the motion may be regarded as taking place approximately in a straight line. Take this line as z-axis, positive toward the right, and choose as origin the point of equilibrium. Let the time t be measured in seconds from the moment in which the pendulum is released. Then the required equation for the lower end of the pendu- lum will be 7. A cork is bobbing up and down owing to the passing water waves. These waves are 4 inches high (i.e. the difference of level between the crest and trough is 4 inches). If seventy of these waves pass in a minute, and we start to count time from one of the instants when the cork has reached its highest position, what equation will describe the motion of the cork approximately ? 8. Show that the following mechanism enables us to convert uniform circular motion into simple harmonic motion. RR' (Fig. 128) is a rod which may slide back and forth in its own direction, its motion being limited by the guides A, B,A', B'. Attached to the rod there is a slot S perpendicular to the rod. A crank C, moving with uniform velocity in a SIMPLE HARMONIC CURVES 231 circle around the point O, is made to fit accurately into the slot S, Every point of the rod RR' will then describe a simple harmonic motion. c B' 9 Fig. 128 9. Show that a point on the piston rod of a steam engine will move to and fro approximately according to the law of simple harmonic motion. 103. Simple harmonic curves. In the equation charac- teristic of a simple harmonic motion, namely, y = a sin (o)^ -f «), let us substitute x in place of t and interpret x and y as the rectangular coordinates of a point in a plane. The curves obtained as a result of plotting such equations, ^= a sin(o)a; + a), are called simple harmonic curves. We may also think of the relation between simple harmonic motion and simple harmonic curves in the following more concrete fashion. At- tach a light pin P (Fig. 129) to one of the prongs of a vibrating tuning fork, and allow it to press lightly against a strip of smoked glass. If this strip of glass is at rest, the pin will make a short straight line upon it. But if the strip be moved with a constant velocity in a direction perpendicular to that of the vibration of the tuning fork, the point P will describe a wavelike curve upon the slide. This curve may be regarded as a record of the motion of the tuning fork. It is easy to see that the record produced in this way by any simple harmonic motion is a simple harmonic curve. The simplest case of a simple harmonic curve is that of the sine curve ^ = sin a:. Fig. 129 232 THEORY OF WAVE MOTION whose form has, by this time, become familiar to the student. (Compare the middle curve in Fig. 130.) It has the form of a wave line with nodes at the points a; = 0, a: = ± tt, x= ± 2 TT, etc., with crests or maxima one unit high above the of the ic-axis, and with troughs or minima one unit deep below the points 3 TT 7 TT 11 TT , a: = — , ^ = "2-' ^= -2"' ®^^-' The length of one complete undulation of the curve is equal to 2 tt. 104. Amplitude. It is clear that the curve (1) y = (^ sin 27, where a is any fixed positive number, is of the same general form as the sine curve. It has the same nodes (points of intersection with the a:-axis), and each of its undulations has the same length as that of the sine curve. Its maxima are above the same points of the rc-axis as those of the sine curve, but they are higher or lower according as a is greater or less than unity, a is called the amplitude of the curve. Figure 130 shows three such curves of amplitude, ^, 1, and 2. It is clear, then, that the depth of the wavelike curve (1) is dependent upon the value of the amplitude. 105. Wave length. The two curves y = sin X and y = sin 2 a; are shown together in Fig. 131. The latter curve has the same general form as the former, but each of its undulations (its wave length) is only half as long. Similarl}^ the curve (1) i/ = sinnx, PHASE CONSTANT 233 where ?^ is a positive integer, is found to be a wave line of the same height as the sine curve, but having n complete undulations between x = and a; = 2 TT. Consequently its wave length X is given by n (2) Fig. 131 Equation (1) represents a simple harmonic curve of 2 TT wave-length \ = — , even if n n is not an integer. For the function sin nx will pass through all of its values jusH once, while nx changes from to 2 tt ; 2 TT that is, while x changes from to . n We may put the wave length into evidence in the equation of the curve, by solving (2) for n and substituting the result- ing value of n in (1). (3) We find n = , and therefore X y = sm- TTX as the equation of a simple harmonic curve of wave length X. If we combine this result with that of Art. 104, we see that (4) y = a sm represents a simple harmonic curve of wave length X and of amplitude a, 106. Phase constant. If we compare the curves y = sin X and y = cos x, we observe that they are identical in form. That is, both are simple harmonic curves of the same wave length and amplitude. They differ only in position. We can, in fact, slide one of these curves along the x-axis in such a way as to 234 THEORY OF WAVE MOTION make it coincide with the other. This, as we have observed before (Art. 78), is the geometrical significance of the equation sin l — -\-x] = cos X. We may therefore dispense with the cosine curve altogether and regard it as a displaced sine curve. More generally, the same thing is true of the curve (1) ^ = sin (2: + a), which coincides with the sine curve if a = and with the cosine curve if « = -5-. Equation (1) represents a sine curve displaced toward the left through a distance of a units. The quantity a is called the phase constant of this curve. If we combine the results of Arts. 104 and 105 with our latest remark, we see that (2) y = a sm (^^— + « 1 represents a simple harmonic curve of wave length X, amplitude a, and phase constant a. This equation may be put into a different form by making use of the addition formula of the sine. (See Art. 79.) For we have r . 