UC-NRLF $B 527 flS2 f 'i- = In % J! "1. 4.^ 4- .Jh ^4^^ -4^ -^ >^ .^^ ^^ ^ Jfc- ^•*- PLANE TRIGONOMETRY COLLEGES AND SECONDARY SCHOOLS BT DANIEL A. MURRAY, B.A., Ph.D. / » . INSTRUCTOR IN MATHEMATICS IN CORNELL UNIVERSITY FORMERLY SCHOLAB AND FELLOW AT JOHNS HOPKINS UNIVERSITY or THf VNIYER8ITY LONGMANS, GREEN, AND CO. 91 AND 93 FIFTH AVENUE, NEW YORK LONDON AND BOMBAY 190X f % nr mRAL j\c<^ Copyright, 1899, fte lONGMANS, GREEN, AND OU ALL EIGHTS EE8ERVED. First Edition, November, 1899. Keprinted, February and September, 1900, September, 1901, PEEFACE. Although there are already many excellent text-books on trigonometry, there appears still to be room for one which shall avoid the extremes of expansion and brevity. Some of the most thorough and scholarly of these contain a great variety of matters which it is impossible to consider in the time usually assigned to this study in school and college. On the other hand, the expla- nations given in many other works are so meagre that the stu- dent is perplexed and bewildered by the new ideas which are so abruptly forced upon him, and the difficulties of the teacher are greatly increased. The manner of presentation adopted in this volume necessitates more reading matter, and, consequently, a somewhat larger number of pages than is found in many of the recent text-books on trigonometry. This has seemed unavoid- able, however, for the general consensus of opinion among those with whom the author has conferred, is that it is essential to explain in some detail the principles of the science, in order that it may be clearly and intelligently understood by an ele- mentary student. With regard to the scope of the book, it may be said that it deals with the subjects considered in the ordinary course in plane trigonometry in colleges and secondary schools. It discusses the topics usually required for teachers' certificates, for entrance to college, and for examinations in trigonometry in the first year of the college curriculum. It treats of all the topics that one who has taken a few months' course in trigonometry may be reasonably expected to know. VI PBEFACE. Careful consideration has been given both to the early diffi- culties and to the possible future needs of the beginner. The book differs somewhat from other text-books on this branch of mathematics both in the arrangement and in the manner of pre- sentation. The oldest and simplest part of trigonometry, namely, the solution of triangles and the associated practical problems, is concluded before the more general and abstract portions of the study are introduced. The first chapters of the book contain little more about trigonometric ratios and angular analysis than is sufficient to enable the beginner to understand clearly the arithmetical part of the science, and its simple practical applica- tions. This arrangement seems to have several advantages. The subject is rendered far less strange at the beginning, and, by means of practical, concrete examples, the student becomes familiar with the trigonometric functions before proceeding to the more general treatment. His progress is thus made easier and more rapid. Teachers who prefer a wider generality of treatment at the outset, however, can select the chapters in a different order from that followed in the text. An endeavour has been made to introduce the several topics in such a way that the pupil may have, from the very start, an intel- ligent idea of each step in advance, as well as of the ultimate purpose of the study. In some cases, especially in Chapter II. (the first chapter on trigonometry), care has been taken to pre- pare the mind of the learner for the reception of new ideas, by the preliminary solution of easy familiar exercises. Throughout the work the author has endeavoured to make each step clear, and thus to prevent the appearance of that puzzled feeling which has such a depressing influence on those entering upon a new study. On the other hand, he has sought to develop independence of mind and the power of mental initiative on the part of the stu- dent. Suggestions as to practical methods of work are frequently PREFACE. Vll introduced, and summaries are made in several places for the purpose of helping the pupil to get a better idea of the subject as a whole. In the practical applications, marked attention has been given to the graphical method of solution, as well as to the method of computation. The former method serves as a check upon the latter, and affords practice in neat and careful drawing. What is perhaps more important, however, is that the students will thus become accustomed to a method which will be used by them in other studies, and which is often employed in practical work by engineers and others. Logarithms are used almost at the beginning of the study as here presented. For this reason, and in order to avoid making a digression later on, an introductory chapter is devoted to a review on logarithms. Examples, simple ones as a rule, are given in the several articles. Questions and exercises suitable for practice and review on the separate chapters are placed at the end of the book instead of at the ends of the chapters. These collections will be found useful, both in the short reviews that may be required on the completion of each chapter and in the larger and more general reviews. Many of the examples have been taken from examina- tion papers set in Great Britain and the United States. Throughout the work there are many historical and other notes ; and an historical sketch is given in the Appendix. It is believed that some knowledge of the historical development of trigonome- try, and of the men of various times and races who have helped to advance the subject, will interest and stimulate those who are entering upon its study. While writing this book, the author has received many valuable suggestions from Mr. J. A. Clark, B.S., of the Ithaca High School, and from several of his colleagues in the departments of mathe- matics and of engineering at Cornell University. He is indebted Vlll PREFACE, ^ to Dr. G. A. Miller and Dr. J. V. Westfall, of the department of mathematics at Cornell University, for their kind assistance in! the revision of the proof-sheets, and to Mr. E. A. Miller, B.S., for his friendly aid in working examples. The drawings have been made by Mr. A. T. Bruegel, M.M.E., formerly instructor in the kinematics of machinery at Cornell University, now of the Pratt Institute, Brooklyn, N.Y. The author uses this opportunity to express his thanks for the pains taken by Mr. Bruegel to make the figures a pleasing feature of the book. D. A. MUERAY. Cornell University, August, 1899. CONTENTS. CHAPTER I. Review op Logarithms. ABT. PAGE 2. Definition of a logarithm 2 3. Properties of logaritiims . ~ 2 4. Common system of logaritlims 4 ] 5. Negative characteristics .5 6, Exercises in logarithmic computation .6 ) CHAPTER II. Trigonometric Ratios of Acute Angles. 8. Ratio. Measure .9 9. Incommensurable quantities. Approximations .... 12 10. Linear measure. Drawing to scale. Direct measurement by means of drawing 15 11. Degree measure. The protractor 18 12. Trigonometric ratios defined for acute angles 20 13. Definite and invariable connection between acute angles and trig- onometric ratios 24 14. Practical problems 26 15. Trigonometric ratios of 45°, 60°, 30°, 0°, 90° 29 16. Relations between the trigonometric ratios of an angle and those of its complement . 32 17. Exponents in trigonometry . 32 18. Relations between the trigonometric ratios of an acute angle . . 33 9. Summary 37 iz CONTENTS. CHAPTER III. Solution of Right-angled Triangles. AKT. PAGE *20. Solution of a triangle 38 21. The graphical method 38 22. The method of computation 39 23. Comparison between the graphical method and the method of computation 40 24. General directions for solving problems 41 0-26. Solution of right-angled triangles .41 26. Checks upon the accuracy of the computation . . . .43 27. Cases in the solution of right-angled triangles . . . .43 CHAPTER IV. Applications involving the Solution of Right-angled Triangles. 28. Projection of a straight line upon another straight line ... 49 29. Measurement of heights and distances 50 30. Problems requiring a knowledge of the points of the mariner's compass 63 31. Mensuration . . 64 32. Solution of isosceles triangles 55 33. Related regular polygons and circles 55 34. Solution of oblique triangles 67 34a. Area of a triangle in terms of its sides 60 346. Distance and dip of the visible horizon . . . . . .01 34c. Examples in the measurement of land 01 35. Summary 63 CHAPTER V. Trigonometric Ratios of Angles in General. 36. Directed lines 65 37. Trigonometric definition of an angle. Angles unlimited in magni- tude. Positive and negative angles 67 38. Supplement and complement of an angle 69 39. The convention of signs on a plane 70 40. General definition of the trigonometric ratios 71 CONTENTS, xi 41. The algebraic signs of the trigonometric ratios for angles in the different quadrants 72 42. To represent the angle geometrically when the ratios are given . 74 43. Connection between angles and trigonometric ratios ... 75 44. Relations between the trigonometric ratios of an angle . . .77 45. Ratios of 90° - A, 90° + A, 180° ^ Ay - A 79 CHAPTER VI. Trigonometric Ratios of the Sum and the Difference op Two Angles. 46. Derivation of the sine and cosine of the sum of two angles when each of the angles is less than a right angle . . . . .85 47. Derivation of the sine and cosine of the difference of" two angles when each of the angles is less than a right angle ... 87 48. Proof of the addition and subtraction formulas for all values of the two angles 89 49. Each fundamental formula contains the others .... 90 50. Ratios of an angle in terms of the ratios of its half angle . . 90 51. Tangents of the sum and difference of two angles, and of twice an angle 92 52. Sums and differences of sines and cosines 93 CHAPTER VII. Solution of Triangles in General. 53. Cases for solution . .97 54. Fundamental relations between the sides and angles of a triangle. The law of sines. The law of cosines 98 54a. Substitution of sines for sides, and of sides for sines . . . 101 55. Case I. Given one side and two angles 101 56. Case II. Given two sides and an angle opposite to one of them . 102 57. Case III. Given two sides and their included angle . . .105 58. Case IV. Given three sides 105 59. The aid of logarithms in the solution of triangles .... 106 60. The use of logarithms in Cases I., II 107 61. Relation between the sum and the difference of any two sides of a triangle. The law of tangents. Use of logarithms in Case III. . 108 Xll CONTENTS. ART. PAOB -62. Trigonometric ratios of the half-angles of a triangle. Use of loga- rithms in Case IV. . . 110 63. Problems in heights and distances . . . . . . .113 64. Summary 115 CHAPTER VIII. LiDE AND Area op a Triangle. Circles connected with A Triangle. 65. Length of a side of a triangle in terms of the adjacent sides and the adjacent angles 116 66. Area of a triangle 117 67. Area of a quadrilateral in terms of its diagonals and their angle of intersection 118 68. The circumscribing circle of a triangle 119 69. The inscribed circle of a triangle 119 70. The escribed circles of a triangle 120 CHAPTER IX. Radian Measure. 71. The radian defined 121 72. The value of a radian 122 73. The radian measure of an angle. Measure of a circular arc . . 123 CHAPTER X. Angles and Trigonometric Functions. 75. Function. Trigonometric functions 128 76. Algebraical note 129 77. Changes in the trigonometric functions as the angle increases from 0° to 360° 131 78. Periodicity of the trigonometric functions 134 79. The old or line definitions of the trigonometric functions . . 135 80. Geometrical representation of the trigonometric functions . . 137 81. Graphical representation of functions 138 82. Graphs of the trigonometric functions 139 83. Relations between the radian measure, the sine, and the tangent of an angle 143 CONTENTS. Xlll CHAPTER XI. General Values. Inverse Trigonometric Functions. ART. PAGE 84. General values 146 85. General expression for all angles which have the same sine . . 146 86. General expression for all angles which have the same cosine . 148 87. General expression for all angles which have the same tangent . 149 88. Inverse trigonometric functions ,; . 151 89. Sum and difference of two anti-tangents. Exercises on inverse functions 152 90. Trigonometric equations 164 CHAPTER XII. Miscellaneous Theorems and Exercises. 92. Functions of twice an angle. Functions of half an angle . .156 93. Functions of three times an angle. Functions of an angle in terms of functions of one-third the angle 157- 94. Functions of the sum of three angles 158 95. Identities .159 q2 i03 96. For an acute angle of 6 radians, cos ^ > 1 , sin ^ > ^ , . 160 4 4 97. One method of computing trigonometric functions .... 161 98. Trigonometry defined. Branches of trigonometry .... 162 APPENDIX. Note A. Historical sketch . 165 Note B. Projection definitions of trigonometric ratios , . . . 169 Note C. On the ratio of the length of a circle to its diameter . .171 Note D. On analytical trigonometry and De Moivre's Theorem . . 176 Questions and Exercises for Practice and Review . . . 181 Answers to the Examples 203 PLANE TRIGONOMETET. CHAPTER I. LOGARITHMS: REVIEW OF TREATMENT IN ARITH- METIC AND ALGEBRA. 1. There is a large amount of computation necessary in the solution of some of the practical problems in trigonometry. The labour of making extensive and complicated calculations can be greatly lessened by the employment of a table of logarithms, an instrument which was invented for this very purpose by John Napier (1550-1617), Baron of Merchiston in Scotland, and described by him in 1614. From Henry Briggs (1556-1631), who was professor at Gresham College, London, and later at Oxford, this invention received modifications which made it more con- venient for ordinary practical purposes."* Every good treatise on algebra contains a chapter on logarithms. This brief introductory review is given merely for the purpose of bringing to mind the special properties of logarithms which make them readily adaptable to the saving of arithmetical work. A little preliminary practice in the use of logarithms will be of advantage to any one who intends to study trigonometry. A review of logarithms as treated in some standard algebra is strongly recommended. * The logarithms in general use are known as Common logarithms or as Briggs' s logarithms, in order to distinguish them from another system, which is also a modified form of Napier's system. The logarithms of this other modified system are frequently employed in mathematics, and are known as Natural logarithms, Hyperbolic logarithms, and also, but erroneously, as Napierian logarithms. See historical sketch in article Logarithms (Ency. Brit. 9th edition), by J. W. L. Glaisher. 1 2 PLANE 'CBIGONOMETBY, [Ch. I 2. Definition of a logarithm. If a^ = 7V; (1 then a; is the index of the power to which a must he raised in order to equal N. For some purposes, this idea is presented in these words : If a^* = Nj then x is the logarithm of N to the base a. The latter statement is taken as the definition of a loganthmf and is expressed by mathematical symbols in this manner, viz. : a; = log^^JV. (2) Equations (1), (2), are equivalent ; they are merely two different ways of stating a certain connection between the three quantities a, X, N. For example, the relations 23 = 8, 5^ = 625, 10-3 = 3-JL^ = . 001, may also be expressed by the equivalent logarithmic equations, log2 8 = 3, logs 625 = 4, logio .001 = - 3. EXAMPLES. 1. Express the following equations in a logarithmic form : 33 = 27, 44 = 256, 112 ^ 121, 93 = 72a 73 = 343, m^ =p, 2. Express the following equations in the exJ)onential form : log2 8 = 3, logs 625 = 4, logio 1000 = 3, loga 64 = 8, log„ P=a. 3. When the base is 2, what are the logarithms of 1, 2, 4, 8, 16, 32, 64, 128, 256 ? 4. When the base is 5, what are the logarithms of 1, 5, 25, 125, 625, 3125 ? -f 5. When the base is 10, what are the logarithms of 1, 10, 100, 1000, 10,000, 100,000, 1,000,000, .1, .01, .001, .0001, .00001, .000001 ? 6. When the base is 4, and the logarithms are 0, 1, 2, 3, 4, 5, what are the numbers ? 7. When the base is 10, between what whole numbers do the logarithms of the following numbers lie : 8, 72, 235, 1140, 3470, .7, .04, .0035 ? 3. Properties of logarithms. Since a logarithm is the index of a power, it follows that the properties of logarithms must be derivable from the properties of indices ; that is, from the laws [. 2.] PROPERTIES OF LOGABITHMS. 3 of indices. The laws of indices are as follows (a, m, n, being any finite quantities) : (1) a^ X a" =^or:+V (2) — = a-» . r^ = a-« = a« ; also, ^ = 1. .-. a" = ll 1 - (3) («'")'» = a'"^ (4) ■^^={a^y = a\ Let M=a"', whence, log^M^m-, (1) and let JV= a% whence, log« N=n.' (2) It follows that MN = »*"+" ; whence, loga ilOT = m + n = loga ilf + loga N, (3) [If P=aP, then log«P = ^, iJf^P = a'"+"+^ ; . whence, log^ MNP = m -\- n -\- p — \og^ M + log« iV + log^ P. ] Also, ^ = ^ = a-«; whence, loga ^ = m - n = loga M - loga N» (4) Also, 3P = (a"*)*" = a'"'' ; whence, log = 8.84948 - 10 = 2.84948» .-. B = 3. Find V742 x .0769. Let B be the value. Then log B = log v'742 x .0769 = log (742 x .0769)^ = ^ log (742 X .0769) = | [log 742 + log .0769]. log 742 = 2.87040 log .0769 = 2.88593 Dividing by 5, 5 1 1.75633 .-. \ogB= .35126 ,-. i2 = 2.245. >'350 X 249 V350 x 249 ^ Let B be the value. Then log i? = ^ (log 456 + log 372 - log 350 - log 249). log 456 = 2.65896, log 350 = 2.54407 log 372 = 2.57054 , log 249 = 2.39620 5.22950 4.94027 4.94027 Dividing by 2, 2 1 .28923 .-. \ogB= .14402 (See Art. 9, Note 1.) .-. i2= 1.395. 8 PLANE TRIGONOMETRY, [Ch. I. 5. Find the value of x in 34* = 19. Since . 34^ = 19, log 34* = log 19, X log 34 = log 19, ^ log 19 1.27875 QQ.oQ 1 ^=i^ = i:53148=-''''''"^^^^y- _ 6. Find the value of (a) ^^, (6) ?^^, (c) ^^^, {d) 267 ^ " 315.2 ^ ' 69.83 ^ ' .4231 7. Find the value of (a) ^^'^ >< ^•^'^\ (&) 8-97 x 6.36 95.83 x 76.49 ^ ^ 674.2 ' ^ ^ 7.84 ' ^ ^ 82.97 8. Find the value of (a) \/63, (6) V630, (c) VoF, (d) v/Tes, (e) V.063, (/) •V/.0063. 9. Find the value of V63.42 x 74.95, V6.35 x 10.87, \/l4.21 x 17.29. 10. Find the value of -J^^-Qx 72- 1\ J 31-21 x 41.7 /4L >'7.81 X 6.95 \ll.39x 15.71 >'73 7 X 85.6 73.4x97.8 3 11. Find the value of 2.5637TT 12. [—rY , b W V X 13. Find x from the equations : (a) 3* = 35, (6) 5* = 70, (c) 10* = 36, ((^) 10* = 127, (c) 10* = 765, (/) 10* = 1364. CHAPTER 11. TRIGONOMETRIC RATIOS OF ACUTE ANGLES. 7. The name Trigonometry is derived from two Greek words wMch taken together mean ^ I measure a triangle/ * At the pres- ent time the measurement of triangles is merely one of several branches included in the subject of trigonometry. The more ele- mentary part of trigonometry is concerned with the calculation of straight and circular lines, angles, and areas belonging to figures on planes and spheres. It consists of tjvo sections, viz. Plane Trigonometry and Spherical Trigonometry. Elementary trigonom- etry has many useful applications, for instance, in the measure- ment of areas, heights, and distances. An acquaintance with its simpler results is very helpful, and sometimes indispensable, in even a brief study of such sciences as astronomy, physics, and the various branches of engineering. Some modern branches of trigonometry require a knowledge of advanced algebra. Their results are used in the more advanced departments of mathe- matics and in other sciences. This work considers only the sim- pler portions of trigonometry, and shows some of its applications. The truths of elementary trigonometry are founded upon geom- etry, and are obtained and extended by the help of arithmetic and algebra. A knowledge of the principal facts of plane geome- try, and the ability to perform the simpler processes of algebra, are necessary on beginning the study of plane trigonometry. Instruments for measuring lines and angles, and accuracy in computation are required in making its practical applications. 8. Ratio. Measure. On entering upon the study of trigo- nometry it is very necessary to have clear ideas concerning the terms ratio and incommensurable numbers as explained in arith- metic and algebra, for these terms play a highly important part * See historical sketch, p. 165. 9 10 PLANE TRIGONOMETRY, ill the subject. The study begins with an explanation of certain ratios which are used in it continually, and most of the numbers that appear in the solution of its problems are incommensurable. If one quantity is half as great as another quantity in magni- tude, it is said that the ratio of the first quantity to the second is as one to two, or one-half. This ratio is sometimes indicated thus, 1:2; but more usually it is written in the fractional form, |. In this example the magnitude of the second quantity is twice that of the first, and the ratio of the second quantity to the first is 2 : 1, or, adopting the more usual style, \, i.e. 2. The ratio of two quantities is simply the number which expresses the magni- tude of the one when compared with the magnitude of the other. This ratio is obtained by finding liow many times the one quantity contains the other, or by finding what fraction the one is of the other. It follows that a ratio is merely a pure number, and that it can be obtained only by comparing quantities of the same kind. Thus the ratio of the length 3 feet to the length 2 inches is -^, i.e. 18 ; the ratio of the weight 2 pounds to the weight 3 pounds is f . But one cannot speak of the ratio of 3 weeks to 10 yards, for there is no sense in the questions : How many times does 3 weeks contain 10 yards ? What fraction of 10 yards is 3 weeks ? When it is said that a line is ten inches long, this statement means that a line one inch long has been chosen for the unit of length, and that the first line contains ten of these units. Thus the 7iumher used in telling the length of a line is the ratio of the length of this line to the length of another line which has been chosen for the unit of length. The measure of any quantity, such as a length, a weight, a time, an angle, etc., is / the number of times the quantity contains \ I or, the fraction that the quantity is of / a certain quantity of the same kind which has been adopted as the unit of measurement. In other words, the measure of a quantity is the ratio of the quantity to the unit of measurement. For example, if half an inch is the unit of length, then the measure of a line 8 inches long is 16 ; if a foot is the unit of length, then the measure of the same line is f ; if a second is the unit of time, then the measure of an hour is 3600 ; if an hour is the unit of time, then the measure of a second is 3 ^nriT' 8.] RATIO. MEASURE. 11 If two quantities have a common unit of measurement, then their ratio is the ratio of their measures. For example, 1 pound being taken as the unit of weight, the ratio of a weight 3 pounds to a weight 7 pounds is -f-, which is also the ratio of the measures 3 and 7. In general, if a quantity P contains m units, and a quantity Q contains n units of the same kind as is used in the case of P, then the ratio quantity P _ m units _ m^ quantity Q n units n The last fraction — is the ratio of the numbers m and n, which n are the measures of the quantities P and Q respectively. EXAMPLES. 1. What is the ratio of each of the following lengths to an inch, viz., 8 in., 2 ft., 3 ft. 6 in., 1.5 yd., 20 yd., a yd., h ft., c in.? 2. What is the ratio of eacli of the following lengths to a yard, viz., 6 yd., 3.75 yd., 8 ft., 2 ft. 6 in., 10 in., 5 in., a yd., & ft., c in. ? 3. What is the measure of each of the following lengths, when a foot is the unit of length, viz., 1.5 mi., 17 yd., 3 yd. 2 ft., 8.5 ft., 2 ft. 6 in., 9 in., 2 in., a yd., h ft., c in. ? 4. What is the measure of each of the following lengths, when 3 in. is the unit of length, viz., 2.5 yd., 1.5 ft., 8 in., a yd., h ft., c in. ? 5. Express the ratio of 2.5 mi. to 10 yd. ; and the ratio of 2\ in. to 3iyd. 6. Compare the ratio of a foot to a yard with the ratio of a square foot to a square yard. 7. What is the unit of measurement in each of the following cases : when the measure of 2 ft. is 4, of 1 yd. is 72, of .5 in. is 4, of 2.5 ft. is .25 ? N.B, The following examples will he used again for purposes of illustra- tion. The student is advised to draw figures neatly and accurately and to preserve the results carefully. 8. In a right-angled triangle the base is 6 ft. and the hypotenuse 10 ft. What is the perpendicular ? Calculate the following ratios, viz. : perpendicular base perpendicular hypotenuse base hypotenuse hypotenuse base base hypotenuse perpendicular perpendicular 12 PLANE TRIGONOMETRY, [C What are these ratios in a triangle whose base is 6 in., and hypotenuse 10 in.? What are they when the base is 6 yd., and the hypotenuse 10 yd. ? When the base is 6 mi., and the hypotenuse 10 mi. ? When the base is 12 ft., and the hypotenuse 20 ft. ? When the base is 3 in., and the hypotenuse 5 in. ? Compare, if possible, the angles in these triangles. 9. In a right-angled triangle whose base is 35 ft. and perpendicular 12 ft., what is the hypotenuse ? For this triangle calculate the ratios specified in Ex. 8. Calculate these ratios for a triangle whose base is 70 yd., and perpendicular 24 yd. Compare, if possible, the angles in these triangles. 10. Calculate these ratios for the triangle whose hypotenuse is 29 ft. , and perpendicular 21 ft. ; for the triangle whose hypotenuse is 2.9 in., and perpendicular 2.1 in. Compare, if possible, the angles in these triangles. 9. Incommensurable quantities. Approximations. If the side of a square is one foot in length, then the length of a diagonal of the square is V2 feet. Thus the ratio of the diagonal to the side is V2, a number which cannot be expressed as the ratio of two whole numbers. Two quantities whose ratio can be expressed by means of two integers are said to be commensurable the one with the other; when their ratio cannot be so expressed, the one quan- tity is said to be incommensurable with the other. For example, the diagonal of a square is incommensurable with the side, and the length of a circle with its diameter.^* The quantities in the examples, Art. 8, are commensurable. Numbers such as V2, V4, VlO are incommensurable with unity, and their values cannot be found exactly. Their values, however, can be found to two, to three, to four, in fact, to as many places of decimals as one please. The greater the number of places of decimals, the more nearly will the calculated values represent the true values of the numbers. In other words, the values of incommensurable num- bers can be found approxiinately ; and the degree of approximation (that is, the nearness to the exact values) will depend only on the carefulness and patience of the calculator. In practical problems there frequently is occasion for the exercise of judgment as to the degree of approximation that is necessary and sufficient. For example, in calculating a length in inches in ordinary engineer- * See Appendix, Note C. 9.] APPBOXIMATIONS. 13 ing work there is no need to go beyond the third place of deci- mals, for engineers are satisfied when a measurement is correct to within -^^ of an inch. As a rule the results obtained in practical problems in mathematics are only approximate and not exact. There are two reasons for this : first, the data obtained by actual measurement can only be approximate, however excellent the instruments used in measuring may be, and however skilled and careful is the person who does the measuring ; second, most of the numbers used in the subsequent computations are incommen- surable. The examples at the end of this article are intended to bring out more clearly the idea of an approximate result. The answers are to be calculated to three places of decimals. It is advisable to compare the values calculated to three places of decimals with the values calculated to two places of deci- mals, and to note the difference between them. The following facts are sup- posed to be known and will be taken for granted. (a) In a right-angled triangle the square of the measure of the hypotenuse is equal to the sum of the squares of the measures of the other two sides. (&) The ratio of the length of any circle to its diameter is a number which is the same for all circles.* The exact value of this ratio is incommensurable and is always denoted by the symbol tt (read;9i).t The approximate values commonly used for tt are 3.1416, 3.14159, fff (i.e. 3.1415929 -.•), ¥ (i.e. 3.142857) ; of these values the last is the least accurate, but it is accurate enough for many practical purposes. (c) The length of a circle of radius r is 2 irr [by (6)] ; and the enclosed area is irr^. Note 1. If a number be calculated to three or more places of decimals, then the closest approximation to, say, two places of decimals is obtained by leaving the number in the second place of decimals unchanged when the number in the third place is less than 5, and by increasing the number in the second place by unity when the number in the third place is greater than 5 or 5 followed by numbers; thus, e.g., 3.72 for 3.724, 3.73 for 3.7261 and * This ratio and facts (c) are considered in Note C, Appendix. The read- ing only requires a knowledge of elementary geometry. t This symbol is the initial letter of periphereia, the Greek word for cir- cumference. Its earliest appearances to denote this ratio are in Jones's Synopsis Palmarioriim Mathesos, London, 1706, and in the Introductio in analysin infinitorum., published in 1748 by Leonhard Euler (1707-1783), a native of Switzerland, who was one of the greatest mathematicians of his time. 14 PLANi: fRIGONOMETRY. [Ch. I 3.7257. When the number in the third place is 5 and this is followed by zeros only, the number in the second place is unchanged if it is even, and is increased by unity if it is odd ; thus, e.g.^ 3.78 for 3.775, 3.78 for 3.785. In a series of calculations the errors made by following this rule tend to balance one another. Note 2. A quantity measured to two places of decimals is correct to the hundredth part of the unit employed, and a quantity measured to three places is correct to the thousandth part of the unit. For example, the length of a circle of 10 feet diameter is 31.4159 . . . feet. For this length 31.416 or 31.42 maybe taken; the former result differs from the true result by less than one-thousandth of a foot, the latter by less than one-hundredth. EXAMPLES. 1. A finds the square root of 3 correctly to two places'of decimals, and B to three. How much closer than A does B come to the exact value of the square root of 3 ? 2. A circle is 50 ft. in diameter. In calculating its length A takes 3.1416 as the ratio of the length of a circle to the diameter, B takes 3.14159, and C takes 22 : 7. What are the differences (in inches) between their results ? 3. The radius of a circle is 49.95 ft. How nearly will a person come to the length of the circle if he assumes the radius to be 50 ft. ? [In this and the following example take tt = 22 : 7.] 4. It is known that the diameter of a certain circle does not differ from 100 ft. by more than 2 in. What will be the outside limits of the error made in calculating the area when the diameter is taken as 100 ft.? 5. Find the difference between the calculations of the numbers of revolu- tions per mile made by a 50-in. bicycle, for tt = 22 : 7 and ir = 3.1416. 6. A lot is 75 ft. by 200 ft. Find the diagonal distance across the lot correctly to within a tenth of an inch. 7. Find the height of an equilateral triangle whose side is 20 yd. 8. The side of an isosceles triangle is 40 ft. and the base is 30 ft. ; find the height. 9. What is the length of the diagonal of a square whose side is 20 ft. ? 10. What is the length of the side of a square whose diagonal is 20 ft. ? N. B. The following examples will be used again for purposes ofillnstra- tion. The student is advised to draw figures and to preserve the results icith those of Exs. 8, 9, 10, Art. 8. 11. (a) In a right-angled triangle the hypotenuse is 12 ft. and'the base is 6 ft. ; calculate the ratios specified in Ex. 8, Art. 8. (6) What are these ratios when the lengths in (a) are taken twice, three times, one-half as great ? Compare, if possible, the angles in these triangles. »Tr™l 10.] DRAWING TO SCALE, 15 12. (a) In a right-angled triangle the base is 8 in. and the perpendicu- lar 12 in. ; calculate the ratios specified in Ex. 8, Art. 8. (&) What are these ratios when the lengths in (a) are taken one-third, and four times as great ? Compare, if possible, the angles in these triangles. 13. (a) In a right-angled triangle the hypotenuse is 35 yd. and the per- pendicular is 15 yd. ; calculate the ratios specified in Ex. 8, Art. 8. (h) What are these ratios when the lengths in (a) are taken four times, six times, one-fifth as great ? Compare, if possible, the angles in these triangles. 10. Linear measure. Drawing to scale. Direct measurement by means of drawings. Various systems of linear nieasurement are described in arithmetic. The system mostly used in English- speaking countries is that in which length is given in miles, yards, feet, or inches. The system which is in common use on the continent of Europe, and which is mainly employed in scientific measurements throughout the world, is the metric system. In this system lengths are given in centimetres, metres, etc., the centimetre being a hundredth part of a metre. A metre is equal to 39.37 . . . inches.* Drawing to scale. It is often desirable to have a drawing on paper which shall serve to give an accurate idea of the relations of certain lines and positions. Maps and architects' plans are familiar examples of such drawings. In a map an inch may represent 1 mile, 10 miles, 100 miles, 500 miles, and so on, accord- ing to the scale on which the map is made ; in a building plan an inch may represent 10 feet, 12 feet, and so on. The operation of drawing on paper lines that shall be a half, a quarter, a tenth, a thousandth, etc., part of the actual length of given lines, is called drawing to scale. In many cases the drawings of objects cannot be made full size ; for instance, the map of a town, the floor plan of a church ; these are drawn to a reduced scale. In other cases the drawings are made larger than the actual objects, for instance, the drawings of the minute things that live in a drop of water, the drawings of the various parts of a flower; * The metric system has the great advantage of being a decimal system. At the present time committees of scientific societies in England and America are working to have the common system replaced by the metric. 16 PLANE TRIGONOMETRY. [Ch. II. these are drawn to an enlarged scale. When a drawing is made to scale, the scale should always be indicated on it. This may be done in various ways. Thus a mere statement may be made, e.g., 1 inch to 10 feet; / or, the scale may be indicated by a fraction which gives the ratia of any line in the drawing to the actual line represented. The scale can also be shown graphically by means of a specially marked line. Both the latter methods are illustrated, for instance, on the map of the Kingdom of Saxony in Tlie Times Atlas : 1:870000. 10 English Miles (69.16 ^ 1° ) Pig. 1. The scale should be expressed fractionally, that is, by expresjy- ing the ratio of a line in the drawing to the actual line repre- sented. Thus in the first example above the scale is 1 : 120 ; in the second the scale is 1 : 870000. When a drawing is made to scale, the distance between tiuo objects can be measured directly, by merely measuring the distance between their corresponding points on the drawing. For instance, if 1 inch represents 120 feet, then 2.5 inches represents 300 feet. Another example : On the map of Saxony referred to above, the distance between Leipzig and Dresden is, approximately, 4i inches, and 4^ inches x 870000 gives about 62 miles as the distance in an air- line between these cities. This method of finding distance can be used in solving many of the problems in trigonometry.* To find the length of the representative line in the drawing when the scale and the actual length are given, is an exercise in simple proportion ; so, also, to find the actual length of a line when the scale and the length of the representative line are known. * This is one of the methods which will be employed in this book in prob- lems involving distance. Proficiency in drawing will be very helpful to the student. 10.] EXAMPLES. 17 EXAMPLES. 1. When an inch represents 10 ft., how long must the lines be that will represent 3 in., 6 in., 1 ft., 2 ft., 5 ft., 15 ft., 7.5 ft., 30 ft., 40ft., 55 ft.? What is the scale ? 2. When an inch represents 5 yd. , how long must the lines be tiiat will represent 2 yd., 4 yd., 7 yd., 11 yd., 3 yd. 2 ft., 4 yd. 1 ft. 8 in. ? What is the scale ? 3. When an inch represents 150 ft., what distances are represented by I in., I in., i in., If in., 2| in., 4.8 in., 5.3 in. ? What is the scale ? 4. When an inch represents 10 mi., what distances are represented by 3 in., 7 in., ^ in., ^ in., 3 J in. ? What lengths on the drawings will represent 7 mi., 18 mi., 25 mi. ? What is the scale ? 