27rx , 2irx . ~l v=a\ sin cos a + cos sm a . Consequently, if we put (3) m = a cos a, w = a sin a, we shall find ,.. . 2 Trrr , 2 7rx (4) y = msm hwcos . Conversely^ any equation of the form (4) has a simple har- monic curve as its graph. WAVE MOTION 235 For if m and n are given, we may compute a and a from (3), thus obtaining the value of amplitude and phase con- stant. Since a is positive, we have, from (3), ^ (5) a= + Vm2 + r^2, tana = — the quadrant of a being determined from the sign of its sine and cosine as given by (3). EXERCISE LVIII Draw the simple harmonic curves which correspond to the equations given in Exs. 1 to 5 : 1. 3^ = 2sin(3a; + ^y 3. y = 3.1 sin (7.6 x + 6.2). 2. y=3cos(2a;-|). 4. y = 2.8sin (1^+72°). « 5. 2/ = sin (a: + tt), y = — sin x. 6. In the general theorj'^ of simple harmonic curves we assumed a to be a positive quantity. Show that this assumption does not, after all, really exclude from consideration those cases in which a is negative. In other words show that a simple harmonic curve, for which a is negative, coincides with another one for which a is positive and whose phase con- stant differs from that of the first curve by tt radians. 7. Write the equations of the curves of Exs. 1 to 5 in the form (4) of Art. 106. 8. A simple harmonic curve is given by the equation y = 2.75 sin ?^^ + 3.76 cos ^^ • ^ 5.76 5.76 Determine its wave length, amplitude, and phase constant. 9. Discuss in the same way the equation y = 3.72 cos (7.52 x)- 2.67 sin (7.52 x). 107. Wave motion. Our use of the word wave^ in con- nection with simple harmonic curves, is not quite in accord- ance with the accepted meaning of this term. Ordinarily when we speak of a wave, a water wave, for example, we mean a peculiar kind of motion. If the cross section of the surface of the water at a given instant is a simple harmonic 236 THEORY OF WAVE MOTION curve, we should properly speak of this curve, not as the wave, but as the instantaneous profile of the wave.* It is characteristic of a wave that this profile is in motion. We shall obtain an excellent idea of wave motion by allowing a simple harmonic curve to glide along the a;-axis with a uniform velocity v. We shall call such a wave a simple harmonic wave, and v its velocity of propagation. Let t denote the time (expressed in seconds), and let us assume that v, the velocity of propagation (expressed in feet per second), is positive, so that the wave advances in the direction of the positive a>axis. Let the simple harmonic curve (1) y = asm{—- + a] be the wave profile at the time t. The profile of the wave at the time t = X) (t seconds earlier) was a curve of the same form as (1), but situated farther toward the left. Therefore its equation can differ from (1) only in the value of the phase constant. Consequently we may assume that (2) y = a sin 1^—- + (^oj is the equation of the wave profile at the time ^ = 0. We wish tp find the relation between a, a^, v, X, and t. The nodes of the wave profile at the time ^ = 0, that is, its intersections with the a:-axis, are obtained from (2) by equating y to zero. These nodes (see Fig. 132), the points Mq, Mq, Mq'\ etc., are infinite in number, and the distance between two consecutive ones is equal to J X or one half of the wave length. One of these nodes, Mq say, will be ob- tained by equating — ^ + «o ^^ ^^^^* Since OMq is the A. abscissa of the point Mq and since for this point — h "o = ^' * The wave profile is the cross section which one would obtain of the surface of the water if it were to freeze suddenly while a wave is passing. WAVE MOTION 237 we shall have (3) Oitfo = -|^ IT After t seconds, the wave profile has moved from its original position MqAqMq' • • • to MAM profile is now given by (1). The nodes of this profile are the points of the curve for which The equation of the wave 2 7rrc + a = JCTT, Fig. 132 where Jc is either equal to zero or to a positive or negative integer. Consequently these nodes are the points of the a;-axis whose abscissas have the values x = Xa IT + 1 k\ where Jc = 0, ±1, ±2, ±3, One of these points is the new position M occupied by the point Mq as a consequence of the motion of the wave profile from MqAqMq^ ... to MAM' •••. Since OM is the abscissa of M^ we shall therefore have (4) OM^-^ + ikX, where A; is a definite positive or negative integer or zero whose precise value remains to be determined. But if V is the velocity of propagation of the wave in feet per second, every point on it moves through a distance of vt feet in t seconds. Therefore (5) OM- OMq = vt. If we substitute in this equation the values (3) and (4) foi OM and OMq, ^ (6) ii:ia_4n+ i^\ = i,^. ^ ^« . 1 Z.X 27r"2^+^~^^ 238 THEORY OF WAVE MOTION This equation must be true for all values of t. In particular it must be true for <= 0. But for ^ = we have a= Wq, so that the equation involves a contradiction unless ^ = 0. Consequently the integer h which appears in equations (4) and (6) must be equal to zero, and we have whence (7) 2irvt If we substitute this value of a in (1), we find (8) y = a^m\-^(x-vf)+aA as the general equation of a simple harmonic wave of amplitude a and wave length \, whose velocity of propagation is equal to v. We may still speak of Uq as the phase constant. It is the phase constant of the wave profile at the time t = 0. The phase constant of the profile curve at any other instant may be computed from (7). If in (8) we assign a fixed value to t, we obtain the equa- tion of the wave profile at that instant. Let us instead assign a fixed value to rr, so that y becomes a function of t alone. In the case of a water wave this would amount to a study of the upward and downward oscillations of a cork floating upon the water. We shall naturally inquire as to the length of time which is required to complete such an oscillation. This time is called the period and may be denoted by T. Clearly T is the time which must be added to t so as to change the argument of the sine function in (8) by ± 2 TT. Now if we increase t hy T without changing x^ this argument changes by , and this will be equal to 2 7rif vT . ^ \ — =1, or r= -. \ V GENERAL HARMONIC MOTION 239 Therefore, if v is the velocity of propagation^ \ the wave lengthy and T the period^ we have the relation (9) X=Tv, Instead of (8) we may now write (10) y = a8m[2^(|- !) + «„] as the general equation of a wave of length \, period T^ ampli- tude a, and phase constant a^. We see finally that a simple harmonic wave has for its profile at any moment a simple harmonic curve^ and that every point upon it oscillates up and down in accordance with the law of simple harmonic motion. EXERCISE LIX If the units of length and time are feet and seconds respectively, com- pute the wave length, period, amplitude, and phase constant of each of the waves represented by the following equations : 1. , = 3sin[2.(|-±) + |]. 3. y = a sin (bx + ct -\- d). 108. General harmonic motion. A simple harmonic motion is the simplest case of an oscillatory phenomenon, and many natural periodic motions may be adequately described as simple harmonic motions. We have already given some ex- amples of such cases. But very frequently the motion, although of an oscillatory character, is more complicated. In the case of a water wave, for instance, we see that the profile of the wave is not a simple harmonic curve, but that there are smaller waves (ripples so to speak) running along the backs of the larger ones, thus complicating the motion. Simple experiments show that the sound waves produced by a tuning fork are very approximately represented by simple harmonic motion; but other musical instruments, such as the 240 THEORY OF WAVE MOTION violin, the piano, the human voice, produce sound waves which resemble the more complicated water waves. A tuning fork which makes 129 oscillations in a second causes a certain simple harmonic motion of the air particles whose period is yjgth of a second and which produces a certain tone usually denoted by C. If the same note is struck on the piano, it is found that the principal part of the motion of the air particles again has y^-gth of a second as its period. But the motion is not simply harmonic. It is a combination of this fundamental motion with one twice as fast, with another three times as fast, and so on. In other words, the motion of the air particles is given by an equation of the form (1) y=.a^^m{^t^a^^-a^^m{-^t^a^ where the period of the first and principal term is T^ that of the second ^ jT, that of the third | T^ etc. This is not the place to discuss details of the theory of sound. Our purpose in entering upon this theory at all was merely to explain one of the many instances in which sums of simple harmonic functions of the form (1) present themselves as indispensable. We wish to learn how the various terms in (1) combine. For that purpose the length of the period jT makes but little difference. We shall therefore put T=2 7r since the formulae will then assume a somewhat simpler appearance. Then (1) reduces to (2) y = «! sin (t + ttj) + ^2 sin (2 ^ + Og) 4- fl^s sin (3 ^ + ctg) +••• . Now each of these terms may be expanded in accordance with the addition theorem (Art. 79), so that ttj sin (t -f ttj) = a^ cos a^ sin t-\- a^ sin a^ cos ^, ^2 sin (2 t + ttg) = ^2 ^os H si^ 2 ^ 4- ^2 sin a^ cos 2 t, etc. GENERAL HARMONIC CURVES 241 Consequently, if we introduce new constants J.^, ^2' "• ' -^v -^2' ••• ^y putting A^ = a-^ sin a^, A^ = ag sin Ogi • * • » j5j = a-^ cos «!, -Bg = ^2 ^^S «2i • • • 9 equation (2) becomes .^ ^ = ^2 cos <4-^ cos 2 f + ^3 cos 3 i+ ••• + B^ sin ^ H- ^2 si^ 2 « + ^3 sin 3 « H . Let us call the fixed point, with which the moving point would tend to coincide if the amplitudes a^ a^<^ etc., of all the simple harmonic motions of (1) were to approach the limit zero, the center of oscillation. We have tacitly assumed so far that the center of oscillation was the origin of coordinates. Let us drop this specialization, and let J Aq be the fixed value to which y would reduce if all of the oscillations were to disappear ; that is, let J A^ be the ordinate of the center of oscillation.* Then we must add ^ A^ to the right member of (3), so that we obtain finally ^A^ y = J Aq -h ^1 cos t-\- A^ cos 2t + A^ cos 3 f + • • • + ^1 sin t-{- B^^in2t^ B^ sin 3 i H as the typical equation of a general harmonic motion. As in the case of a simple harmonic motion, we may make a graphical record of this motion. Suppose, for instance, that the motion to be in- vestigated is the vibration of a metallic wire. We attach a light pen to the wire so as to enable it to write upon a strip of smoked glass. If the glass be left at rest while the wire is caused to vibrate, the pen will merely describe a straight line upon the smoked glass. If, however, the glass be moved with a rapid uniform motion at right angles to the direction of vibration of the wire, there will appear as record a wavelike curve. This curve will belong to the class considered in the next article. 109. General harmonic curves. If we put x in place of t in equation (4) of Art. 108, we find (Vi 2^ = 2 ^0 + ^1 ^^s ^ + ^2 ^^s 2x-\- A^ cos 3 a; 4- ••• H- B^ sin a; + -^2 sin 2 a; + ^3 sin S x + • • • . * The reason for denoting this quantity by ^ Aq rather than Aq will appear later. (See Art. 112.) 242 THEORY OF WAVE MOTION The curves which are obtained as a result of plotting an equation of this form are called general harmonic curves and are capable of an extraordinary variety of forms. In fact it can be shown, by methods involving the integral calculus, that an infinite series of the form (1) may be found to repre- sent almost any continuous curve, and even extensive classes of discontinuous curves *. In this book, however, we are con- cerned only with sums of the form (1) involving a finite number of terms and the curves represented by them. The name harmonic curves will be understood to apply only to such curves. We proceed to discuss an example. Let us plot the curve whose equation is (2) y = sin a? + sin 2 ar. We begin by drawing the two familiar curves (3) y^ = sin x and 7/2 = sin 2 x, the two dotted curves of Fig. 133. From these curves it is easy to con- struct the curve (2). For we see from (2) and (3) that y = 2/1 + 2/2 for every value of x. If then we find the ordinate of each of the two dotted curves for a given value of x, their algebraic sum will be the ordinate of a point on the required curve. The resulting curve is in- dicated in Fig. 133 by a full line. A few points of this curve may easily K /;''\ \"^N /' ^N p. 