5. What are the scales when 1 in. represents 100 ft., ^ in. represents a mile, I in. represents 20 ft., ^^ in. represents 15 yd., 1 in. represents 1 mi., 10 mi., 100 mi. ? 6. Draw to a scale 1 : 240 (20 ft. to the inch) the circles in Exs. 2, 3, 4, 5, Art. 9. 7. On a map in Baedeker's Guide to Paris the distance between the nearest corners of the Eiffel Tower and Notre Dame Cathedral is 7f in. What is the distance between those points, the map being drawn to a scale 1 : 20000 ? 8. Make the comparison of angles asked for in Exs. 8, 9, 10, Art. 8 ; Exs. 11, 12, 13, Art. 9. Suggested Exercises. Make drawings to scale of the floor plan of a dwelling house, of some other building, of some grounds. Find the distances between various points, such as diagonally opposite corners, by making measurements in the drawing and applying the scale. Compare the results obtained in this way with the results obtained by other methods. Other methods that may be used are : (1) making an off-hand estimate of the dis- tance ; (2) actually measuring the distance by "pacing off" or by using a rule or tape line ; (3) making a computation. Let the student, from his own experience, form a judgment as to which of the four methods referred to is the easiest, and which the more exact. Find the air-line distances be- tween places by measuring the distances between them on maps. Several maps may by used so as to have a variety of scales. Note. The word scale also has another meaning in drawing and meas- urement. Engineers and draughtsmen use various kinds of rules called scales. The faces of these rules contain different numbers of divisions to an inch, one 10 divisions, one 20, one 30, and so on ; and generally, one inch on each face is subdivided so that a small fraction of an inch may be set off or read. Some paper scales are on the protractor inserted in this book. 18 PLANE TRIGONOMETRY. [Ch. I » 11. Degree measure. The protractor. It has been seen geometry that: (1) When one line is perpendicular to another line, each of the angles made at their intersection is a right angle ; (2) All right angles are equal to one another. In some geomet- rical propositions angles are compared, and one angle is shown to be greater or less than another. But geometry, with the excep- tion of a few cases, does not show by exactly how 7nuch the one angle is greater or less than the other. In order to show this, measurement is necessary ; and in order to measure, a unit angle of measurement must be chosen. The unit of angular magnitude which is generally used in practical work is the angle that is one- ninetieth part of a right angle. This unit angle is called a degree. All degrees are equal to one another, since all right angles are equal to one another. Each degree is subdivided into 60 equal parts called minutes, and each minute is subdivided into 60 equal parts called seconds. Hence comes the following table of angular measure : 60 seconds = 1 minute, 60 minutes = 1 degree, 90 degrees = 1 right angle. The magnitude of an angle containing 37 degrees and 42 minutes and 35 seconds, say, is written thus : 37° 42' 35", read 37 degrees, 42 minutes, 35 seconds. This system of measurement is sometimes called the rectangular system^ sometimes tlie sexagesimal system. In this chapter only acute angles, that is, angles which contain between 0° and 90°, are considered. Chapter V. considers angles of all magnitudes. Note 1. An angle 1° is subtended by 1 in. at a distance 4 ft. 9.3 in., and by 1 ft. at a distance 57.3 ft. An angle 1' is subtended by 1 in. at a distance 286.5 ft., and by 1 ft. at a distance 3437.6 ft., about two-thirds of a mile. An angle 1" is subtended by 1 in. at a distance of nearly 3| mi., by 1 ft. at a distance a little greater than 39| mi., by a horizontal line 200 ft. long on the other side of the world, nearly 8000 mi. away. These facts can be verified later. See Ex. 3, Art. 83. Note 2. Another system of angular measurement was advocated by Briggs and other mathematicians (see Art. 1), and was introduced in France at the time of the Revolution. In this system, which is a decimal o^ie and called the centesimal system., a right angle is divided into 100 equal parts called grades., each grade into 100 equal parts called minutes, and each minute into 100 equal parts called seconds. It has not been generally adopted, on account of the immense amount of labour that would be necessary in order to change the mathematical tables computed for the other system. |] THE PROTRACTOR. 19 Note 3. The sexagesimal system (from sexagesimus, sixtieth) was in- vented by the Babylonians, who constructed their tables of weights and measures on a scale of 60. Their tables of time (1 day = 24 hr., 1 hr. = 60 min., 1 min. = 60 sec.) and circular measure have come down to the present day. It has been suggested that their adoption of the scale of 60 is due to the fact that they reckoned the year at 360 days. " This led to the division of the circumference of a circle into 360 degrees, each degree representing the daily part of the supposed yearly revolution of the sun around the earth Probably they knew that the radius could be applied to the circumference as a chord six times, and that each arc thus cut off contained 60 degrees. Thus the division into 60 parts may have suggested itself . . . . Babylonian science has made its impress upon modern civilization. Whenever a surveyor copies the readings from the graduated circle on his theodolite, whenever the modern man notes the time of day, he is, unconsciously perhaps, but unmis- takably, doing homage to the ancient astronomers on the banks of the Eu- phrates." — Cajori, History of Elementary Mathematics^ pp. 10, 11. Note 4. Another system of angular measure is described in Chapter IX. See Art. 71. The protractor. The protractor is an instrument used for meas- uring given angles and laying off required angles on paper. Pro- tractors are of various kinds, of which the semicircular and the full-circled are the most common. The degrees are marked all round the edge. A paper protractor is inserted in this book for use in solving problems.* In order to draw a line that shall make a given angle with a given line at a given point, proceed as follows : Place the centre of the protractor at the given point and bring its diameter into coincidence with the given line, keep- ing the semicircle on the side on which the required line is to be drawn; prick off the required number of degrees with a sharp pencil or fine needle. The line joining the point thus fixed and the given point, is the line required. In order to measure a given angle with the protractor, place the centre at the vertex of the angle, and place the diameter in coincidence with one of the boundary lines of the angle; the number of degrees in the arc intercepted between the boundary lines of the angle is the meas- ure of the angle. * A horn protractor costs about 25 cents, and a small metal one about 50 cents. One who is neat and handy can make a paper protractor. 20 PLANE TBIGONOMETRY. [Ch. II. Note. Before proceeding further, the student should be able to draw with ease a right-angled triangle, having been given: (a) The hypotenuse and a side ; {h) the two sides about the right angle ; (c) the hypotenuse and one of the acute angles ; (d) one of the sides about the right angle and the opposite angle ; (e) one of the sides about the right angle and the adja- cent angle. It is here, taken for granted that these problems have been con- sidered in a course in plane geometry or in a course of geometrical drawing. EXAMPLES. N. B. The student is advised to do Exs. 1-6 carefully, and to preserve the results, for they will soon be required for purposes of illustration. 1. Draw to scale the triangles considered in Exs, 8, 9, 10, Art. 8, and Exs. 11, 12, 13, Art. 9, and measure the angles. 2. Make drawings, on two different scales, of a right-angled triangle whose base is 20 ft. and adjacent acute angle is 55°. In each drawing meas- ure the remaining parts and thence deduce the unknown parts of the original triangle. In each drawing calculate the ratios specified in Ex. 8, Art. 8. 3. Same as Ex. 2, for a right-angled triangle whose hypotenuse is 30 ft. and angle at base is 25". 4. Same as Ex. 2, for a right-angled triangle whose base and perpendicular are 30 ft. and 45 ft. respectively. 5. Same as Ex. 2, for a right-angled triangle whose hypotenuse is 60 ft. and base is 45 ft. 6. Same as Ex. 2, for a right-angled triangle in which the base is 50 ft. and the angle opposite to the base is 40°. 7. What angles of a whole number of degrees can easily be constructed geometrically without the aid of the protractor ? Make the constructions. 12. Trigonometric ratios defined for acute angles. The ratios referred to at the beginning of Art. 8 will now be explained so far as acute angles are concerned. (Before proceeding, the student Via. 2, 12.] THE TBIGONOMETRIC RATIOS. 21 should glance over the work on Exs. 8-10, Art. 8 ; Exs. 11-13, Art. 9 ; Exs. 1-6, Art. 11.) Let A be any acute angle. In either one of the lines containing the angle take any point P and let fall a perpendicular FM to the other line. The three lines AP, AM, MP, can be taken by twos in three different ways, and hence six ratios can be formed with them, namely : MP AM MP AM AP^ AP AP AP' am' MP AM MP It is shown in Art. 13 that each of these ratios has the same value as in Fig. 2, no matter where the point P is taken on either one of the lines bounding an angle which is equal to A. For the sake of convenience of reference, each one of these six ratios is given a particular name with respect to the angle A. Thus ; MP — — - is called the sine of the angle A ; AP ^ ' AM — -— is called the cosine of the angle A ; AP ^ ' MP ' ■ is called the tangent of the angle A ; AM AM is called the cotangent of the angle A : MP AP — — is called the secant of the angle A ; AP is called the cosecant of the angle A. ■ ay These six ratios are known as the trigonometric ration of the angle A. According to the definition of a ratio (Art. 8) they are merely numbers. Eor brevity they are written sm A, cos A, tan A, * In Chapter V. the trigonometric ratios are defined for angles in general. The definitions givo^iri this article will be found to follow immediately from those given in Art. 40. 22 PLANE TRIGONOMETRY. [Ch. J(| cot A, sec A, cosec A (or esc A).* Thus tan A is read ^'tangent J.," and means "the tangent of the angle ^." The giving of names in (1) may be regarded as defining the trigonometric ratios. Definitions (1) may be expressed as follows : AP ' AM ' AM 4^ = cos ^, 4^= cot A, ^ =cosec A, AP ' MP ' MP ' . . (2) These definitions can be given a slightly different form which is more general, and, accordingly, more useful in applications. In any right-angled triangle AMP (Fig. 2), M being the right angle, with reference to the angle A let MP be denoted as the opposite side, and AM as the adjacent side. Then these defini- tions take the form : — * The term sine first appeared in the twelfth century in a Latin translation of an Arabian work on astronomy, and was first used in a published work by a German mathematician, liegiomontanus (1436-1476). The terms secant and tangent were introduced by a Dane, Thomas FincJc (1561-1646), in a work published in 1583. The term cosecant seems to have been first used by Bheticus, a German mathematician and astronomer (1514-1576), in one of his works which was published in 1596. The names cosine and cotangent were first employed by Edmund Gunter (1581-1626), professor of astronomy at Gresham College, London, who made the first table of logarithms of sines and tangents, published in 1620, and introduced the Gunter's chain now used in land surveying. The abbreviations sin, tan, sec, were first used in 1626 by a Flemish mathematician, Albert Girard (1590-1634), and those of cos, cot, appear to have been earliest used by an Englishman, AVilliam Oughtred (1574-1660), in his Trigonometry, published in 1657. Tliese con- tractions, however, were not generally adopted until after their reintroduction by Leonhard Euler (1707-1783), born in Switzerland of Dutch descent, in a work published in 1748. They were simultaneously introduced in England by Thomas Simpson (1710-1761), professor at Woolwich, in his Trigonom- etry, published in 1748. [See Ball, A Short History of Mathematics, pp. 215, 367.] When first used these names referred, not to certain ratios con- nected with an angle, but to certain lines connected with circular arcs sub- tended by the angle. This is explained in Art. 79, which the student can easily read at this time. See Art. 80, Notes 2, 3. 12.1 THE TRIGONOMETRIC BATIOS, 23 ^h. sill A = opposite side, " hypotenuse, . adiacent side, cos A = —^ i hypotenuse, .^-—— ^ tan A = opposite side, V /{ adjacent^ side, ^ cot A - ^^3^^^^^ side, opposite side, see^= hypotenuse, adjacent side, cosec^= t»yP°te""^e, opposite side, [The word perpendicular is sometimes used instead of opposite sidCy and base instead of adjacent side.^ It is necessary that these definitions be thoroughly memorized. rx EXAMPLES. N.B. The student is requested to preserve the work and results of these Exs. for purposes of future reference. 1. In AMP (Fig. 2) give the trigonometric ratios of angle APM. Note what ratios of angles A and P are equal. 2. In Figs. 45 a, 45 b, Art. 46, give the trigonometric ratios of the various acute angles. 3. Find the trigonometric ratios of the acute angles in the triangles in Exs. 8-10, Art. 8 ; Exs. 11-13, Art. 9; Exs. 2-6, Art. 11. 4. In a triangle PQB right-angled at §, the hypotenuse PB is 10 in. long, and the side QB is 7. ]Find the trigonometric ratios of the angles P and B. Note what ratios of P and B are equal. 5. For each of the angles in Ex. 4, and for each of any three of the angles in Ex. 3, calculate the following, and make a note of the result. [Let x de- note the angle whose ratios are being considered.] (1) sin ic cosec x (2) cos x sec x (3) tan x cot x (4) sin2 X + cos2 x (5) sec^ x — tan2 x (6) cosec^ x - cot^ x (7) tana; - sinx cosx (8) cot X — sin X 6. Make the same calculations for angle A in Fig, 2, Art. 12. 24 PLANE TRIGONOMEfUT. [Ch. II. 13. Definite and invariable connection between (acute) angles and trigonometric ratios. It is important that the following prin- ciples be clearly understood: (1) To each value of an angle there corresponds hut one value of each trigo7iometric ratio. (2) Tujo unequal acute angles have different trigonometric ratios. (3) To each value of a trigonometric ratio there corresponds hut one value of an acute angle. (1) In Fig. 2, Art. 12, from any point ^S' in AB draw ST per- pendicular to AL. Let angle B (Fig. 3) be equal to A, and from any point G in one of the lines containing angle B draw GK perpendicular to the other line. Then, by definition (3), Art. 12, sin A (in AMP) = ^, ^inAimAST) ST AS' sm B (= sm ^) = — . But the triangles AMP, AST, BKG, are mutually equi- angular. Hence the sides about the equal angles are propor- tional, and - MP^ST^KG AP AS BG' Therefore all angles equal to A have the same sine. In like manner, these angles can be shown to have the same tangent, secant, etc.* (2) Let EAL and EALi be any two un- equal acute angles, placed, for convenience, so as to have a common vertex A and a common boundary line AR. From any point P on AL draw PM perpendicular ^ to AR. Take AP^ = AP, and draw P,M, perpendicular to AR. Then * In Euclid's text on geometry, the properties of similar triangles are con- sidered in Bk. VI. Pupils who study Euclid and have not reached Bk. VI. can be helped to understand these properties by means of a few exercises like those referred to in Ex. 3, Art. 12. TABLES. 25 sm HAL = -—-, sill RAL^ = ^ ^- But M^P, > MP, and ^Pj = .4P; hence sin RALi > sin RAL. In a similar manner the other ratios can be shown to be respec- tively unequal. Ex. In this construction AP\ is taken equal to AP. Why does this not affect the generality of the proof ? (3) This property follows as a corollary from (1) and (2). The trigonometric ratios for angles from 0° to 90° are arranged in tables. In some tables the calculations are given to four places of decimals, in others to five, six, or seven places. There are also tables of the logarithms of the ratios (or of the logaritlims in- creased by 10),* which vary in the number of places of decimals to which the calculations are carried oufc. The student is advised to examine a table of the trigonometric ratios at this time. A good exercise will consist in finding the logarithms of some of the sines, tangents, etc., adding 10 to each logarithm, and comparing the result with that given in the table of Logarithmic sines, tan- gents, etc. [What are denoted as Natural sines and cosines in the tables, are merely the actual sines and cosines, which have been discussed above ; the so-called Logarithmic sines and cosines are the logarithms of the Natural sines and cosines with 10 added.] A book of logarithms and trigonometric ratios is the principal help and tool in solving most of the problems in practical trigonom- etry; and hence, proficiency in using the tables is absolutely necessary. The larger part of the numerical answers in this book have been obtained with the aid of a five-place table. Those who use six-place or seven-place tables will reach more accurate results. EXAMPLES. 1. Compare each of the ratios of BALi with the corresponding ratio of BAL. 2. Suppose that the line AB (Fig. 4) revolves about ^ in a counter-clock- wise direction, starting from the position AM: show that, as the angle MAL * These are usually called Logarithmic sines, tangents, etc. 26 PLANE TRIGONOMETRY. [Ch. U. 11 increases, its sine, tangent, and secant increase, and its cosine, cotangent™' and cosecant decrease. Test this conclusion by an inspection of a table of Natural ratios. 3. Find by tables, sin 17° 40', sin 43° 25' 10", sin 76° 43', sin 83° 20' 25", cos 18° 10', cos 37° 40' 20", cos 61° 37', cos 72° 40' 30", tan 37° 40' 20", tan 79° 37' 30", cot 42° 30', cot 72° 25' 30". Log sin 37° 20', Log sin 70° 21' 30", Log cos 30° 20' 20", Log cos 71° 25', Log tan 79° 30' 20", Log cot 48° 20' 40". 4. Find the angles corresponding to the following Natural and Logarithmic ratios : sine = .15327, sine = .62175, sine = .82462, sine = .84316, cosine = .85970, cosine = .61497, cosine = .84065, cosine = .60165. tangent = .42482, tangent = .60980, tangent = 1.6820, tangent = 2.4927, Log sine = 9.79230, Log sine = 9.94215, Log cosine = 9.96611, Log cosine = 9.74743, Log tangent = 9.82120, Log tangent = 10.37340. 14. Practical problems. The problems in this article are in- tended to help the learner to realize more clearly and strongly the meaning and the usefulness of the ratios which have been defined in Art. 12. The student is earnestly recommended to try to solve the first three problems below witliout help from the book. He will find this to be an advantage, whether he can solve the prob- lems or not. If he can solve them, then he will have the pleas- urable feeling that he is to some extent independent of the book ; and he will thus be encouraged and strengthened for future work. Should he fail to solve them, he will have the advantage of a closer acquaintance with the difficulties in the problems, and so • will observe more keenly how these difficulties are avoided or overcome. Throughout this course the student will find it to be of immense advantage if he will think and study over the subject- matter indicated in the headings of the articles and make some kind of an attack on the problems before appealing to the book for help. If he follows this plan, his progress, in the long run, will be easier and more rapid, and his mental power more greatly improved than if he is content merely to follow after, or be led by, the teacher. or author. 14.] PUACncAL PROBLEMS. 2T EXAMPLES. " 1. Construct the acute angle whose cosine is |. What are its other trigo- nometric ratios ? Find the number of degrees in the angle. The definition of the cosine of an angle shows that the required angle is equal to an angle in a certain right-angled triangle, namely, the triangle in which " the side adjacent to the angle is to the hypote- nuse in the ratio 2:3." Thus the lengths of this side and hypotenuse can be taken as 2 and 3, 6 and 9, 200 and 300, and so on. Taking the lengths 2, 3, (these numbers being simpler and, accordingly, more conven- ient than the others), construct a right-angled triangle AST which has side AS = 2, and hypotenuse AT =Z. The angle A is the angle required, for cos A = ^. V5 Now /S'T = V32-22 Hence, the other ratios are sin ^ = ^ = .7454, tan ^ = ^ 3 ' 2 sec J. = - = 1.5000, cosec^ 2.2361. 1.1180, cot ^ = -^ = .8944, V5 3 1.3416. 2 ' V5 The measure of the angle can be found in either one of two ways, viz. : (a) by measuring the angle with the protractor ; (6) by finding in the table the angle whose cosine is | or .6667. The latter method shows that A = 48° 11' 22". [Compare the result obtained by method (a) with the value given by method 2. A right-angled triangle has an angle whose cosine is f , and the length ^ of the hypotenuse is 50 ft. Find the angles and the lengths of the two sides. By method shown in Ex. 1, construct an angle A whose cosine is |. On one boundary line of the angle take a length AG to represent 50 ft. Draw GK per- pendicular to the other boundary line. Fig. 6. cos^ AK AG Cos^ = f = .6666 ..., .-. ^ = 48°11'22^ .• -B = 90 - ^ = 41° 48' 38", = .6666-., sin^=^^ 3 = 50 X .6666 . . KG V5 *' " AG~ 3\ = 33.333 ..., .-. /iG^ = :^ X 50 = 37.27 (Ex. 1) 28 PLANE TRIGONOMETRY. [Ch. II. • The problem may also he solved graphically as follows. Measure angles A, G, with the protractor. Measure AK^ KG directly in the figure. 3, A ladder 24 ft. long is leaning against the side of a building, and the foot of the ladder is distant 8 ft. from the building in a horizontal direction. What angle does the ladder make with the wall ? How far is the end of the ladder from the ground ? Graphical method. Let AC represent the ladder, and BC the wall. Draw AC^ AB, to scale, to represent 24 ft. and 8 ft. respectively. Measure angle ACB with the pro- tractor. Measure BC directly in the figure. Method of computation. BC = ^AG^ - AB^ = V676 - 64 = V6l2 = 22.63 ft. sin ^05 = ^ = A = .33333, .-. ^0J5 = 19° 28' 16". 4. Find tan 40° by construction and measurement. With the protractor lay off an angle SAT equal to 40°. From any point P in J. T draw PB perpendicu- larly to AS. Then measure AB., RP, and substitute RP the values in the ratio, tan 40° = AR Compare the result thus obtained with the value given for tan 40° in the tables.* 5. Construct the angle whose tangent is f . Find its other ratios. Meas- ure the angle approximately, and compare the result with that given in the tables. Draw a number of right-angled, obtuse-angled, and acute-angled triangles, each of which has an angle equal to this angle. 6. Similarly for the angle whose sine is | ; and for the angle whose co- tangent is 3. 7. Similarly for the angle whose secant is 2^ ; and for the angle whose cosecant is 3^. 8. Find by measurement of line« the approximate values of the trigo- nometric ratios of 30°, 40°, 45°, 50°, 55°, 60°, 70° ; compare the results with the values given in the tables. * The values of the ratios are calculated by an algebraic method, and can be found to any degree of accuracy that may be required. 15.] thigonometric hatios of 45°. 29 If any of the folloiving constructions asked for is impossible, explain why it is so. 9. Construct the acute angles in the following cases : (a) When the sines are ^, 2, ^, f ; (&) when the cosines are |, ^, 3, .3; (c) when the tangents are 3, 4, |, ^ ; (d) when the cotangents are 4, 2, |, .7 ; (e) when the secants are 2, 3, |-, |, 4 J ; (/) when the cosecants are 3, 2.5, .4, f. 10. Find the other trigonometric ratios of the angles in Ex. 9. Find the measures of these angles, (a) with the protractor, (&) by means of the tables. 11. What are the other trigonometric ratios of the angles : (1) whose sine is-; (2) wnose cosine is -; (3) whose tangent is -; (4) whose cotangent b b b ■ is-; (5) whose secant is -; (6) whose cosecant is - ? b ^ ^ & ^ ^ b 12. A ladder .32 ft. long is leaning against a house, and reaches to a point 24 ft. from the ground. Find the angle between the ladder and the wall. 13. A man whose eye is 5 ft. 8 in. from the ground is on a level with, and 120 ft. distant from, the foot of a flag pole 45 ft. 8 in. high. What angle does the direction of his gaze, when he is looking at the top of the pole, make with a horizontal line from his eye to the pole ? 14. Find the ratios of 45°, 60°, 30°, 0°, 90°, before reading the next article. 15. Trigonometric ratios of 45°, 60°, 30°, 0°, 90°. The ratios of certain angles which are often met will now be found. A. Ratios of 45°. Let AMP be an isosceles right-angled tri- angle, and let each of the sides about the right angle be equal to a. The angle ^ = 45°, and^P=aV2. *. sin 45° = sin ^ = MP Fig. 10. ^P aV2 V2 By using the same figure it can be shown that cos 45° =- V2' sec 45° tan 45° = 1, cot 45° = 1, V2, cosec 45° = V2. so PLANE TlitGONOMETMV. [Ch. It The sides of triangle AMF are proportional to 1, 1, V2. Hence, in order to produce the ratios of 45° quickly, it is merely necessary to draw Yig. 10 ; from this figure the ratios of 45° can ^B J, be read off at once. B, Ratios of 30° and 60°, Let ABC be an equilateral ^ triangle. From any vertex B draw a perpendicular BD to the opposite side AC. Then angle DAB = 60°, an- gle ABD = 30°. A [\ /wt° V ovrN \ / A\ A /C\ 4^ a ^D G 1 Fig 11. Fig, 12. liAB = 2a, then AD = a, and DB =-V4.a^-d' = aVS. .-. sin 60° = sin DAB = ^ = ^ = 2^, AB 2a 2 By using the same figure it can be shown that cos 60° = 1, tan 60° = V3, cot 60° "'■ 2 V3' sec 60° = 2. cosec 60° = Also, Similarly, cos 30° sin 30° = sin ABD = — AB 2 a V3 V3 tan 30' 1 V3' sec 30° = V3' cot 30° = V3, cosec 30° = 2. In ADB the sides opposite to the angles 30°, 60°, 90°, are respectively proportional to 1, V3, 2. Hence, in order to produce the ratios of 30°, 60°, at a moment's notice, it is merely necessary to draw Fig. 12, from which these ratios can be imjuediately read off. C. Ratios of 0" and 90° read now. The algebraical note, Art 76, may he 15.] TBIGONOMETRIC RATIOS OF 0° AND 90°, 31 Let the hypotenuse in each of the right-angled triangles in Fig. 13 be equal to a. MP Bin MAP = cos MAP = AP' AM AP' It is apparent from this figure that if the angle MAP approaches zero, then the perpendicular MP approaches zero, ^ and the hypotenuse AP approaches to an equality with AM; so that, finally, if MAP = 0, then MP = 0, and AP = AM. Therefore, when MAP — 0, it follows that : sin 0° = - = 0, a cos 0° = - = 1, tan 0° = - = 0, a cot 0° = ^ 00. sec 0° = - a 1, cosec 0° = - = 00 As MAP approaches 90°, AM approaches zero, and MP ap- proaches to an equality with AP. Therefore, when MAP = 90°, it follows that : sin 90° = - = 1, a cos 90° = - = 0, a tan 90" = 00, sec 90° = ^=00, cot 90° = 5 = 0, cosec 90° = - = 1. a a EXAMPLES. N, B. Read the first few lines of Art. 17 before attacking the problems. Find the numerical value of 2. sec2 30° + tan3 45°. 4. cos 0° sin 45° + sin 90° sec2 30°. 6. 3 tan3 30° sec^ 60° sin2 90° tan2 45°. 8. 2 sin^ 30° tan^ 60° cos^ 0°. 9. X cot8 45° sec2 60° = 11 sin2 90° ; find x. 10. x(cos 30° + 2 sin 90° + 3 cos 45° - sin2 60°) = 2 sec 0° - 6 sin 90° ; find x. 1. sin 60° + 2 cos 45°. 3. sinS 60° + cot8 30°. 5. 4 cos2 30° sin2 60° cos2 0°. 7. 10 cos* 45° sec6 30°. -L-^ b2 PLANE TRIGONOMETRY, [Ch. II. 16. Relations between the trigonometric ratios of an acute angle and those of its complement. When two angles added together make a right angle, the two angles are said to be complementary ^ and each angle is called the complement of the other. For example, the acute angles in a right-angled triangle are complemen- tary ; the complement of ^ is 90° — ^ ; the complement of 27° is 63°. Ex. 1. What are the complements of 10°, 12° 30', 47°, 56° 27', 35° ? Ex. 2. What angles are complementary to 23°, 42°, 51°, 78°, 86° ? In Fig. 2, Art. 12, the angle APM is the complement of the angle A. Now, sinP = 4|; tanP = ^, secP = ^, AP' MP' MP' cosP = ^, cotP = i^, cosecP = 4^. AP AM AM Comparison of these ratios with the ratios of A in (2), Art. 12, shows that cos A = sin P, tan A = cot P, sec A = cosec P, sin A = cos P, cot A — tan P, cosec A = sec P. These six relations can be expressed briefly : Each trigonometric ratio of an angle is eqnal to the corresponding co-ratio of its complement. Ex. Compare the ratios of 30° and 60° ; of 0° and 90°. 17. Exponents in trigonometry. When a trigonometric ratio has an exponent, a particular way of placing the exponent has been adopted. For example, (since) 2 is written sin^jc. There is no ambiguity in the second form, and the advantage is apparent. Thus, cosmic, tan^a;, sec* a;, represent or mean (cosec) 3, (tanx)^ (seccc)*. There is one exponent, however, which must not be written with the brackets removed. This exception is the exponent — 1. Thus, for example, (cos ic)-i, which means , must never be written cos-i x. The reason cosx for this is that the symbol cos-i x is used to represent something else. This symbol denotes the angle whose cosine is x, and is read thus, or is read " the 16-18.] BELATIONS BETWEEN RATIOS. 33 anti-cosine of x," '.'the inverse cosine of x," "cosine minus one x." The number — 1 , which appears in cos-i x, is not an exponent at all, but is merely part of a symbol. Suppose that (a) " the sine of the angle J. is f." The latter idea can also be expressed by saying (&) "J. is the angle whose sine is f " ; or, more briefly, by saying, (c) " ^ is the anti-sine of f." The two ways, (a), (c), of expressing the same idea can be indicated still more briefly by equations, viz., sin ^ = f , A = sin-i f . Thus, (sinic)-i and sin-^a; mean very different things; for (sincc)-i is , which is a number, and sin-i x is an angle. sinic Note. The symbols sin-ia;, cos~ia;, •••, are considered in Art. 88. Ex. Express sin -4 = |, cos jc = |, tan C = 4, sec ^ = 9, cosec A = ^,m the inverse form. 18. Relations between the trigonometric ratios of an acute angle. [N.B. Some relations between these ratios may have been noticed or discovered by the student in the course of his preceding work. If so, they should now be collected, so that they can be compared with the relations shown in this article. ] Some of the preceding exercises have shown that when one trigonometcic ratio of an angle is known, the remaining Jive ratios can be easily determined. This at least suggests that the ratios are related to one another. In what follows, A denotes any acute angle. A. Reciprocal relations between the ratios. Inspection of the definitions (3), Art. 12, shows that : (a) sin A = -, cosec A = -: -, or, sin A cosec A = l; cosec A sm A (b) cos A = -, sec A = -, or, cos ^ sec ^ = 1 ; sec A cos A (c) tan^ = -, cotJ. = 7j or, tan-4cot^ = l. cot -4 tan 4 (1) 34 PLANE TBIGONOMETBT. [Ch. II. B. The tangent and cotangent in terms of the sine and cosine. In the triangle AMP (Fig. 2, Art. 12), MP AM ^^^ * _ MP _AP unA ^ _^ ^ _ AM _ ^lP_ cos^ (9\ m\ ^P ^P O. Relations between the squares of certain ratios. In the triangle AMP (Fig. 2, Art. 12), indicating by mP the square of the length of MPj MP^ + AM^ = AP, On dividing each member of this equation by -, E, F, by the algebraic method. Verify the results by the method used in Art. 14. 8. Find by the algebraic method the ratios required in Exs. 1, 5-7, 10, 11, Art. 14. 9. Express all the trigonometric ratios of an angle A in terms of : (a) cos A ; (&) tan A ; (c) cot A ; (d) sec A ; (e) cosec A. Arrange the results and those of Ex. 2, neatly in tabular form. Prove the following identities : 10. (sec2 ^ - 1) cot2 ^ = 1 ; cos y1 tan yl = sin ^ ; (1 - sin^ A) sec^ A = l. 11. sin2 6 sec2 6 = sec^ ^ - 1 ; tan2 6 - cot^ 6 = sec2 6 - cosec^ 6. io 1 I 1 _i. si n ^ I cos yl _ ., . sec2^ cosec2^ cosec ^ sec^ (tan^ + sec^)2 = l±4^. ^ ^ 1-sin^ 18. sec2 A + cosec2 A = tan2 ^ + cot2 ^ + 2 ; 1 + tan2 A _ sin2 A . cosec A _ ^^^ ^^ 1 + cot2 A cos2^ ' cot ^ + tan J. ^^ cos^ ^ sin^ =sin^ + cos^; 1 — tan A 1 — cot ^ = sec J + tan A ; sec* A — sec2 A = tan* ^ + tan 2 A. sec A — tan A 19.] BUMMABY, 37 Solve the following equations : 15. 2 sin e = 2 - cos d. 16. tan ^ + cot ^ = 2. 17. tan ^ + 3 cot ^ = 4. 18. 6 sec2 - 13 sec 61 + 5 = 0. 19. 8 sin2 ^ - 10 sin ^ + 3 = 0. 20. sin ^ + 2 cos 6 = 2.2. 19. Summary. In this chapter important additions have been made to the knowledge concerning angles that one gained in geometry. A process of measuring angles has been introduced. The close connection between angles and the ratios of lines has been emphasized. It has been shown that each (acute) angle has, associated with it, a definite set of six numbers, called trigonometric ratios ; and it has been seen that the sets of num- bers are different for different angles. It has also been shown that the seven quantities (namely, the angle and the six numbers) are so related, that, if one of the seven be given, then the remaining six can he detennined. A few applications to the measurement of lines and angles have been made in some of the preceding articles. The next two chap- ters are taken up with a formal treatment of such applications. It should be stated, however, that any one who understands the contents of this chapter is in possession of all the principles which will be used in the next two chapters, and can proceed directly to the solution of the problems given there. The student is recom-") mended to attack some of the exercises in Chapters III., IV., \ before reading the explanations given in the text. Attention may again be given to the first part of Art. 14. N.B. Questions and exercises suitable for practice and review on the sub- ject-matter of Chapter II. will be found at pages 182, 183. CHAPTER 111. SOLUTION OF RIGHT-ANGLED TRIANGLES. Before the solution of right-angled triangles is entered upon, a few remarks will be made on the solution of triangles in general. Some of the ideas expressed in Arts. 20-24 are applicable to prac- tical problems throughout the book. 20. Solution of a triangle. Every triangle has three sides and three angles. These six quantities are called the parts or elements of a triangle. Sometimes one or several of the parts of a triangle are known ; for instance, the three sides, two angles and a side, two sides, one side, three angles, and so on. In such cases the questions arise: Can the remaining parts be found or deter- mined ? and, if so, by what method shall this be done ? The pro- cess of deducing the unknown parts of a triangle from the known, is called solving the triangle, or, the solution of the triangle. This Chapter and Chapter VII. are concerned with showing, in detail, methods of solving triangles. There are two methods which can be used to find (only approximately, in general) the unknown parts of a triangle when some of its parts are given. These methods are: (a) The graphical method; (6) The method of computation, 21. The graphical method. This method consists in draiving a triangle which has angles equal to the given angles, and sides proportional to, and thus representing the given sides, and then measuring the remaining parts directly from, the drawing. For example, a triangle has two sides whose lengths are 10 ft., 5 ft., and the included angle is 28° 30' ; the third side and the other angles are required. The graphical solution is as follows : Construct a triangle QPR having two sides, PQ, PB, representing 10 ft., 5 ft., respectively, on some con- 88 20-22.] METHOD OF COMPUTATION. 39 venient scale, and with their included angle, QPB, equal to 28° 30', as shown in Fig. 14. Measure the angles PBQ, PQB with the protractor; measure the side PQ and, by reference to the scale, find the length represented by PQ. [The results thus obtained may be compared with those obtainable by the method of com- putation explained in Arts. 54, 57. The latter results are P = 128° 26' 46", Q = 28° 3' 14", 7^^ = 6.092.] The conditions necessary and sufficient for constructing a tri- angle, and the methods of drawing triangles that satisfy given con- ditions, are shown in plane geometry and in geometrical drawing. It is obvious that the grajjhical method can be employed only when the values of the parts given are consistent loith one another, and when the parts given are sufficient in number to determine a definite triangle. For instance, suppose that one is asked to lind the remaining parts of a triangle one of whose sides is 10 inches long. In this case as many unequal triangles as one please, can be constructed, all of which will satisfy the given condition. Again, a given side and a given angle are insufficient data on which to proceed to find the remaining parts of a triangle, for there is an infinite number of unequal triangles which can have parts equal to the given parts. So also the method fails if three angles be given; for an infinite number of unequal triangles can be drawn whose angles are equal to the given angles. Again, let it be required to find the angles of a triangle whose sides are 10 feet, 40 feet, GO feet. Such a triangle is impossible, since the length of one side (GO feet) is greater than the sum of the lengths of the other two. One more instance : let two given angles be 85° and 105° and the included side be 40 inches ; this triangle is im- l)ossible, since the sum of the two given angles is greater than two right angles. 