'i N +a; FiQ. 133 * Such series are usually called Fourier's series in honor of the great mathe- matical physicist who first stated, and in part proved, the above theorem. The first rigorous proof was furnished much later by Dieichlbt. TRIGONOMETRIC INTERPOLATION 243 be obtained by inspection. For x = 0, y^ and 7/^ are both zero, and there- fore also y = y^-{- y^ = O. Consequently the point is on the curve. For X = 7r/2, yi = 1 and y^ = 0, so that y=l and the curve passes through the point A. For values of x between 7r/2 and tt, y^ is negative so that 2/1 + 2/2 ^^i^l b^ l^ss ^^^'^ ^r Consequently the full line curve in this in- terval lies below the corresponding portion ^jB of the curve y^ = sin x. For a certain value of x in this interval (determined by the equation sin a: -f sin 2 a; =: 0) ^^ and y^ will be numerically equal but opposite in sign, so that at that point the curve y = sin x -f sin 2 a: will cross the a;-axis. This is the point L of Fig. 133. For any value of x we may obtain y^ and y^ by measurement from the two dotted curves. If we form the sum of these two quantities with due regard to sign, we find the corresponding ordinate of the required curve.* EXERCISE LX Plot the following harmonic curves : 1. y = 2 sin X + sin 2 x. 3. y = sin a: -|- cos 2 x. 2. y = sin x + ^ sin 2 a:. 4:. y = 5 sin x + sin 4 x. 110. Harmonic analysis or trigonometric interpolation. We have 'seen how a number of simple harmonic curves may be compounded into a single general harmonic curve. It often happens that a curve is given, actually drawn out on paper, as for instance in the case of a self-recording barometer or thermometer. If the curve is of a periodic character, the question arises whether it may be regarded as a harmonic curve, that is^ whether it is possible to compound it out of a number of simple harmonic curves by the method of Art. 109. And if so, the problem presents itself to actually find the component simple harmonic curves. The process of solving this problem is known as harmonic analysis and is of great importance in many branches of pure and applied mathematics. Let us suppose that the given curve is periodic, so that it * MiCHELSON and Stratton have devised a machine for performing mechan- ically the operation of combining a number of simple harmonic curves. This machine is also capable of performing the inverse operation discussed in Art, 110. For this reason it has been called a harmonic analyzer. 244 THEORY OF WAVE MOTION consists of an infinite number of equal pieces, and let the length of one of these pieces, the wave length, be equal to X. If X is less than 2 tt, we may, by a process of magnification or stretching, replace the given curve by another one similar to it whose wave length is just equal to 2 tt. If we can solve our problem for this second curve of wave length 2 tt, it will be easy to solve the corresponding problem for the original curve. If X is greater than 2 tt, a process of compression enables us to reduce the problem again to the case of a curve of wave length 2 tt. We may therefore assume that the wave length of the given periodic curve is equal to 2 tt without reducing, by this assumption, in any essential fashion the general applicability of our results. The ob- ject of this assumption is merely to simplify the resulting formulae. Let PqP^P^ '" (Fig. 134) be the given curve of wave length OA = 2 tt, and let us divide OA into an odd number, +; J n^ -^^ A V2 2/3 ?A \ X, X. A Vo / ^' X2 Xa X, \ f Po f \ % Ve A t i" ^ +« ^a Fig. 134 say 2 m 4- 1, of equal parts. Not counting A^ there will then be 2m-\-l points of division; namely, 0, Xj, X^, •" X^^^. In Fig. 134 we have made 2m + l = l. At these points of division we construct the ordinates OPq., X-jP^^ JL2-P2' *** of the curve. Let y^^ y^, Vi-^"' Vim ^^ these ordinates, eacl with its proper sign prefixed. On account of the periodic character of the curve, the ordinate at A will be the same as that at 0. This is the reason that we did not count A as TRIGONOMETRIC INTERPOLATION 245 one of the points of division. If we had included A^ we should really have been counting twice. We can always .find a harmonic curve involving terms in x^ 2x, Sx, "' mx, Vvhich passes through the 2 m + 1 points Pq, Pj, Pg' ••• ^2m' ^^^^ *^® general equation of such a harmonic cuiive is » .^ . ^ = ^ Aq + A^ cos x-\- A^ cos 2x-\- • • • + J.^ cos mx -\- B^sin X + B2sin2x-\' • • • -|- B^ sin mx, and therefore contains 2m + l coefficients Aq, ^j, ••• ^, 5j, ••. P^, which may be determined in such a way as to make the corresponding curve (1) pass through the 2 m + 1 given points. In fact, the curve (1) will pass through the point Pq, if the value of y obtained from (1), for a: = 0, is equal to the ordinate i/q of the given point Pq ; that is, if (2) I/O = 1^0 4- A + ^2+ - +-^m. 2 TT The abscissa of P, is OX. = . Theref c ^ ^ 2m+l (1) will pass through Pj, if the value of y obtained from (l),for:.= -l^, 2 m-h 1 point Pj ; that is, if 2 TT The abscissa of P, is OX. = . Therefore the curve ^ ^ 2m+l irough Pj, if (1), for X = - — — — , is equal to the ordinate «/, of the given 2 m-h 1 27r . . „„ 2.27r (3) ^1 = 1 ^0 + ^1 cos — - + A cos 2m+l ^ 2w+l W • 2 TT + ... + J.^cos 2m-\-l 27r . „ . 2.27r H-^jSin- — ^^ + P2sin 2m+l ^ 2m+l W . 2 TT + ... +^„sin 2m + l In the same way we find that the curve (1) will pass through the points Pg, Pg, ••. P^m ^^ ^^® following additional equations are satisfied ; 1 246 THEORY OF WAVE MOTION ^ A , A 47r ,4 2 47r , 2m-}-! ^ 2m + l ^ 2m + l , , E» . m • 4:7r (4) 2w + l' 2w4-l _ 2m4- 1 2w + l W • 2 WTT + ... H-^^sin 2w+l Now the equations (2), (3), and (4) are together 2 m + 1 in number, and involve 2m +1 unknown quantities, viz., the quantities, Aq^ A^ • - - A^^ B^, . . . B^. The coefficients of these quantities (sines and cosines of known angles) are known numbers, and the quantities y^, y^, ... ^^m') ^^^ ordinates of. the given points Pq, Pj, ... Pgm' ^^® known. We may there-, fore, in general, solve these 2 m + 1 equations for the 2 m + 1; unknown quantities and then substitute the values obtained, for Aq, ^j, .•• A^, B^, ... B^ in (1). The harmonic curve, obtained by plotting (1) will then pass through the given points. If the given curve is fairly smooth, and if the points Pq Pj, ... P^m be taken close enough together, that is, if m b< chosen large enough, we may find by the above method harmonic curve which may be regarded as replacing th< original curve everywhere with any desired degree of aj proximation.* The following example will illustrate this method. Let y he a, pei odic function of x, of period 2 tt, such that the vajues of y which corre spoud to X = 0°, 72°, 144°, 216°, 288° * A more precise formulation of this statement must be left for a much latej stage of the student's mathematical career. TRIGONOMETRIC INTERPOLATION 247 are respectively y=Q, +2, + i - i, - 2. Our general theory tells us that we may find an expression of the form y = \Aq-{- ^icosa; + ^2 cos 2 a: + ^1 sin a: + ^2 sin 2 x, which assumes the five values assigned to y for the five given values of x. Moreover we have the following five equations for the five unknown co- efficients Aq, Ai, A2, Bi, B2'. Q = IAq + Ai + A2, 2 = ^ ^0 + ^1 cos 72° + ^2 cos 144° + Bi sin 72° + B2 sin 144°, I = ^ ^0 + ^1 cos 144° + ^2 cos 288° + Bi sin 144° + B2 sin 288°, - 1 = ^ ^0 + ^1 cos 216° + .42 cos 72° + Bx sin 216° + B^ sin 72°, - 2 = I ^0 + ^1 cos 288° + ^ 2 cos 216° + ^1 sin 288° + B2 sin 216°. Since we have sin 216° = sin (360° - 144°) = - sin 144°, cos 216° = cos (360° - 144°) ^ cos 144°, sin 288° = sin (360° - 72°) = - sin 72°, cos 288° = cos (360° - 72°) = cos 72°, the above equations may also be written as follows : (1) J ^0 + ^1 + ^2-0, (2) \ Ao + ^1 cos 72° + ^2 cos 144 + 5^ sin 72° + ^3 sin 144° = 2, (3) ^ ^ + ^ 1 cos 144° + ^ 2 cos 72° + B^ sin 144° - B^ sin 72° = |, (4) \Aq + A^ cos 144° + ^2 cos 72° - Bi sin 144° + B2 sin 72° = - 1, (5) \Aq + Ai cos 72° + Ai cos 144° - Bx sin 72° - Bi sin 144° = -2. From (2) and (5) we find by addition (6) ,\Aq + Ax cos 72° + A2 cos 144° = 0, and similarly from (3) and (4), (7) ^ ^0 + ^1 cos 144° + J 2 cos 72° = 0. From (1) we have ^4o =- ^1 - ^2 which, substituted in (6) and (5), gives (8) Ax (cos 72° - 1) + ^2 (cos 144° - 1) = 0, Ax (cos 144° - 1) + ^2 (cos 72° - 1) = 0. If we multiply both members of the first of these equations by cos 72° —1, those of the second by — (cos 144° — 1), and add, we find (9) ^i[(cos 72° - 1)2 - (cos 144° - 1)2] = 0. From the table of natural functions, we find to two decimal places cos 72° = 0.31, cos 144° = - cos 36° = - 0.81. 248 THEORY OF WAVE MOTION Therefore cos 72^ - 1 = - 0.69, cos 144° - 1 = - 1.81, so that the coefficient of u4j in (9) is not equal to zero. Consequently we conclude from (9) that A^ = 0. According to (8) and (1), we must then have also ^.j = 0, ^^ = 0. If now we put Af^= A^ = A^^ 0, in (1) to (5), these five equations reduce to the following two : B. sin 72° + B„ sin 144° = 2, rio) 1 -T- 2 ^ ^ B^ sin 144° - B^ sin 72° = |. From these equations we eliminate first B^ and then B-^^ , giving (sin2 72° + sin2144°)5i = 2 sin 72° + | sin 144°, ^ ^ (sin2 72° + sin2 144°) ^3 = 2 sin 144° - \ sin 72°. From the table of natural sines we find, correct to two decimal places, sin 72° = 0.95, sin 144° = sin 36° = 0.59, so that sin2 72° + sin2 144° = 0.90 + 0.35 = 1.25. Consequently, equations (11) become 1.25^1= 1.90 + 0.30 = 2.20, 1.25i?2 = 1-18 -0.48=: 0.70; whence finally (12) ^1=1.76, ^2 = 0.56. Since we have already found Afy = A^= A^-= 0, the function which we were seeking is (13) y = 1.76 sin a; + 0.56 sin 2 a;. In order to check our result we may substitute the five given values of X in (13) and verify that the corresponding values of y are actually those which were originally given. That equation (13) gives y = for a: = is obvious. For x = 72° and for x = 144°, we find from (13) y = 1.76 X 0.95 + 0.56 x 0.59 = 1.67 + 0.33 = 2.00, and y = 1.76 X 0.59 + 0.56 x (- 0.95) = 1.04 - 0.53 = 0.51, respectively, checking to within one unit of the last decimal place computed. To check the other two pairs of values requires no additional computation. In this example, the values of y iov x = 144° = 180° — 36° and for aj= 216° = 180° + 36° were numerically equal but opposite in sign. The values of y iov x= 72° = 180° — 108° and for x = 288° = 180° + 108° were also numerically equal TRIGONOMETRIC INTERPOLATION 249 but opposite in sign. It is owing to this circumstance that J.Q, J-i, and A^ turned out to be all three equal to zero. Having become aware of this fact, we may abbreviate our work very much by at once equating Aq, A^, A^, etc., to zero, whenever the given values of the function are numerically equal but opposite in sign for those among the given angles which differ numerically by the same amount from 180°. Similarly, if the given values of the function y are equal numerically and in sign for all of those among the given angles x which differ numerically by the same amount from 180°, the expression for 7/ will contain no sine terms ; that is, ^j, B^t -Sg, etc., will all be equal to zero. EXERCISE LXI 1. Find a periodic function, involving only the angles x and 2 x, which assumes the values 2/ = 0, + 2, + 1, - 1, - 2, for X = 0°, 72°, 144°, 216°, 288°, respectively. Compute the coefficients to two decimal places. 2. Find a periodic function, involving only the angles x and 2 x, which assumes the values 2/ = + 2, + 1, - i, - i + 1, for x = 0°, 72°, 144°, 216°, 288°, respectively. Compute the coefficients to two decimal places. 