22. The method of computation. This method is applicable in precisely the same cases in which the preceding method can be employed ; namely, in the cases in which the parts given are con- sistent with one another, and afford conditions sufficient to enable one to construct a definite triangle. This will be fully apparent later, when the various cases will be treated in detail. One of 40 PLANE TlilGONOMETRY, [Cii. III. the principal purposes of this book is to show the different methods of computation applicable to various sets of given condi- tions. Oyie of the principal objects of a student who is taking a first course in trigonometry should he to acquire facility and, above all, accuracy in using these methods of computation. 23. Comparison between the graphical method and the method of computation. The experience gained in some of the exercises in the preceding chapter has probably shown the student that he can attain much greater accuracy by using the method of computation than by using the graphical method. The accuracy of the results obtained by the latter method depends upon the carefulness and skill with which the figures are drawn and measured; in the other method, accuracy depends upon the care and patience em- ployed in performing arithmetical work. While the results attainable by the gr.aphical method, even in the case of skilled persons with excellent drawing instruments, are not as accurate as the results attainable by the other method, yet they are often accurate enough for practical purposes. When the computations required are long and complicated, the graphical method is much the more rapid of the two. There are several reasons why it is advisable for the learner to use the graphical method, as well as the method of computation, in solving problems in this course. The first reason is that the former method icill serve as a check upon the latter. With or- dinary care the graphical method very quickly gives a fair approximation to the result. This result will sometimes show that there is an error in the result obtained by computation. A little error in arithmetic may yield a quantity which is ten times too great or too small ; but this can be detected at once if the other method has also been used.* A second reason for using the graphical method is that this method incidentally provides train- ing in neat, careful, and accurate drawing ; this training will not only be a benefit in itself, but will be of very great advantage in other studies, and especially in the applied sciences. A third * The results can also be tested by methods of computation, which will be sliown in due course. 23-25.] GENERAL BIEECTIONS. 41 reason is that the pupil will gai7i some knowledge and experience of a method that is used in other subjects, for instance, in physics and in mechanics, and that is extensively employed by engineers in solving problems in v/hich the computations required by the other method may be overwhelmingly cumbrous. 24. General directions for solving problems. A third method of approximating to the magnitude of lines and angles may be men- tioned here, for it has often to be employed in practical life. In this method the student may suppose that he possesses neither measuring instruments, drawing materials, nor mathematical tables, and thereupon he may give an off-hand estimate concerning the magnitudes required. This method also serves as a check, by showing when great arithmetical blunders are committed. The pupil is advised to use all three methods in working each practical problem in this course, and to do so in the following order: (1) Make an off-hand estimate as to what the magnitude required may he, and write this estimate down; (2) Solve the problem by the graphical method; (3) Solve the problem by the slower but more accurate and reliable method of computation. There may be some interest found in comparing the results obtained by these three methods. The exercise in judging linear and angular magnitudes afforded by the first method, the practice in neat and careful drawing neces- sary in the second, and the training in accurate computation given by the third, will each afford some benefit to the learner. 25. Solution of right-angled triangles. Let ABC be a right- angled triangle, C being the right angle. In ^b what follows, a, b, c, denote the lengths of the sides opposite to the angles A, B, C, respectively. The sides and angles of ABC are connected by the following relations: „ „ ' r . . (Geometry) 42 PLANE TEIGONOMETBT. [Cu IIL . (Definitions (3), Art. (3), (4) sin^=-=cos5; (5), (6) cos^ = - = sin^ c (7), (8) tan^ = -=cot5; (9), (10) cot^ = - = tan5. Note. To these may be added : sec ^ = - = cosec B, cosec A = ~ = secB. b a These relations, however, are not needed, and are rarely used in solving triangles, for they are equivalent to (3)-(6), and few tables give secants and cosecants. See Art. 18, Note 2. Equation (1) shows that no other element of the triangle can be derived from the two acute angles only. Each of the remain- ing equations, (2)-(10), involves three elements of the triangle, and at least two of these elements are sides. Hence, in order that a right-angled triangle be solvable, two elements must be known in addition to the right angle, and one of these must be a side. If any two of the elements involved in equations (2)-(10) are known, then a third element of the triangle can be found therefrom. Hence the following general rule can be used in solving right-angled triangles : When in addition to the right angle, any two sides, or one of the acute angles and any 07ie of the sides, of a right-angled triangle are Tcnown, and' another element is required, write the equation involving the required element and two of the known elements, and solve the equation for the required element. Eor example, suppose that a, c are known, and that A, B, b are required. In this case, sin.4 = ^, B c 90 -A, b' = Vc' The quantities A, B, b can be found from these equations. If an error has been made in finding A, then B will also be wrong. Hence it is advisable to check the values found by seeing whether they satisfy relations differing from those already employed. 26, 27.] CHECKS. CASES. 43 For example, check formulas which may be used in this case are: - = tan B, -= cos A. a c 26. Checks upon the accuracy of the computation. As already pointed out, large errors can be detected by means of the off-hand estimate and by the use of the graphical method. The calculated results in any example can be checked or tested by employing rela- tions which have not been used in computing the results, and examining whether the newly found values satisfy these relations. An instance has been given in the preceding article. The student is advised not to look up the answers until after he has tested his results in this way. Verification by means of check formulas is necessary in cases in which the answers are not given. The test- ing of the results also affords practice in the use of formulas and in computation. When a check formula is satisfied it is highly probable, but not absolutely certain, that the calculated results are correct. 27. Cases in the solution of right-angled triangles. All the possible sets of two elements that can be made from the three sides and the two acute angles of a right-angled triangle are the following : (1) The two sides about the right angle. (2) The hypotenuse and one of the sides about the right angle. (3) The hypotenuse and an acute angle. (4) One of the sides about the right angle, and an acute angle. (5) The two acute angles. (This case has already been referred to.) Some examples of these cases are solved. The general method of procedure, after making an off-hand estimate and finding an approximate solution by the graphical method, is as follows : First : Write all the relations (or formulas) which are to he used in solving the problem. Second: Write the check formulas. Third: In making the computations arrange the ivork as neatly as possible. 44 PLANE TRIGONOMETRY. [Cu. III. This last is important, because, by attention to this rule, the work is presented clearly, and mistakes are less likely to occur. The computations may be made either with or without the help of logarithms. The calculations can generally be made more easily and quickly by using logarithms. Note 1. Relations (3), (4), Art. 25, may be written: a = csinj., a = c cos B. 'J'hese relations may be thus expressed : A side of a right-angled triangle is equal to the product of the hypotenuse and the cosine of the angle adjacent to the side. A side of a right-angled triangle is equal to the product of the hypotenuse and the sine of the angle opposite to the side. Note 2. Relations (7), (8), Art. 25, may be written: a = & tan ^, a = & cot B. These relations may be thus expressed : A side of a right-angled triangle is equal to the product of the other side and the tangent of the angle opposite to the first side. A side of a right-angled triangle is equal to the product of the other side and the cotangent of the angle adjacent to the first side. EXAMPLES. i. In the triangle ABC, right-angled at C, a = 42 ft., 6 = 56 ft. Find B ' the hypotenuse and the acute angles. I. Computation without logarithms. [Four-place tables. ] tan ^ = ^ = ^ = .7500. .-. A = 36° 62'. 2. b 56 B=QO-A. .♦. ^ = 53° 7'.8. c = Va2 -f- 62 ^ V1764 + 3136. .-. c = 70 ft. Check : a=:ccosB = 70 x cos 53° 7'.8 = 70 x .6000 = 42 ft. II. Computation with logarithms. Given : a = 42 ft. To find :* A = 6 = 56 ft. B = c = Formulas : tan A = -. (1) B = 90°-A. (2) (3) Checks : tan B = — a a 62 sin^ ^' =z(c-H6)(c-6). » This is to be filled after the values of the unknown quantities have been found. It is advisable to indicate the given parts and the unknown parts clearly. 27.] EXAMPLES. 45 Logarithmic formulas : log tan ^ = log a — log b. [See Note 6] , log c = log a — log sin A. log a = 1.62325 log a = 1.62325 log& = 1.74819 log sin ^ = 9.77815 -10 .-. log tan^ = 9.87506 - 10 .-. log c = 1.84510 .-. ^ = 36° 52' 12" .-. c = 70 .-. 5 = 53° 7' 48" The work can be more compactly arranged, as follows : Checks : log a = 1 .62325 log tan B = 10. 12494 - 10 log 6 = 1. 74819 . •. J5 = 53° 7' 48" , log tan A = 9.87506 - 10 c + b = 126 .-. ^ = 36° 52' 12" c-b= 14 .-. 5 = 53° 7' 48" log (c + 6) = 2.10037 log sin A = 9.77815 - 10 log (c - 6) = 1.14613 log c = 1.84510 .-. c = 70 •. log a2 = 3.24650 .-. loga = 1.62325 Note \. The latter form is preferable when all the parts of a triangle are required. Note 2. If there is difficulty in calculating log a — log sin A in the second form, wnce log sin A on the edge of a piece of paper and place it immediately beneath log a. Note 3. The formula tan _B = - can be used instead of (2). A check then is A -\: B = 90°. Instead of (3), one of the following formulas can be used, viz. b a b sin 5' cos 5' cos J. There is often a choice of formulas that can be used in a solution. Note 4. In every example it is advisable to make a complete skeleton scheme of the solution^ before using the tables and proceeding with the actual computation. In the last exercise, for instance, such a skeleton scheme can be seen on erasing all the numerical quantities in the equations that follow the logarithmic formulas. Note 5. Time will be saved if all the logarithms that can be found at one place in the tables, be written at one time. Thus, for example, in the preceding exercise find log sin A immediately after A has been found. 46 PLANE TRIGONOMETRY. LCh. III. Note 6. The logarithmic formulas can be written on a glance at the formulas such as (1), (2), (3). The writing of the logarithmic formulas may be dispensed with when the student has become familiar with calcu- lation by logarithms. A glance at the original formulas will show how the logarithms are to be combined in the computation. 2. In a triangle ABG right angled at C, c = 60 ft., 6 = 50 ft. ; find side a and the acute angles. B ^.^ I. Computation without logarithms. a cos^=- = — = .8333. .-. ^ = 33°33^75. c 60 B = 90° - A. .'. B - 56° 26'. 25. A b~50Ft. a a = c sin ^ = 60 X .5528 = 33.17 ft. ^^^' I''- Check : a = 6 tan ^ = 50 x .6635 = 33.17. II. Computation with logarithms. Given : c = 60 ft. To find : A - 6 = 50 ft. B = Formulas : cos A= — a = c B = 90°- A. Checks: a^=c^-h'^={c-\-h){c- a = 6 tan A. -6). a = c sin A. , Logarithmic formulas : log cos A = log h — log c. (If necessary.) log a = log c + log sin A. log6 = 1.69897 (1) log tan ^ = 9.82173 -10 (6) log c = 1.77815 (2) .-. loga= 1.52070 (7) .-. log cos A = 9.92082 - 10 (3) = (l) + (6) = (l)-(2) c-hb = 110 .-. A = 33° 33' 27" c-b=: 10 .-. B = 56° 26' 33" log (c + b) = 2.04139 log sin ^ = 9.74255 -10 (4) log(c-6)=l .-.log a = 1.52070 (5) .-. log a2 = 3.04139 -(2) + (4) .-. loga = 1.52070 .-.a = 33.10 Note. There is a slight difference between the results obtained by the two methods. This is due to the fact that the calculations have been made with a four-place table in one case, and with a five-place table in the other. A four-place table will give an angle correctly to within one minute ; a five- place table will give it correctly to within six seconds, and sometimes, to within a second. Ex. Make the computation I. with a five-place table. 27.] EXAMPLES. 47 3. In a triangle right angled at C, the hypotenuse is 250 ft., and angle A is 67° 30'. Solve the triangle. I. Computation without loganthms. ^ = 90° - ^ = 90° - 67° 30' = 22° 30'. a = c sin ^ = 250 X sin 67° 30' = 250 x .9239 = 230.98. 5 = c cos ^ = 250 X cos 67° 30' = 250 x .3827 = 95.68. Checks : a^ = (fi — h^, or a = b tan A. II. Computation with logarithms. Given : c = 250 ft. A = 67° 30'. To find : B = a = 6 = Formulas : J5 = 90^ - A. a = c sin A. b = G cos A. Checks : a^ = c'^-b^ = (c-}-b)ic 6). Logarithmic formulas log a = log c + log sin A. log 6 = log c + log cos A. .'. B = 22° 30' log c = 2.39794 log sin ^ = 9.96562 -10 Jogcos^ = 9.58284- 10 .-.log a = 2.36356 .-.log 6 = 1.98078 .-.a = 230.97 .-. b = 95.67 c + 6 = 345.67 c-b = 154.. 33 log (c + 6) = 2.53866 log(c- &) = 2.18845 .-. log a2 = 4.72711 .-.log a = 2.36356 \ 4. In a triangle ABC right angled at C, 6 = 300 ft. and ^ = 37° 20'. Solve the triangle. I. Computation without loganthms. B = 90-A = 90°- 37° 30' = 52° 40'. b 300 c = = 377.3. cos^ .7951 a = 6 tan u4 = 300 X .7627 = 228.8. Checks : a^ = c^ — b"^, a = c sin A. 11. Computation with logarithms. Given Formulas : B 37° 20'. 300 ft. 90° - A. cos^ a = b tan A. To find Checks 48 PLANE TRIGONOMETRY, [Ch. III. .'.B = 52° 40' log 6 = 2.47712 log cos ^ = 9.90043 -10 log tan A = 9.88236 - 10 .-. log c = 2.57669 .♦. loga = 2.35948 .-. c = 377.3 .-. a = 228.8 c + 5 = 677.3 c-h= 77.3 log (c + 6) = 2.83078 log (c- 6) =1.88818 /. log a2 = 4.71896 /. log a = 2.35948 I N.B. Check all results in the following examples. belong to a triangle ABC which is right angled at C. The given elements From the given elements solve the following triangles : 6. c = 18.7, a = 16.98. 6. a = 194.5, b = 233.5. 7. c = 2934, A = 31°- 14' 12''. 8. a = 36.5, B = 68° 52' 9. a = 58.5, b = 100.5. 10. c = 45.96, a = 1.095. 11. c = 324, A = 48° 17'. 12. b = 250, A = 51° 19' 13. c = 1716, ^ = 37° 20' 30". 14. a = 2314, b = 1768. 15. b = 3741, ^ = 27° 45' 20". 16. c = 50.13, a = 24.62. Solve Exs. 17-24 by two methods, viz. : (1) with logarithms ; (2) without logarithms. ^ = 62° 40'. 18. c=9, a = 5. & = 7.5. 20. c = 15, ^ = 39°40^ 5 = 71° 20'. 22. c = 12, a = 8. 5 = 42° 30'. 24. a = 8, 6 = 12. 17. a = 40, 19. a = 4.5, 21. c = 12, 23. 6 = 15, N.B. Questions and exercises suitable for practice and review on the subject-matter of Chapter III. will be found at page 183. CHAPTER IV. APPLICATIONS INVOLVING THE SOLUTION OF RIGHT-ANGLED TRIANGLES. Some practical applications of trigonometry will now be given. It is not necessary that all the problems be solved, or all the articles be considered, before Chapter V. is taken up. 28. Projection of a straight line upon another straight line. If from a point P a perpendicular PO be drawn to the straight line STj then is called the projection of the point P upon the line ST. If perpendiculars be drawn from -two points A, By to Fig. 20. a line LR, and intersect LP in M, N, respectively, then MN is called the projection of AB upon LP. Let I be the length of AB, and let a be the angle at which the two lines AB, LP are inclined to each other. Through A draw AD parallel to LP. Then Projection = MN= AD = AB cos DAB = I cos a. That is, the projection of one straight line upon another straight line is equal to the product of the length of the first line and the cosine of the angle of inclination of the two lines. Note. The projection discussed here, is orthogonal (i.e. perpendicular) projection. If a pair of parallel lines AMi^ BNi^ not perpendicular to Zi?, be drawn through J., B^ then M^Ni is an oblique projection of AB on LB. 49 _ 50 PLANE TBIGONOMETRT. [Ch. IV. EXAMPLES. In working these examples use logarithms or not, as appears most con- venient. Check the results. 1. A ladder 28 ft. long is leaning against the side of a house, and makes an angle 27° with the wall. Find its projections upon the wall and upon the ground. 2. What is the projection of a line 87 in. long upon a line inclined to it at an angle 47° 30' ? 3. What are the projections : (a) of a line 10 in. long upon a line inclined 22° 30' to it ? (6) of a line 27 ft. 6 in. long upon a line inclined 37° to it ? (c) of a line 43 ft. 7 in. long upon a line inclined 67° 20' to it ? (d) of a line 34 ft. 4 in. long upon a line inclined 55° 47' to it ? 29. Measurement of heights and distances. There are various instruments used for measuring angles. The sextant can be used for measuring the angle between the two lines drawn from the observer's eye to each of two distant objects. Horizontal and vertical angles can be measured with a theodolite or engineer's transit. When great accuracy is not required, vertical angles can be measured by means of a quadrant. When an object is above the observer's eye, the angle between the line from the eye to the object, and the horizontal line through the eye and in the same vertical plane as the first line, is called the angle of elevation of the object, or simply the elevation of the object. When the object is below the observer's eye, this angle is called the angle of depression of the object, or simply Hie depression of the object. Fig. 21. Note. In ordinary work engineers get angular measurements exact to within one minute, and in the best ordinary work to half a minute. In very particular work, like geodetic survey, they can get measurements exact to five seconds. For ordinary engineering work five-place tables are gen- erally used ; four-place tables are used in some kinds of work. See Art. 11, Note 1, Art. 27, Ex. 2, Note. >.] HEIGHTS AND DISTANCES. 51 EXAMPLES. A few simple examples are given here ; others ivill be given later. 1. At a point 150 ft. from, and on a level with, the base of a tower, the angle of elevation of the top of the tower is observed to be 60°. Find the height of the tower. Let AB be the tower, and P the point of observa- tion. By the observations, ^lP=150ft., APB = e()°. ^P=^Ptan60°=150xV3 = 150x 1.7321=279.82 ft. 2. In order to find the height of a hill, a line was measured equal to 100 ft., in the same level with the base of the hill, and in the same vertical plane with its top. At the ends of this line the angles of elevation of the top of the hill were 30° and 45°. Find the height of the hill. Let P be the top of the hill, and AB the base line. The vertical line through P will meet AB produced in C. AB = 100 ft., C^P= 30°, OPP=45° j the height CP is required. Let BG = x^ and GP=y. In triangle CAP, CP, ^ AG' CP tan 30°; l-^-lOO Ft Fig. 23. in GBP, BC = tan 45°. Hence, and y. — = tan 30° = .57735, a; + 100 ^ = tan 45° = 1. X (1) (2) From (2), x = y. Substitution in (1) gives 2/=(?/ + 100)x.67735. .-. 2/(1 -.57735) =57.735. .-. CP^ 2/ = ^^:^= 136.6 ft. ^ .42265 3. A flagstaff 30 ft. high stands on the top of a cliff, and from a point on a level with the base of the cliff the angles of elevation of the top and bottom of the flagstaff are observed to be 40° 20' and 38° 20', respectively. Find the height of the cliff. 52 PLANE TRIGONOMETRY, [Cn. IV. Let BP be the flagstaff on the top of the cliff BL, and let C be the place of observation. i?P = 30 ft., LGB = 38° 20', LCP = 40° 20'. Let CL = x, LB = y. In LCB, In LCP, LB = tan 38° 20' tan 40° 20' Hence, on division, On solving for y, LG y- = .7907. X LP LG y±M = .8491. X y ^7907 y + 30 8491* LB = y = 4SiQ.\?> it. 4." At a point 180 ft. from a tower, and on a level with its base, the eleva- tion of the top of the tower is found to be 65° 40. 5'. What is the height of the tower ? 5. From the top of a tower 120 ft. high the angle of depression of an object on a level with the base of the tower is 27° 43'. What is the distance of the object from the top and bottom of the tower ? 6. From the foot of a post the elevation of the top of a column is 45°, and from the top of the post, which is 27 ft. high, the elevation is 30°. Find the height and distance of the column. 7. From the top of a cliff 120 ft. high the angles of depression of two boats, which are due south of the observer, are 20° 20' and 68° 40'. Find the distance between the boats. 8. From the top of a hill 450 ft. high, the angle of depression of the top of a tower, which is known to be 200 ft. high, is 63° 20'. What is the distance from the foot of the tower to the top of the hill ? 9. From the top of a hill the angles of depression of two consecutive mile-stones, which are in a direction due east, are 21° 30' and 47° 40'. How high is the hill ? 10. For an observer standing on the bank of a river, the angular elevation of the top of a tree on the opposite bank is 60° ; when he retires 100 ft. from the edge of the river the angle of elevation is 30°. Find the height of the tree and the breadth of the river. 11. Find the distance in space travelled in an hour, in consequence of the earth's rotation, by an object in latitude 44° 20'. [Take earth's diameter equal to 8000 mi.] 12. At a point straight in front of one corner of a house, its height subtends an angle 34° 45', and its length subtends an angle 72° 30' j the height of the house is 48 ft. Find its length. 30.] THE MARINER S COMPASS. 53 30. Problems requiring a knowledge of the points of the Mariner's Compass. The circle in the Mariner's Compass is divided into 32 equal parts, each part being thus equal to 360° -r- 32, i.e. The points of division are named as indicated on the tigure. Hi' Fig. 25. It will be observed that the points are named with reference to the points North, South, East, and West, which are called the cardinal points. Direction is indicated in a variety of ways. For instance, suppose C were the centre of the circle ; then the point P in the figure is said to bear E.N.E. from C, or, from C the hearing of P is E.N.E, Similarly, C bears W.S.W. from P, or, the bearing of C from P is W.S.W. The point E.N.E. is 2 poii;ts North of East, and 6 points East of North. Accordingly, the phrases E. 22i° N., N. 67i° E., are sometimes used instead of E.N.E. EXAMPLES. 1. Two ships leave the same dock at 8 a.m. in directions S.W. by S., and S.E. by E. at rates of 9 and 9| mi. an hour respectively. Find their distance apart, and the bearing of one from the other at 10 a.m. and at noon. 2. From a lighthouse L two ships A and B are observed in a direction N.E. and N. 20° "W. respectively. At the same time A beai's S.E. from B. If LA is 6 mi., what is LB ? 54 PLANE TRIGONOMETRY. [Ch. IV. 31. Mensuration. Let ABC be any triangle, and let the lengths of the sides opposite the angle A, B, C be denoted by a, 6, c, respec- tively. From any vertex C draw CD at right angles to the opposite side AB. It has been shown in arithmetic and geometry, that the area of a triangle is equal to one-half the product of the lengths of any side and the perpendicular drawn to it from the opposite vertex. [In (1) A is acute, in (2) A is obtuse.] area ABC (Tig. l)=iAB' DC; = iAB- AC sin A; = ibc sin A. area ABC (Fig. 2) = iAB'DC; = iAB'ACsmCAD; = ibc sin (180 - A). It will be seen in Art. 45, that sin (180 — A) = sin A. Hence, the area of a triangle is equal to one-half the product of any two sides and the sine of their contained angle. EXAMPLES. 1. Find the area of the triangle in which two sides are 31 ft. and 23 ft. and their contained angle 67° 30'. area = §l-><^ X sin 67° 30' = 31 x 23 x .92388 ^ 339 37 f ^^ 2 2 2. Find area of triangle having sides 125 ft., 80 ft., contained angle 28° 36'. 3. Find area of triangle having sides 125 ft., 80 ft., contained angle 151° 25'. [Draw figures carefully for Exs. 2, 3.] 4. Find area of parallelogram two of whose adjacent sides are 243, 315 yd., and their included angle 35° 40'. 32.J ISOSCELES TiiiANGLES. 55 5. Find area of parallelogram two of whose adjacent sides are 14, 15 ft., and included angle Ib"^. 6. Find area of triangle having sides 40 ft., 45 ft., with an included angle 28° 57' 18". 7. Write two other formulas for area ABC, similar to that derived above. Also, derive them. 32. Solution of isosceles triangles. In an isosceles triangle, the perpendicular let fall from the vertex to the base bisects the base and bisects the vertical angle. An isosceles triangle can often be solved on dividing it into two equal right-angled triangles. EXAMPLES. 1. The base of an isosceles triangle is 24 in. long, and the vertical angle is 48° ; find the other angles and sides, the perpendicular from the vertex and the area. Only the steps in the solution will be indicated. Let ABC be an isosceles triangle having base AB = 24 in., angle C = 48°. Draw CD at right angles to base ; then CD bisects the angle ACB and base AB. Hence, y in the right-angled triangle ADC, AD — \ AB = 12, / ACD = I ACB = 24°. Hence, angle A, sides AC, / DC, and the area, can be found. / 2. In an isosceles triangle each of the equal sides is s~-i2-"- 363 ft., and each of the equal angles is 75°. Find the p « 28 base, perpendicular on base, and the area. 3. In an isosceles triangle each of the equal sides is 241 ft., and their included angle is 96°. Find the base, angles at the base, height, and area. 4. In an isosceles triangle the base is 65 ft. , and each of the other sides is 90 ft. Find the angles, height, and area. 6. In an isosceles triangle the base is 40 ft., height is 30 ft. Find sides, angles, area. 6. In an isosceles triangle the height is 60 ft., one of equal sides is 80 ft. Find base, angles, area. 7. In an isosceles triangle the height is 40 ft., each of equal angles is 63°. Find sides and area. 8. In an isosceles triangle the height is 63 ft., vertical angle is 75°. Find sides and area. 33. Related regular polygons and circles. The knowledge of trigonometry thiis far attained, is of service in solving many 56 PLAKB TJUQONOMETHY. [Ch. IV. problems in which circles and regular polygons are concerned. Some of these problems are : (a) Given the length of the side of a regular polygon of a given number of sides, to find its area; also, to find the radii of the inscribed and circumscribing circles of the polygon ; (b) To find the lengths of the sides of regular polygons of a given number of sides which are inscribed in, and circumscribed about, a circle of given radius. J^IQ. 29. Eor example, let AB (Fig. 29) be a side, equal to 2 a, of a regu- lar polygon of n sides, and let C be the centre of the inscribed circle. Draw CA, CB, and draw CD at right angles to AB. Then D is the middle point of AB. 360° 180° By geometry, angle ACD = ^ ACB = i It IV Also, by geometry, Tangle DAO = ^(^^^^0^ = f^)^*^"-! Hence, in the triangle ADC, the side AD and the angles are known ; therefore CD, the radius of the circle inscribed in the polygon, can be found. On making similar constructions, the solution of the other problems referred to above will be apparent. The perpendicular from the centre of the circle to a side of the inscribed polygon is called the apothem of the polygon. EXAMPLES. 1. The side of a regular heptagon is 14 ft. : find the radii of the inscribed and circumscribing circk\s ; also, find the difference between the areas of the heptagon and the inscribed circle, and the difference between the area of the heptagon and the area of the circumscribing circle. 34.] OBLIQUE TRIANGLES. 67 2. The side of a regular pentagon is 24 ft. Find quantities as in Ex. 1. 3. The side of a regular octagon is 24 ft. Find quantities as in Ex. 1. 4. The radius of a circle is 24 ft. Find the lengths of the sides and apo- thems of the inscribed regular triangle, quadrilateral, pentagon, hexagon, heptagon, and octagon. Compare the area of the circle and the areas of these regidar polygons ; also compare the perimeters of the polygons and the circumference of the circle. 5. For the same circle as in Ex. 4, find the lengths of the sides of the circumscribing regular figures named in Ex. 4. Compare their areas and perimeters with the area and circumference of the circle. 6. If a be the side of a regular polygon of n sides, show that i?, the 180° radius of the circumscribing circle, is equal to ^ a cosec ; and that r, the 180° ^ radius of the circle inscribed, is equal to A a cot n 7. If r be the radius of a circle, show that the side of the regular inscribed 180° polygon of n sides is 2 r sin ; and that the side of the regular circum- 180° ^ scribing polygon is 2 r tan . n 8. If a be the side of a regular polygon of n sides, B the radius of the cir- cumscribing circle, and r the radius of the circle inscribed, show that area of polygon = i naP' cot — = 1 nija sin '^ = nr^ tan — • n n n 34. Solution of oblique triangles. Since an oblique triangle can be divided into right-angled triangles by drawing a perpendicular from a vertex to the opposite side, it may be expected that know- ledge concerning the solution of right-angled triangles will be of service in solving oblique triangles. This expectation will not be disappointed. An examination, which it is advisable for the stu- dent to make before proceeding farther, will show that all the sets of data from which a definite triangle can be drawn are those indicated in (l)-(4) below. The ability to make the following geo- metrical constructio7is is presupposed : (1) To draw a triangle on being given two of its angles and a side oppo- site to one of them ; (2) To draw a triangle on being given two of its sides and an angle oppo- site to one of them ; (3) To draw a triangle on being given two of its sides and their included angle ; (4) To draw a triangle on being given its three sides. 58 PLANE TRIGONOMETRY, [Ch. IV. In what follows, only the steps in the solutions will be indi- cated. The examples that are worked may be saved, so that the amount of labor required by the method of solution shown here can be compared with the amount required by another method which will be described later. There are four cases in the solution of oblique triangles ; these cases correspond to the four problems of construction stated above. Case I. Given tioo angles and a side opposite to one of them. o c Fig. 30. In ABG let A, B, a be known. Angle C and sides 6, c are re« quired. From G draw CD at right angles to AB or AB produced. In triangle CBD, angle CBD and side CB are known. .-. BD and DC can be found. Then, in triangle ACD^ side DG and angle A are known. .'.AG and AD can be found. Side AB = AjD-BD when B is obtuse, smd AB = AD + DB when B is acute. Angle G=1S0°-(A + B). Another method of solution is given in Art. 55. 1. Ex. 1, Art. 55. 3. Ex. 2, Art. 60. EXAMPLES. 2. Ex. 2, Art. 55. 4. Other Exs. in Arts. 65, 60. Case II. Given two sides and an angle opposite to one of them. N.B. The first part of the text in Art. 56 should be read at this time. Let (Fig. 30) AG, BG, angle A be known. [In a certain case, as shown in Art. 56, two triangles can be drawn which satisfy the given conditions.] From G draw CD at right angles to AB or AB produced. OnLiqUM TMtANGLES, 59 In ACB, AC and A are known. .-. AD, DC, angle ACD, can be found- Then, in BCD, BC and CD are known. .-. BD, angle DBC, can be found. In one figure, AB = AD- BD, angle ABC = 180° - CBD. In other figure, AB = J.Z> + DB. In both figures, angle ACB = 180° - (CAB + ^5(7). Another' method of solution is given in Art. ^Q, EXAMPLES. 1. Ex. 1, Art. 56. 2. Ex. 2, Art. 56. 3. Ex. 1, Art. 60. 4, Other Exs. in Arts. 66, 60. Case III. Given two sides and their included angle. In ABC let b, c, A be known. Side a, B, C are required. From C draw CD at right angles to AB or AB produced. In ACD, AC and angle A are known. .-. CD and AD can be found. Then, in triangle CDB, CD is now known, and BD=AD—AB or ^5 - AD. .-. Angle C5i> can be found. Angle ABC = 180° - CBD in figure on the left. Angle ACB = 180° -{A + B). Another method of solution is given in Art. 57. EXAMPLES. 1. Ex. 1, Art. 57. 2. Ex. 2, Art. 67. 3. Ex. 1, Art. 61. 4. Other Exs. in Arts. 57, 61. 6, Ex. in Art. 21. Case IV. Given the three sides. In ABC let a, 6, c be known. The angles A, B, C are required. From any vertex C draw CD at right angles to AB or AB pro- duced. CD'=b'-AD'', (1) also, CD' = a^ - DjB^ = a^ - (c - ADf. .-. 62 _ ^/>2 ^a'-(c- ADf. .%^Z> = ^' + ^'-< (2) 2c ^ '^ 60 PLANE TRIGONOMETRY, [Ch. IV. Also, DB = AD — c (one figure), or c — AD (other figure). Hence, in ACD, AC, AD are known. .-. A can be found. Also, in CDB, CB, DB are known. .-. B can be found. C = 1S0°-(A + B), Another method of solution is given in Art. 58. EXAMPLES. 1. Ex. 1, Art. 58. 2, Ex. 2, Art. 58. 3. Ex. 1, Art. 62. 4. Other Exs. in Arts. 58, 62. 5. Solve some of the problems in Art. 63 by means of right-angled triangles. 34 a. The area of a triangle in terms of the sides. (See Fig. 30.) From (1), (2), Case IV., Art. 34, CT2 _ ^2 f h' + c^ - ay ^ 4 c'h' - Qf- + c^ - a^ \ 2c J 4c2 _ [2 c6 + ?>' + c^ - a^M2 ch - (b' + c^ - a')1 4c2 _ [(6 + cY - g^ [«' - (^ - c)n 4c2 ^ (g + 6 + c)(- g + 5 + c)(a - & + c)(a + 6 - c) 4c2 Let a + b + c = 2s; then 2(s — g) = g4-& + c — 2g = — g + 6+c. Similarly, 2 (s — b) = a — b -\- c; 2 (s — c) = a + b — c. 2 I Then CD = - Vi, DiD can be computed. Now area ABCD = trapezoid D^DCCi + trapezoid CiCBBi — triangle ADDi — triangle ABBi. The areas in the second member can be computed ; it will be found that area ABCD = 26 acres. Note. Sometimes the bearing and length of one of the lines enclosing the area is also required. These can be computed by means of the latitudes and departures of the given lines. The formulation of a simple rule for doing this is left as an exercise to the student. 2. In Ex. 1, deduce the length and bearing of DA from the lengths and bearings of AB, BC, CD, 3. A surveyor starts from A and runs 4 chains S. 45° E. to 5, thence 5 chains E. to (7, thence 6 chains N. 40° E. to D. Find the distance and bear- ing of A from D ; also, the area of the field ABCD. Verify the results by going around the field in the reverse direction, and calculating the length and bearing of BA from the lengths and directions of AD, DC, CB. 4. A surveyor starts from one corner of a pentagonal field, and runs N. 25° E. 433 ft., thence N. 76° 55' E. 191 ft., thence S. 6° 41' W. 539 ft., thence S. 25° W. 40 ft., thence N. 65° W. 320 ft. Find the area of the field. Deduce the length and direction of one of the sides from the lengths and directions of the other four. 5. From a station within a hexagonal field the distances of each of its corners were measured, and also their bearings ; required its plan and area, the distances in chains and the bearings of the corners being as follows : 7.08 N.E., 9.57 N. iE., 7.83 N.W. by W., 8.25 S.W. by S., 4.06 S.S.E. 7°E., 5.89 E. by S, 3i°E. 35. Summary. Chapter II. was concerned with defining and investigating certain ratios inseparably connected with (acute) angles, and attention was directed to the tables of these ratios and their logarithms. In Chap. III. it was shown how these definitions and tables can be used in finding parts of a right- angled triangle when certain parts are known. In Chap. IV. the knowledge gained in Chap. III. was employed in the solu- tion of some of the many problems in which right-angled triangles appear. In Art. 34 it has been seen that this knowledge can serve for the solution of oblique triangles. It follows, then, that it can serve for the solution of problems in which oblique tri- angles appear, and, accordingly, for the solution of all problems 64 PLANE TRIGONOMETRY. [Cn. IV. invoh)iny the measurement of straight lines only. Consequently, the student is now able, without any additional knowledge of trigonometry, to solve the numerical problems in Chaps. VII., VIII. It is thus apparent that even a slight acquaintance with the ratios defined in Chap. II. has greatly increased the learner's ability to solve useful practical problems. Oblique triangles can sometimes be solved in a more elegant manner than that pointed out in Art. 34. In order to show this, further consideration of angles and the trigonometric ratios is necessary. Consequently, in Chap. V. some important additions are made to the idea of a straight line and the idea of an angle ; the trigonometric ratios are defined in a more general way, namely, for all angles, instead of for acute angles only, and the principal relations of these ratios are deduced. Chapter VI. treats of the ratios of two angles in combination. While it is necessary to con- sider these matters before proceeding to the solution of oblique triangles given in Chap. VII., it should be said that the know- ledge that will be gained in Chaps. V., VI., VII., is necessary and important for other purposes besides the solution of triangles. In fact, the latter is one of the least important of the results obtained in these chapters. N.B. Questions and exercises suitable for practice and review on the sub- ject-matter of Chap. IV. will be found at page 184. CHAPTER V. TRIGONOMETRIC RATIOS OF ANGLES IN GENERAL. 36. Directed lines. Let MN be a line unlimited in length in the directions of both M and N. Suppose that a point starts at P and moves along this line for some given distance. In order to mark where the point stops, it is necessary to know, not only this distance, but also the direction in which the point has moved from P. This direction may be indicated in various ways ; by saying, for instance, that the point moves toward the right from P, or toward the left from P; that the point moves toward N, or toward M; that the point moves in the direction of JSf, or in the direction of M; and so on. Mathematicians, engineers, and others F P B CD I I I I I M N riG. 31. have agreed to use a particular method (and this practically comes to the adoption of a particular rule) for indicating the two oppo- site directions in which a point can move along a line, or in which distances along a line can be measured. This convention, or rule which has been adopted for the sake of convenience, is as follows : Distances measured along a line, or along parallel lines, in one direction shall he called positive distances, and shall be denoted by the sign +/ distances measured iyi the opposite direction shall be called negative distances, and shall be denoted by the sign — . The convenience of this custom, fashion, or rule, will become apparent in the examples that follow.* In Fig. 31 let distances * Advances in mathematics have often depended upon the introduction of a good custom which has at last been universally adopted and made a rule. Thus, for example, the custom of using exponents to show the power to which a quantity is raised, which was first introduced in the first half of the sixteenth century, and made gradual progress until its final establishment in the latter half of the seventeenth century, has beeu of great service in aiding the advances of algebra. 65 66 PLANE TRIGONOMETRY, [Ch. V, measured in the direction of JVbe t2^Q\i positively ; then distances measured in the direction of M will be taken negatively. On directed lines the direction in which a line is measured, or in which a point moves on a line, is indicated by the order of the letters naming the line. Thus, for example, if a point moves from B to (7, the distance passed over is read BC. In this reading, the starting point is indicated by the first letter B, and the stopping point, by the last letter C. After the same fashion, CB means the distance from C to B. If, for instance, there are 3 imits of length between B and (7, then ^(7 = + 3, 05 = - 3. EXAMPLES. 1. Suppose a point (Fig. 31) moves from P to J5, thence to C, thence to Z>, thence to F. Let the number of units of length between P and i?, B and O, C and Z>, F and D, be 2, 3, 2, 10, respectively. The point starts at P and stops at F ; hence the distance from the starting point to the stopping point is PF. In this case the point's trip from P to i^ is made in several steps as indicated above. That is, on properly indicating the lines passed over, PF=PB-hBC + CD + DF = 2 + 3 + 2-10 [•.• FD = + 10, then DF = - 10.] = -3. This shows that the final position of the moving point is three units to the left of P. This example also shows one great convenience of the rule of signs in measurement^ namely, that by attending to this rule and to the proper naming of the lines passed over by a moving point, one immediately obtains the result of the successive movements. ' Note. In the following examples, in lines that lie east and west, let measurements toward the east be taken positively ; in lines that lie north and south, let measurements toward the north be taken positively. 2. A man travelling on an east and west line goes east 20 mi., then east 16 mi., then west 18 mi., then east 30 mi. What is his final distance from the starting point ? [Draw a figure, and indicate the successive trips by letters.] 3. A man travelling on an east and west line goes west 20 mi., then east 10 mi., then east 25 mi., then east 30 mi., then west 45 mi. Do as in Ex. 2. 4. A man travelling on a north and south line goes north 100 mi., then south 60 mi., then south 110 mi., then north 200 mi., then north 15 mi., then south 247 mi. Do as in Ex. 2. 37.] TRIGONOMETBIC RATIOS OF ANGLES. 67 37. Trigonometric definition of an angle. Angles unlimited in magnitude. Positive and negative angles. In books on plane geometry a plane angle is defined in various ways, namely, as the inclination of two lines to one another, which meet together, but are not in the same direction; or, as the figure formed by two straight lines drawn from the same point; or, as the amount of divergence of two lines which meet in a point, or would meet if produced; or, as the opening between two straight lines which meet ; or, as the difference in direction of two lines which meet ; and so on. In these definitions an angle is always regarded as less than two right angles. A definition according to which angles are less restricted, is adopted in trigonometry. Trigonometric definition of an angle. The angle between two lines which intersect is the amount of turning which a line revolving about their point of intersection makes, when it begins its revolution at the position of one of the two lines and stops in the position of the other line. Thus, for example, the angle between OX and OQ is the amount of turning which is made by a line OP revolving about when OP starts re- volving from the position OX and stops its revolution at the position OQ. The line OX at which the revolution begins, is called the initial line; the line OQ at which the revolution ends, is called the terminal line ; when the turning line OP has reached the terminal position OQ, OP is said to have described the angle XOQ. Let YOTi be at right angles to XiOX. When OP has revolved until it lies in the position Y, it has described a right angle, or 90° ; when it has revolved until it lies in the position OXi, it has described two right angles, or 180° (this is usually termed "a straight angle " or " a flat angle '') ; when OP keeps on turning until it is in the position Fj, it has described three right angles, or 270°; when OP has again reached the position OX, that is, when it has made one complete revolution, it has described four right angles, or 360°. 68 PLANE TRTGONOMETBT, [Ch. V. Angles unlimited in magnitude. Now OP may start revolving from OX, make one complete revolution, continue to revolve, and then cease revolving when it has again reached the position OQ. This is indicated in Fig. 33. Or, OP may make two complete revolutions before it comes to rest in the position OQ ; or, it may make three revolutions, or four, or as many as one please, before ceasing its revolution at the position OQ. An an- gle of 360° is described each time that OP makes a complete revolution, and OP can make as many revolu- tions as one please. According to the trigonometric definition of an angle, therefore, angles are unlimited in magnitude. Moreover, when this definition of an angle is adopted, the same figure can represent an infinite number of different angles* Any two of these angles differ from each other by a whole number of complete revolutions. For instance, Fig 34 may represent 60°, 360° -f 60° or 420°, 2 . 360° + 60° or 780°, 3 • 360° + 60° or 1140°, '",n- 360° + 60°, in which n denotes any whole number. A7iy ttvo of these angles i differ by a multi2:)le o/360°. Angles which have the same initial and terminal lines may be called coterminal angles. Positive and negative angles. The revolving line OP (Fig. 32) may revolve about in the same direction as that in which the hands of a watch revolve, or it may revolve in the opposite direc- tion. The following convention (see Art. 36) has been adopted for the sake of distinguishing these two opposite directions : When the turning line revolves in a counter-clockwise direction, the angles described are said to be positive, and are given the plus sign; when the turning line revolves in a clockwise direction, the angles described are said to be negative, and are giveii the minus sign. Thus, for example. Fig. 34 represents the angles -f 60°, — 300°; further, this figure represents the angles 60° ± 7i • 360°, in which n denotes any whole number. The angle —300° is included in these angles, for, on putting —1 for n, there is obtained 60°— 360°, i.e. — 300°. (Negative angles are also unlimited in magnitude.) Initial Line Fia. 34. 38.] SUPPLEMENT AND COMPLEMENT. 69 As in the case of lines, the sign of an angle can he denoted by the order of the letters used in naming the angle. Thus XOQ denotes the angle formed by revolving OX toward OQ, and QOX denotes the angle formed by revolving OQ toward OX. Accordingly, qox= -XOQ. Quadrants. In Fig. 32, XOY, YOX^, X^OT^, Y,OX, are called the first, second, third, and fourth quadrants, respectively. When the turning line ceases its revolution at some position be- tween OX and Y, the angle described is said to be an angle in the first quadrant; when the final position of the turning line is between Y and OXj, the angle described is said to be in the sec- ond quadrant; and so on for the third and fourth quadrants. For example, the angles 30°, - 345°, 395°, 725° are all in the first quad- rant ; the angles — 60°, 340°, 710° are all in the fourth quadrant ; the angle — 225° is in the second quadrant, and the angle 225° is in the third quadrant. Note. While all acute angles are in the first quadrant, all angles which are in the first quadrant are not acute. EXAMPLES. Note. When it is necessary, the number of revolutions and their direction may be indicated on the figure in the manner shown in Fig. 34. Lay off the following angles with the protractor: In the case of each angle name the least positive angle that has the same terminal line. Name the quadrants in which the angles are situated. In the case of each angle name the four smallest positive angles that have the same terminal line. 1. 137°, 785°, 321°, 930°, 840°, 1060°, 1720°, 543°, 3657°. 2. _ 240°, - 337°, - 967°, - 830°, - 750°, - 1050°, - 7283°. 3. _ 470 ^ 230° + 37°, 420° - 470° + 210° - 150°, 230° - 47° + 37°, 230° + 37° - 47°. 38. Supplement and complement of an angle. TJie supplement of an angle is that angle which must be added to it in order to make two right angles, or 180°; the complement of an angle is that angle which must be added to it in order to make one right angle, or 90°. Thus, if A be any angle, then supplement of angle A = 180° — A, complement of angle A = 90° — A. 70 PLANE TRIGONOMETRY, [Ch. V. EXAMPLES. 1. What are the complements and supplements of 40°, 227°, — 40° ? complement of 40°= 90°- 40°= 50°; supplement of 40° = 180° - 40° = 140°. complement of 227° = 90° - 227° = - 137° ; supplement of 227° = 180° - 227° = - 47°. complement of - 40° = 90° - (- 40°) = 130° ; supplement of - 40° = 180° - (- 40°) = 220°. 2. By means of a figure verify the results obtained in Ex. 1. 3. What are the complements of - 230°, 150°, - 40°, 340°, 75°, 83°, 12°, - 295°, - 324°, 200°, 240°, - 110°, - 167° ? 4. What are the supplements of the angles in Ex. 3 ? 6. Verify the results in (3), (4), by drawing figures. 39. The convention of signs on a plane. Articles 36, 37 contain statements of the conventions adopted regarding the algebraic signs to be given to distances measured on parallel straight lines, and to angles described by the revolution of a turning line. A figure, such as Figs. _^__ 32, 35, will be frequently used in the j -^ articles that follow. In this figure, ! OX is the initial line, the turning P4* line revolves about O, and YOY^ is at right angles to X^OX. The fol- PiG. 35. lowing convention has been adopted re- garding the lilies which will be used : Horizontal lines measured in the direction of X are taken positively ; Horizontal lines measured in the direction of Xi are taken negatively ; Vertical lines measured upward are taken positively ; Vertical lines measured downward are taken negatively. The distance of points, such as Pj, P2, P3, Pi, from X^X, is always measured /rom X^X toward the points. Any turning line (or oblique line) as OP is measured positively -P. X, 40.] GENERAL DEFINITION OF RATIOS. 71 from toward the end of the turning line which lies in the direction of X from O when the turning line coincides with the initial line. Thus a distance + 3 on OP will terminate at T, dis- tant 3 units from 0, and a distance — 3 on OP will terminate at Ti, distant 3 units from 0, but in the direction opposite to the former. This is sometimes briefly expressed in the words: the turning line carries its positive direction with it in its revolution, 40. General definition of the trigonometric ratios. The remarks in this article apply to each of the four figures below. In each figure, O is the point about which the angle is described, OX is the initial line, and OP is the terminal line. The first figure represents any angle in the first quadrant; the second figure represents any angle in the second quadrant ; the third figure, any angle in the third quadrant ; and the fourth figure, any angle m the fourth quadrant. In each figure the angle will be called A. X M X Fig. 36. Let P be any point in OP, the terminal line of any angle A. From P draw PM at right angles to the initial line OX, or to the initial line produced in the negative direction. In each figure, OM is the distance measured along X^OX from the point to the foot of the perpendicular MP, and MP is the distance from XiOX to the point P. Following are the definitions of the trigo- nometric ratios ; these definitions apply to the angles represented in Fig. 36, and, accordingly, to all angles whatsoever. ^Particular attention should be j^aid to the order of the letters used in naming the lines, for this order iiidicates the direction in which the line is measured. See Art. 36.] * 72 PLANE TBIGONOMETRT, [Ch. V. MP The ratio -— ■ is called the sine of the angle A, The ratio -— — - is called the cosiyie of the angle A. MP The ratio -— — : is called the tangent of the angle A, The ratio •— —- is called the cotangent of the angle A. MP OP The ratio -- — is called the secant of the ansrle A, OM OP The ratio -— — is called the cosecant of the angle A, MP These definitions may be briefly stated : (1) '•»-=^- — ^- ^--W,- A = H~' cot A = ^^» cosec A = OP MP MP Inspection will show that the definitions of the trigonometric ratios for acute angles given in Art. 12, are in accordance with these general definitions. N.B, The projection definitions of the trigonometric ratios are given in Note 13, Appendix. 41. The algebraic signs of the trigonometric ratios for angles in the different quadrants. Figures 36 show that if the angle A is in the first, second, third, fourth quadrants, then the algebraic sign of MP is +, +, — , — , respectively, and the algebraic sign of OM is +, ^, — , 4-, respectively. As stated in Art. 39, OP is always taken positively. Hence, on paying regard to the algebraic signs of OM, MP^ OP, in the several quadrants, it will be seen that the ratios of the angles in these quadrants are posi- tive or negative, as indicated in the fc^lowing table : 41.] ALGEBRAIC SIGNS OF RATIOS. 73 Quadrant, I. II. III. IV. Sine + + Cosine + — . — + Tangent .... + — + - Cotangent .... -f — + — Secant . . * . -1- — — + Cosecant .... + + — — Fig. 37. EXAMPLES. The student is advised to preserve his work on these examples. If he regards his results attentively, he will probably discover some useful facts, and he able to deduce some useful theorems^ concerning angles in general. Any preceding results, such as those in Art. 15, may be used as an aid in solving these exercises. 1. Find the ratios of 945°. -^i ^ "1 ^ ^ 945° = 2 X 360° + 225°. .•• OP, the terminal line of angle 945°, has the position shown in Fig. 37. For this position of the terminal line, 03/ and ilfP are both negative. As shown in Art. 15, the lines OM, MP, OP, in this figure are respectively proportional to 1, 1, y/2. These are indicated on the figure with their proper algebraic signs. It is immediately apparent that sin 945°=--^, cos 945° = - -i^, tan 945° =+1, cot 945° =+1, \/2 \/2 sec 945° = - \/2, cosec 945° = - \/2. 2. Construct, and find the ratios of, 420°, 780°, 1140°. 3. Construct, and find the ratios of, 120°, 480°, 240°, 600°, - 60°, 300°, 660°, -720°. 4. Construct, and find the ratios of, 150°, 410°, 210°, -150°, 330°, -390°. 5. Construct, and find the ratios of, 45°, 765°, 135°, - 225°, 225°, 585°, -405°, 1035°. 6. Construct, and find the ratios of, - 754°, 487°, - 245°. 7. Compare the ratios of 90° - 30°, 90° - 60°, 90° - 45°, 90° - 135°, 90° - 240°, 90° - 300°, with the ratios of 30°, 60°, 45°, 135°, 240°, 300°, respectively. 74 PLANE TRIGONOMETRY, [Ch. V. 8. Compare the ratios of 90° + 30°, 90° + 60°, 90° + 45°, 90° + 135°, 90° + 240°, 90° + 300°, with the ratios of 30°, 60°, 45°, 135°, 240°, 300°, respectively. 9. Compare the ratios of 180° - 30°, 180° - 60°, 180° - 45°, 180° - 135°, 180° - 240°, 180° - 300°, with the ratios of 30°, 60°, 45°, 135°, 240°, 300°, respectively. So, also, the ratios of — 30°, — 60°, — 45°, etc. 10. Are any general relations indicated by the results of Exs. 7, 8, 9 ? If so, state these relations. Try to prove them. 42. To represent the angles geometrically when the ratios are given. In constructing the angles in tins article it is necessary to bear in mind that, according to the definitions given in Art. 40 : When MP is positive, it can he drawn in the Jirst and second quadrants; When MP is negative, it can be drawn in the third and fourth quadrants; When OM IS positive, it is to be drawn in the direction OX; When OM is negative, it is to be drawn in the direction OXi ; and OP is to be taken positively. EXAMPLES. 1. Represent by a figure the angles which have sines equal to f . Calcu- late their other ratios. Let A denote an angle whose sine is | ; i.e. let sin^ = f. But sin ^ = :^ (Art. 40). Hence, iJfP:OP = 3:4; and if OP = 4, then MP = 3. Now, MP can be drawn . positively in both the first and second quadrants. Hence the problem amounts to finding a point in the first quadrant and a point in the second quadrant, each at a distance 4 from and a distance 3 from XiOX. The result is indicated in Fig. 38. The student can make the construction for himself. The angles having sines equal to f, accord- ingly, include all the angles which have OP for a terminal line, and all the angles which have OPi for a terminal line. By Art. 37 each of these two sets of angles consists of an infinite number of angles, any two of which differ from one another by a whole number, positive or negative, of complete revolutions. A general algebraic expression which includes all these angles, iti deduced in Art. 85. X, M, 42-43.] GEOMETRIC REPRESENTATIOK. 75 Figure 38 shows that for angles having OP for a terminal line, cosine is ■ — , tangent is — -, etc. ; and that for angles having OPi for a terminal line, 4 _ V7 ^-7 Q cosine is , tangent is — — , etc. Since the given sine is positive, it is 4 VV apparent that the angles required, must be in the first and second quadrants. (See Art. 41.) 2. Bepresent by a figure all the angles which have tangents equal to — f . Let A denote an angle whose tangent is — | ; i.e. let tan A--^. prhis may be written, i-^ or — -1 But tan A = ^^^ (Art. 40). Hence, if MP = 3, then OM = -i, and if MP = - 3, then 031 = 4. When 3IP is positive and 03/ is negative, OP can lie p-v only in the second quadrant. When 1 ^v MP is negative and OM is positive, 3| >s. OP can lie only in the fourth quadrant. j _ _ ^s ^ ilfi Figure 39 represents the angles. The x M -4 student can make the construction for himself. Thus, the angles whose tan- gents are equal to — f , consist of the set of angles, infinite in number, which ^^^' ^' have OP for a terminal line, and the set of angles, infinite in number, which have OPi for a terminal line. 3. Calculate the other ratios of the angles in Ex. 2. 4. Represent geometrically all the angles whose cosine is |. Calculate their other ratios. 5. So, also, when the cosine is — |. r 4 14 4-1 6. So, also, when the tangent is |. Note. - = -^—~ = — -• 7. So, also, when the sine is — f . 8. So, also, when the secant is f. 9. So, also, when the secant is — |. 10. So, also, when the cosecant is — 2 ; |. N.B. The student is now strongly recommended to delay the reading of the next article until after he has reviewed the properties stated in Art. 13, and, if possible, determined what are the correct corresponding statements for angles in general. 43. Connection between angles and the trigonometric ratios. For the same terminal position of the revolving line OP each of the MP ratios, — — , etc., in (1) Art. 40, is always the same, no matter 76 PLANE TRIGONOMETRY, [Ch. V. where the point P is taken on the revolving line. Thus, for example, let any other point Pi (Fig. 40 a) be taken on the ter- X^ M Ml O X X^M^M O X Fig. 40a. Fig. 40b. minal line, and let PiMi be drawn perpendicular to X^OX. Thtv. ^—- — ^ \ That is, the sine of any angle whose terminal line OP OPi is OP, has a fixed definite value. The same can be shown for the other ratios. Hence, as already shown in Art. 13 for angles between 0° and 90^ (1) To each angle there corresponds but one value of each trigono- metric ratio. Now let OP revolve a little from OP into the position OPi (Fig. 40 b). For convenience keep OPi equal to OP. Draw PiMi at right angles to X^OX, Then, MP the sine of the angle whose terminal line is OP — — — , MP and the sine of the angle whose terminal line is OPi = — ^— ^» OPi Since MiP^ is not equal to MP, it follows that these two sines are unequal. Hence, the sine of an angle changes when the angle changes. The same can be shown for the other ratios. Hence, (2) TJie ratios of an angle change ivhen the angle changes. The variation in the ratios as the angle increases, is discussed in Art. 77. Ex. 1. In the above, OPi is taken equal to OP. Why does this not affect the generality of the deduction ? Ex. 2. Trace the changes in the trigonometric ratios as the turning line revolves from 0° to 360°. Compare your results with those of Art. 15 (7, and those given in the table at the end of Art. 77. It has been shown in Art. 37 that an infinite number of angles have the same terminal line. Jt follows that in each of the 44.] RELATIONS OF TRIGONOMETRIC RATIOS. 11 MP figures in Art. 40, -—— is tlie sine of an infinite number of angles. The same is true in the case of the other ratios. Moreover, the geometrical solutions in Art. 42 show that there are two sets of angles correspo7iding to each given ratio, and that each set is infinite in number J and has a particular terminal line. Hence, (3) To each value of a trigonometric ratio there corresponds an infinite number of angles. Note. The student will see, by turning to Arts. 84-87, that all angles which have the same sine, can he given in a simple formula, and that the same fact is true in the case of each of the other ratios. The deduction of these formulas, while easily possible at this place, is postponed in order to permit the early completion of the solution of triangles. 44. Relations between the trigonometric ratios of an angle. The relations between the trigonometric ratios of an acute angle were set forth in Art. 18. It will now be shown that these relations also hold for the ratios of any angle. A. Inspection of the definitions (1), Art. 40, shows the reciprocal relations, namely : sin ^ cosec ^ - 1 ; cos^sec^ = l; tan ^ cot 4 = 1. (1) B. In each of the figures in Art. 40, MP OM t,r, 1-^P-OP slnA. ^^. .OM OP ^A *^'* OM OM cos A' *"*'* MJP MP sin a' OP OP (2) C. In each of the figures in Art. 40, On dividing both members of this equation by OP , OM , MP , in turn, and following the same process as that adopted in Art. 18, it results that sm^A + cos^A = l; sec^^ = 1 +tan2^; cosec^ JL = 1 + cot^ ^. (3) From the first of relations (3) it follows that cos -4 = ± Vl — sin^ A, 78 PLANE TRIGONOMETRY. ICu. V. This shows that, corresponding to a given sine, there are two cosines which are numerically equal, and opposite in algebraic sign. Ex. 1, Art. 42, illustrates this. This is also manifest in the table of signs in Art. 41. As indicated in this table, the sine is positive in the first and second quadrants, and then the cosine is positive and negative, respectively ; the sine is negative in the third and fourth quadrants, and then the cosine is negative and positive, respectively. The signs of the remaining ratios corre- sponding to a given sine will be apparent on a short geometrical inspection, or by a glance at this table of signs. When any single ratio is given, there is an ambiguity as to the signs of some of the other ratios. Thus, to take another instances-it follows from the second of (3) that I tan A = ± Vsec^ A — 1. The secant of A is positive in the first and fourth quadrants^ and then the tangent is positive and negative respectively; the secant of A is negative in the second and third quadrants, and then the tangent is negative and positive respectively. The double sign which appears in these relations was referred to in the examples. Art. 18. The student is advised to review and work the examples, for angles in general, in Art. 18. EXAMPLES. 1. Given that sin ^ = | ; find the other ratios of A by means of the relations shown in this article. [In Ex. 1, Art. 42, this problem is solved geometrically ; here it will be solved algebraically. '\ ±V7. ... ._ 1 _ 4 . cos ^ = ± V 1 — sin"'^ A = — ; sec A 4 cos A ^ ^^7 cosec^=-J-=i; tanyl = ^iil^ = -J-; cot A = -^— = ±2^. sin. A 3 cos^ ±Vl tan A 3 Since the given sine is positive, the corresponding angles are in the first and second quadrants. Hence the double values of the calculated ratios are paired as follows : 46.] BATIOS OF 90'' - A, ETC. 79 sin A cos A tan A cot A sec A cosec ^ 3 + V7 3 V7 4 4 4 4 V7 3 V7 3 3 -V7 3 -V7 4 4 4 4 V7 3 V7 3 Find the other ratios algebraically , and verify the results geometrically^ when : 2. cos ^ = — |. 3. tan ^ = f . 4. sec A = A. 5. cosec A = — b. 6. sin ^ = — f 7. cos A = ^. 8. tan J. = — 3. 9. cot A = l. Find the other ratios algebraically, and verify the results geometrically, when angle J. satisfies the following p«iVs o/conrfiYiows: . / 10. sin ^ = i and tan A = 11. tan ^1 = VS and sec ^ = — 2. V3 \/5 12. cos J. = — I and sin ^ = + — — 13. sin ^ = — f and tan A = \, o Give a geometrical^lution of the following trigonometric equations : 14. sin A = l, " 15. cos ^ = - f 16. tan ^ = 4. 17. sec A = 5. Namoi^^Rur least angles, and also the four least positive angles, that satu^PPI^quations : 18. sin^=— . 19. tan^=\/3. 20. cos^= — —» 21. cot^=-\/3. \/2 V2 45. Ratios of 90° - A, 180° - A, 90° -{-A, -A, compared with the ratios of Ay A being any angle. The student may have suspected, from Exs. 7-10, Art. 41, that there is a close connec- tion between the ratios of an angle A on the one hand, and the ratios of the angle — A and of angles differing from A and — A by multiples of 90°, on the other. He may have discovered already what the connection is. This connection, which is set forth in this article, is interesting in the study of angles, and has an important bearing on the construction of trigonometric tables, and on the solution of triangles. In each figure in this article OP is the terminal line of the angle A, and OP^ is the terminal line of the related angle which is under consideration ; for the purpose of easy comparison, OPi is always taken equal to OP] MP, M^P^, are the perpendiculars drawn from the initial line to P, Pj, respectively. The deductions 80 PLANE TRIGONOMETRY, [Ch. V. made in the simplest case, namely, when A is an angle in the first quadrant, are true for all angles. The student is advised to con- sider only the simplest case, when first he considers the subject of this article, and then to draw the figures and make the deduc- tions for himself, in the cases in which angle A is in the second, third, and fourth quadrants, respectively. Note. A compound angle, 90° — A^ for instance, can be described by- revolving the turning line forward through 90'', and then backward through an angle equal to A; or, these steps may be taken in a reverse order, namely, by revolving the turning line backward through an angle pqual to A^ and then forward through 90°. Similarly, for the compound yr;l<^s 90° + J, 180° ± A, etc. For the sake of clearness of construction, it is better not to take the terminal line of A nearly midway between XiOXand Y\OY. A. Ratios of 90° - A. Describe the angles A, 90° — A. Let OP, OPi, be the terminal lines of A, 90° — A, respectively. In Figs. FiQ. 41a. Fig. 41t). Fig. flOr Fig. 41d. 41 a, 41 b, 41 c, 41 d, A is an angle in the first, second, third, and fourth quadrants, respectively. Take OP^ equal to OP, and draw MP, M^P^, at right angles to the initial line. In the triangles M^OPi, MOP, (in each figure) the angles at M^, M, are right angles, angle M^OPi = angle 0PM, and OPi = OP. Hence these two triangles are equal, and OMj^ = MP, M,P, = CM. The figures also show that, for A in each quadrant, OJ/j, MP have the same algebraic sign, and JfiA, OM have the same sign. Hence, for all angles A, M,P, OM OP sin (90° - A) 1^ 1 _ OM^ MP cos A cos (90 -^) = ^ = _ = sm A; 46.] RATIOS OF 180" -A. 81 tan (90° - -'')=^=^'="^^ cot (90° - -->'^r"r-'^' sec (90° - -^'-Ji-ip-'""^^ osec (90° • Hence, the ratio of any angle is the same as the co-ratio of its complement. Compare with Art. 16. The relations for tangent, secant, cotangent, cosecant, can also be deduced from those of sine and cosine by means of Art. 44 (1), (2). Thus, for example, tan (90° - ^) = «-Hli5^^^ = ^ = cot A ^ ^ cos (90°-^) sin^ B. Ratios of 180° - A. Describe the angles A, 180° — A. Let OP, OPi be the terminal lines of A and 180° — A respectively. In Figs. 42 a, 42 b, 42 c, 42 d, A is an angle in the first, second, third, fourth quadrants, respectively. Take OPi equal to OP, and draAV MP, M^P^, at right angles to the initial line. In the two triangles OM^P^, OMP, in each figure, the angles at M^, M, are Fia. 42a. Fia. 42b. right angles, angle M^OPx = angle MOP, OPi= OP. Hence, 0M^= OM, and M^P^= MP. In each figure, OM^, OM have oppo- site algebraic signs, and M^Pi, MP, have the same sign. Hence, for all angles A, sin (180» -A) = ^^=^^AuA; /^OAO ^\ 0M^ — OM M cos (180° - ^) = ^ = —^^ = - cos A. 82 PLANE TRIGONOMETRY. [Ch. V. So also, tan (180° - ^) = - tan ^ ; cot (180° - A) = - cot ^ ; sec (180° — A) = — sec J. ; cosec (180° — A) = cosec A. The last four relations can be deduced by means of the figures, or by means of relations (1), (2), Art. 44. Hence, any ratio of an angle is equal in magnitude to the same ratio of its supplement; the sines of supplementary angles have the same algebraic sign, and so have the cosecants; the other ratios of supjjlementary angles have opposite signs. C. Katios of 90° + A. Describe the angles A, 90° + A. Let OPy OPi, be the terminal lines of A, 90° + A, respectively. In Figs. 43 a, b, c, d, A is an angle in the first, second, third, fourth quadrants, respectively. Take OPi = OF, and draw MP, M^P^, at right angles to the initial line. In the two triangles, OMiP^, OMP, (in each figure) the angles at Mu M, are right angles, Fia. 43a. Fig. 43b. Fia. 43c. angle JfiOPi = angle 0PM, OPi=OP. Hence, M^P^= OM, and OM^ = MP. In each figure M^P^, OM, have Ihe same alge- braic signs, and OM^, MP, have opposite signs. Hence, for all angles A, MyPi ^ OM OP sin (90° + A) OP cos A'y COS (90° + ^) = I*: MP = — sin A. OP So also, tan (90° + ^) = - cot ^ ; cot (90° -h ^) = - tan ^ ; sec (90° 4- ^) = - cosec A ; cosec (90° + A) = sec A. These four relations can be deduced from the figures, or by means of (1), (2), Art. 44. D. Ratios of - A. Describe the angles A, — A. Let OP, OPif be the terminal lines of A, — A, respectively. In Figs. 44 a, 45.] RATIOS OF —A, 83 b, c, d, A is in jhe first, second, third, fourth quadrants, respec- tively. Take OPi = OP, and draw PM, PiM^, at right angles to the initial line. In the two triangles, OM^P^, OMP, (in each figure) the angles at Mi, M, are right, angle M^OPi = angle MOP, Fig. 44a. Fia. 44b, Fig. 44c. OPi=OP. Hence, MiPi = MP, OM=OMi. In each figure, 03fi, OM, have the same sign, and MiP^, MP, have opposite signs. Hence, for all angles A, sin (—A) M,P, OP, MP OP = — sin A ; , .. OMi OM . eos(-A) = ^ = — = cosA. So also, tan (— A) = — tan A; cot (— ^) = — cot ^ ; sec (— ^) = sec ^ ; cosec (— A) = — cosec A. The last four relations can be deduced from the figures, or by means of (1), (2), Art. 44. Ex. 1. Show that sin (180° + A)=- sin A, cos (180° + ^) = - cos ^, tan (180° + J.) = tan^, etc., when A denotes an angle in any one of the four quadrants. Ex. 2. Deduce the relations between -each of the following angles and angle A, viz. 270° - A, 270° + A, 360° - A, 360° + A, n • 360° ± A, n being any whole number. By means of the relations shoivn in this article, the ratios of any angle can be expressed in terms of the ratios of an angle between 0° and 45°. Thus, for example. 84 PLANE TRIGONOMETRY. [Ch. V. sin 700° = sill (360° + 340°) = sin 340° = sin (- 20°) = - sin 20° ; tan 975° = tan (2 • 360° + 255°) = tan 255° = tan (180° + 75°) = tan 75° = cot 15° ; cosec (- 1160°) = - cosec 1160° = - cosec (3 • 360° + 80°) = - cosec 80° = - sec 10°. .-. sin 700° = - .34202 ; tan 975° = 3.7321 ; cosec (1160°) = - sec 10° = -^=J-- = -=J- = - 1.015. ^ ^ cos 10° .98481 This property, and the property that the ratio of an angle is the co-ratio of its complement, account for the arrangement and extent of the trigonometric tables. EXAMPLES. 1. Express the ratios of the angles in Exs. 1-7, Art. 41, in terms of ratios of angles between 0° and 45°. Also find the ratios. 2. Do likewise for the angles in Exs. 1, 3, Art. 38. Also find the ratios. 3. Do likewise for the angles in Exs, 1, 2, 3, Art. 37. Also find the ratios. N.B. Questions and exercises suitable for practice and review on the subject-matter of Chap. V. will be found at page 186. CHAPTER VI. TRIGONOMETRIC RATIOS OF THE SUM AND DIFFERENCE OF TWO ANGLES. N.B. Another way of making the derivations shown in Arts. 46-48 is given in Note B of the Appendix. The method of projection, as it is called, used in Note B, is preferred by many. 46. Derivation of the sine and cosine of the sum of two angles when each of the angles is less than a right angle. In this article and the following one, careful regard must be paid to the dii-eo- tions in which lines and angles are measured, and to the order of the letters used in measuring them. See Arts. 36, 37, 40. To deduce sin (A + B) and cos (A + B). Let A and B be twA angles each of which is less than a right angle. Let the turning line revolve from the initial line OX, and about describe the pI / /L n-h i- ...-V Im\\ M N X Fig. 45a. M O N Fig. 45b. angle XOL equal to A, and then revolve forward from the posi- tion OL and describe the angle LOT equal to B. Thus, angle XOT=A + B. [In Fig. 45 a, ^ + J5 is less than 90° ; in Fig. 45 h, A-\- B \^ greater than 90°.] Take any point P on OT, the ter- minal line of the angle {A-{- B), and draw PQ at right angles to OL, the terminal line of the angle A. From P, Q, draw PM, QN, at right angles to the initial line; and through Q draw VQR parallel to OX and intersecting MP in U. 85 86 PLANE TRIGONOMETBY. [Ch. VI. fA . T^ ■ vni:. ^P NQ + BP NQ- RP ^OQsm^^QPsinFQP (Definitions, Art. 40.) Now, by definitions in Art. 40, and by Art. 45, ^ = cos QOP= cos i5; S?= sin QOP= sin J5; sin VQP = sin (180° - PQR) = sin PQR = cos RQO = cos XOQ = cos A, /. sin iA + B)= sin AcosB -^ cos A sin B, (1) cos(^ + P)=cosXOP = ^=^^^ ^ 0^t9)QE ^0N QR OP, OP OP ^ OQcos^ QPco^VQP OP OP ' (Definitions, Art. 40.) Now, by definitions in Art. 40, and by Art. 45, 0^ = ,,,B, S^=sinJB; OP 'OP ' cosFQP=cos (180°-PQP)= -cosPQP= -sinPQO= -sin A .'. cos (^ + B) = cos ^ cos J8 - sin A sin J5. (2) EXAMPLES. 1. Sin 75° = sin (30° + 45°) = sin 30° cos 45° + cos 30° sin 45° = 1 1 ■ \/3 1 ^ 1+V3 2 ' V2 2 * V2 2V2 * 2. Find cos 75° by putting 30° + 45° for 75° and using formula (2). 3. Deduce the sine and cosine of 15° from the results in Exs. 1, 2. 47.] SIN (A-B) AND COS {A-B). 87 - 4. Find sin 90°, cos 90°, by putting 90° = 30° + 60°. Also by putting 90° = 45° + 45°. Also by putting 90° = 75° + 15°. 5. Find sin 120°, cos 120°, by putting 120° = 60° + 60° ; 120° = 90° + 30° ; 120° = 75° + 45°. 6. Find sin 150°, cos 150°, by putting 150° = 75° + 75° ; 150° = 90° + 60°. 7. Find sin 135°, cos 135°, by putting 135° = 75° + 60° ; 135° = 90° + 45°. 8. Given sin x = f , sin ?/ = |, x and y both in the first quadrant ; find sin (x + y), cos(x + y). 