111. Theorems leading to the general solution of the problem of trigonometric interpolation. We have shown in Art. 110 that the problem of trigonometric interpolation may be re- duced to that of solving a system of 2 m + 1 equations of the first degree with 2m-\- 1 unknowns. But we can accomplish much more than this. We shall derive elegant and con- venient formulae for the solutions of these equations, en- abling us to find the values of the coefficients Ak and B^ by a direct and simple process. But before we can do this, we must prepare the way by proving some theorems necessary for this purpose. 250 THEORY OF WAVE MOTION We begin by proving that the following formula (1) sina + sin(a + + sin(« + 2 04- ••• -{-s'm(a + mt) / . mt\ . (m-\- l)t sm( a + —]sm^ — ^|— ^ . t sm- published by Euler * in 1743, is true for all values of a and t, provided that sin - is not equal to zero. Proof. Let us denote the sum in the left-hand member of (1) by 8^. If we multiply ^^ by 2 sin -, we shall have A 2 s„ sin - = 2 sin a sin - + 2 sin (a + t) sin - A ''A A + 2 sin (a + 2 t) sin - + ••• +2 sin (a + mt) sin |. A A Every term in the right member of this equation contains a product of two sines, and may therefore be expressed as a difference of two cosines by means of formula (4) of Art. 82 ; that is, sin a sin yS = I [cos (a — yS) — cos (a + /S)]. We find, in this way, 2 8^ sin - = cos {a — 1 ~ ^^^ (^ + i A + COS {a 4- \ — ±l^ . t sin- 252 THEORY OF WAVE MOTION a formula which may also be obtained directly by a process strictly analogous to the one employed for the proof of equation (1). Formula (2) is also due to Euler. If in equations (1) and (2) we put a = 0, we find sin — sin ^^ — - — — (3) sin f+sin 2 ^+ sin 3 ^H +^\nmt = and sm^ cos -- sin -^^ — ' ^ 2 2 (4) l-|-cosf + cos2f/H-cos3fH \-co^mt = -^ * sin- Let us subtract J from both members of (4). We obtain the formula mt . (m 4-1)^ cos — sin -^^ ' — — 1 2 2 1 - + cos 1 4- cos 2 ^ H h cos mt = ^ ' t A sin- o mt . (m-^l)t . t . f , 1\^ , . t . t ' 2 cos — sin ^ ^ sm - sin [m-i--Jti- sin- - sin- 2 sin - 2 sin - 2 2 where we have made use of one of the equations '(4) of Art< 82. The numerator of the last fraction obviously reduces ti its first term, so that we find finally sin ^^ -! — ^ 1 2 (5) - +cosf+cos 2*+cos St-i — -\-cos mt = r 2 2 sin - a formula which was known to Snellius in 1627. The angle t which occurs in all of these equations ma^ have any value whatever excepting only those values foi which sin - is equal to zero ; that is, excepting the valueJ Li 2 Ajtt, where h is an integer. We shall now apply these f ormuh TRIGONOMETRIC INTERPOLATION 253 to the particular case when ^ is a commensurable fractional part of the entire circumference, so that n where both h and n are positive integers and where n>l. We shall further put m = n—\. We may express these assumptions more concretely as fol- lows. Let us divide the circumference of the circle into n equal parts, where n> 1. Then -^ will be the angle sub- tended by one of these parts. The smallest multiple of this angle which is equal to a complete circumference is of course the nth.. Therefore the angles iz, 2?^, 3-^,...(ri-l)^ n n n n are all distinct, and we propose to find the sum of their sines as well as of their cosines. We consider next the angle 2 = and all of its multiples up to (w — 1) and calcu- n n n late the sum of their sines and of their cosines. In general, we consider the angle k ^^^ = - — - and its multiples 2 — —, n n n S- — —'" (ri — 1) — — and compute the sum of their sines and n n of their cosines. According to (3) we have (putting t = and m = n — l') sm h sm 2 \- sm 3 h • • • -I- sm (w — 1) n n n n . (n— l)k7r . J sm -^^ '^ — sin fCTT n . kir sm — n a formula which will be valid unless sin — is equal to zero; 254 THEORY OF WAVE MOTION that is, unless k is an integral multiple of n. Excluding this case for a moment, we see that the right member is equal to zero since sin kir occurs as a factor in the numerator and since the sine of any integral multiple of ir is equal to zero. Consider now the excluded case when ^ is a multiple of n. Then every term on the left member is individually equal to zero. We see therefore that (b) iSj^ = sm f- sin 2 f- sin 3 h • • • n n n + sin {n — 1) = 0, n for every value of h^ whether k is divisible by n or not. We may prove in the same way, making use of equation (4), that ^r7^ /> -I , 2 kTT , o 2 kir , o 2 ^TT , (7) Ch=l-hcos |-cos2 |-cos3 (- ••• n n n + cos (n — 1) = 0, n ifk is not a multiple of n. If A; is a multiple of n, all of the angles , 2- , etc., n n are integral multiples of 2 tt, so that each of the cosines which appears in OJ. is equal to unity. There are n terms in Cj,. Therefore (8) Ok= 1 + cos h cos 2 h ... 4-cos (n—l) =n, n n n if k is a multiple of n. 2 kir 2 /tt Let us now consider two angles of the form and , n n and denote by Oki the sum (9) Cki—i^-\- cos cos h cos 2 cos 2 n n n n + ••• + cos (n— 1) cos (n — 1) ^ n n TRIGONOMETRIC INTERPOLATION 255 A product of two cosiaes may be expressed as a sum by means of formula (3) of Art. 82 ; that is, cos a cos y8 = |^[cos (a — /3) + cos (a + y8)] . Therefore we may write, if we apply this formula to each term of C^i^ [_ n n J + J COS 2—^ ^— + cos2 ^ ^ ^ [_ n n ^ + + I rcos (n - 1) ^(^ - ^)" + cos (n - 1) ^(^ + 0^ Now the sum of all of the terms in the first column may be equated to \Ck-i\i we again make use of the notation 0^ de- fined by equations (7) and (8). Similarly we observe that the sum of the terms in the second column is ^ C^^i. Therefore (10) (7« = KC.-,+ <7w)- Consider now the expression, analogous to (7^^, (11) ^Ski = sin sin f- sin 2 sin 2 1- • • . n n n n + sin {n — 1) sin (n — 1) • n n Since we have (see Art. 82) sin a sin y8 = 1^ [cos (a — /3) — cos (a + y8)], we find ^ki H[. ^« = — ^ if k—l is not divisible by n and k + 1 is divisible by w, Sf^i = il k — I and k -}- 1 are both divisible by n. (^Sk, Oi) = for all values of k and l. TRIGONOMETRIC INTERPOLATION 257 112. General solution of the problem of trigonometric inter- polation. We are now ready for the solution of the problem of trigonometric interpolation. We divide the circumference into 2 m H- 1 equal parts by means of the angles — 2 TT 47r 6 TT 4 mir ' 2m + l' 2m+1' 2m + l'"'' 2m + l' and denote the corresponding values of the function y by ^0' VV Vv ^3' ••• V'hm respectively. The function y = \A^-\- A-^ cos x-{- A^ cos 2x-\- • • • + ^,„ cos mx + j5j sin 2; + ^2 si^^ ^ ^ + •*• + -^m sin wa; will actually assume these values, if the 2 w + 1 quantities Aq, J.J, ..., J.