47. Derivation of the sine and cosine of the difference of two angles when each of the angles is less than a right angle. The construction and derivation are veiy similar to that made in the preceding article. To deduce Hln(A-B) and cos(^-jB). Let A and B be two angles each of which is less than a right angle, and let A be the greater. Let the turning line revolve from the initial line OX, ^ jL /T ^< %^/ 1 /\^/ fp- y k t. 7\ N M FIG. a J. and about describe the angle XOX equal to A, and then revolve backward from the position OL and describe the angle LOT equal to - B. Then angle XOT=A-B. Take any point P on OT, the terminal line of the angle (A — B), and draw FQ at right angles to OL, the terminal line of the angle A. From P, Q, draw PM, QN, at right angles to the initial line ; and through P draw liPV parallel to OX and intersecting NQ in R. ^ OQ sin A PQ sin VPQ OP OP ' (Definitions; Art. 40.) 88 PLANE TRIGONOMETRY. [Ch. VL Now, by definitions in Art. 40, and by Art. 45, -^ = cos QOP = cos (- B) = cos B; _PQ^QP^^ qOP= sin(-B)= - sin B: OF OP ^ ^ ^ / sin VPQ = sin (180° - QPR) = sin QPE = cos BQP = cos A. .'. sin ( 4 - J5) = sin ^ cos ^ - cos A sin B. (3) cos(^-B) = cosXOP = §f=Mi5^=g|_g ^ OQcos^ PQ cos VPQ Qp OP ' (Definitions, Art. 40.) Now, by definitions in Art. 40, and by Art. 45, — ^ = cos B, and — — ^ = — sin Bj as shown above ; cosFPQ=cos(180°- QPP)= -cos QPE=-sinBQP= -sin A. .'. cos (A- B) = cos ^ cos J5 + sin A sin B, (4) If B is greater than A, then the formula, sin (B — A) = sin B cos A — cos jB sin A, can be deduced as above. Since sin(^ -B) = - sin(B - A), then sin ( J. — 5) = sin A cos 5 — cos A sin ^. 7i^ is shoivn in Art. 48 that the formulas (1), (2), (3), (4), are true for all values of A and B. These formulas are called the addition and suhtrojction formulas or theorems in trigonometry. They are of such great importance, and so many thorems can be deduced by means of them, that they are called the fundamental for- mulas of trigonometry.* They should be memorized. Note. Arts. 48, 49, may be omitted, if deemed advisable, until after the solution of triangles is completed. Art. 48 can also be shown geometrically. * Adrian Romanus (1561-1625), professor of mathematics and medicine at the University of Louvain, was the first to prove the formula for sin(^ + B). The formulas for cos(^ ± B) and sin(^ — 7?) were given by Pitiscus (1561-1613), a German niatliematician and astronomer, in his Trig- onometry published in 1595. 48.] ADDITION AND SUBTRACTION FORMULAS. 89 EXAMPLES. 1. Derive sin 15°, cos 15°, on putting 60° - 45° for 15°. 2. Derive sin 15°, cos 15°, on putting 45° - 30° for 15°. 3. Find sin (x — ?/), cos (x — y) in tlie cases in Ex. 8, Art. 46. 48. Proof of addition and subtraction formulas for all values of A and B. These formulas have been proved in Arts. 46, 47, for values of A and B which are less than a right angle. In Art. 45 c it has been shown that for any angle, say X, cos X = sin (90° + X), sin X = - cos (90° + X). Hence, cos ^ = sin (90° + A), sin ^ = - cos (90 °+ A), cos(^+i3) = sin(90°+^+jB), sin(.4+^)=-cos(90°+^4--S). The substitution of these values for cos Aj sin A, cos {A + B), sin {A + B), in (1), Art. 46, gives - cos (90°+^+ J5) = -cos (90°+ J.) cos 5+sin (90°+u4) sin 5; .♦. cos(90°+^+^)=cos(90°+^)cos5-sin(90°+^)sinJ3. (1) The substitution of the same values in (2), Art. 46, gives sin (90° + ^ + ^) = sin (90°+ A) cos B + cos (90° + A) sin B. (2) Hence, formulas (1), (2), Art. 46, are true when one of the angles is increased by a right angle. In a similar way, these formulas can be shown to remain true when one of the angles in (1), (2), of this article is increased by a right angle. It is thus evident that the formulas are true, no matter how many right angles are added to either one or both of the angles. It can easily be shown that sin J. = cos (J.— 90°), cos J. = — sin (^—90°). Then, in the same way as that just emplo'yed, it can be shown that the formulas (1), (2), Art. 46, hold when eitlier one or both of the angles is diminished by integral multiples of 90°. Hence, formulas (1), (2), Art. 46, are true for angles in any quadrant, that is, for all angles. In a similar Avay, formulas (3), (4), Art. 47, can be shown to be universally true. 90 PLANE TUIGONOMETRY, [Ch. VI. 49. Each fundamental formula contains the others. From any one of the four fuildamental formulas, the remaining three can be derived. Thus for example : In (1) Art. 46, change A into 90° -^; then sin (90° - ^ + 5) = sin (90° - A) cos B + cos (90° - A) sin B, From this, sin (90°- J. - B)f i.e. cos {A-B)= cos ^ cos i? + sin A sin B. In (1) Art. 46, change B into (— -S); then sin {A — B)= sin ^ cos (— 5) + cos ^ sin (— B) = sin A cos B — cos A sin B. In (1) Art. 46, change A into (90° + A) ; then sin (90° + -4 + 5) = sin (90° + A) cos J5 + cos (90° + A) sin B, whence, cos {A + B)= cos ^ cos 5 — sin A sin B. Ex. 1. From formula (2), Art. 46, derive the other three fundamental formulas. Ex. 2. So also, from formula (3), Art. 47. Ex. 3. So also, from formula (4), Art. 47. 50. Ratio of an angle in terms of the ratios of its half angle. In this article and Arts. 51, 52, a few deductions will be made from the addition and subtraction formulas, which have been shown to be true for all angles. These deductions are necessary for the explanations concerning triangles, as well as useful for other purposes. More ample opportunity will be afforded later for working exercises involving the use of these formulas. The fundamental formulas may be brought together: sin {A-\- B)- sin A cos B + cos vl sin B. (1) sin {A- B) = sin ^ cos B - cos A sin -B. (2) cos(^ + B) = cos A cos B - sin A sin B. (3) cos(^ - B) = cos ^ cos B + sin A sin J5. (4) 49-50.] SIN 2 A, COS 2 A. 91 Let B = A; tlien, from (1), sin (A + A)= sin A cos A -\- cos A sin A j that is, sin 2 ^ = 2 sin A cos A. (5) Similarly, from (3), cos 2 ^ = cos*^ A - sin^^. n (6) Since cos^ A + sin^ J. = 1, it follows that cos2^ = l-2sin2^. (7) and cos 2 4 = 2 cos^ A- 1. (8) In formulas (l)-(8), A, B, denote any angles whatsoever. These formulas occur so often, and are so useful, that it is well to translate them into words. Thus, sine sum of any two angles = sin j^rs^ • cosine second + cosine first • sine second sine difference of any two angles = sine j^Vs^ • cosine secoyid — cosine first • sine second cosine sum of any two angles = cosine first • cosine second — sine j^Vs^ • sine second cosine difference of any two angles = cosine first • cosine second + sine first • sine second Since A is one-half of 2^4, formulas (5)-(8) can be translated as follows : sine any angle = 2 sine half-angle • cosine half-angle, cosine any angle = (cosine half-anglef — (sine half-angle)'"^, = 1 — 2 (sine half-angley, = 2 (cosine half-angle^ — 1. EXAMPLES. 1. Find cos 221° from cos 45°. 2 cos2 22|° = 1 + cos 45° by (8) ; ^ V V2/ 2V2 2x 1.4142 .-. cos22i° = .9239. 92 PLANE TRIGONOMETRY. [Ch. VI. 2. Express cos 4 x in terms of sin x and cos x. cos4ic = 2cos2 2x - 1 = 2(2cos2x- 1)2- 1 = 8 cos'* x - 8 cos2 cc + 1. 3. Deduce sin 30°, cos 30°, from cos 60°. 4. Deduce sin 75°, cos 75°, from cos 150°. [Logarithms may be helpful.] 5. Deduce sin 22|°, from cos 45°. 6. Deduce sin 15°, cos 15°, from cos 30°. 7. Express cos 6 x, sin 6 x, in terms of ratios of 3 x. 8. Express cos 3 x, sin 3 x, in terms of ratios of f x. 9. Express sin f a*, cos | x, in terms of ratios of f x. 10. Express cos 6 x, sin 6 x, in terms of ratios of 12 x. 11. Express cos 3 x, sin 3 x, in terms of ratios of 6 x. 12. Express sin | x, cos | x, in terms of ratios of | x. 13. Show that sin (w + 1)^ + sin (w - 1) J. = 2 sin nA cos A, and cos (n + V)A + cos {n — 1)^ = 2 cos nA cos A ; and hence, express sin 2 ^, cos 2 ^, in terms of sin A^ cos ^. 51. Tangents of the sum, and difference of two angles, and of twice an angle. Let Ay B, be any two angles. It is required to find tan {A + B) and tan {A — B). tan (A-\-B) = ^^^ (^ + B) _ sin ^ cos -S + cos A sin B cos {A + B) cos ^ cos ^ — sin ^ sin B On dividing each term of the numerator and the denominator of the second member by cos A cos B, there is obtained tan(^ + B)=: tan^ + tang , (1) 1 - tan ^ tan ^ In the same way it can be shown that tan(^-B)^ *^»/-;f ^ . (2) 1 + tan ^ tan B Formula (2) can also be deduced from (1) by changing B into -B. If B = A, then (1) becomes tan2^ = ^iM.^. ^ (3) 1 - tan^ A 51-52.3 SUM OF TWO SINES. 93 Formulas (1), (2), (3), can be translated into words, as follows : tangent sum any two angles = ^ ^ ; 1 — product of tangents tangent difference any two angles = difference of tangents 1 + product of tangents tangent^angle^ ^fg;^;;;^^, EXAMPLES. 1. tan P = 2, tan § = f Find tan (P+ Q), tan (P - Q). t tan(P+§)=:-l±l^ = 7; tan (P - ^) = Azii^ = i. 2. Find tan 75° by means of tan 4^°, tan 30°. 3. Find tan 15° by means of tan 60°, tan 45°. 4. Find tan 22|° from tan 45°. 5. Find tan 37|° from tan 75°. 6. Derive cot iA±B) = ^ot^cotP:Fl .^^ 2 ^ = co^^jzi. cot J5 ± cot ^ 2 cot A 52. Sums and differences of sines and cosines. The set of formulas (l)-(4), Art. 50, can be transformed into two other sets which are very useful. From (1), (2), (3), (4), Art. 50, on addi- tion and subtraction, there is obtained : sin(^ + B)+sin(^-.B)=2sin^cosJ5. (1) sin (A + B)- sin (A-B)=2 cos A sin B. (2) cos (A + B)-{- cos iA-B)=2 cos A cos B. (3) cos iA + B)- cos (^ - ^) = - 2 sin ^ sin B, (4) If A + B = P, and A — B= Q, then 2 ^ = P+ Q, and ^ = i (P^+ Q), 2B==P- Q, and JS = i(P- Q). 9^ PLANE TRIGONOMETBY. [Ch. VI. Substitution of these values of A, B, in (l)-(4) gives siu r+s'mQ= 2 sin ^-^ cos ^^- (5) sin P- sin mC. .•.2U sin C (2) Ex. 1. Explain why the circumscribing circle of a triangle depends only 'upon one side and its opposite angle. Ex. 2. Derive the law of sines by drawing a perpendicular from ^ to J5C. Ex. 3. Derive 2B = — ^, 2B = — ^, by means of figures, sin A sin B II. The law of cosines. An expression for the length of the ^ side 'of a triangle in terms of the cosine of the opposite angle and the lengths of the other two sides, will now be deduced. The angle A is acute in Fig. 49 a, obtuse in Fig. 49 b, right in Fig. 49 c. From C draw CD at right angles to AB. The direc- tion AB is taken as positive. In Figs. 49 a, 49 6, In Fig. 49 a, in Fig. 49 b, BC' = DC' + DB", DB = AB-AD) DB=DA-\-AB = -AD-\- AB. Hence, in both figures, BC = DC + (AB - ADy DC + AD- + AB' -2AB' AD. 100 PLANE TBIGONOMETRY. [Ch. VII. In Fig. 49 a, AD = AC cos BAC; in Fig. 49 6, AD = ^(7 cos BAC (Art. 40). Also, DO' + Aff=ACf. Hence, in both figures, W = J^O' + Zb' -2 AC- AB cos ^ ; tliat is, a2 := 62 4- c2 - 2 6c cos ^. (3) This formula also holds for Fig. 49 c ; for there, cos ^=: cos 90° = 0. Similar formulas for 6, c, can be derived in like manner, or can be obtained from (3) by symmetry : 62 = c2 + a2-2cacos^, c" = o? -\-h'' -2ah(tosC. (3') These formulas can be expressed in words : In any triangle, the square of any side is equal to the sum of the squares of the other two sides minus tivice the product of these two sides multiplied by the cosine of their included angle. Note. In Fig. 49 at, A is acute and cos A is positive ; in Fig. 49 b, A is obtuse and cos A is negative. Hence formula (3) shows that in Fig. 49 a, a^ is less than 6^ + c^, and that in Fig. 49 b, a'^ is greater than b^ + c^. In Fig. 49 c, a2 = 62 + c2. -<« Relation (3) may be expressed as follows : oosA = ^±^, (4) and similarly for cos B, cos O. Ex. Derive the formulas for 6^ and for c*. Each of the relations (1), (3), (3'), involves four of the six ele. ments of a triangle. If any three of the elements in any one of these relations are known, then the fourth element can be found by solving the equation. Inspection shows that relations (1) are serviceable in the solution of Cases I., II., Art. 53, and that relations (3), (3'), are serviceable in the solution of Cases III., IV., Art. 53. The student is advised to try to work some of the examples in Arts. 55-58 before reading the text of the articles. (See Arts. 20-24, 34.) 54 a, 55.] SOLUTION OF TRMNGLES:. CASE I- 101 54 a. Substitution of sines for sides, and of sides for sines. Since — ^ = — — = ^ , the sines of the opposite angles can be sub- sin A sin B sin C stituted for the sides of triangles, and vice versa, when they are involved homogeneously in the numerator and denominator of a fraction, or in both members of an equation. Thus, on putting — ^ = — ^ = — ^ = x, it follows that a = x sin A^ sin A sin B sm G b = xsmB, c = x sin C. a^ ax sin A a sin ^ Then, for example, 6 + c ic sin ^ + ir, sin (7 sin JB + sin G EXAMPLES. 1. Show that in any triangle ^±^ = ^ ^^-^) , c sm I C a + 5_ sin^ + sin ^ _ 2 sin |(yl + ^)cosK^ — -^) _ cos^(^ — -S) c "~ sin C 2 sin I C cos ^G sin ^ O ' for sin l(A + B)= cos | O, since 1(A + B) + ,IG = 90°. 2. Derive two other relations similar to that in Ex. 1. 3. Show that 3 qg 4. 2 62 _ 3sin2^ + 2sin2^ _ 3 sin^ A + 2 sin^ B _ 3 sin^ ^4 + 2 sin^^ .g abc ~ a sin ^ sin O 6 sin ^ sin O c sin J. sin J? 55. Case I. Given one side and two angles. In triangle ABC, suppose that A, B, a are known ; , it is required to find C, b, c. In this case (see Fig. 47 a, Art. 54), C=1S0°-(A + B)', , whence b = .^ , • sin 5 : sm^ _ •, whence c = -: — - • sin O. sin C sm A sm A b c b^ -{- c^ — 2bc cos A, — = -: — -, the result in sm B sm C Ex. 1, Art. 54 a. Other checks will be discovered later. h a sin 5 sin J.' c a 102 :p^ANK TMIGONOMETBT. [Ch. VII. EXAMPLES. 1. Solve the triangle PQB, given : PQ = 12 in., Q = 40°, P = 75°. B = 180° - (P + ^) = 180° - (40° + 75°) = 65°. &\nQ sini? " sinP sini? Solution : B PB BQ = sin JK 12 sin 65 12 • sin Q, ;.sin40°, X . 6428, P^ = -^.sinP, sinP 12 sin 65° 12 . sin 75° X 9659, .9063' ' .9063 = 13.24 X • 6428, = 13.24 x 9659, = 8.51 in. = 12.8 in. 2. In ABC, A = 50°, B = 75°, c = 60 in. Solve the triangle. 3. In ABC, A = 131° 35', B = 30°, b = ^ ft. Find a. 4. In ^PO, B = 70° 30', C = 78° 10', a = 102. Solve the triangle. 5. In ABCy B - 98° 22', C = 41° 1', a = 5.42. Solve the triangle. 56. Case II. Given two sides and an angle opposite to one of them. In the triangle ABC let a, h, A be known, and G, Bj c be required. The triangle will first be constructed with the given b G / A / a / / c. Fig. 50. Fig. 51. b ^C \/ / I c \ s/- P;\ A D FlO. 50. ^^'B M 56.] SOLUTION OF TRIANGLES : CASE It. 103 elements. At any point J. of a straight line LM, unlimited in length, make angle MAG equal to angle A, and cut off AG equal to h. About (7 as a centre, and with a radius equal to a, describe a circle. This circle will either : (1) Not reach to LM, as in Fig. 50. (2) Just reach to LM, thus having LM for a tangent, as iu Fig. 51. (3) Intersect LM in two points, as in Figs. 52, 53. Each of these possible cases must be considered. In each figure, from G draw GD at right angles to AM] then GD = b sin A. In case (1), Fig. 50, GB < GD, and there is no triangle which can have the given elements. Hence, the triangle is impossible when a GD', that is, a>b sin A. If a > b, then the points B, B^, in which the circle intersects LM, are on opposite sides of A, as in Fig. 52, and there is one triangle tvhich has three elements equal to the given elements, namely, ABG. Jf a a, then sin B>1, and, accordingly, B is impossible and there is no solution. If, in equation (1), bsin A = a, then sin ^ = 1, and B = 90°. The con- sideration of the trigonometric equation (1) leads, therefore, to the same results as the preceding geometrical investigation. Checks: A-{-B-\- (7=180°, and, as in Case I. Other checks will be found later. EXAMPLES. 1. Solve the triangle STV, given : ST = 15, VT =\2, S= 52°. sin V _ smS . ^T VT ' .78801 8T sin V sin 52° = .065668. 15 12 12 .-. sin F= 15 X .065668 = .98502. .-. F= 80° 4' 20", or 180° - 80° 4' 20", i.e. 99° 55' 40". Both values of V must be taken, since the given angle S V^ V EiG. 54. is opposite to the smaller of the given sides. The two triangles correspond- ing to the two values of Fare STV, S2'V\, Fig. 54, in which /SFr= 80° 4' 20", >S'Fir=99°55'40". In STVi In STV angle STV^ = 180° - (^ + SV^T) angle STV= 180° -(6^+ 8VT) = 28° 4' 20". = 47° 55' 40". SVx _VxT . smSTVi sin/S' SV VT . sin STV smS* .47059 -.78801 -^^•^^^• '^^ = 15.228. .74230 .-. SVi = lAl. .-. >S'F=11.3. The solutions are : Fi = 99° 55' 40" ■ F=80° 4' 20" T=28° 4' 20" > r= 47° 55' 40" /S'Fii^7.17 ^F=11.3 In the ambiguous case, care must be taken that the calculated sides and angles are combined properly. 57.] SOLUTION OF TRIANGLES : CASES III., IV. 105 2. Solve ABC, given : a = 29 ft., 6 = 34 ft., ^ = 30° 20'. 3. Solve ABC when a ^ 30 ft., & = 24 ft., 5 = 65°. 4. Solve ABC when a = 30 in., 6 =z 24 in., ^ = 65°. 5. Solv^ ABC when a = 15 ft., & = 8 ft., J5 = 23° 25'. 57. Case III. Given two sides and their included angle. In the triangle ABC, a, b, C are known, and it is required to find A, B, c. In this case, c can be determined from the relation c^ = a^ -^b^ — 2ab cos (7, Art. 54 ; angle A can be^ determinedrfromihe relation sin A _ suiC . a c angle B can be determined from the relation ^ + 5+ (7=180°, or from ^=^HL^. b c Checks: ^a^ = W -\- c^ — 2 be cos A, b^ ==a^ 4- c- — 2 ac cos A^ th e result in Ex. 1, Art. 54 a ; other checks will be found later. EXAMPLES. 1. In triangle PQIi, p = 8 ft., r = 10 ft., Q = 47°. Find q, P, B. q2 —p2 ^ ^2 _ 2pr cos Q = 64 + 100 - 2 X 8 X 10 X .6820 = 54.88. .-. q = 7.408. sin P = P^rL9. = 8j^^7314 ^ ^g^g^ ._ p ^ ^^o ^q, q 7.408 sin B = ri^HLS = ^» X -7314 ^ gg^g ._ ^ ^ g^o 50/. g 7.408 2. Solve ylPC, given : a = 34 ft., & = 24 ft., C = 59° 17'. 3. Solve ABC, given : a = 33 ft., c = 30 ft., B = 35° 25'. 4. Solve J^/ST, given : r = 30 ft., s = 54 ft., r=46°. 6. Solve PQB, given : p = 10 in., g = 16 in., B = 97° 54'. 58. Case IV. Three sides given. If the sides a, b, c are known in the triangle ABC, then the angles A, B, C can be found by means of the relations (3), Art. 54. Checks: Kelations (1), Art. 54: A + B-^C^180°. Other c'hecks will be shown later. 106 PLANE TRIGONOMETUr, [Ch. VII. EXAMPLES. 1. In ABC, a = 4, 6 = 7, c = 10 ; find ^, B, C. cos A = yL±^J^ot ^ 49 + 100-16 ^ 133 ^ ^^q^. .-. A = 18° 12'. 2 6c 2 X 7 X 10 140 cos^ = ^'^ + ^'-^' = ^QQ + ^^-^^^g^ = .8375. ...^ = 33° 7' 30". 2 ca 2 X 10 X 4 80 cos o^ «^ + &^ - c^ ^ 16 + 49 - 100 ^ -35 ^ _ ^^^^ .^ = 128°40'52". 2x4x7 56 Angle C is in the second quadrant since its cosine is negative. Check: 18°12'+33°7'30" + 128°40'52"=180°0'22"., The discrepancy is due to the fact that four-place tables were used in the computation. Had five-place tables been used, the discrepancy would have been less. 2. In PQB, p= 9, g = 24, r = 27. Find P, Q, B. 3. In RS2\ r = 21, s = 24, « = 27. Find B, S, T. 4. In ABC, a = 12, & = 20, c = 28. Find A, B, C. 6. In ABC, a = 80, & = 26, c = 74. Find A, B, G. 6. Solve Ex. 1, using five-place tables. 59. The aid of logarithms in the solution of triangles. It was pointed out in Art. 6 that an expression is adapted for logarithmic computation when, and only when, it is decomposed into factors. In Cases I., II., Arts. 55, 56, the expressions used in solving the triangle can be computed with the help of logarithms. On the other hand, the side opposite to the given angle in Case III., Art. 57, and the angles in Case IV., Art. 58, are found by evalu- ating expressions which are not adapted to the use of logarithms. Other relations between the sides and angles of a triangle will be found in Arts. 61, 62. By these relations the computations in Cases III., IV., can be made both without and with logarithms. These relations are useful not merely for purposes of computation ; they are important in themselves, and valuable because many im- portant properties of triangles can be deduced from them. The explanations given in Arts. 55-57 are presupposed in Arts. 60-62. The general directions to be observed in working the problems are as follows: 59,60.] USE OF LOGARITHMS IN CASES /., //. 107 1. Write down all the formulas wliicli will he used in the compu- tation. 2. Express these formulas in the logarithmic form. [As soon as the student perceives that this step does not afford any additional assistance, it may be omitted. See Art. 27, Ex. 1^ Note 6.] 3. Mahe a skeleton scheme, and arrange the arithmetical work neatly and clearly. The skeleton schemes in the worked examples that follow, are apparent when the numbers are omitted.* Checks: The various formulas can serve as checks on the re- sults of one another. The relations derived in Exs. 1, 2, Art. 54 a, are also useful as checks. 60. The use of logarithms in Cases I., II. An example worked out, will give suf&cient explanation. EXAMPLES. 1. In ABC, given : a = 447, To find : B = b = 576, C = A = 47° 35'. c = (Write the results here.) Since a < &, there may be two sohitions. Construc- tion shows there are two sohitions. Formulas : sin ABC = -sin A = sin ABiC. a ACB=180°-{A+ABC). ACBi = lSO°-(A-]-ABi^C). AB a sin^ sin ACB. ABi a sin^ sin^C^i. .*. log sin ABC = log & + log sin ^ — log a = log sin ABi C ; log AB = log a -f log sin ACB — log sin A ; log ABi = log a + log sinACBi — log sin A. * Cologarithms are not used in the solutions in the text. In extensive computations the use of cologarithms is favoured by many computers ; but it seems best for beginners in trigonometry first to become accustomed to the obvious and direct method of working with logarithms. 108 PLANE TRIGONOMETRY, [Ch. VIL log a = 2.65031 log 6 =2.76042 logsin^ = 9.86821 -10 •. log sin B = 9.97832 - 10 .-. ABC = 72° 2' 45" .-. ACB = 60° 22' W log sin ACB = 9.93914 - 10 .-. log^^ = 2.72124 .-. AB = 526.3 and ^5iC=107°57'15" .-. ACBi = 24^ 2T i'o" log sin ACBi = 9.61710 - 10 .-. log ^i?i = 2.39920 .-. ^i;i = 250.7 In obtaining log J.^, for instance, log sin ^C^ may be written on the margin of a slip of paper, placed under log a, the addition made, log sin A placed beneath, and the subtraction made. Solve the triangle ABC, when the following elements are given : 2. A = 63° 48', B = 49° 25', a = 825 ft. Z. B = 128° 3' 49", O = 33° 34' 47", a = 240 ft. 4. ^ = 78° 30', b = 137 ft., a = 65 ft. 5. a = 275.48, b = 350.55, B = 60° 0' 32". 6. c = 690, a = 464, A = 37° 20'. 7. a = 690, & = 1390, A = 21° 14' 25". 61. Relation between the sum and difference of any two sides of a triangle. The Law of Tangents. Use of logarithms in Case III. In any triangle ABC, for any two sides, say a, 6, a _ sin A b sin B b sin ^ — sin B a -f- 6 sin A + sin B [By equation (1), Art. 54.] [By composition and division.] ^pos^A±B)m^^A-:B)^ [Art. 52, Formulas (5), (6).] 2sm:L(^A + B)cos^(A-B) ^ ' \ J>\ J ^ « + ^ tanliA + B) [Art. 44, A, B.] (1) That is, the difference of any trvo sides of a triangle is to their sum as the tangent of half the difference of their opposite angles is 61.] LAW OF TANGENTS, CASE III, 109 to the tangent of half their sum. This is sometimes called the law of tangents. Kow A-\-B=lSO°-G, and, consequently, i{A-\-B)=90° - C O Hence, tan ^(A + B) = cot — , and, accordingly, relation (1) may be written tan^(JL-B) = |^cot|C. 2 (2) Formulas for b, c, and c, a, similar to the formulas for a, b in (1), (2), can be derived in the same way as (1), (2), have been derived. These formulas can also be written down immediately, on noticing the symmetry in formulas (1), (2). Ex. Write the formulas for sides b, c and c, a. Derive these formulas. Case III, In a triangle ABC^ a, b, (7, are known, and c, B, Ay are required. Here, ^ (A-{- B)= 90° — ^Cj also, ^(A — B) can be found by (2). Hence, A and B can be found ; for J. = |(J. + J5) + i(J.-^), and B = ^(A + B)~-L{A-B). The side c can then be found by (1), Art. 54. (In using (1), (2), write the greater side and the greater angle first, in order that the difference may be positive.) Fonnulas (1), (2), can also be used as a check in the cases discussed in the preceding aii:icles. Other checks will be shown in the next article. 1. In triangle ABC, Given EXAMPLES. h = 472, Find : B = c = 324, C = ^ = 78°40^. a = Formulas : tan ^ (-B — C) = cot I A. b + c B = l{B+C)+KB- C). C=KB+ C)-l(B- C). h sin A csin^ Checks : A + B+ C = shown in ihe next article. sin B sin C 180°, formulas in preceding articles, and formulas 110 PLANE TRIGONOMETRY, [Ch. VII. log tan KB - O) = log(6 - c) + log cot ^ ^ - log(6 + c), log a = log & + log sin A — log sin B ; or = log c + log sin A — log sin C. h = 472 log(6 - c) = 2. 17026 log 6 = 2.67394 c = 324 log(& + c)= 2.90091 log sin ^ = 9.99145 - 10 A = 78° 40' log cot ^ ^ = 10.08647-10 log sin B = 9.95223 - 10 6-c = 148 .-. log tan K^-C7)= 9.35582-10 .-. log a = 2.71316 6 + c = 796 .-. i^B- C')= 12° 47' 1" .-. a = 516.6 I ^ = 39° 20' KB+ Q = 50° 40' .-. ^ = 63° 27' 1" .-. C= 37° 52' 59" Check : ^ + 5 + C = 78° 40' + 63° 27' 1" + 37° 52' 59" = 180°. Note. Formulas (1), (2), are adapted to logarithmic computation; but the computations can be made without the aid of logarithms. 2. Solve ABC, given b = 352, a = 266, C = 73°. 3. Solve PQR, given j? = 91.7, q = 31.2, R = 33° 7' 9". 4. Solve ABC, given a = 960, b = 720, O = 25° 40'. 5. Solve ABC, given b = 9.081, c = 3.0545, Jl = 68° 14' 24". 6. Solve Exs. 1, 5, Art. 57, using the formulas of this article, without logarithms. 62. Trigonometric ratios of the half angles of a triangle. Use of logarithms in Case IV. In any triangle ABC, cos A = ^'+J^'-< [Art. 54 (4).] Now l-cos^ = 2sin2i^, \ and 1 + cos ^ = 2 cos^ \ A. [Art. 50 (7), (8).] Also.l-cos^=l-^^^+^^-^^ = ^^^-(^^ + ^^-^ ' 2 6c 2 he _ a^-(b^ + c^- 2 be) ^ a'-jb- cf 2 be 2 be 2 6c .'. 2sin'»^^=:(^-^ + ^)(^+^-") (1) 62.] USE OF LOGARITHMS IN CASE IV. Ill Also,l + cos^ = l+ ^^^ = ^26^ ^ _ (6 + cf - g^ 2 6c ., 2cos^A = (^±^±f^^±^^^^ (2) Let a4-& + c = 2s; then 2(s — c) = (a + 6 + c) — 2c = a + & — a Similarly 2 (s — 6) = a — & + c, 2 (s — a) = — a + 6 + c. The substitution of these values in (1) and (2) gives o . 21 . 2(s-b)'2(s — c) o« 21 A 2s'2(s-a) 2sm^A = ^ br '^ 2cosH^ = ^^. .«. sm4^ = ^^-^^^^-^) ; (3) cos4^=5l^^. (4) Since tam' J ^ = sin' ^ ^ -s- cos' | A, it follows that ten''U = ^''-^H^-"> - (6) sm^ 2 \ s(8-a) Note. By geometry, b + c>a. Hence, — a + & + c>0, and, accord- ingly, s — a is positive. Similarly, s — b, s — c, are also positive. Therefore, the quantities under the radical signs are positive. The positive sign must be given to the radical, for A is less than 180°, and consequently J A lies between 0° and 90° 112 PLANE TRIGONOMETRY. [Ch. VII. Similar formulas hold for i B and \ C. They can be deduced in the same manner as those for i A ; or, they can be written immediately, from the symmetry apparent in the formulas (3)-(5). The student is advised to derive the similar formulas for ^ B, i O, viz. : sin2i5 = (£j=i^K£j=^; cosH^ = ^^^-^=-^; (3') sinHC7= ^'"^^^'~^^ - cosHO = '^'~'^ ' (4') ^ ah ^ ah ^ ^ ^ s(s-b) ' tanHO=^'"'^^^'~^>» ^ s(s-c) (5') Formula (5) can be given a more symmetrical form; for, on multiplying and dividing its second member by (s — a), ^ s(s-ay ' whence tan ^ 4 = -1-^ /(8-a)(s-&)(g-c). 2 s-a\ s If y.^l(s-a)(8-b)(8-c)^ (6) (7) then t&nlA^—^ — (S) 2 s-a "- ^ Similarly, tan 45 = —^, tan 4 (7=-^. (8') s — 6 s — c When all the sides are known, the angles can be found by means of formulas (3)-(5') or, by (7)-(8'). When all the angles are required,, the tangent formulas are better, since fewer loga- rithms are required than in (3), (4), (3'), (4'). It will be shown in Art. 69 that r is the radius of the circle inscribed in the triangle. 63.] HEIGHTS AND DISTANCES. 113 1. In triangle ABC, a A, B, C. Formulas : EXAMPLES. 25.17,6 = 34.06,0 = 22.17. Find r=V^ a)(s — b)(s — c tan ^A = tan i C ; lau f o = a s — s — c :. log r = i [log (s - a) + log (s - 6) + log (s - c) - log s]. log tan ^ J. = log r - log (s - a); log tan | ^ = log r log tan I O = log r — log {s — c). A C -^ 22.17 B Fia. 60. log(s-6); Check : a = 25.17 b = 34.06 c = 22.17 2s = 81.40 s = 40.70 -a =15.53 -b= 6.64 - c = 18.53 Check : A + B-i- C = 180^ logs = 1.60959 log(s-a) =1.19117 log (s- ft) =0.82217 log (s-c) =1.26788 .-. logr2 = 1.67163 .-. logr = 0.83582 log tan ^^= 9.64465-10 i^ = 23°48'28" log tan 1 B = 10.01365 - 10 ^ 5 = 45° 54' log tan 10= 9.56794-10 ^C = 20° 17' 35" .-. A = 47° 36' 56", B = 91° 48', C = 40° 35' 10" A + B + C= 180° 0' 6". 2. Solve ABC, given a = 260, b = 280, c = 300. 3. Solve ABC when a = 26.19, b = 28.31, c = 46.92. 4. Solve PQB, given p = 650, q = 736, r = 914. 5. Solve EST, given r = 1152, s = 2016, t = 2592. 6. Solve Exs. 1, 4, Art. 58, using formulas (3)-(8'), without logarithms. 63. Problems in heights and distances. Some problems in heights and distances have been solved in Art. 29 by the aid of right-angled triangles. Additional problems of the same kind will now be given, in the solution of which oblique-angled tri- angles may be used. It is advisable to draw the figures neatly and accurately. The graphical method should also be employed. EXAMPLES. 1. Another solution of Ex. 2, Art. 29. In the triangle ABP (Fig. 23), AB = 100 ft., 5^P= 30°, PBA = 180° — 45° = 135°. Hence the triangle can be solved, and BP can be found. When ^P shall have been found, then in the triangle CBP, PPis known and £P = 45° ; hence CP can be found. The computation is left to the student. 114 PLANE TRIGONOMETRY, [Ch. VII. 2. Another solution of Ex. 3, Art. 29. In the triangle CBP (Fig. 24), BP = 30 ft. , BGP = 40° 20' - 38° 20^ = 2°, PBC = 90° + ZC^ = 128° 20'. Hence CBP can be solved and the length of CB can be found. When CB shall have been found, then, in the triangle LCB^ angle = 38° 20', CB is known, and hence LB can be found. The computation is left to the student. 3. Find the distance betw^een two objects that are invisible from each other on account of a wood, their distances from a station at which they are visible being 441 and 504 yd., and the angle at the station subtended by the distance of the objects being 55° 40'. 4. The distance of a station from two objects situated at opposite sides of a hill are 1128 and 936 yd., and the angle subtended at the station by their distance, is 64° 28'. What is their distance ? 5. Find the distance between a tree and a house on opposite sides of a river, a base of 330 yd. being measured from the tree to another station, and the angles at the tree and the station formed by the base line and lines in the direction of the house being 73° 15' and 68° 2', respectively. Also find the distance between the station and the house. 6. Find the height of a tower on the opposite side of a river, when a horizontal line in the same level with the base and in the same vertical plane with the top is measured and found to be 170 ft., and the angles of elevation of the top of the tower at the extremities of the line are 32° and 58°, the height of the observer's eye being 5 ft. 7. Find the height of a tower on top of a hill, when a horizontal base line on a level with the foot of the hill and in the same vertical plane with the top of the tower is measured and found to be 460 ft. ; and at the end of the line nearer the hill the angles of elevation of the top and foot of the tower are 36° 24', 24° 36', and at the other end the angle of elevation of the top of the tower is 16° 40'. 8. A church is at the top of a straight street having an inclination of 14° 10' to the horizon ; a straight line 100 ft. in length is measured along the street in the direction of the church ; at the extremities of this line the angles of elevation of the top of the steeple are 40° 30', 58° 20'. Find the height of the steeple. -^ 9. The distance between the houses C, Z>, on the right bank of a river and invisible from each other, is required. A straight line AB^ 300 yd. long, is measured on the left bank of the river, and angular measurements are taken as follows : ^50= 53° 30', CBD=^5° 15', CAD=ii7°, i>^5=58°20'. What is the length CD ? 10. A tower CD, C being the base, stands in a horizontal plane ; a hori- zontal line AB on the same level with the base is measured and found to be 468 ft.; the horizontal angles ^^C, ABC, are equal to 125° 40', 12° 35', respectively, and the vertical angles CAD, CZ?Z>, are equal to 30° 20', 11° 50', respectively. Find the height of the tower and its distances from A and B. 64.] SUMMABY. 115 11. A base line AB 850 ft. long is measured along the straight bank of a river; O is an object on the opposite bank; the angles BAC, ABC, are observed to be 63° 40', 37° 15', respectively. Find the breadth of the river. 12. A tower subtends an angle a at a point on the same level as the foot of the tower and, at a second point, h feet above the first, the depression of the foot of the tower is j3. Show that the height of the tower is h tan a cot ^. 13. The elevation of a steeple at a place due south of it is 45°, and at another place due west of it the elevation is 15°. If the distance between the two places be a, prove that the height of the steeple is a(\/3 — 1) ^2^2. 14. The elevation of the summit of a hill from a station ^ is a ; after walking c feet toward the summit up a slope inclined at an angle ^ to the horizon the elevation is 7. Show that the height of the hill above A is c sin a sin (7 — jS) cosec (7 — a) ft. 64. Summary. The preceding discussions on the solution of triangles have shown that a triangle may be solved in the following ways : I. By the graphical method. [Arts. 10, 14, 21-24.] II. If the triangle is right-angled, it can be solved, either with or without logarithms, by the methods shown in Arts. 25-27. III. If the triangle is oblique, it can be divided into right- angled triangles, each of which can be solved by either of the methods II. [Art. 34.] IV. The triangle, whether right-angled or oblique, can be solved without using logarithms, by means of formulas (1), (3), Art. 54; (1) or (2), Art. 61 ; (3)-(8), Art. 62. V. The triangle, whether right-angled or oblique, can be solved with the use of logarithms, by means of formulas (1), Art. 54; (1) or (2), Art. 61 ; (3)-(8), Art. 62. Checks : Any formula not employed in the computation can be employed as a check ; that is, as a test for the correctness of the result. Two things are necessary on the part of one who wishes to do well in the solution of triangles : (1) The formulas referred to above should be clearly under- stood and readily derived. (2) The arithmetical work required should be done accurately, N.B. Questions and exercises suitable for practice and revieuo on the subject-matter of this chapter will be found at pages 189-193. CHAPTER VIII. SIDE AND AREA OF A TRIANGLE. CIRCLES CON- NECTED WITH A TRIANGLfe-^-':^'" 65. Length of a side of a triangle in terms of the adjacent sides and the adjacent angles. In this proof, regard is paid to the con- ventions about signs, described in Arts. 36, 37. Let ABC be any triangle. From A draw AD perpendicular to BC, or BG pro- duced. The positive direction of BC is in the direction of V. [At the first reading, only Fig. 61 a may be regarded.] B DC Fig. 61a. V B C Fig. 61a BO = BD-hDC, = -GD + BD, = -CA cos VGA + BA cos VBA. [Art. 40.] But VGA = 180° - AGB. .-. cos VGA = cos (180° - AGB) = - cos AGB. [Art. 45.] .-. BG = GA cos AGB + BA cos GBA ; i.e. a = 6 cos C + c cos B, (1) Therefore^ in any triangle each side is equal to the sum of the products of each of the other sides by the cosine of the angle which it makes with the first side. When (7 is a right angle, (1) reduces to a = c cos B. Example. Write the corresponding formulas for b and c. Derive these formulas. 116 65, 66.] AREA OF A TRIANGLE. 117 66. Area of a triangle. Suppose that the area of a triangle ABC is required. Let the length of the perpendicular DC from A D B Fig. 63a, C to AB, «or AB produced, be denoted by p, and let the area be denoted by S. The following cases may occur : I. One side and the perpendicular on it from the opposite angle known J say (c, p). S = \cp. [By geometry.] ^ (1) II. Two sides and their included angle Jcnoiun, say, 6, c, A. (See Figs. 62 a, 62 b.) S=:^cp = iC' AC sin BAO. [Art. 40.] . •. S = lbc sin A. [Compare Art. 31.] (2) III. Three sides known. /S = ^ 6c sin ^ = i &c • 2 sin 1 ^ cos I A, be (s-b)(s-c) Ri^ \ ha \ ho. ' be ^ be .'. S = Vs{s — a) (s — b) (s - c), [Art. 50.] [Art. 62.-] (3) That is, the area of a triangle is equal to the square root of the product of half the sum of the sides by the three factors formed by subtracting each side in turn from this half sum. See Art. 34 a for another derivation qf this formula. IV. One side and the angles known, say, a, A, B, G. a sin JB ^ ab sin O. S = Now b = 1 a^ sin O sin B sin J. 2 sin^ Example. Write and also derive the similar formulas in b and c. (4) 118 PLANE TRIGONOMETRY, [Ch. VIIL EXAMPLES. 1. Find the areas of the triangles in Exs. 1-5, Art. 61. 2. Find the areas of the triangles in Exs. 1-5, Art. 62. 3. Find the areas of the triangles in Exs. 2, 3, Art. 60. 67. Area of a quadrilateral in terms of its diagonals and their angle of intersection. Area^J5C'i)=area^i)0 + area ABC. %^ Area AD(7=area ALB + area QLB =i AL ' LD sin ALD A B +i GL ' LD sin CLD [Art. m (2).] fig. 63. =^{AL+LC)DL smALD, (since sin CLD=smALD) =^ AG ' DL sin ALD. Similarly, area ABG =^ AG - BL sin BLG= ^ AG - BL sin ALD. .-. area ABGD = i AG(DL+LB) sin DLA=^ AG - BD sin DLA. .'. area of a quadrilateral is equal to one-half the product of the two diagonals and their angle of intersection. EXAMPLES. ^ 1. Find the area of a quadrilateral whose diagonals are 108, 240 ft. long, and inclined to each other at an angle 67° 40'. Find the sides and angles of a parallelogram having these diagonals. 2. So also when the diagonals are 360, 570 ft. long, and their inclination is 39° 47'. 3. The diagonals of a parallelogram are 347 and 264 ft., and its area is 40,437 sq. ft. Find its sides and angles. 4. Solve an isosceles trapezoid, knowing the parallel sides a =682. 