^, 5j, ^2^ "'1 ^m ^1*6 chosen so as to satisfy the 2 m 4- 1 equations 3/0 = 2 A + -^1 + A + ••• + -^m. 2 TT ^i=2A + Acos- — -+ ... +J.^cosm (1) 2m + l "* 2m + l + Asin- — -+ ••• +^mSinm- — -, 2mH-l 2mH-l ^2m = 2 A + A cos — - + ... + An cos m 2m+l 2m+l + A sin 7^ —- + ••• 4-^^sin w- — , 2m-\-l 2 m-^1 all of which may be expressed by the single equation (2) yv = i A + A cos —- + ••• + A cos m 2m+l 2m+l 2 I/TT . . T^ . 2 Z^TT H- ^1 sin 4- ••• + B^ sm m 2m + l 2w-hl if we think of v as assuming in succession the 2 w + 1 values 1^ = 0, 1, 2, 3, ..., 2m. Let r be any integer between and m, let us multiply Vq oy 1, ^1 by cos — -, y^ by cos 2 -— , ..., y^ by m + 1 772 + 1 258 THEORY OF WAVE MOTION cos V -, ..., «/2^ by cos 2 m —^ -, and add. If we m + 1 2m + 1 make use of equations (1) and the notations introduced in Art. Ill, we find (3) y^+y^00S^^+ ... +2,,„C0S2™^|^ From Art. Ill, (17), we know that all of the quantities (^Sk^ Ci) are equal to zero. Since r was chosen as an integer between and w, r can be divisible by 2?7i + 1 (which num- ber corresponds to the n of equations (15), (16), (17) of Art. Ill), only if r = 0. In that case we shall have, accord- ing to (15) of Art. Ill, Or= 0^ = 71=- 2m + 1, and according to (16), Art. Ill, ^10 ~ ^20 = ••• = Wio ~ ^» since none of the numbers 1, 2, •••, m are divisible by 2m-\-l. Consequently, equation (3) reduces to (4) ^0 + yi + ^2 + - + y2m = Mo(2 ^ + 1) in the case r=0. If r>0, Cr=0. Moreover, (7^ will be zero for all values of k for which neither k — r nor k -\-r is divisible by 2 w -h 1. But ^ as well as r can at most be equal to m, so that neither k — r nor ^ + r is ever as large as 2m-\-l. The only case therefore in which one of these numbers can be divisible by 2 m -hi is when k — r. Thus we have, according to (16), Art. Ill, Ckr = iov k different from r, ^ _ 2m-hl ^ TRIGONOMETRIC INTERPOLATION 259 Consequently equation (3) gives, for r > 0, (5) ^'o + j/icos^— ^ + y,cos2 2^^^+... o 2r7r 2m + l . + Vo^ cos 2 m = — ! — A.^ ^2'" 2w + l 2 and we notice that equation (4) may be thought of as included in (5) for r = 0.* We find therefore the formula 2 7-7r . ('^ ^^=2^^+^^^^^rirTi+^^'"'' 2w + l + Vorr. cos 2 m - (r = 0, 1, 2, ...,7/i), enabling us to compute A^, A^^ ••-, J.^ in terms of the given quantities ?/o,«/i, -", y^^. It remains to find a corresponding formula for B^. In order to do this we return to equations (1), multiply them m order by 0, sm- -^ sin 2- -,..., sinzw -, and add. This gives ^ism- -— + ?/2Sin2- —}■ + ••• + y2mSm2w — 2 m -\- 1 2 m -\- 1 2m + l = i^.S', + A,(S,, Ci) + A(^.. ^2) + - + An(^., 6U + A^lr + ^2^2r H ^^ ^m^mr' All of the terms in the right member of this equation reduce to zero except B^S^^^ which, according to Art. Ill, (17), be- comes equal to B^ ^ — Consequently we find a) ^.= 2m + l Vi sm- + v„ sin 2 -f- • • • ^^ 2m + 1^^ 2m +1 • o 2r7r 1 + ^2.sm2m^--^J (r=l, 2, 3, ...,w). * It was for this reason that the notation 5 ^0 > rather than Aq , was chosen in Art. 108 for the constant term of the harmonic fmiction. 260 THEORY OF WAVE MOTION Equations (6) and (7) furnish the complete solution of the problem of trigonometric interpolation. The coefficients of the harmonic function 1/ = IAq-\- A^Gosx-{- A^cos2x-\ — -\-A^Gosmx + B^ sin X -{- B^ sin 2 X -^ • '• -{- B^ sin mx which assumes the arbitrarily/ assigned values for the 2 m -{-1 equidistant values of the argument — ^ TT 4 TT imir ^~ ' 2m+l' 2w + l' ' 2m+ 1' are given hy equations (6) and (7). EXERCISE LXII 1. Solve the problem treated in detail in Art. 110 by the method of Art. 112. 2. Solve the problems of Exercise LXI by the general method of Art. 112. INDEX References are to pages Abscissa Accuracy, degree of, in cal- culation of five-place tables .... of measurements .... Addition theorems . 184, 185, Ambiguous case of oblique tri- angles Ill- Amplitude 228, Analysis, Harmonic .... Angle, betw^een directed lines cardinal 145, general horizontal inclined initial side of measurement of .... natural measure of . . . negative of elevation or depression . positive quadrant of an standard position of . . . subtended terminal side of .... vertical Arc, length of functions of 5 Area of a triangle . . . 84, 91 Aristotle ^ Aryabhata 155, Auxiliaries S and T . . 61, Axis of a coordinate system rBall . •. Base of a system of logarithms Bearing 138 37 59 1-5 212 114 232 243 204 173 133 73 73 135 3 165 134 73 134 145 136 74 135 73 165 152 ,92 47 192 198 138 47 43 74 Briggs 47 Biirgi 38 Cajori 39 Center of oscillation .... 241 Chaining 74 Characteristic, of a loga- rithm 47, 49 Checks 30, 70 Chord 155 Cof unctions 21 Cologarithm 54 Compass 5, 74 Complementary angles, func- tions of 20, 169 Composition, of displacements, velocities, forces and vectors 126 Coordinates 138, 139 Cosecant, definition of . . 14, 140 graph of 163 line representation of . . . 154 Cosecant Curve 163 Cosine, definition of . . 14, 87, 140 graph of 161 line representation of . . . 152 Cosine Ciurve 161 Cosines, law of 85 Cos (aiP) 184,186 Cos A ± cos 5 188 Cos 2 a 189 Cotangent, definition of . . 14, 140 graph of 163 line representation of . . 153 Cotangent Curve 163 Cot (a ± p) 185, 186 Crelle's Tables 38 Crest of a wave 232 261 262 INDEX Decimal places, number of . . 34 Departure 75 Descartes 130, 159 Difference of latitude ... 75 Difference of two angles, func- tions of 186 Difference of two cosines . . 188 of two sines 188 Directed lines 201 line segments 202 Direction cosines 209 Dirichlet 242 Displacements 126 Distances, measurements of . 1 Double angles, functions of . 189 Eichhorn slide rule ... 66, 95 Equations, trigonometric 222-225 Errors, gross 68 small 70, 71 Eudemus 7 Euler 250 Extraction of roots by loga- rithms 55 Fermat 159 Fink 101 Forces 127 Form ratios of a triangle . . 