7 metres, c = 1242.6 metres, and the non-parallel equal sides b = d = 986.4 metres. Find the angles, the area, the lengths and angle of inclination of the diagonals. ___._.._ _ __ __.^J 67-69.] CIRCLES CONNECTED WITH A TRIANGLE. 119 68. The circumscribing circle. Let the radius of the circle described about a triangle ABC be denoted by R. It has been shown (see equation (2), Art. 54) that 2 sin ^ 2 sin B 2 sin C ^ ^ That is, the radius of the circumscribing circle of aiiy triangle is equal to half the quotient of any side by the sine of the opposite aiigle. From (2), Art. m, sin^ first of equations (1), gives 2S_ be ' Substitution of this in the It = abc 69. The inscribed circle. Let the radius of the circle inscribed in a tri- angle ABC be denoted by r. Join the centre and the points of contact L, M, N. By geometry, the angles at L, M, ]^ are right angles. Draw OA, OB, OC Area 500+ area 00^ + area AOB = area ABC. (2) iar-^i br + i I.e. cr = V s(s sr=:S. i){s — b)(s — c), or S. .-. r=\ (s - a)(s - b)(s s (3) That is, the length of the radius of the inscribed circle of a triangle is equal to the number of units in its area divided by half the sum of the lengths of its sides. See reference in Art. 62. Note. Formula (8), Art. 62, can be readily derived from Fig. 64. By geometry, AN = MA, BL = NB, CM = LC. NO Now But NO = r, and AN = (iAN tan ^ A = tan BA = AN BL+ CM)-{BL-VLC)=s .'. tan \ A= . 120 PLANE TEtGONOMETRY, [Ch. VIIL I.e. 70. The escribed circles. An escribed circle of a triangle is a circle that touclies one of the sides of the triangle and the other two sides produced. Let r^ denote the radius of the escribed circle touching the side BC opposite to the angle A. Join the centre Q and the points of contact L, M, N. By geometry, the angles at L, My iV are right angles. Draw QA, QB, qa Area ABQ + area CAQ — area BCQ = area ABO. '-' iKG + irJ)-iraa = S, ... ^(c + b-a)r, = S; (s — d)r^ = S, S ra = Similarly, . S 8 — a S S— G Other interesting relations between the sides, angles, and related circles, of a triangle, are indicated in the exercises in the latter part of the book. EXAMPLES. 1. Find the radii of the circumscribed, inscribed, and escribed circles of some of the triangles in Arts. 55-58. 2. Find the radii of these related circles of some of the triangles in Exs. 1-3, Art. 66. N.B. Questions and exercises suitable for practice and review on the subject-matter of this Chapter will he found at pages 193, 194. CHAPTER IX. RADIAN MEASURE. 71. The radian defined. The system of measuring angles with a degree as the unit angle, was described in Art. 11. Since the time of the Babylonians this system has been the common practical method employed. Another method of measuring angles was introduced early in the last century. This method is used to some extent in practical work, and is universally used in the higher branches of mathematics. It is employed, on account of Fig. 66. its great convenience, in the larger and more important part of what is now called trigonometry, namely, the part which is not concerned with the measurement of lines and angles, but which pursues investigation of the properties of the quantities that, so far in this book, have been called the trigonometric ratios. A very little knowledge of the trigonometric ratios is sufficient for the solution of triangles. The more detailed and extended study of angles and their six related numbers, constitutes part of what is sometimes called Higher Trigonometry, but, more generally. Analytical Trigonometry. This subject is a large one, and has close connections with many other branches of modern mathematics. 121 122 PLANE TRIGONOMEmr, [Ch. IX. The system of angular measurement now to be described, is sometimes referred to as the theoretical system of measurement. In this system the unit angle is the angle which at the centre of a circle subtends an arc equal in length to the radius. This unit angle is called a radian. Thus, if a circle with any radius be de- scribed about as a centre, and an arc AB he taken equal in length to the radius, then the angle AOB is a radian. 72. The value of a radian. In order that a quantity may be used as a unit of measurement, it must have a fixed value ; that is, using the customary mathematical phrase, it must be a constant quantity. The proof that a radian has a fixed value, or is a con- stant quantity, depends upon two geometrical facts, viz. : (a) In the same circle two angles at the centre are in the same ratio as their intercepted arcs. (6) The ratio of a circumference of a circle to its diameter is the same for all circles. [See Art. 9 (6).] For the proof of (a), reference may be made to any plane geome- try ; for instance, to Euclid VI., 33.* The proof of (b) is not con- tained in all geometries ; for instance, Euclid does not give it.f Accordingly, an outline of such a proof and the calculation of tt are given in Note C of the Appendix. This note should now be studied by those whose course in plane geometry has not included * The truth of theorem (a) can easily, by an inductive method, be made evident to students who have not proved the theorem in plane geometry. Thus, on taking angles which are twice, three times, four times, one-half, one- third, etc., of a given angle, it can be seen that their respective arcs bear the same relations to one another. t Euclid lived about 323-283 b.c. Archimedes (287 ?-212 b.c), the greatest mathematician of antiquity, measured the length of the circle and the area contained by it, and also measured the surface of the sphere. He showed that the ratio of the circle to its diameter lies between ^^ and -2^. In 1794 a French mathematician, Adrien Marie Legendre (1762-1833), pub- lished his Elements of Geometry, in which the works of Euclid and Archi- medes on elementary geometry are blended together. The elementary text- books now in use on the continent of Europe and in the United States, are written mainly on Legendrean lines ; the geometrical text-books generally studied throughout the British Empire, are editions of Euclid's Elements. 72, 73.] THE BADlAN. 123 the measurement of the circle. Theorems (a) and (h) are assumed in what foHows. T -c^- oi' tJi^ radian AOB arc AB rT> / n n In Fig. 60, -J-T77 -. = r^ . . , [By (a).] 4 ri^/ii angles circumference of circle 1 2 .*. the radian = -— x 4 right angles = - X right angle. (1) Since all right angles are equal, and since each radian is a fixed 2 fraction, namely, -, of a right angle, it follows that all radians IT are equal. It will be remembered that the unit in the common practical system is one-ninetieth of a right angle. From (1), A radian = i^° (2) IT - — 5:55! — = 57° IV 44 ".81 approximately * 3.14159... = 206265" approximately. Ex. "With a protractor lay off an angle approximately equal to a radian. Compare it with angle 60°. An angle 60°, at centre of a circle, is subtended by a chord equal in length to the radius ; a radian is subtended by an arc equal in length to the radius. 73. The radian measure of an angle. Measure of a circular arc. The radian measure of an angle is the ratio of the angle to a radian. [See Art. 8.] For instance, if an angle A is twice a radian, then its radian measure is 2 ; if an angle B is two-thirds of a radian, then its radian measure is J. This is expressed thus : A = 2 radians = 2'' ; ^ = | radians = f. (3) Here, r is used as the symbol for radians just as ° is used as the symbol for degrees in 23°. In general discussions the radian * The value of the radian has been calculated by J. W. L. Glaisher to 41 places of decimals of a second. \_Proc. Land. Math. Soc, Vol. IV. (1871-73), pp. 308-312.] 124 PLANE TRIGONOMETRY. [Ch. IX. measure of an angle is often expressed by Greek letters ; thus, the angles a, jS, 6, cf), etc., contain a, p, 6, cjy, etc., radians. In these cases the symbol r is usually omitted, but it is always understood that the radian is the unit of measurement. If the circular arc subtended by an angle is equal in length to twice the radius, then the radian measure of the angle is obviously two; if the arc is one-half t\iQ length of the radius, then the angle contains half a radian. Tlie radian measure of an angle may he given a second definition, which depends on Theorem (a). Art. 72. Let AOP, Fig. QQ>, be any angle, and AOB be a radian. Describe a circle with any radius OA, equal to r, about the vertex O as a centre. Let arc AB be equal to the radius, and draw OB. Then angle AOB is a radian, by the definition in Art. 71. j^^le^OP^ar^ [Th. (a). Art. 72.] (4) radian AOB arc AB ^ ^ ^' ^ ^ ^ angle AOP arc AP zt.. I.e. —-^ — = — (5) the radian the radius That is, the number of radians in an angle, or the radian measure of an angle, is the answer to the question : how many times does any circular arc subtended by it, contain the radius % Thus, for example, the radian measures of the angles which subtend circular arcs equal in length to 2, 3, 1.5, .825 radii are 2, 3, 1.5, .825, respec- tively. The circular arc subtended by 360° = 2 Trr ; 2 ttT hence, radian measure of 360° = = 2 ir, r and radian measure of 180° = tt. (6) This shows that an angle 2 it radians is described each time that the revolving line makes a complete revolution. Relation (6), namely, 180° = IT radians, (7) connects the two systems of angular measurement. By means of (7), an angle expressed in the one system can be expressed in the other. The word radians is usually omitted from (7), but is always understood. Relation (7) may also be deduced directly 73.] RADIAN MEASURE OF AN ANGLE. 125 from (1). Just as angles are considered as unlimited in mag- nitude, so arcs are considered as unlimited in length. Note 1. The term circular measure is often used for radian measure, Ex. 1. Express 30° in radian measure. Since 180»_.., r-^;^; TT "6' Ako,30»--- 3.14159'... 30° = T%^ = 77' Also, 30° = ^ = "-""7" = .52359^ .... 180 g > g g Tlie term radians and the symbol for radian is usually omitted from the second members of these equations, but is always understood to be there. Ex. 2. Express 45°, 60°, 135°, 210°, 300°, 330°, 270°, 225°, - 75°, 63°, 27°, — 33°, — 150°, in radian measure, (a) as fractions of tt, (&) numerically, on putting TT = ^1^. Ex. 3. Express the angle ^^ tt in degrees. Here, "t^tt" means "^^^^ x -^i^ radians.''^ Since tt = 180°, i^ tt = /^ x 180° = 162°. Ex. 4. Express the angles — , -, ^, J, J, |7r, ^tt, 10 tt, 4 7r, Stt, 2 3 4 6 5 6 TT, f TT, f TT, in degrees, and their complements and supplements, in radians. Ex. 6. Express the angles — | tt, — 5 tt, — f tt, — ii tt, — 25 tt, in degrees. Ex. 6. Express 2'' (2 radians) in degrees. Since tt = 180°, TT .-. 2'- = 2 X — = 114° 35' 29.6" approximately. TT Ex. 7. Express — , 4'', 3% — , — , 5'-, lO**, — , in degrees. 2 3 5 10 Measure of an arc. Since, by (5), subtended circular arc _ ^^^^^^^ ^^ ^^^i^,,^ j^ ^j^^ ^„glg_ radius then len^h of arc = radius x number of radians in the angle. If a denote the length of any arc AP, r the radius, 6 the radian measure of angle AOPj then a = r9. (8) 126 PLANE TRIGONOMETRY, [Ch. IX. In words : The length of any circular arc is equal to the product of the radius and the radian measure of its subtended central angle. For example, the arc of 360° = 2 tt radii, arc of 180° = tt radii, etc. These arcs are usually referred to as the arcs 2 7r, tt, etc.; but it is always understood that the radius is the unit of measurement. The symbol tt, which always denotes the incommensurable num- ber 3.14159 •••, can thus be used in three connections in trigo- nometry : (1) With other numbers, as a number simply. (2) With reference to angles; in which case it denotes an angle containing tt radians, i.e. 3.14159 ••• radians. (3) With reference to arcs; in which case it denotes an arc containing 3.14159 radii. This is an arc subtended by a central angle of tt radians. The expression 180° = tt does not mean 180° = 3.14159 ... ; it means 180° = 3.14159 radians. The expression ^^ arc tt " does not mean arc 3.1416 ; it means "arc of 3.1416 radii.^^ In any particular instance, the context will show to what tt refers, whether to angle or arc. It is evident from the second definition of radian measure that, like the trigonometric ratios, the radian measure of an angle is also a ratio of one line to another, namely, the ratio of the subtended circular arc to its radius. Note 2. If the radius he taken as unit lengthy then, by (8) or (5), the number of units of length in the arc is the same as the number of radians in the angle. EXAMPLES. 8. What is the radian measure of the angle which at the centre of a circle of radius 1^ yd. subtends an arc of 8 in.? Also express the angle in degrees. Let 6 denote the radian measure of the angle. Then Since arc 8 in. _ 8 _ 4 rad 1.5 yd. 64 27' TT = 180°, M^ = l«^ •• (/?)'■ = iV X 22 X 180° = 8° 29' approximately. 73.] EXAMPLES. 127 9. Give the trigonometric ratios of TTTTTTTr » c >, e — , — , — , — , TT, fTT, — #7r, — iTT, — # T. 6 4 3 2 10. Find the numerical values of (a) sin2 - + cos^ | tt + tan2 — , 6 3 (6) 3sinf Trcos-i^Trtan^^TT, (c) 2 sin -2/ tt cos -2/ tt tan ^/ tt. 11. Find the number of radians (a) as fractions of tt, (b) numerically (on putting w = ^-^), in each interior and exterior angle of the following regu- lar polygons: pentagon, hexagon, heptagon, octagon, decagon, dodecagon, quindecagon. 12. Find the number of radians and the number of degrees in the follow- ing angles subtended at the centres of circles : (1) arc 10 in., radius 3.5 in.; (2) arc } ft., radius 2 ft.; (3) arc 1 mi., radius 7920 mi.; (4) arc 250 mi., radius 8000 mi.; (5) arc 10 yd., radius 10 mi.; (6) arc ^ mi., radius 10 ft. 13. What are the radii when an arc 10 in. in length subtends central angles containing 1, 2, 4, 6, 8, 12, 15, 20, |, ^, f, f , radians respectively ? 14. What are the radii when an arc 10 in. in length subtends central angles containing 1°, 2°, 3°, 16^ 28°, 120°, 30', 20', 10', 10", 20", 45", re- spectively ? 15. In a circle whose radius is 10 in., what are the lengths of the arcs subtended by central angles containing 1, 4, 7, 8, 12, .5, .375, .125, radians respectively ? 16. In the circle in Ex. 15, what are the lengths of the arcs subtended by central angles containing 2°, 25°, 48°, 135°, 250°, 30', 45', 30", 50", respec- tively ? 17. What are the areas of the circular sectors in Exs. 13, 15 ? [See Note C, 5.] N.B. Questions and exercises suitable for practice and review on the subject-matter of this Chapter will be found at pages 194, 195. CHAPTER X. ANGLES AND TRIGONOMETRIC FUNCTIONS. 74. Chapters II., V., contain little more about the trigonometric ratios than is needed in the. solution of triangles. In this and the following chapters a further study of these ratios is made. Although the results of this study are not applicable to such ordinary practical uses as the measurement of triangles, heights, and distances, yet they are very interesting in themselves, and help to give a better and fuller understanding of the connection between angles and trigonometric ratios. These results are also useful in further mathematical work, and in the study of various branches of mechanical and physical science. In reading Chap- ters X., XI., acquaintance will be made, or renewed, with some important general ideas of mathematics. 75. Function. Trigonometric functions. If a number is so re- lated to one or more other numbers, that its values depend upon their values, then it is a function of these other numbers. Thus the circumference of a circle is a function of its radius ; the area of a rectangle is a function of its base and height ; the area of a triangle is a function of its three sides. Note. The values of such expressions as 2 a; — 5, x'^ — 4 x + 7, logiox, 2*, depend upon the values given to x. These expressions are, accordingly, functions of x. A function of x is usually denoted by one of the symbols /(x), F(x), 0(x), etc., which are read "the /-function of x," "the i^-func- tion of X," "the Phi-innction of x," etc. The trigonometric ratios of an angle depend upon the value (i.e. magnitude) of the angle. On this account the trigonometric ratios are very often called the trigonometric functions. They are also frequently called the circular functions. The trigonometric (or circular) functions include not only the 128 74-76.] ALGEBBAICAL NOTE, 129 six functions previously discussed, namely, sme, cosinej tangent, cotangent, secant, cosecant, but also three others, viz, : versed sine of J. = 1 — cos A, written vers A, coversed sine of ^ = 1 — sin A, written covers A, suversed sine of ^ = 1 + cos A, written suvers A. The versed sine is used not unfrequently ; the latter two are rarely used. EXAMPLES. Find the remaining eight trigonometric functions when : 1. sin ^ = .3. 2. cos J. = .4. 3. tanj. = — 3. 4. cot ^ = .7. 5. sec J. = — 3. 6. cosec A = .8. 7. vers ^ = 1.5. 8. vers A = .5. 9. Show that ^^^^ ^ ^ - ^f ^' ^ = tan A. 1 — vers A 10. Show that cos d vers ^ (1 + sec 6) = sin^ 6. 76. Algebraical note. It will be useful to have an idea of the meaning of the word limit as used in mathematics. In the geometrical series the sum of 2 terms is 1|, of 3 terms is If, of 4 terms is 1|, of 5 terms is l|f . The sum of the series varies with the number of terms taken ; and the greater the number of terms taken, the more nearly does their sum approach 2. It is stated in arithmetic and algebra that the sum of an infinitely great num- ber of terms of this series is 1 -4- (1 — |), i.e. 2. This simply means that, by making the number of terms as great as one please, the sum can be made to approach as nearly as one please to 2 ; or, in other words, the greater the number of terras taken, the more nearly does their sum approach the value 2. This idea is expressed in mathematics in slightly different language: "The limit of the sum of this series is 2." In geometry (see Note C) it is shown that if a regular polygon be inscribed in a circle, the length of the perimeter of the polygon approaches nearer and nearer to the length of the circle as the number of the sides of the polygon is increased ; also the area of the polygon approaches nearer and nearer to the area of the circle. The length of the circle is said to be the limit of the length of the perimeter of the inscribed polygon, and the area of the circle is said to be the limit of the area of the polygon, as the number of its sides is indefinitely increased. Definition. If a varying quantity approaches nearer and nearer to a fixed quantity (or given constant)^ so that the difference between the two quantl- 130 PLANE TlilGONOMETRY. [Ch. X. ties may become, and remain, as small as one please, then the fixed quantity- is called the limit of the varying quantity. The following algebraic principles are required in some of the articles that follow : I (a) If the numerator of a fraction is finite, and its absolute value either remains constant or increases while the denominator decreases, then the absolute value of the fraction increases. Thus, if in ^, a either remains a ^ constant or increases while x decreases, then - increases ; e.g. _a ^ _a_^ _a_^ ^^^^ values 10a, 100a, 1000a, .... Tiy T7(7 1000 It is also evident that the smaller x becomes, the larger does - become ; X and that - approaches an exceedingly great value when x approaches zero. X In other words, when the number x approaches the value zero as its limit, then the number - approaches an inconceivably great value as its limit. In X mathematics numbers of the latter kind are each denoted by the word infinity and by the symbol oo ; that is, the symbol co denotes any number which is greater than any number that can be assigned. The principle just stated, may be briefly expressed : If x = 0, then - = 00, X in which the symbol = indicates the following reading : If x approaches zero as a limit, then - approaches infinity as a limit. The same idea is also ex- pressed thus : Limit a a; = 0x = °°' i.e. the limit of -, when x is zero, is infinity. X (b) It is also evident that as x increases, - decreases (a remaining ^ a finite); and that when x approaches an infinitely great value, - approaches zero. That is, if x = cx>, then ^ = 0; or, ^^^^^^ = 0. (c) If X approaches zero, then - approaches zero, so long as a does not] approach zero ; that is, ^ Limit ^ _ rt 77.] CHANGES IN TRIGONOMETRIC FUNCTIONS. 131 77. Changes in the trigonometric functions as the angle increases from 0° to 360^ For convenience the revolving line will be kept constant in length in the following explanations. The student should try to deduce the changes in the functions for himself, especially after reading about the changes in the sine. Change in sin ^ as ^ increases from 0° to 360°. If OP be any position of the revolving line, then sin XOP MP op' Now OP is kept the same in length, say length a, as XOP in- creases from 0° to 360°. Hence, in order to trace changes in the sine as the angle changes, it is necessary to consider only the changes in MP. Let the angle be denoted by A. When ^ = 0, OP coincides with OB, and MP = 0. •. sinO° = - = 0. a As OP revolves from OX to O Y, MP increases in length and is positive. When A = 90°, OP coincides with 0(7, and MP = a. .'. sin 90° = - = 1. Hence, as the angle A increases from 0° to 90°, its sine increases from to 1. As OP revolves from F to OXi, MP decreases in length and is positive. When A = 180°, OP coincides with OBu and MP = 0. .-. sin 180° = - 1= 0. Hence, as the angle A increases from 90° to 180°, its sine decreases from 1 to 0. ^ As OP revolves from OXi to Yi, MP increases in length and is negative ; i.e. MP really decreases. 132 PLANE TRIGONOMETRY. [Ch. X. When A = 270°, OF coincides with OCi, MP = a and is negative. .-. sin 270-^ = -^^ = -1. a Hence, as the angle A increases from 180° to 270°, its sine decreases from to - 1. As OP revolves from O Ti to OX, MP decreases in length and is negative ; i.e. MP really increases. When A = 360°, OP coincides with OB, and MP = 0. .*. sin 360° = - = 0. Hence, as the angle A increases from 270° to 360°, its sine increases from - 1 to 0. If OP continues to revolve, then the sine again undergoes the same changes in the same order, and does so during each suc- cessive revolution. Change in cos^ as A increases from 0° to 360°, In Fig. 67, cos XOP = -^—. Hence, in order to trace the changes in the cosine as the angle changes, it is necessary to consider only the changes in OMj since OP is kept at a constant length a. When A = 0°, OP coincides with OB, and OM = a. .'. cos 0° = - = 1. As OP revolves from OX to F, OM decreases in length and is positive. When A = 90°, OP coincides with OC, and OM =0. .-. cos 90° = - = 0. a Hence, as the angle A increases from 0° to 90°, its cosine decreases from 1 too. As OP revolves from OF to OXi, Oilf increases in length and is negative, i.e. Oi¥^ really decreases. When A = 180°, OP coincides with 0-Bi, OM = a, and is negative. .-. cos 180°==^ = -!. a Hence, as the angle increases from 90° to 180°, its cosine decreases from to - 1. On proceeding in the same manner, the student will discover that : As A increases from 180° to 270°, cos A increases from — 1 to ; As A increases from 270° to 360°, cos A increases from to 1. If OP continues to revolve, then the cosine again undergoes the same changes in the sf'^ne order, and does so" during each succes- sive revolutioT" 77.] CHANGE IN TAN A, 133 Change in tan JL as A increases from 0° to 360°. MP In rig. 67, tan XOP = . Hence, in order to trace the changes in the tangent as the angle changes, it is necessary to consider the changes in MP and OM. When A = 0°, OP coincides with OB, MP=0, OM=a. .-. tan ^ = -= 0. As OP revolves from OX to F, MP increases and OM decreases, and both are positive ; hence, tan A increases. When A = 90°, OP coincides with 00, MP = a, OM=:0. As OP revolves from OF to OXi, MP decreases and is positive, OM in- creases in length and is negative ; hence, tan A decreases in magnitude and is negative ; i.e. tan A really increases. [When OP passes at OF from the first quadrant into the second, the value of the tangent changes from + co to — 00, for Oilf changes its sign from + to — .] When A = 180°, OP coincides with 07?i, MP=0, 0M = - a. .-. tanl80°= — = 0. — a Hence, as A increases from 90° to 180° tan A increases from — oo to 0. On proceeding in the same manner the student will discover that : As A increases from 180° to 270°, tan A increases from to + co ; As A increases from 270° to 360°, tan A increases from — oo to 0. If OP continues to revolve, then the tangent again undergoes the same changes in the same order, and does so during each successive revolution. In the same way as above, the student can trace the changes in sec A, cosec A, cot A, as A increases from 0° to 360°. The changes in these functions can also be deduced from the results obtained for sin A, cos A, tan A, and the relations sec A = 7, cosec A = -: — -, cot A = cos A sin A tan A The results are collected in the following table : * *This method of indicating the changes in the trigonometric functions i ; that given in Loney's Plane Trigonometry, p. 67. 134 PLANE TRIGONOMETRY. [Ch. X. In the second quadrant the sine decreases from 1 to cosine decreases from to r In the first quadrant the sine increases from -1 cosine decreases from to 1 to 1 tangent increases from - cotangent decreases from secant increases from - -Qo to to -QO to -00 -1 tangent increases from cotangent decreases from secant increases from to 00 to 1 to 00 00 cosecant increases from 1 to 00 cosecant decreases from 00 to 1 v In the third quadrant the sine decreases from to -1 In the fourth quadrant the sine increases from — ! to cosine increases from - -1 to cosine increases from to 1 tangent increases from cotangent decreases from secant decreases from - to 00 to -1 to 00 — 00 tangent increases from cotangent decreases from secant decreases from — 00 to to ooto -00 1 coseoaiUt increases from - -co to -1 cosecant decreases from -1 to — OQ Note. It should be observed that the algebraic sign of each function changes when the function passes through either of the values zero and infinity. Ex. Trace the changes in the versed sine as the angle changes from 0° to 360°. 78. Periodicity of the trigonometric functions. It has been seen in Arts. 40-44 that all angles coterminal with XOP have the same ratios. That is, the same ratios as XOP has, are obtained each time that the revolving line returns to the position OP, no matter how many complete revolutions in the positive or negative X^ X direction it may make in the meantime. In the Fio. 68. last article it was pointed out that the sine, for instance, always goes through all its changes (the cycle of changes, namely, to 1, 1 to 0, to — 1, — 1 to 0) in the same order when the turning line revolves from the position OX through the angle 360° or 2 tt. According to the opening remarks of this article, all angles which differ by any integral multi- ple (positive or negative) o/ 2 tt radians have the same sine. These facts are expressed mathematically by saying : The sine is a periodic function, and the period of the sine is 2 ir. 78,79.] PERIODICITY OF FUNCTIONS, 135 Similar considerations show that the cosine, secant, cosecant, are periodic functions, and that each of them has a period 2 tt. The tangent and cotangent, however (Art. 77), go through all their changes, while the angle increases by 180° or ir radians. Hence, the period of the tangent and cotangent is tt. Note 1. These properties may be expressed as follows, w denoting any- positive or negative whole number, and x being any angle : sin X = sin(2 imr + x), cos x = cos(2 mir + x), and similar for sec x, cosec x ; tan X = tan(m7r + x), cot x = cot(?7i7r + x). Note 2. (Algebraic.) When a function /(x) has the property that f{x) = f(x + k), in which x can have any value and A; is a constant, the function /(x) is said to be a periodic function. If k is the least quantity for which this equation is true, then k is called the period of the function. If f{x) = f{x + k), then f(x) = f(x + nk), n being any positive or nega- tive whole number. For f(x + k)=f(x + k + k)=f{x + 2 A;), and so on. Also, since f(x)=f(x + k) for all values of x, this equation holds when X — A; is put for x; that is, f(x—k) = f{x). Similarly, /(x-2A:) = f{x — k)=f(x)^ and so on. Note 3. It has been shown above that each of the trigonometric func- tions has but one period, namely, tt, in the case of the tangent and cotangent, and 2 TT in the case of each of the other functions. Hence, the trigonometric functions are singly periodic functions. Functions which have more than one period appear in some branches of higher mathematics. For instance, certain functions called elliptic functions have two periods, and, accordingly, are said to be doubly periodic. See Questions on Chap. X., Ex. 16. 79. The old or line definitions of the trigonometric functions. The trigonometric functions were formerly considered as belonging to arcs rather than to angles, and were certain lines related to these arcs. Let APB be a circle described with any radius R aboufc as a centre. Let OA, OB, be at right angles to each other, and let AP be any arc having A for its initial point. Draw OP; from P draw PM at right angles to OA ; through A draw a tangent AT to meet OP pro- Em. 69. duced in T; through B draw a tangent C7\ to meet OP produced in T^ ; from P draw PM^ at right angles 136 PLANE TRIGONOMETRY. [Ch. X. to OB. The lines MP, AT, OT, AM, were called respectively, the sine, tangerit, secant, cover sed sine, of the arc AP; and M^P, BT^, OTi (the sine, tangent, secant, of the complementary arc PB) were called respectively, the cosine, cotangent, cosecant, of the arc AP. These definitions are expressed in words as follows : The sine of an arc is a straight line drawn from one extremity of the arc perpendicular to the radius passing through the other extremity. The tangent of an arc is a straight line touching the arc at one extremity, and limited by the radius produced through the other extremity. The secant of cm arc is thG straight line joining the centre of the circle, and the further extremity of the tangent drawn at the origin of the arc. The sine, tangent, and secant of the complement of an arc are called the cosine, cotangent, and cosecant of that arc. Since the arc measures the angle at the centre (the number of degrees in this arc is the same as the number of degrees in the subtended angle), these lines were also called the sine, cosine, • ••, of the central angle AOP measured by the arc AP. These lines were known as the trigonometric lines. Note. By " the length of a line " is meant the number of units of length which it contains. The lengths of these lines depend on the length of the radius of the circle., as well as on the magnitude of the central angle subtended by the arc. Hence it was necessary to specify the radius when the functions were discussed. This inconvenience has led to the adoption of the ratio definitions. In Fig. 69 let R denote the length of the radius. Then, on using the ratio definitions, sin ^0P = ^, tan ^0P = —, sec ^0P = ^. R R R Hence, the ratio definitions of the trigonometric functions can be derived from the line definitions by dividing the lengths of the lines by the length of the radius. Jf the length of the radius is unity, then the lengths of the lines used m the line definitions are equal to the ratios in the ratio definitions. This suggests a geo- metrical or graphical method of representing the trigonometric functions, which is shown in the next article. 80.] GEOMETRICAL BEPBESENTATION. 137 80. Geometrical representation ef the trigonometric functions. Let a circle of radius equal to unity be drawn. This circle is called a unit-circle. Let the construction described in Art. 79 be made for each of the angles AOP^, AOP^, AOP^, •••. In any circle the lines M^P^, M^P^, M^P^, "•, are proportional to, and hence represent the sines of these angles, the lines AT^, ATo_, AT^, "-, represent the tangents of these angles, the lines 0J\, OT^, OT^, .••, represent the secants of these angles, ^^^' '^^^ and so on for the other ratios. In the unit-circle, however, the measures of these lines, the radius being the unit of length, are the very same numbers as the respective ratios mentioned. In the unit- eircle also, the linear measure of the arc is the same as the radian measure of the angle which it subtends. [See Art. 73, Note 3.] Suggested Exercises. (1) By means of the lines on the unit-circle, trace the changes in the trigonometric functions as the angle changes from 0° to 90°. Compare the results with those of Art. 77. (2) For particular values of the angle AOPi, measure the lengths of the related lines on the unit-circle, and compare the results with the values given in the tables of natural sines and tangents. Note 1. The origin of the terms circular functions, tangent., secant^ is apparent from Art. 79. Note 2. The name sine comes from the Latin word sinus, which was the translation of the Arabic word for this trigonometric function. The Arabic word for the sine resembled a word meaning an indentation or gulf. Note 3. In trigonometry the Greeks used the ivhole chord FiQ instead of the /iaZ/-chord or sine. For example, Ptolemy, the celebrated astronomer who flourished about 125-151 a. d., gave a table of chords in Book L of the Almagest, his work on astronomy. The Hindoos, on the other hand, always used the half-chord or sine. The Arabian astronomer, Al Battani (or Alba- tagnius) (877-929), in his work The Science of the Stars, like the Hindoos determined angles "by the semi-chord of twice the angle," i.e. by the sine of the angle, taking the radius as unity. The translation of this work into Latin in the twelfth century introduced the word sine into trigonometry. The Hindoo sine was finally adopted in Europe in preference to the Greek chord in the fifteenth century. [See Art. 12, foot-note.] 138 PLANE TBIGONOMETRY. [Ch. X. 81. Graphical representation of functions. Graphical representation. The different values which a varying quan- tity takes, are often represented by means of a curve. Many illustrations can he given of the graphical representation of various things whose values can be denoted by means of numbers. For example, the curve in Fig. 71 shows the record of the barometer at Ithaca from May 22 to May 29, 1899. 29.7 " " ~ " "" 29.5 29.1 29.3 29.2 29.1 29.0 28.9 >■ > ,^^ - «» _ -' S *" s > *«^ ^^> H EIGHT OF Bt ROV.ETER > ^ ""X^ AT ITHACA, N.Y. MAY 22 TO MAY 29 1899 - -r 28.7 2a6 llllllllllll L Inches 29.7 29.6 29.5 29.4 29.3 29.2 29.1 29.0 12 3C9123C9123C9123 9123C9123C9123C9123C9123C9123C9123C91236 9123C9123C912 M. P.M. Al. P.M. M. P.M. M. P.M. M. P.M. M. P.M. M. P.M. M. Monday Tusiday Wednesday Thursday Friday Saturday Sunday Monday May 22 May 23 May 2i May 25 May 26 May 21 May 2S May 29 Fig. 71. In this figure an hour is represented by a certain length, and the lengths rep- resenting hours are measured along a horizontal line. Each inch of height of the barometer is also represented by a certain length. At the points cor- responding to the successive times perpendiculars are drawn-, the lengths of which represent the heights of the barometer at the respective times. (In the figure the position of the horizontal line marked 29, represents the upper ends of heights of 29 inches.) The smooth curve drawn through the extremities of the perpendiculars is the barometric curve or curve of barometric heights for the period May 22 to May 29, 1899. This curv^ will give to most persons a clearer and more vivid idea of the range and variation of the height of the barometer during this period than a column of numbers of inches of heights is likely to give. If the scales used in representing the hours and the heights of the barometer were changed, then the curve would be somewhat altered, but its general appearance would remain the same. The graph of a function. The graph of a function of as, say /(a;), is obtained in the following way: Take a horizontal line XiOX; choose a point 0, from which, distances representing the different values of x are measured along the line ; measure positive values of x toward the right from 0, and negative values toward the left. At particular points of XiOX, at convenient distances apart, draw perpendiculars to represent the values of f(x) at the respective points. Draw the perpendiculars upward from XiOX when the values of f(x) are positive, and downv/ard when these values are negative„ The smooth curve drawn through the extremities of these perpendiculars is the graph of /(x). The nearer the perpendiculars Sl-82.] GRAPHS OF TRtGONOMETHlC FUNCTIONS. 139 are to one another, the better is the graph. For example, the function 2 x is represented (for certain values of x) by Fig. 72, the function ^x^, by Fig. 73, the function Vx, by Fig. 74. The pupil is advised to construct these graphs by following the method just described above. Graph of V^ Fig. 74. Exs. Draw the graphs for 3 oc, ^ jc, 4 a; + 5, 4 a; — 5, ^ x^, i x^. v25 — x'^. Note. The notion of representing a function by a curve is the funda- mental notion in algebraic geometry, or, as it is usually termed, analytic geometry. This geometry was invented, in the form in which it is now known, by the philosopher and mathematician, Ren6 Descartes (1596-1650), and first published by him in 1637. This article may be regarded as a short lesson in the subject. 