98 Foster 63 Fourier 242 Framework of a calculation . 69 Fuller's slide rule 65 Function, definition of . . . 13 graph of 156-163 Functions, of acute angles . 14 of cardinal angles . . . 147 of complementary angles 20, 169 of double angles . . . . 189 of a general angle . 137, 140 of half angles 190 of obtuse angle .... 89 of small angles or angles near a right angle ... 60 of sum or difference . 184, 186 of supplementary angles . 89 of symmetric angles . . . 167 Grade 12 Gradient 12 Graphic method ..... 8 Graphs, of functions, some of whose values are given . 156 of simple algebraic func- tions 158 of the trigonometric func- tions 159-163 Greek alphabet 132 Gross errors 68 Gunter . 62 Gunter's scale ...... 61 Half angle formulae .... 95 Half angle, functions of . . 190 Harmonic analysis 243 Harmonic curves, simple . . 231 general 241 Harmonic motion, simple . . 226 general 239 Heights and distances . . . 118 Hero 92 Hero's formula 92 Hipparchus 192 Horizontal angles, lines and planes 73 Identities, involving a single angle 149 involving several angles . 185 Inaccuracies of measurements . 1 Inclined lines, planes and angles 73 Index of Refraction . . . . 129 Index laws 41 Indices, fractional, negative and zero 40 Inskip's Tables 31 INDEX 263 Interpolation, in use of tables. 26 trigonometric 243 Inverse functions 217 Latitude 78 Law, of cosines 85 of sines .... 84, 96, 212 of sines generalized . . . 212 of tangents 101 Level 5 Level chaining 80 Leveling 74 Limits of variation of the func- tions 147 Limits i^^ and *^ ... 195 e e Line representation of func- tions 151 Lines, directed 201 Line segments, directed . . . 202 Logarithms, arrangement and use of the tables . . 51-53 base of a system .... 43 calculation with .... 47 characteristics of . . . 47, 49 common 47 definition of 43 extraction of roots by . . 55 mantissa of 47, 48 of numbers 47-56 of trigonometric functions 57 properties of .... 44, 45 Logarithmic calculations in- volving negative numbers 56 Logarithmic scale 61 Longitude 78 Mannheim 64 Mannheim's slide rule ... 63 Mantissa .(.... 47,48 Mariner's compass .... 75 Measurement of lines and angles 1, 3 Menelaus 192 Meridian 74 Method of mathematical induc- tion 170, 184 Michelson 243 Mobius 213 MoUweide 105 MoUweide's equations . . . 104 Napier 38 Nasir Addin 84 Natural functions 25 Natural unit of circular meas- urement 165 Navigation 75 Negative angles 134 line segments 202 Newton 105 Object of trigonometry ... 10 Oblique triangles, ambiguous case of 111-114 area of 84, 91, 92 solution of 107-131 theory of 82-106 Ordinate 138 Origin of arcs, primary and secondary 151, 152 Origin of coordinates .... 138 Oscillation, center of ... . 241 Oughtred 62 Parallax 198 Parallelogram of forces . . . 127 Periodicity of trigonometric functions 147 Period of a simple harmonic motion 228 of a wave motion .... 238 Phase 229 Phase constant ... 228, 233, 239 Pitiscus 192 Plane sailing 118 Plane, surveying .... 73^ 118 Plumb line 72 264 INDEX Plutarch 7 Points of the compass ... 75 Pothenot's problem .... 123 Principal values of inverse func- tions 219 Products of sines and cosines . 187 Profile of a wave 236 Projection, definition of . . . 201 theorems on 207 of a broken line .... 208 Proportional parts, tables of . 53 Protractor 3 Ptolemy 87, 192 Pythagoras, generalized theo- rem of 85 Quadrantal formulae . . 171-173 Quadrants, the four .... 145 Radian 165 Radius, of circumscribed circle 96 of escribed circle .... 97 of inscribed circle .... 94 Radius vector 139 Ratios, trigonometric functions defined as 15 Reflection of light 128 Refraction of light .... 129 Relations, between functions of a single angle . . .18, 148 between functions of com- plementary angles . . 20, 169 between functions of supple- mentary angles .... 89 between functions of sym- metric angles 167 Rheticus . 192 Right triangles, of unfavorable dimensions 80 solution by logarithms . . 67 solution by natural func- tions ........ 29 S, the auxiliary Secant curve . . . 61, 198 . . 163 Secant, definition of ... . 14 graph of .163 line representation of . . 154 Sector, area of 166 Segment, area of 166 Sense of a line segment . . . 202 Signs of coordinates .... 138 of trigonometric functions . 145 Sine, definition of . . . 14, 83, 140 graph of 161 line representation of . . 152 origin of name 155 Sine curve 161 Sines, law of . . . 84, 96, 212 Sin (a ± P) 184, 186 Sin 2 a . 189 Sin ^ ± Sin 5 100 Skeleton form of a computation 69 Slide rule 62 Slope 12 Small angles, functions of . . 60 Smoley's tables 31 Snellius 125, 130, 252 Solution of oblique triangles 107-131 of right triangles ... 29, 67 Spirit level 5 Squares, table of 30 Standard position of an angle . 136 Stratton 243 Subtended angle 74 Subtraction formulae .... 186 Sum of two angles, functions of 184, 185, 212 Sum of two sines or cosines . 188 Supplementary angles, func- tions of 89 Symmetric angles, functions of 167 T, the auxiliary .... 61, 198 Tables, explanation of . 25, 30, 57 Tangent, definition of . . 14, 140 graph of 163 line representation of . . 153 Tangent curve 163 I INDEX 265 Tangents, law of 101 Tan (a ± P) 185, 186 Thacher's slide rule .... 65 Thales of Miletus 7 Theodolite 4 Transit 4 Trigonometric functions, defini- tions of ... . 14, 89, 140 table of logarithms of . . 57 table of values of ... . 25 Trigonometric interpolation . . 243 Trigonometry, object of . . . 10 Trough of a wave 232 Variation of functions . . . 146 Vector 127 Velocities 127 Velocity of propagation of a wave 236 Vernier 5 Vertical lines, planes and angles 72,73 Vieta 87, 101 Von Opel 105 Wave length 232 Wave motion ...... 235 Zero, division by 142 ¥ m UNIVERSITY OF CALIFORNIA LIBRARY This book is DUE on the last date stamped below. ocr ^Q '% 5 MAY,, 5 1948 230ct'5Cr ' ft t950 I ll^pr52RL LD 21-100ot-12,'46(A2012s16)4120 'T!^^ W^l^^ R.£C'D LD '25Feb'58ji' REC'D FtB 21 1! LD i9'i8 FEB 9 1959 \ 1 6 1968 3 » l-j; ^ UNIVERSITY Ol^ CALIFORNIA LIBRARY *wtT«f?rr^^iTflf?l?f?gt H ff * niVfff ^l