82. Graphs of the trigonometric functions. Graph of sin 0. In order to draw the graph of sin 9 take dis- tances, measured from along the line X^OX, to represent the number of radians in the angle 0. At points (not too far apart) on X^OX draw perpendiculars to represent the sines of the angles corresponding to these points. The smooth curve drawn through the extremities of these perpendiculars will be the graph of the sine. Thus, for example, let a radian be represented by a unit length, and let the ratio unity be also represented by a unit length. Then (see Fig. 75) angle tt {i.e. 180°) is represented by OM^ = 3|. The perpendiculars at and M^ are zero, since sin = and sin7r = 0. Erect perpendiculars equal to •••, sin 30°, sin 45°, sin 60°, sin 90°, •••, for instance, at the points corresponding to •••, 7r TT TT TT G' 4' 3' 2' ., (i.e. .-., 30°, 45°, 60°, 90°, ••.,) respectively. Do the 140 PLANE TRIGONOMETRY, [Ch. X. same at points between L and M^, and draw the smooth curve OQ\Mi through the extremities of the perpendiculars. The suc- cessive perpendiculars from tt to 2 tt are the same in length as those from to tt, but negative. From 2 tt to 4 tt the values of the sine are repeated in the same order as from to 2 tt. Hence, the graph of the sine can be obtained by merely successively reproducing the double undulation OG1M1G2M2, as indicated in Fig. 75. This is called the curve of sines, sine curve, or sinusoid. Note 1. The unit circle (Art. 80) will be of service in drawing the graphs of the sine and the other trigonometric functions. For, if the scales for radians and ratios be those adopted above, then the horizontal distances from will be equal to the lengths of the arcs (Art. 73, Note 2), and the lengths of the perpendiculars will be the lengths of the lines in the line defini- tions (Art. 79). Note 2. If tt radians {i.e. 180°) be represented by a length different from that adopted in Fig. 75, then the graph of the sine will differ somewhat from Fig. 75, but its main features will be the same as in that figure. Figures 76 and 77 show portions of the graph of sin 6 when tt is represented on two other scales, while the sin- {i.e. 1) is represented by a unit length. Hence the curve of sines, or the sinusoid, may be defined as the curve in which hori- zontal distances measured on a certain line are proportional to an angle, and the perpendiculars to this line are proportional to its sine. Ex. Draw the graphs for cos 6, tan 6, cot 0, sec 6, cosec 6. Graph of cos 0. On using the same scales for radians and ratios as those adopted in Fig. 75, the graph of cos 6 takes the form shown in Fig. 78. It is the same as the graph of sin in Fig. 75 would be if and the other points in XiOX were all moved a distance 1 tt toward the right. This might have been expected, since the sine of an angle is equal to the cosine of its comple- ment. The values of the sine and the cosine alike range from + 1 to - 1. Graph of tan 0. On using the same scales for radians and ratios as have been adopted in Fig. 75, the graph of tan takes the form shown in Fig. 79. Graph of sec 0. On using the same scales for radians and ratios as have been adopted in Fig. 75^ the graph of sec takes the form shown in Fig. 80. 62.] GRAPHS OF SIN 6 AND COS 6. 141 \ s / CO f ^1- \ \ 8 00* 142 I*LANE TRIGONOMETRY, [Ch. X. ft I i 1 1 1 / /j 2 / 1 / y ! 37r ^^ 1 1/ ^OJL 77 A 37r / /- |57r / X Graph of tan Fig. 79. Xy^ ^.E. Graph of sea d Fig. 80u 83.] RELATIONS BETWEEN 6, SIN 6, TAN 6. 143 EXAMPLES. 1. Draw the graph for cot d, using the scales adopted in Fig. 75. 2. Draw the graph for cosec 6, using the scales adopted in Fig. 75. 3. Construct various graphs for sin 0, cos d, tan 0, cot d, sec 6, cosec d, by varying the scales used in representing radians and ratios. 4. Draw graphs for : (a) sin X + cos x, (b) sin x — cos x, (c) sin 2 x, (d) cos 2 x. 83. Relations between the radian measure, the sine, and the tan- gent of an acute angle. A. If e be the radian measure of an acute angle, then sin L±JL. (1) 1 1 — 7nn 1 - nm ' In a similar manner it can be shown that tan-i m - ten"! n = tan > ~^^ =i^. (2) 1 ' inn 30.] EXAMPLES, 153 EXAMPLES. Kl. Find tan-i2 ± tan-i|. (Compare Ex. 1, Art. 51.) tan-i 2 + tan-i i = tan-i ^ + ^ = tan-i 7 = w . 180° + 81° 52' 11".5. 1 — 2 .^ tan-i 2 - tan-i 4 = taii-i ^ ~^ = tan"! 1 = ?i7r + -. ' 1 + 2. i ^4 By the tables, taking acute angles only, tan-i 2 = 63° 26' 4". 3, tan-i i = 18° 26' 6", the sura is 81° 52' 10". 3, and the difference is 44° 59' 58". 3. The slight discrepancy between the results obtained by the two methods is due to the fact that the angles found by the tables are only approximately correct. In the following examples test or verify the result in the manner shown in Ex. 1. 2. Find tan-i 7 ± tan-i 3. 3. Find tan-i 2 + tan-i . 5. 4. Find tan-i | + tan-i \. 5. Find tan-i 3 + tan-i 2 + tan-i.6. (Suggestion. Find tan-i 3+tan-i 2, then combine the result with tan-i .6.) 6. Find2tan-il.5, 2tan-i3, 2tan-i2, 3tan-i.2. 7. Show that 2 tan-i m = tan-i -^-B- . Show that 2 ^ = tan-i { ^^^"^ \ . l-m2 Vl-tan2^/ 8. Show that 4 tan~i \ — tan-i -^\^ = ^ when the angles are between 0° and 90°. ^ 9. Find sin (sin~i \ + sin-i \) when the angles are between 0° and 90°. 10. When the angles are between 0° and 90°, show that : (a) sin (sin-i m ± sin"i n) = wVl — n^ ± ny/1 — w^. (Suggestion. Let x = sin-i ?w, y = sin""i «.) (6) cos (sin-i m ± sin-i n) = Vl — n^ Vl — m^ =F ww. (c) sin (sin-i m ± cos-i n) = mw i Vl — m* Vl — n^. (d) cos (sin-i m ± cos"i w) = ?z Vl — m^ + m Vl — w^. 11. Find sin (sin-i | + sin-i f ), cos (sin~i | — cos-i |), sin (cos-i f — cos-i j%), sin (tan~i 4 — cos-i |), tan (sec-i 3 — sin-i |), (a) when the angles are between 0° and 90°, (6) when this restriction is not imposed. 12. Two lines, AB, AC, intersect a horizontal line at B, C, making angles whose tangents are f, f. Find the angle BAC. 13. Two lines, LM, LN, make angles whose tangents are |, 2, with a horizontal line. Find the angle MLN. 154 PLANE TRIGONOMETBT, [Ch. XI. 90. Trigonometric equations. Trigonometric equations have appeared in many of the preceding articles. When an angle, 6 say, is the unknown quantity in a trigonometric equation, the complete solution is the general value of which satisfies the equor tion. For- example, if a be an angle whose sine is s, then the solution of the equation, sin B=zs, that is, of ^ = sin-^s, is 6 = nir -\- {— ly a, n being any integer. EXAMPLES. (See the definition of principal value in Art. 88.) V| 2 * 1. Solve the equation cos d = The principal value of 6 is -. Hence the complete solution is = 2 wtt -jl ^, 6 6 n being any integer. 2. Solve the equation (See Ex. 2, Art. 86.) ^ = tan -11. = n7r+I. (See Ex. 1, Art. 87.) The principal value is -. 3. Solve the equation sin x cos x=— \ V3. . .♦. sin X Vl-sin^x = - ^Vs. .-. sin2 x{l~ sin2 ic ) = j%. .-. sin* X - sin2 a; + ^^ = 0. .-. (sin2 ic - |) (sin2 aj - ;^) = 0. .-. sin2x = f ; sm'^x = \. Whence (a) sin a; = ± V| ; (6) sin a; = ± ^. The given equation shows that sinx and cosx have opposite algebraic signs. Hence, x can only be in the second and fourth quadrants. .-. In (a), a; = 120°, 300°, etc., its general value is n . 180 - 60°, where n is any positive integer. In (&), x = 150°, 330°, etc. ; its general value is « • 180° - 30°, n being any positive integer. 4. Solve the equation sin 5 ^ + sin ^ = sin 3 d. ^2 sin 3 e cos 2 6]=: sin 3 0. .-. sin 3 ^(2 cos 2 ^ - 1) = 0. ~ .-. (a) "sin 3^=0, (6) 2 cos 2 - 1 = 0. From (a), 3 ^ = 0°, 180°, etc. ; the general value of 3 ^ is rnr {n being any integer). .'. e = 0°, 60°, etc. ; the general value of 3 ^ is ^. 3 From (&), Cos 2 ^ = J. .-. 26 = ± 60°, etc. ; its general value is 2 nv ± ^. o /. 6 = ± 30°, etc. ; its general value is nr ± -• 6 90.] EXAMPLES. Find solutions of these equations ; 5. 3(sec2^ + cot2^)=: 13. 7. seco; + tanx = 2. 9. sec2 X — tan x = 3. 11. 2sin X + 5 cosx — 2. 13. 4sin^cos2^= 1. 15. 3 (tan2 q ^ cot2 d) = 10. 6. cot ^ — tan ^ = 2. 8. sec2x + tanx = 7. 10. cos — cos 7 ^ = sin 4 6. 12. sin 2 ^ + sin 4 ^ = V2 • cos 6. 14. tan*^ - 4 tan2 J. + 3 = 0. 16. cos-i X — sin-i x = cos-i x \/3. N.B. Questions and exercises suitable for practice and review on the subject-matter of Chapter XI. imll be found at pages 197-199. CHAPTER XII. MISCELLANEOUS THEOREMS AND EXERCISES. 91. Chapters II.-VIII. were devoted to the oldest and the simplest application of trigonometry; namely, the measurement of triangles. Angles and the trigonometric functions connected with angles were more fully discussed in Chapters IX.-XI. This chapter does not introduce any new principles. Most of its articles may be regarded as exercises on the relations shown in Chapters II.-VIII., and more especially on the properties an- nounced in Arts. 44, 50-52. The articles just mentioned should be reviewed. Some of the results in the exercises in this chapter are useful and important; but the student should direct attention mainly to the methods whereby the results are . obtained, so that he can proceed quickly and confidently to the solutions of similar exercises. These solutions require a ready and an accurate knowledge of (that is, an intelligent familiarity with) the for- mulas deduced in the earlier chapters. It is on this account, per- haps, that such exercises are regarded with favor by examiners. 92. Functions of twice an angle. Functions of half an angle. Relations (6)-(8), Art. 60, (3), Art. 51, give the sine, cosine, and tangent of twice an angle in terms of the functions of the angle. On rearranging (7), (8), Art. 50, there is obtained, ^U,A = ^ll=JfAA^ eoiA = yl ^ + 'l^^^ (1) whence, tan ^ = ™4 =a/M^4 (2) COS A '1 + cos 2^. On putting I x for A, these relations take the forms («) sinix=^l^l|^, (6) cosix=V^^^' (c) tan ^ a; = Ji_ZL22i^ . (3) >< 1 + cos X In (1), (2), (3), angles A and x denote any angles. 156 91-93.] FUNCTIONS OF THEEE ANGLES. 157 EXERCISES. 1. Account for the ambiguity of the radicals in (1), (2), (3). 2. Find sin 45°, given tliat cos 90° = 0. From (3) a, sin 45° ^J l-^os90" ^ _1_. V2 3. Find sin 22° 30', cos 22° 30', tan 22° 30' by means of (1), (2). Com. pare the values vv^ith those given in the tables. 93. Functions of three times an angle. Functions of an angle in terms of functions of one-third the angle. To express tan S A in terms of tan A. Let A denote any angle. tan3^ = tan(2^ + ^) ^ . . z ^— r + tan A tan 2A-\- tan A 1 — tan^ A {X t ni \ l-tan2^tan^ ^_ 2tan2^ l-tan^^ ... tan 3 ^ = BtanA-Un^A^ qx 1-3 tan2 A ^ On putting x for ZA, (1) becomes ^^^^^^3tanJ^^-tan3j^, 1 -3tan2ix To express sin 3 A in terms of sin A, sin 3 ^ = sin (2 ^ + J.) = sin 2 ^ cos J. + cos 2 ^ sin A [Art. 50, (1)] = 2 sin J. cos2 ^ + (1 — 2 sin^ A) sin A = 2 sin ^(1 - sin2 ^) + (1 - 2 sin2 A) sin A. .-. sin 3 ^ = 3 sin ^ - 4 sin^ A. (2) In a similar way, cos 3 A can be expressed in terms of cos A. COS 3 -4 = 4 cos^ ^ - 3 COS A. (3) EXERCISES. 1. Derive formula (3). 2. On substituting x for 3^1, write (2), (3). 3. Express formulas (1), (2), (3), and the results of Ex. 2, in words. 4. Assuming the value of sin 30°, calculate sin 90°. 158 PLANE TRIGONOMMBT. [Ch. Xll. 5. From cos 30°, derive cos 90° ; from tan 30°, derive tan 90°. 6. Derive sin 180°, cos 180°, tan 180°, from sin 60°, cos 60°, tan 60°, respectively. 7. Derive sin 75°, cos 75°, tan 75°, from sin 25°, cos 25°, tan 25°, respec- tively, as given in the tables. 8. Derive sin 37° 30', cos 37° 30', tan 37° 30', from the ratios of 75°. 94. Functions of the sum of three angles. i5Ilii±*2£A+tanO 1 — tan A tan B ^ tan A + tan B . ^ 1 ^ • tan G 1 — tan A tan B (1) _ tan^+tan JB+tanO— tan^tanjBtan (7 1 —tan A tan ^— tan B tan (7— tan C tan A Cor. 1. It Az=B = C, (1) reduces to (1), Art. 93. CoR. 2. If A + B + C = 180°, then tan(^ + B + C) = 0, and, accord- ingly, the numerator of (1) is equal to zero. Hence, if A, B, G, are the three angles of a triangle, tan ^ + tan ^ + tan C = tan A tan B tan C. (2) CoR. 3. If A-\- B + C = 90°, then tan(^ + JB + C) = oo, and, accord- ingly, the denominator of (1) is equal to zero. Hence, tan ^ tan ^ + tan 5 tan C + tan a tan ^ = 1, when A + B + = 90°. (3) EXERCISES. 1. Show that sin(^ + B -]-C) = smA cos J5 cos C + cos A sin S cos C + cos A cos J5 sin C — sin A sin B sin C. If A + B + C = 180°, the first member is zero. Division of the second member by cos A cos B cos C will give relation (2) above. 2. Show that cos(^ + J5 + O) = cos ^ cos J5 cos C — cos A sin J5 sin O — sin A cos i? sin C — sin A sin B cos C. What does this become when A-{-B + C= 180° ? If A + B + 0= 90°, the first member is zero. Division of the second member by cos A cos B cos C will give relation (3) above. 94-95.] IDENTITIES. 159 3. It A + B-\-C=: 180°, prove that A Ti C COS J. + COS ^ + COS C = 1 + 4 sin — sin — sin -• 2 2 2 Since A + B + G=1S0°, A±A = QO°-^- COS ^ + COS J5 + cos O = 2 cos ^^^ cos ^-^ + cos C [Art. 52 (7)] = 2 sin - cos ^-^ +1-2 sin2 ^ [Art. 50 (7)] = l + 2sin^fcos^^:^-sin^'\ 2V 2 2^ = l + 2sin^fcos^^l:^-cos^^il^l 2V 2 2 y = 1 + 2sin - .2sin-sin- [Art. 52 (8)] = l + 4sin^sin:5sin^. 2 2 2 4. If ^ + ^ + = 180°, prove that sin -4 + sin 5 + sin C 5. li A-\-B-\-0- 180°, prove that ABC sin ^ + sin 5 + sin C = 4 cos — cos — cos — • 2 2 2 cos ^ + cos JS — cos = — 1 + 4 cos — cos — sin -. 2 2 2 6. Also, that sin (J. + B) sin (5 + C) = sin A sin C 7. Also, that sin2 A + sin2 B + sin2 O = 2 + 2 cos ^ cos £ cos C 8. If ^ + ^ + O = 90°, show that sin 2 ^ + sin 2 5 — sin 2 (7 = 4 sin A sin B cos C. 9. Find tan 4 ^4, tan 5 A^ tan 6 J^, tan 7 ^, in terms of tan A. 95. Identities. In the following exercises it is required that the first member be changed into the second member. When it is difficult to do this, help is sometimes afforded by taking some steps in changing the second member into the first. The direct steps to be taken from the first member to the second may be indicated by this means. No general directions can be given con- cerning the making of these transformations. The two following suggestions, however, are frequently useful : (a) Since sin^^+ cos^^=l, unity can be substituted for the first expression, and the first expression can be substituted for unity. ip (6) The change of tan a;, cot x, sec ic, cosec x, into their values ih terms of the sine and cosine, is sometimes helpful. The examples in Art. hi belong to this class. 160 PLANE TRIGONOMETRY. [Ch. XII. EXERCISES. 1. Show that ^-^^^^^ = tan2 A. 1 + cos 2 ^ 1 - cos 2 ^ ^ 1 -(l-2sin2^) ^ sin^^ _ ^^^^ ^ l + cos2^ l+(2cos2^-l) cos2^ 2. Show that tan^ ^ ^ 1 - cos 2 ^ ^ 1 + cos 2 ^ tanM = ^^^'^ = ^^^ - cos 2^) ^ 1 - cos 2^ cos2^ 1(1 + cos 2^) l + cos2^ Note. The fact that tan^ A = , suggests that the numerator in Ex. 1 cos2^ be expressed in terms of the sine, and the denominator in terms of the cosine. In Ex. 2, the plan of transformation is more obvious. Prove the following identities : 3. _^^^!^_=sec2 5. 2-sec2J5 4. l-2sin2(45°-^)=sin2A 5. cos2 A -f sin2 ^ cos 2 J5 = cos2 B + sin2 B cos 2 A. 6. 1 + cot 2 ^ cot e = cosec 2 d cot d. 7. 4 sin A sin (60° + ^) sin (60° -A)= sin 3^. 8. cos 5 = cos(3 ^ + 2 ^) = 16 cos^ 0-20 cos^ + 5 cos 6, 9. sin 5 = 16 sin^ 0-20 sin^ + 5 sin 0. 10. tan(45^ + A)- tan (45° - ^) = 2 tan 2 A. 11. cos* 5 - sin* 5 = cos 2 5. j2^ sinl^^^cos3^^2sin2^-l. sin u4 + cos ^ 13. sinx + sin2x ^^^^^^ 1 + cos ic + cos 2 x 14. 4(cos« X + sin^ ic) = 1 + 3 cos^ 2 x. 15. sin 4 ^ = 4 sin A cos^ A — A cos A sin^ ^. 16. cos 4 ^ = 1 - 8 cos2 ^ + 8 cos* A. 96. For an acute angle of radians, cos 6 > 1 , sin 8 > 9— — • 2 4 By Art. 60, (7), cos (9 = 1 - 2 sin^ t by Art. 83, sin ^ < ^- Hence, cos ^ > 1 — 2 ( | ] , i.e. cos > 1 - 1- \ By Art. 50, (5), 2"~2 --""2-2 --2C-^'4) sin0 = 2sin^cos? = 2tan^cos2^=2tan^ 96, 97.] COMPUTATION OF TRIGONOMETRIC FUNCTIONS. 161 But by Art. 83, tan^>^, and sin^<^. I '-©•!"■ Hence sin ^ > -— 1 1 — f ^ ) 1 ; t'.e. sin 9 > - ^. 97. One method of computing the trigonometric functions. A method of computing the trigonometric functions of angles which are in an arithmetic progression having the common difference Z)", will now be shown. sin in + 1) D" + sin (n - 1) i)" = 2 sin nZ)" cos B'\ [Art. 52, (5).] .-. sin (n + 1) D" = 2 sin nD" cos D" - sin {n - 1) U\ (1) Also cos i)" = Vl-sin^Z)". Hence, if the sines of the angles D", 2 D", 3D", up to nD" be known, then sin {n + 1) D" can be computed by formula (1). The other functions can be derived from the sine. The functions for angles from 0° to 45° will serve for the angles from 45° to 90°, since the ratio of an angle is the co-ratio of its complement. When the functions have been computed for angles up to 30°, the computations for angles greater than 30° can be made more easily. For, if A is an angle less than 30°, sin (30° + ^) 4- sin (30° - ^) = 2 sin 30° cos A = cos A. ' sin (30° + ^) = cos ^- sin (30°-^). (2) Similarly, cos(30°-f ^)=cos(30° -^)~sinA (3) In formula (1) suppose that D" = 10", and let its radian meas- ure be denoted by B, Then sin 10" <^, >^-j- [Arts. 83, 96.] Since 180° = tt, 10" = ^^ = 3.1415926535 ^ 000048481368 .•• radians. 180 X 60 X 60 64800 .-. sin 10" <. 00004848..., > [.00004848 ••.- ^ (.00004848 ...)3]. Hence, to 12 places of decimals, sin 10" = .000048481368. From this, sin 20" can be found by (1) ; then sin 30", then sin 40", and so on. The functions of several angles can be found independently of the method just shown. Formulas involving these angles, and 162 PLANE TRIGONOMETRY. [Ch. XII. Euler's and Legendre's verification formulas, may be ased to test the accuracy of the tables. The latter formulas are (see Exs. 7- 10, Ch. XII.), sin(36°+^)-sin(36°-^)-sm(72°+^) + sm(72°-^) = sin^ (4) cos(36°H-^)+cos(36°-^)-cos(72°+^)-cos(72°-^)=cosA(5) EXERCISES. 1. Test the tables of natural sines and cosines by means of formulas (4), (5), taking A equal to 4°, 10°, 15°, and other values. 2. Assuming the func- tions of 1° as known, calculate the sines of 2°, 3°, 4°, 6°, 6°, by formula (1). 3. By means of formulas (2). (3), calculate the sines and cosines of 33°, 37°, 41°, 47°, 53°, 67°, and other angles. 98. Trigonometry defined. Branches of trigonometry. Before concluding this text-book it may be well to indicate to the student the relation of the part of trigonometry treated in the preceding pages to the subject as a whole, and also to try to give him a little idea of another branch of trigonometry ; namely, analytical trigonometry. In Chapters II.-IX., plane angles, the solution of plane tri- angles, and applications connected therewith were discussed. This is what is usually known as plane trigonometry. The study of solid angles, the solution of spherical triangles, and the as- sociated practical applications, constitute spherical trigonometry. These branches of mathematics are founded on geometrical con- siderations, and may be looked upon as applications of algebra to geometry. Pure mathematics is sometimes regarded as consisting of two great branches ; namely, geometry and analysis. Analysis includes algebra, infinitesimal calculus, and other subjects which employ the symbols, rules, and methods of algebra, and do not rest upon conceptions of space. (Geometrical ideas may be used in analysis, however, for the sake of exposition and illustration, and, on the other hand, algebra may be employed in expounding the principles of geometry.) Since the eighteenth century, trigo- nometry has also been treated as a branch of analysis.* *The meaning of the word " analysis" thus used in mathematics, should not be confounded with the ordinary meaning of the word, or with the meaning attached to the term *' analysis " in logic. 98.] ANALYTICAL TRIGONOMETRY. 163 Analytical (or algebraical) trigonometry treats of the general relations of angles and their trigonometric functions without any reference to measurement. It discusses, among other things, the development of exponential and logarithmic series, the connections between trigonometric and exponential functions, the expansions of an angle and its trigonometric functions into infinite series, the calculation of ir, the summation of series, and the factorization of certain algebraic expressions. The properties stated in formulas, (l)-(3) Art. 44, (l)-(8) Art. 50, (l)-(8) Art. 52, (l)-(3) Art. 93,, are analytical properties, and can he derived without the did of geom- etry. Analytical trigonometry includes hyperbolic trigonometry ; that is, the treatment of what are called the hyperbolic functions. While the trigonometric functions may be defined and discussed on a geometrical basis, as done in this book (and this is the easiest way for beginners), it may be stated that they can also be defined and their properties deduced on a purely algebraic basis. It is beyond the scope of this work to show this, but the student may obtain a little light on the subject by reading Notes A and D. It may be stated further, that, under certain restrictions, some of the most important theorems and properties found in analytical trigonometry can be derived easily in an elementary course in the infinitesimal calculus. It has been pointed out that the trigono- metric functions can be defined in a purely geometrical manner, and in a purely algebraic manner ; they can also be given defini- tions depending on the infinitesimal calculus, and their properties deduced therefrom. Finally, it may be said that trigonometry is merely a brief chapter in the modern Theory of Functions, and may be defined as the science of singly periodic /mictions (see Art. 78). For a treatment of trigonometry, either as a part of algebra, or, as "an elementary illustration of the application of the Theory of Functions," see Lock, Higher Trigonometry ; Loney (Part II.), Analytical Trigonometry ; W. E. Johnson, Trea- tise on Trigonometry, Chaps. XII.-XXII. ; Casey, A Treatise on Plane Trigonometry ; Levett and Davison, Elements of Plane Trigonometry (Parts II., III., Eeal Algebraical Quantity, Com- plex Quantity); Hay ward, Vector Algebra a7id Trigonometry; Hobson, A Treatise on Plane Trigonometry ; Chrystal, Algebra, Part I., Chap. XII. ; Part II., Preface, and Chaps. XXIX., XXX. APPENDIX, NOTE A. HISTORICAL SKETCH. The most ancient mathematical writing known at the present time is an Egyptian papyrus preserved in the British Museum. It is the work of Ahmes, an Egyptian priest who lived at least seventeen hundred years b.c, and is believed to have been founded on older works dating as far back as 3400 B.C. The treatise is concerned with practical mathematics, and merely, gives rules for making geometrical constructions and determining areas. The area of an isosceles triangle is obtained by taking the product of half the base and one of the sides. The area of a circle is found by deducting from the diameter one-ninth of its length, and squaring the remainder- -a proceeding which is equivalent to taking tt = 3.1604 ••-. The ancient Greeks brought geometry to a high state of perfection, but showed little aptitude for algebra and trigonometry. They were not inclined to be satisfied with approximate results, and regarded the practical applica- tion of mathematics as degrading to the science. Trigonometry was invented to supply practical needs, and its development, in the earlier stages, was due to men of the Egyptian, the Hindoo, and the Semitic races. Astronomy was one of the studies most cultivated by the ancients, but astronomy could not advance, or even become a science, without the aid of trigonometry, Hipparchus of Nicsea in Bithynia, the greatest astronomer of antiquity, who flourished about 160-120 e.g., is regarded as the founder of trigonometry, which he developed solely as a necessary part of astronomy. Moreover, trigonometry continued to exist, for the most part, merely as a handmaid of astronomy for over eighteen hundred years. On this account, the theorems of spherical trigonometry were developed earlier than those of plane trigonometry. Of the writings of Hipparchus, all but one have been lost ; but it is known that he constructed a table of chords, which serves the same purpose as a table of natural sines. Hero of Alexandria, who flourished some time between 155 and 100 b.c, and is supposed to have been a native Egyptian, found the area of a triangle in terms of its sides, and placed 166 166 PLANE TRIGONOMETRY. engineering and land-surveying on a scientific basis. Ptolemy, a native of Egypt, the records of whose observations cover the period 127-151 a.d., wrote the Syntaxis Mathematica (called the Almagest by the Arabs), a work founded on the investigations of Hipparchus. This was regarded as a kind of astronomical Bible for thirteen hundred years, until the Ptolemaic theory, namely, that the sun, planets, and stars revolve around the earth, was shown to be erroneous by Copernicus and Galileo. The Almagest is divided into thirteen books. Book I. treats of plane and spherical trigonometry, contains a very accurate table of chords, probably derived from Hijjparchus, and shows the method of forming the table. It develops spherical before plane trigonometry, and does not give the solution of plane triangles. " Whereas the Ptolemaic system (of astronomy) was . . . overthrown, the theorems of Hipparchus and Ptolemy, on the other hand, will be, as Delambre * says, forever the basis of trigonometry." t Whatever advance was made in trigonometry during the thousand years after Ptolemy, was due to the Hindoos and Arabs. The Hindoos had tables of the half-chords, or sines, and found that the arc equal in length to the radius contained 3438'. Aryabhatta (476-530 a.d. ?) wrote a work containing sections on astronomy, spherical and plane trigonometry. This contained tables of natural sines of the angles in the first quadrant at intervals of 3f°, the sine being defined as the semi-chord of twice the angle. Pe gave 3.1416 as the value of tt. Other writers were Brahmagupta, born 598, and Bhaskara, about 1150, who gave some trigonometric formulas. The Hindoos knew how to solve plane and spherical right triangles. During the period of the Dark Ages in Europe, the sciences of the Greeks and Hindoos were preserved, and, to some slight extent, improved by the Arabs. The latter studied trigonometry only for the sake of astronomy. The term sine is due to the celebrated Arabian astronomer Al Battani (Albategniiis), a native of Syria, who died about 930 a.d. Another Arabian astronomer, Abii H Wafa (940-998), a native of Persia, was the first to intro- duce the tangent of the arc into the science ; he calculated a table of tangents. Among the Western Arabs, to whom the development of the subject is in- debted, were Ihn Yunos of Cairo (died 1008), and Gabir ben AJlah, who was born at Seville and who died at Cordova in the latter part of the eleventh century. The latter wrote an astronomy in nine books, the first of which is devoted to trigonometry ; he also contributed to the advancement of spherical trigO]iometry. The next stage in the history of trigonometry is marked by the introduc- tion of the Arabian works into Europe, and the development of the arithmet- ical part of the subject, especially the calculation of tables. This was largely ' *Jean Baptiste Delambre (1749-1822), a French mathematician who de- rived important formulas in spherical trigonometry. t Ency. Brit., Art. Ptolemy. J APPENDIX, 167 the work of German astronomers, and chiefly of Begiomontanus and Mheticus. Georg Purbach (1423-1461), professor of mathematics and astronomy at the University of Vienna, wrote a table of natural sines computed for intervals of ten minutes, which was published in 1541. Begiomontanus (John Mtiller) (1436-1476), a native of Franconia, who was one of the greatest mathemati- cians that Germany has ever produced, in conjunction with Purbach made a translation of the Almagest, which was published in 1496. In this he sub- stituted sines for chords, and gave a table of natural sines. He reinvented the tangent, and made a table of natural tangents for all degrees of the quad- rant ; this was published in 1490. In 1464 he wrote his De Triangulis, which was the earliest modern systematic exposition of plane and spherical trig- onometry. This was printed in 1533, and a second edition appeared in 1561. The only functions introduced were sines and cosines. Copernicus (1473- 1543), born in Prussia, wrote a short text-book on the subject about 1500, which was published in 1542. Bheticus (Georg Joachim) (1514-1576), a native of the Tyrol, professor of mathematics at Heidelberg, constructed tables (published in 1596) which are the basis of those still in use. He intro- duced secants and cosecants, and found the values of sin 2 6, sin 3 in terms of sin 6, cos 6. Hitherto the trigonometric functions had been considered as lines related to circular arcs. Rheticus was the first who constructed the right triangle and used the ratio definitions which depend directly on the angle. These definitions were not adopted, however, and, although intro- duced two hundred years later by Euler in 1748, they did not come into common use until after the middle of the present century. Pitiscus (1561- 1613), professor of mathematics at Heidelberg, made important corrections in and additions to the tables of Rheticus. His trigonometry, published in 1599, contained formulas for cos (A± B), sin (A — B). Adrian Bomanus (1561-1625), a Belgian mathematician, professor at the University of Lou vain, first found the formula for sin (yl + B). Franrms Vieta (1540-1603), the greatest French mathematician of the sixteenth century, extended the tables of Rheticus. He made one of the earliest attempts to find the value of tt by means of infinite series, and was the first who made any considerable appli- cation of algebra to trigonometry. In his work, Ad Angulares Sectiones, he gave formulas for sin nd, cos nO, in terms of sin 6, cos d. John Napier (1550-1617) discovered the important formulas in spherical trigonometry which are commonly called Napier's Analogies. His invention of logarithms greatly lessened the arithmetical work necessary in astronomy and trigonom- etry, and thus ushered in a new era in the history of these sciences. Edmund Ounter (1581-1026), professor of astronomy at Gresham College, London, gave the first tables of logarithms of sines and tangents. He first used the terms cosine, cotangent, cosecant. Albert Girard (1590-1634), a Flemish mathematician, published a trigonometry in which the contractions sin, tan, sec were used. William Oiightred (1575-16(50), an English mathematician, wrote a trigonometry, published in 1657, containing abbreviations for sine, 168 PLANE TRIGONOMETRY. cosine, but they did not come into general use until Euler reintroduced them nearly a century later.* Thus far, trigonometry had been confined to the bounds set by the ancients, namely, to expressing the relations between the sides and angles of plane and spherical triangles, to the solution of triangles, and to the calculation of tables. Trigonometry had been founded on geometrical conceptions, and was regarded mainly as an appendage of geometry and astronomy. In the seventeenth and eighteenth centuries, however, a new branch of the subject, namely, analytical trigonometry, was created, chiefly by the genius of De Moivre and Euler. In the new development of the science, the symbols, rules, and methods of algebra were employed, and geometrical conceptions were disregarded. [See Art. 98 and Note D.] The older trigonometry still retains its position as a necessary department of applied mathematics. The modern analytical (or algebraical) side of the subject, however, has been so highly developed since the middle of the eighteenth century, and its results are so much employed in other branches of mathematical and physical sci- ence, that it, may be regarded as the larger and more important part of trigonometry. The new development began with the discovery and investigation of expo- nential, logarithmic, and trigonometric series. The chief investigators of infinite series were : John Wallis (1G16-1703), professor of geometry at Oxford ; James Gregory (1638-1675), professor of mathematics at Edin- burgh ; Nicolaus Mercator, died 1687, a native of Holstein, who settled in England; Isaac Newton (1642-1727); Gottfried William Leibnitz (1646- 1716). Several of these series greatly simplified the calculation of tt ; some of them were obtained by means of the infinitesimal calculus invented by Newton and Leibnitz. Before 1669, Newton obtained the series for the arc in powers of the sine, and the series for the sine and cosine in powers of the arc. In 1670, Gregory discovered the series for the arc in powers of the tangent, and the series for the tangent and secant in powers of the arc ; Leibnitz discovered the first of these independently in 1673. * " To England falls the honour of having produced the earliest European writers on trigonometry." (Cajori, History of Mathematics, p. 135.) Thomas Bradwardine (1290 P-1349) , archbishop of Canterbury, Richard of Wallingford (1292?-1336), abbot of St. Albans, John Mauduith (about 1310), fellow of Merton College, Oxford, who were mathematicians and astronomers, left writings containing trigonometry and tables drawn from Arabic sources. The earliest English books in which spherical trigonometry is used, are those of Thomas Digges (died 1595), one of the foremost English mathematicians of the sixteenth century. The earliest book in which plane trigonometiy is introduced, is a -work published by Thomas Blundeville in 1594. J APPENDIX, 169 John Bernoulli (1667-1748), a native of Switzerland, originated the idea of trigonometric functions, and treated trigonometry as a branch of analysis. He was tlie first to obtain real results by using the symbol V— 1. Abraham de 3Ioivre (1667-1754), a French Huguenot who settled in London, did much to advance analytical trigonometry, by his use of (so-called) imaginary quantities, and the discovery of the great fundamental theorems, which are called by his name. (See Note D.) Johann Heinrich Lambert (1728-1777), a native of Alsace, developed de Moivre's theorems, introduced the functions called hyperbolic sine and cosine, and showed their connection with the hyperbola. He also found that ir is incommensurable. Modern trigonometry is indebted most of all to Leonhard Euler (1707-1783), a native of Switzerland. In his Introductio in Analysin Injinitorum, published in 1748, he system- atized and generalized what was then known about algebra and trigonometry. He discussed the expressions of functions in series, and treated trigonometry as a branch of analysis. The latter was effected by regarding trigonometric functions, not as straight lines belonging to arcs, and thus depending on the radius of a circle, but as ratios, and thus as functions of the angle only. He reintroduced the abbreviations now used. [This was done simultaneously in England by Thomas Simpson (1710-1761), professor of mathematics at Woolwich, in his trigonometry, also published in 1748.] Euler first showed the connection between exponential and trigonometric functions (see Note D), and discovered many of their analytical properties.* Since the time of Euler, analytical trigonometry has benefited by the immense advances made in the theory of functions of complex quantities ; that is, quantities of the form x + V^H! y. It is now coming to be regarded, more properly and more logically, as an elementary chapter in the modern theory of func- tions. See Chrystal, Algebra^ Part II., p. vii.t NOTE B. 1. Projection definition of the trigonometric ratios. [Supplementary to Art. 40.] In Fig. 20, Art. 28, MN is the projection of AB on LB, and NM is the projection of BA on LB. In naming the projection, the points obtained by projection are taken in the same order as the corresponding points in the original line. It is apparent that, for any line, the projections upon a series of parallel lines are equal. For instance, AD = MN, Fig. 20. This may also be seen by drawing a series of lines parallel to LB and pro- jecting AB upon them. * The first English book in which trigonometry received an analytical treatment was that of Robert Woodhouse (1773-1827), professor at Oxford, which was published in 1809. t The principal sources from which this historical sketch has been drawn, are Hobson, Article Trigonometry (JEncyclopcedia Britannica, Oth edition), Ball, A Short History of Mathematics^ Cajori, A History of Mathematics, 170 PLANE TRIGONOMETRY. Suppose that in Fig. 36, Art. 40, YOY^ be drawn at right angles to X^OX. Then OM is the projection of the turning line OP upon OX, MP is equal to the projection of the turning line OP upon OY. In two cases in Fig. 36, the projection of OP on OX is in the direction opposite to OX, that is, it is negative ; in two cases, the projection of OP on F is opposite to the direction of Y, that is, it is negative. The definitions. Art. 40, may now be stated as follows : .i^^^proJ-OPonOr ^^^ j^^Vm^OPonOY ^ OP proj. OP on OX sec-4=- OP proj. OP on OX OP ^^^^^p^oj^OPonOX. eot^=PMJ>P^OX eosec^=- OP proj. OP on O r proj. OP on F These differ from (1), Art. 40, merely in the fact that names are given to OM and MP. The properties shown in Chap. V. follow from these definitions. 2. Theorem on projection. The projection of one side of a polygon upon any straight line is equal to the algebraic sum of the projections of the other sides. s K L P FiQ. 91. Let PQRS be any polygon. Draw parallel lines P/), Qq, Bi\ Ss, from its vertices to any straight line LK. Then it is apparent that i.e. ps = pq + qr + rs ; proj. PS = proj. PQ + proj. QR + proj. RS. In Fig. 91, ?*s, the projection of RS, is negative. This proposition is true whether the projection be oblique or orthogonal. The theorem may also be stated thus : The projection of a broken line upon a straight line is equal to the projec- tion of the line, drawn from the initial point to the terminal point of the broken line. Thus, the projection of the broken line PQRS upon any straight line is equal to the projection of PS upon the same line. APPENDIX. 171 3. The sine and cosine of the sum of two angles. [Supplementary to Art. 46.] Let the construction be made as indicated in Art. 46. Then cin^^a. PN _ Pr03» Oi^ on or _ proj. 0^ on or proj.QPonOr [Theorem in (2).] _ proj. OQ on or OQ proj. ^P on Or QP OQ 'op QP ' OP = sui J. cos P + sin VQP sin B = sin A cos B + cos A sin B. coc CA I B^ = P^^J- ^^ ^" ^^ = P^^-^- OQ^^O^ \ P^^oJ- QP o^^ ^-^ ^ ^ OP OP OP _ pro3.0(>onOX 0^ proj. QP on OX QP OQ 'op QP 'op = cos A cos B + cos VQP' sin B = cos A cos B — sin J. sin B. in the projection proof of the addition formulas for the sine and cosine, A and B can have any magnitudes, positive or negative. The formulas for sin(yl — B), cos(^ — jB), can also be derived by substituting — B for + B In the addition formulas. NOTE C. [Supplementary to Arts, 9, 72.] ON THE LENGTH AND AREA OF A CIRCLE. 1. The main purpose of this note is to outline a method of approximating to the value of ir ; that is, to the ratio of the length of a circle to its diameter. This method depends only on elementary geometry.* There are simpler and more expeditious methods of finding rr, but they require a greater knowledge of mathematics than beginners in trigonometry generally possess. By the methods of elementary geometry, as shown in the texts of Euclid and others, regular polygons of 3, 4, 5, 6, 15 sides can be inscribed in, and circumscribed about a given circle. Moreover, inscribed and circumscribing regular polygons of 2, 4, 8, 16, •••, times each of those numbers of sides can also be constructed by successively bisecting the arcs subtended by the sides, and joining the consecutive points of division. This process can evidently be * A section on the mensuration of the circle is given in many geometries. Reference may be made to the geometries of Beman and Smith (Ginn & Co,)^* Gore (Longmans, Green, & Co.), Phillips and Fisher (Harpers), and others. 172 PLANE TRIQONOMETRT, carried on until tlie inscribed and circumscribing polygons have an infinitely great number of sides ; that is, regular polygons of 3.2'*, 4.2«, 5.2", 15.2" sides, n being any positive integer, can be inscribed in, or circumscribed about, a given circle. 2. Outline of a proof of the theorem that the lengths of circles are pro- portional to their diameters. (a) The length of a circle is gi-eater than the perimeter of an inscribed polygon, and is less than the perimeter of a circumscribing polygon of any finite number of sides. (&) As the number of sides of a regular polygon inscribed in, or circum- scribed about, a circle is increased, the length of the perimeter of the polygon approaches nearer and nearer to the length of the circle. In other words, by increasing the number of sides, the difference between the length of the perimeter of the polygon and the length of the circle may be made as small as one please, and this difference approaches zero when the number of sides approaches infinity. (c) Let any two circles be taken, and let the radii be iJ, r. Let AB be a side of a regular polygon of n sides inscribed in the circle having centre O ¥iQ. 93. and radius jB, and let ah be the side of a regular polygon of n sides inscribed in the circle having centre o and radius r. Let P denote the perimeter of the first polygon, p that of the second ; let C denote the length of the first circle, and c that of the second. Then P=G-D, p = c-d, where D, d may each be made smaller than any assignable quantity by mak- ing the number of sides, ?i, infinitely great. The polygons are similar, since they are regular and have the same num- ber of sides. Hence, by geometry, P^OA^B, p oa r ' that is, ^-^ = R. c — d r . From this, rC -rD-Bc- Rd\ Whence, rC - i2c = rD - i?d. APPENDIX. 173 Now, let n become infinitely great. Then the second member becomes smaller than any assignable quantity, since r, J?, each remains finite, and d, Z>, each approaches zero. Hence, when n is infinitely great, rC -Ec = 0. (1) rrom(l), ^ = ^, and 1 = 1 (2) The first of equations (2) may be expressed in words : lengths of circles are to one another as their radii. According to the second equation, the length of the first circle is to its radius as the length of the second circle is to its radius. But these are any two circles. Hence, the ratio of the length of a circle to its radius, and, consequently, to its diameter, is constant. The ratio of the length of a circle to its diameter, which ratio is denoted by tt, will now be approximately determined. [See Arts. 9 (6), (c), 72.] 3. The formulas used in this determination of ir are deduced in problems Af B, that follow : A, Given the radius of a regular inscribed polygon, to compute the side of a similar circumscribing polygon. Let AB be the side of the inscribed polygon, and OC = B, the radius of the circle ; let LM be a side of the similar circumscribing polygon. Let LMhe obtained by producing OA, OB, to intersect the tangent drawn at C, the middle point of the arc AB. The triangles LCO, AEO, are similar. Hence, LC ^ PC AE Oe' OCxAE BxAE .'. LC = LM OE E X AB OE OE In the right-angled triangle OAE, OE Vo^^^aF=\ B2 .'. LM = 4 2B X AB V4 B2 - AB^ = lV4i22 2 AB\ (1) J5. Given the radius and the side of a regular inscribed polygon, to com- pute the side of the regular inscribed polygon of double the number of sides. In Fig. 93, let AB be the side of a regular inscribed polygon of n sides. Draw AC ; then ^C is the side of a regular inscribed polygon of double the number of sides, namely, 2 n sides. It is required to express AC in terms of the radius B and the side AB. Produce CO to D and draw DA. The triangles ACD^ AGE^ are similar, since the angles DAC^ AEC, are equal, 174 PLANE TRIGONOMETRY. both being right angles, and the angle ACE is common to both triangles. Hence, CD : AC = AG: CE. .-. AC^= CD' GE^ CDi^CO - E0)=1 R{B - E0)= R{2 B -2 EO). But EO =Vo^^ - ae' =yR2 - ^ = I V4 i?2 - AB". AC^ = B(i2 B --^4. B^ - AJ^). . AC = \'i?C2 B-^4: B^ - AB^) (2) 4. To determine approximately the ratio of the circumference of a circle to its diameter. If the radius is 1, the length of the circle is 2 tt, and the length of the semicircle is tt. Hence the length of the semi- perimeter of each inscribed and circumscribing regular polygon, is an approxi- mate value of TT, and approaches nearer and nearer to tt, the greater the number of sides in the polygon. The side of the inscribed square of the circle of radius 1 is V2 ; its semi-perimeter is 2.8284271. Successive appli- cations of (2), Art. 3, give the sides of the inscribed polygons of 8, 16, 32, — sides, and successive applications of (1) give the sides of the similar circum- scribing polygons. The successive semi-perimeters are obtained by taking one-half the product of the length of a side and the number of sides in the polygons. The results of the computation are given in the following table. The table also gives the results when the initial polygon taken, is the inscribed hexagon. The figures in bold type show the approximations. Lengths of semi-perimeters of regular inscribed and circumscribing poly- gons of circle of radius = 1. OF Sides. Inscribed. Circumscribing. OF Sides. Inscribed. CiRCUMSCKIBlNG. 4 2.8284271 4.0000000 6 3 3.4641016 8 3.0614675 3.3137085 12 3.1058285 3.2153903 16 3.1214452 3.1825979 24 3.1326286 3.1596599 32 3.1365485 3.1517240 48 3.1393502 3.1460862 64 3.1403312 3.1441184 96 3.1410319 3.1427146 128 3.1412773 3.1422236 192 3.1414524 3.1418730 256 3.1416138 3.1417504 384 3.1415576 3.1416627 512 3.1415729 3.1416321 768 3.1415838 3.1416101 1024 3.1415877 3.1416025 1536 3.1416904 3.1415970 2048 3.1415914 3.1415951 4096 3.1415923 3.1415933 8192 3.1415926 3.1415928 APPENDIX. 175 5. Area of a circle. Area of a circular sector. The area of a circum- scribing polygon of a circle is equal to one-half the product of the lengths of the perimeter and the radius. When the number of sides of the polygon increases indefinitely, the perimeter of the polygon approaches the length of the circle as its limit, and the area of the polygon approaches the area contained by the circle as its limit. Hence, area of circle = \ length of circle x length of radius ; i.e. area of circle = lx2irBx B = iriJ'^. Since the area of a sector of a circle has the same ratio to the area of the circle that the arc of the sector has to the length of the circle, area of circular sector = \ length of arc x length of radius. Hence, if 6 is the radian measure of the angle of the sector, area sector = \Bdx B = fiJ^e. [For example, see Art. 73, Ex. 17.] 6. Historical Note. The problem to find a square whose area is equal to that of a given circle, which is commonly known as "squaring the circle," or "the quadrature of the circle," has long been of interest to mathematicians and others. Since the area of a circle is one-half the radius by the length of the circle, and the ratio of the length of the circle to the diameter is a con- stant, it follows that "squaring the circle " comes to determining this ratio. The ancient peoples used 3 as the value of tt ; see 1 Kings vii. 23, 2 Chron. iv. 2. Ahmes used 3.1604; Archimedes showed, by the method described above, and by successively inscribing and circumscribing regular polygons of 6, 12, 24, 48, 96 sides, that -k is between 3ff and 3f. Ptolemy used 3^^, and Aryabhatta, 3.1416. Adrian of Metz, in 1527, by using poly- gons up to 1536 sides, showed that the ratio is between f|^ and |^|. By taking the mean of the numerators for a new numerator, and the mean of the denominators for anew denominator, he obtained the value ff|, which is correct to six places of decimals. In 1579, Vieta by using polygons of 32,316 {i.e. 6 X 216) sides, got the value of tt correctly to ten places. His method is not the same as that of Archimedes. (Professor Newcomb has remarked that the value of tt to ten places of decimals would give the circumference of the earth correctly to within a fraction of an inch, if the diameter were accurately known.) In 1593, Adrian Roman us of Louvain computed tr to 15 places of decimals and Ludolph van Ceulen (d. 1610), a German residing m Holland, calculated it to 35 places. Hence tt is often referred to in Germany as " the Ludolphian number." The discoveries of trigonometric series (see Note A) made the work of com- puters easier and more mechanical. In 1699, Abraham Sharp (1651-1742), 176 PLANE TRIGONOMETRY, found IT to 71 places, and in 1706, John Machin, died 1751, professor of astronomy at Gresham College, London, extended the value to 100 places. Fautet de Lagny (1660-1734) carried it, in 1719, to 127 places ; Baron Georg Vega (1756-1802), in 1794, to 136 places; Z. Dase of Vienna, in 1844, to 200 places; William Rutherford (1798 ?-1871), Royal Military Academy, Woolwich, in 1853, to 440 places ; Richter, in 1854, to 500 places ; and W. Shanks, in 1873, to 707 places of decimals. The laborious calculations of the " TT-computers " have neither theoretical nor practical value. About 1761 Lambert showed that tt is incommensurable, and in 1882, F. Lindemann, in Freiburg, showed that it is transcendental, that is, it cannot be a root of any algebraic equation with integral coefficients. See article, " Squaring the circle," Ency. Brit., 9th edition. N.B. The ratio w is often calculated approximately by means of Gregory's series (discovered in 1670) and certain identities, namely, tan-ix = X -^x^ + ^x^ — ^x"^ H toco, ^ = 4tan-i^ — tan-i^^, , 4 - = tan-i I + tan-i I + tan-i |, - = 4 tan-i | - tan-i ^ + tan-i ^^ in which the principal values of the inverse tangents are taken. The student is advised to verify these identities, and to find an approximate value of TT by means of Gregory's series ; also to verify the remark made above concerning the circumference of the earth. Also to show that the method of Ahmes (see Note A) for finding the area of a circle is equivalent to taking IT = 3.1604.... NOTE D. DE MOIVRE'S THEOREM, AND OTHER RESULTS IN ANALYTICAL TRIGONOMETRY. [In what follows i denotes V— 1.] 1. De Moivre's Theorem. For all values of n, positive and negative, integral and fractional, (cos 6 + 1 sin 0)** = cos nB + i sin wO. (a) When n is a positive integer. (cos dy + i sin ^i)(cos 62 + i sin ^2) = COS dx cos 02 — sin di sin $2 + i (sin di cos 62 + cos di sin ^2) cos(^i + 62) + i sin(^i + ^3). (1) APPENDIX, 111 On multiplying each member of (1) by cos dz + i sin ^3, there is obtained, (cos di + i sin d{) (cos ^2 + i sin ^2) (cos 6^ + i sin ^3) ={cos(6'i + 62) + i sin(^i + ^2)} (cos dz + i sin ^3) ={cos(^i + ^2) cos ^3 — sin(^i + d^) sin ^3} + i{sin(^i + 62) cos dz + cos(^i + 6^) sin ^3} = cos (^1 4- ^2 + ^3) + *■ sin(^i + ^2 + ^3). In a similar way, the product of four or more such factors can be found. Thus, for n factors, (cos di + i sin ^1) (cos 62 + i sin ^2) •••(cos dn + i sin ^„) = cos(^i + ^2 + — + dn) + i sin(^i + ] then = —^ and (3) becomes (0 . . 0N" . . cos - + 1 sm - = cos + I sm 0. n nj On transposing and taking the nth root of each member of this equation, there is obtained (cos + i sin 0)" = cos - + i sin -• (6) n n (The second member of (5) is one of the n roots of the first member.) n (d) When n is a fraction, — • p 1 1 (cos d -\-i sin 0)<' = [(cos d + i sin ^)p]« = (cos;?^ + i sin pd)9 = cos^d + i sin ^ ^. (6) (The second member of (6) is one of the q roots of the first member.) 178 PLANE TRIGONOMETRY. For all the roots of the first members of (5), (6), see one of the works referred to in Art. 98. For a geometrical representation of the factors considered above and the results (l)-(6), and for a proof of these results, see Hobson, Flane Trigonometry, Chap. Xlll. ; Chrystal, Algebra, Part I. , Chap. XII. 2. Following are some of the theorems proved in analytical trigonometry, which will be met by those who read only a little farther in mathematics : COSa;=l- — + ^^ to 00. (1) sinx = a;-^ + |i to oo. (2) 01 O I Expansions (1) and (2) were first shown by Newton in 1669. If e" denote the series 1 + x + — + ^ + ••• to oo, (3) 2 1 31 pix j_ p—ix . pix p—ix then cos x = ^—Ll — sin x = — (4) 2 2i ^ ^ Formulas (4) were first given by Euler. The expansions (1), (2), (3), are also derived in works on the differential calculus. They are convergent for all finite values of x. Either (1), (2), or (3), (4), may be taken as definitions of the sine and cosine. Hyperbolic functions. The hyperbolic sine and cosine of x, denoted by sink x, cosh x, may be defined in either one of the following ways, namely, rr2 y4 cosh ic = l+ — + — +... to 00 2! 41 sinhx =x-h — + — + ...to 00 31 61 (5) and coshx = ^^i^-^, sinhic = ^^ — ^ (6) 2 2 A geometrical definition may also be given to the hyperbolic functions. In this definition, they are related to the hyperbola in a manner analogous to a way in which the trigonometric (circular) functions are related to a circle. It may be said that the formulas or definitions (l)-(C) may be applied to all numbers x, real, pure imaginary, or complex; i.e. quantities of the form a + 6 V— 1. (When x is real in (l)-(4), it denotes the radian measure of the angle.) See Chrystal, Algebra, Part II., Chap. XXIX.* APPENDIX. 179 EXERCISES. 1. Substitute 1 for x in the series (3), and thus deduce 2.71828 as an approximate value of e. 2. (a) Write the series for e»"* and e-^ by substituting ix and — ix for X in (3). (&) Then find the value of cos x in (4), and compare the result with (1). (c) Then find the value of sin x in (4), and compare the result with (2) . 3. Using formulas (4), show that cos^aj + sin^aj = 1. 4. Substitute ix for x in (1) and (2), and compare the results with (5). Substitute ix for x in (4), and compare the results with (6). Each substitu- tion shows that cos {ix) = cosh x, and sin (ix) = i sinh x. 6. Show by means of the formulas (l)-(4) that the cosine of an angle of magnitude zero is unity, and that the sine of such an angle is zero. 6. By means of (1), (2), find approximate values of sin 10°, cos 10°, sin 15°, cos 15°, sin 20°, cos 20°, sin 30°, cos 30°. [First, express the angles in radian measure.] * Also see McMahon, Hyperbolic Functions, (Merriam and Woodward, Higher Mathematics, Chap. IV., pp. 107-168). QUESTIONS AND EXERCISES FOR PRACTICE AND REVIEW. 3>'sin45° + sin30° * ,^. 1 + cot 60° _ / 1 + cos 30° l-cot60° U-cos30° (c) tan2 60° - 2 tan2 45°= cot2 30° - 2 sin2 30° - f cosec2 45°. 6. The sine of an angle defined as a ratio being less than unity, explain why the tabular logarithms of the sines of angles are expressed with whole numbers as characteristics. Given log tan 18° = 9.51178, show what the tabular logarithm of cot 18° must be. 7. (a) Given log 2 = .30103, log 3 = .47712, find log sin 60° and log tan 30°. (6) Given log 5 = .69897, find the logarithmic sine of 30°, and the logarithmic cosine of 45°. 8. Compute the trigonometric ratios of yl in a right triangle ABC{C= 90°), when b = ^c. 9. Construct the following right-angled triangles : (a) ABC, in whicli, C = 90°, c = 5, cot J. = I ; (&) when one of the legs is 3, and the sine of the adjacent acute angle is f ; (c) hypotenuse 4, and sine of one of the acute qu:estions and exercises, 183 angles f ; (d) = 90°, sin ^ = f , 6 = 7; (e) C = 90°, cosec ^ = f , b = 10 ; write the values of sin A, cos ^, tan A ; (/) (7 = 90°, cos^ = |, a = 9. 10. In the triangle ABC, C = 90°, tan B = if. If AB = 510 ft., find AC. 11. In ABC, C= 90°, J5(7 = 10 ft., tan 5 = 1.05 ; find the other sides. 12. The string of a kite is 250 ft. in length. How high is the kite above the ground when the string, supposed stretched quite tight, makes with the ground an angle whose tangent is ^^ ? 13. ABC is an isosceles triangle, right-angled at (7; Z> is the middle point of AC. Prove that DB divides the angle B into two parts whose cotangents are as 2 : 3. 14. (a) Given L. cos 20° = 9.97 and L. cot 20° = 10.44 ; find each of the other logarithmic ratios of 20°. (b) Given L. sin 40° = 9.808, L. tan 40° = 9.924 ; find log cot 40°, log cos 40°, log sec 40°, log cosec 40°. 2 Vab sin — 15. If tan e = ;— ?- , find d when a = 5, 5 = 2, C= 120°. a ~ b 16. Calculate sin3 23° x V27.268 -4- 2 cos2 48°. 17. Find x in the equations : (a) x sin 74° = 235 tan 37° cos 63°, (&) x^ cos 39° = 47.5 sin2 46° sec2 64°. 18. Solve (sin 8° + cos 8°)2^ = 2 sin 16° (tan 32°)^ CHAPTER III. 1. State what parts of a right plane triangle must be given that it may be constructed, and show how a right triangle may be solved, in each of the four possible cases. 2. Derive the formulas for computing B, a, and c of a right triangle when C = 90°, and A and b are given. Also find a formula that shall include only the required parts. 3. In ABC, C = 90°, & = 22 ft., and sin A = .42. Find a, c, sin B, and the area. 4. In ABC, C = 90°, cosA = ^, c = 40 ft. Find the values of cos B, cot B, a, b, and the area. 5. Solve the following right-angled triangles by (1) making an ofE-hana estimate, (2) measuring on a drawing made to scale, (3) computing without logarithms (four-place tables), (4) computing with logarithms. Check the results by computation. If a solution is impossible, explain why it is so. (Each triangle is denoted by ABC, and (7 = 90°.) (i.) a = 45, & = 62 ; (ii.) a = 685, 5 = 34° 47' 25"; (iii.) c = 560, a = 310 ; (iv.) c = 327,6 = 450; (V. ) c = 520, A = 36° 40' 20" ; (vi. ) 6 = 720, B = 61° 24' 30" ; (vii.) c = 425, B = 32° 45' 35" ; (viii.) a = 11524, 6 = 35976 ; (ix.) a = 67213, 6 = 75324 ; (X.) c = 35421, 6 = 23462. 184 PLANE TBIGONOMETBY. 6. Two sides of a triangle are as 5 : 9, and the included angle is a right angle. Find the other angles. 7. Find the acute angles of a right-angled triangle whose hypotenuse is six times as long as the perpendicular let fall upon it from the opposite angle. CHAPTER IV. 1. (a) Derive the formula for the area of a right triangle in terms of (i.) an angle and its opposite side, (ii.) an angle and its adjacent side. (&) One side of a triangle is seven times another, and the included angle is a right angle. Find the other angles. 2. Show how an isosceles triangle may be divided into right triangles, and how it may be solved by aid of these right triangles when the following elements are given : (a) Base and vertical angle, (&) base and side, (c) side and vertical angle, (d) base and perpendicular from vertex on the base. Discuss any other possible cases. 3. Solve the isosceles triangles (a) whose base is 126 ft., and vertical angle is 127° ; (6) whose base and perpendicular on it from the vertex are each 721.34 yd. 4. (a) Find the area of a regular octagon the side of which is 26 yd. (&) Find the side of a regular pentagon inscribed in a circle whose radius is 43 ft. (e) If a regular pentagon and a regular decagon have the same perimeter, prove that their areas are as 2 : Vb. v/ 5. (a) At 120 ft. distance, and on a level with the foot of a steeple, the angle of elevation of the top is 62° 27' ; find the height, (h) From a cliff 330 ft. high the angle of depression of a boat at sea is 40° 35' 25" ; how far is the boat from the foot of the cliff ? 6. When the altitude of the sun is 30° the length of the shadow cast by Bunker Hill monument is 381 ft. What is the height of the monument ? 7. The angles of depression from the top of a tower 48.6 ft. high to two points, on a level with its base and in line with the tower, are 45° and 30° respectively. Find the distances of each point from the other and from the top of the tower. 8. A pole 40 ft. high is erected at the intersection of the diagonals of a square courtyard. When the sun's altitude is 43° 40', the shadow just reaches a corner of the yard. Find the length of the side of the square. 9. (a) When the altitude of the sun was 67° 30' 45" the length of the shadow of a perpendicular pole was 73.4 ft. Find the length of the shadow when the sun's altitude is 35°. (&) The shadow of a tower is observed to be half the known length of the tower, and some time after to be equal to the full length. How much will the sun have gone down in the interval ? qu:estions and exercises, 185 10. A flagstaff which leans to the east is found to cast shadows of 198 ft. and 202 ft., when the sun is due east and west respectively, and his altitude is 7°. Find the length of the flagstaff and its inclination to the vertical. 11. What angle will a flagstaff 24 ft. high, on the top of a tower 200 ft. high, subtend to an observer on the same level with the foot of the base, and 100 yds. distant from it ? 12. Looking out of a window with his eye at the height of 15 ft. above the roadway, an observer finds that the angle of elevation of the top of a telegraph post is IT*^ 18' o5", and that the angle of depression of the foot of the post is 8° 32' 15". Calculate the height of the telegraph post and its distance from the observer. 13. A man in a balloon, when it is one mile high, finds the angle of depression of an object on the level ground to be 35° 20', then after ascending vertically and uniformly for 20 min., he finds the angle of depression of the same object to be 65° 40'. Find the rate of ascent of the balloon in miles per hour. 14. A man observes the elevation of a mountain top to be 15°, and after walking 3 mi. directly toward it on level ground, the elevation is 18°. Find his distance from the mountain. 15. From a boat the angle of elevation of the highest and lowest points of a flagstaff, 30 ft. high, on the edge of a cliff are observed to be 46° 12' and 44° 13'. Determine the height of the cliff and its distance. 16. The angles of elevation of the top of a tower, observed at two points in the horizontal plane through the base of the tower, are tan-i | and tan-i _5^ j the points of observation are 240 ft. apart, and lie in a direct line from the base. Find the height of the tower. 17. A person standing due south of a lighthouse observes that his shadow cast by the light at the top is 24 ft. long ; on walking 100 yd. due east he finds his shadow to be 30 ft. Supposing him to be 6 ft. high, find the height of the light from the ground. 18. An observer is 384 yd. due south of a point from which a balloon ascended ; he measures a horizontal base due east, and at the other extremity finds the angle of elevation to be 60° 15'. Find the height of the balloon. 19. A surveyor starts from A and runs 766 yd. due east to J5, thence 622 yd. N. 20° 30' E. to O, thence 850 yd. N. 41° 45' W. to D, thence S. 42° 36' W. to E. Find the distance and bearing of A from E, and determine the area of the field ABODE. 20. A surveyor runs 253 yd. N.E. by E., thence N. by E. 212 yd., thence W.N.W. 156 yd., thence S.W. by S. 210 yd., thence to the starting- point. Find the bearing and distance of the starting-point from the last station, and determine the area of the field which the surveyor has gone around. 186 PLANE TRIGONOMETRY. CHAPTER V. 1. Define and illustrate angle, negative angle, complement of an angle, supplement of an angle, quadrant, angle in the third quadrant. 2. Define and illustrate the six trigonometric ratios. Find the greatest and least values that each of them can have. Arrange in tabular form the algebraic signs of the trigonometric ratios of an angle in each quadrant. 3. Explain how the trigonometric ratios of an angle of any magnitude, positive or negative, can be found, {a) by means of tables which give these ratios for angles up to 90° only, (&) by means of tables which give these ratios for angles up to 45° only. 4. Prove that if two angles have the same sine, and also any of the other five trigonometric ratios (with one exception) the same, they will differ by a multiple of 360°. 5. State and prove the chief relations which exist between the trigono- metric ratios of any angle A. 6. Express the trigonometric ratios of 90° — A, 90° + A, 180° — A, 180°+^, 270°-^, 270° + ^, 360° -yl, -A, in terms of the trigono- metric ratios of A. t. Name three pairs of trigonometric ratios such that the product of each pair shall equal 1 ; one pair, the sum of whose squares shall equal 1 ; two pairs, the difference of whose squares shall equal 1. 8. Compare the trigonometric ratios of any angle (a) with those of its complement, (6) with those of its supplement. 9. Prove that sin ^ = cos ^ tan A ; sec^ ^ = 1 + tan^ A ; cot A = cosec A cos A; sin2 ^ _|. cos^ A = l; sin ^ = tan ^ : VT+ tan^ e ; cos X = Vcosec^ x — l: cosec x. 10. (a) Express the following trigonometric ratios in terms of trigono- metric ratios of positive angles not greater than 45° : sin 237°, cos (— 410°), tan 2000°, cot (-137°), sec 445°, cosec (- 650°), sin 185°, tan 267°, sec 345°, cos 87°, cot ( — 19°) ; (6) by means of the tables give the numerical values of these ratios. 11. Find, without the use of trigonometric tables, the numerical values of cos 1410°, tan (-1260°), cosec (- 1710°), tan 225°, cot 1035°, cosec 210°, cos 1600°, sin 1665°, tan (-1665°), all the trigonometric ratios of -1125° and 930°. 12. Construct the angles : (a) whose secant is 3, (6) whose tangent is V2 4- 1, (c) whose cotangent is |. Find the other ratios of these angles. 13. (a) Find sin A, cot A, when cos ^ = - j%, and A < 180°. (6) Find the other ratios of A and x when cot A = -^ and cos x = — |. (c) Find the other ratios of A when cos ^ = — ^, and A lies between 540° and 630°. QUESTIONS AND EXEttCISES. 187 ^_^ ((2) Find the trigonometric ratios of 180° + 6 and 270° — ^, given tan 6 = \. (e) Given sec a; = — |, and x in the third quadrant ; find the value of sin X + tan x _ cos X + cot X 14. Do Ex. 9, Art. 18, J. being any angle. Explain the ambiguities in the algebraic signs. If A is an angle in the third quadrant, express cos A, tan A, cot A, sec J., cosec A in terms of sin A. 15. (a) If sec A = n tan A, find the other ratios of A. (6) If 2 sec = tan a + cot a, find tan and cosec d. (c) Solve x^ cot 108° = 128° sin 72° cos 18°. 16. Prove the identities : sin^ 6 + cos^ 6 = (sin d + cos 6) (1 — sin 6 cos 6); cos-* A — sin* ^ = 1 — 2 sin^ A ; sin x (cot x + 2) (2 cot ic + 1) = 2 cosec x + 5 cos X ; sec2 5 — cos^ B = cos^ i^ tan'^ B + sin^ J5 sec^ B ; cos^^ ^ + sin^ A = 1 — 3 cos2 A sin2 ^ ; cos^ aj + 2 cos* x sin^ x + cos^ x sin* x + sin^ x = 1. 17. (a) Find the value of x not greater than two right angles which will satisfy the equation 4\/3 cot x=7 cosec x— 4 sin x. (&) Likewise, in the case of the equation sinx+cosxcotx=2. (c) Likewise, in tan*x— 4tan2x+3=0. (d) If 1 + sin2 ^ = 3 sin 6 cos 6, find tan 6. (e) Find the least positive value of A that satisfies the equation 2\/3 cos^^ = sin J.. (/) Find all the angles between 0° and 500° which satisfy the equation 4 sin^ ^ = 3. (g) If 2 cos J. + sec ^ = 3, what is the value of ^ ? (h) Find A when tan^ A + cosec2 ^ = 3. CHAPTER VI. 1. (a) Write the values of cos(^+J5), cos(A — B), sm(A + B)^ sin(^ — J5), tan (J. + B), tan (A — B) in terms of the trigonometric ratios of A and B. (6) Deduce these values, (c) Express them in words. 2. (a) Express in terms of the trigonometric ratios of A each of the following: sin 2 J., cos 2^ (three different forms), tan 2 J., cot 2 A (6) Derive these expressions. 3. (a) Show that sin ^ + sin 5 = 2 sin A±^ cos ^ ~ ^ . (&) Show ^ 2 2 ^t~Z that cos 2 ^ + cos 2 5 = 2 cos {A + B) cos (^A — B). (c) State and derive ^ an equivalent expression for the difference of two sines ; (d) for the differ- ence of two cosines. 4. Show how to find cos^J. when cos J. is known. Explain the ambiguity in the result. Determine the sign of the result when A is an angle in the third quadrant. Find the cosine of 112° 30'. [From cos 225°.] * . 5. Prove 2 cos ~ = — Vl + sin^ — Vl — sin J. if J. is between 270° and 360°. ^ 6. Derive an expression for each of the following : sin 3 ^ in terms of sin A^ cos 3 A in terms of cos A^ tan 3 J. in terms of tan A. [Suggestion : 3^=2^ + ^. See Art. 93.] 188 PLANE TBIGONOMETRY. 7. (a) If tan A= and tan .5= , prove that tan (^-5) = .375. 4-V3 4+\/3 1 3 (h) If sin A = — ■::z and cos B = -, find the value of tan(^ + B). (c) Find VlO 5 tan (J. + B), given that sin A = —, sin 5 = — . 17 13 8. (a) If tan^ = ^, show that sin^=-l^o' sin 2 J. = i«^<«i::i^. (6) If tan A = K prove that J«^ + a/^^ = J-^2^. 9. (a) Find sin 45°, and thence deduce the ratios of 22° 30'. (6) Prove that tan 67° 30' = 1 + V2. (c) Deduce the ratios of 67° 30', (i.) from ratios of 45°, 22° 30', (ii.) from ratios of 135°. 10. (a) Given sin 30° = ^ and cos45° = |\/2; find sin 15°, cos 75°. (6) Given sin 30° = | ; find the numerical values of the other ratios of 30° ; thence derive the ratios of 15°, thence derive the ratios of 75°, 105°, 165°, 195°. (c) Prove the following : tan 15° + tan 75° = 4, cos 15° . cos 75° = .25, sin 105°+ cos 105°= cos 45°, tan 15° (tan 60°- tan 30°) = tan 60°+ tan 30°-2. 11. (a) Express sin8^ + sin2J. as a product. (6) Express as a sum or difference : (i.) 2cos^cosJ5, (ii. ) 2 sin 50° cos 20°. (c) Prove without using tables that (i.) sin 70° - sin 10° = cos 40°, (ii.) cos 20° + cos 100° + cos 140° = 0. Verify by the tables. 12. Show that: (1) cot^ cot 5cos(^+B)=cos ^ cos 5(cot^cotS--l); (2) cos {A + B) cos A + sin (A + B) sin ^ = cos J5 ; (3) cos ^ - sin ^ = V2 cos(A + 45°) ; (4) 2 cos^ x-2 sin^ x = cos 2 x(l + cos2 2 x) ; (5) cos2 A + siii2 ^ cos 2 5 = cos2 B + sin^ ^ cos 2 ^ ; (6) cos2 ^ - cos ^ cos (60^ + A) + sin2 (30° - ^) = .75 ; (7) tan ^ ^ = sin ^ : 1 + cos 6 ; (8) cos (135° + A) + sin(135° - J) = ; (9) cosec 2 ^ + cot 2 ^ = cot ^. 13. Prove that: (1) sin a; + sin y ^_ . . . ^2) tan^ = ^^ cos X- cosy ^^ ^^' ^^ 2 ' 2sin^-sin2^ ^^^ ^^^ _ 8 cot ^ 2 sm ^ + sm 2 ^ ^ ^ \ « y v j ^^^^ ^ - 3 (4) gg^+^^°)=sec2^-tan2^- (5^ cos2 J g-cos2 vl ^ sin 2^-sin 2 g_ cos(^-45°) ' ^ ^ sin2^+sin2^ cos2^ + cos2i? tan(^-^); (6) sec2^-itan2^sin2^=-^"iiA±iH^. cot2 A — tan2 A a 14. (a) Find values of d not greater than 180°, which satisfy cot ^=tan -. (6) Give all the positive angles less than S60°, which satisfy the equation sin 2 ^ = V3 cos 2 A. 16. Show that the value of sin(n + l)5sin(n-l)B+cos(« + l)J5cos(n-l).B is independent of n. 4 QViJSTlONS AND EXERCISES. 189 ^ 16. The cosines of two angles of a triangle ABC are | and i|, respec- tively ; find all the trigonometric ratios of the third angle without using tables. Verify the results by means of the tables. 17. Two towers whose heights respectively are 180 and 80 ft., stand on a horizontal plane ; from the foot of each tower the angle of elevation of the other is taken, and one angle is found to be double the other ; prove that the horizontal distance between the towers is 240 ft., and show that the sine of the greater angle of elevation is .6. CHAPTER VII. 1. In a triangle ABC, show that (1) sin(^ + B)= sin C, (2) cos(^ + B) = -cosO, (3)sin^-±^ = cos-^, (4) cos "^ + ^ = sin ^. 2 2 2 '2 2. (a) State and prove the Law of Sines for the plane triangle. (6) State and prove the Law of Cosines for the plane triangle, (c) If the sines of the angles of a triangle are in the ratios of 13 : 14 : 15, prove that the- cosines are in the ratios of 39 : 33 : 25. 3. (a) Prove that in ABC, 6 + c : 6 - c = tan ^ (5 + (7) : tan |(i? - C) = cot \ ^ :tan 1{B—C). (h) Write and derive the expressions for the cosine of an angle of a triangle, and the cosine and the sine of half that angle, in terms of the sides of the triangle, (c) In the triangle ABC derive the for- mulas expressing tan^^, tan ^ i?, tan 1(7, in terms of a, 6, c. {d) Prove that in any triangle ABC, sin ^ = — \/s(s — a){s — b)(s — c). be 4. (a) Show how to solve a triangle when the three sides are given, (i.) without logarithms, (ii.) with logarithms. Derive all the formulas necessary. (&) Do the same when two sides and their included angle are given, (c) Do the same when two angles and a side are given. 5. (a) Explain carefully, and illustrate by figures, the case in which the solution of a triangle is ambiguous, {b) Write formulas for a complete solution and check, of a triangle, when two sides and an angle opposite to one of them are given. How many solutions are there ? Discuss fully all cases that may arise, (c) Given the angle A, and the sides a and & of a triangle ABC, determine whether there will be one solution, two solutions, or no solution, in each of the following cases : (i.) A < 90°, a > 6, (ii. ) -4 < 90°, a = b, (iii.) A < 90°, a < 6, (iv.) A > 90°, « > &, (v.) 4 > 90°, a = b. 6. Show by the trigonometric fornuilas that the angles of a triangle can be found when the ratios of the three sides are given. Give the geometrical explanation. 7. Show by the trigonometric formulas that the other two angles of a tri- angle can be found when the third angle, and the ratio of the sides contain- ing it, are known. Give the geometrical explanation. 190 PLANE TRIGONOMETRY, 8. Assuming the law of sines for a plane triangle, prove that a -{■ b :c = cos l(A- B): sin ^ O, and a - 6 : c = sin ^ (^ - £) : cos ^ C. 9. (a) UA:B:C = 2:S:4, prove 2 cos ^ = ^^^- (&) If 2 a = ft + c, B C 1 2 prove tan — tan — = -. (c) If a, b, c, the sides of a triangle, be in arithmetical AC B progression, prove that 2 sin — sin — = sin—. [Suggestion : Put 2 6 = a + c.] 10. [In each of the examples in Ex. 10, J., J5, C, denote the angles, a, 6, c, the sides of the triangle.] Solve the following triangles (1) by making an estimate, (2) by the method of construction, (3) by computation, without using the logarithms of the trigonometric ratios (four-place tables), (4) by computation. Using logarithms (five-place tables), (5) by dividing some of the oblique triangles into right-angled triangles. Check the results by com- putation. When a solution is impossible, or ambiguous, explain why it is so. (I) a = 753, 6 = 621, c = 937 ; (2) a = 9, & = 17, c=14; (3) a = 1236.5, b = 1674.8, c = 2532.7 ; (4) a = 30, & = 42, c = 36 ; (5) a = 621, b = 237, c = 325 ; (6) a = 1237, b = 1014, A = 39° 42' ; (7) a = 1114, b = 1345, A = 46° 54' 20" ; (8) c = 832, b = 694, B = 54° 47' 30" ; (d) a = 1020, b = 240, B = 70° 25' ; (10) c = 794, b = 832, B = 65° 30' 20" ; (11) c = 230, a = 950, C = 63° 47' ; (12) a = 237, c = 452, C = 37° 49' ; (13) a = 420, c = 337, C = 42° 46' ; (14) a = 452, b = 624, C = 37° 23' ; (15) a = 1237.4, c = 1941.6, B - 23° 41' 20"; (16) 6 = 237.41, c = 556.82, ^ = 85° 45' 35"; (17) ^ = 37° 41', B = 49° 32', c = 385.9 ; (18) B = 47° 21' 30", C = 81° 49' 45", 6 = 374.26. 11. (a) If 6 : c = 11 : 15, and A = 37° 40', find B and C. (6) If one side of a triangle be five times the other, and their included angle be 64°, find the remaining angles. 12. (a) In ABC, if a : 6 : c = 8 : 7 : 5, find the angles. (6) The sides of a triangle are proportional to the numbers 4, 5, 6. Find the least angle. 13. Given a = 2 6, C = 120°, find A, B, and the ratio c : a. 14. (a) Two adjacent sides of a parallelogram are respectively equal to 12 and 20 in., and a diagonal is equal to 25 in. Find the angles of the parallel- ogram, the other diagonal, and the area. (6) The sides of a quadrilateral taken in order are 8, 10, 16, 18, and one diagonal is 18. Find its angles and area. 15. A ladder 52 ft. long is set 20 ft. in front of an inclined buttress, and reaches 46 ft. up its face. Find the inclination of the face of the buttress. 16. A privateer is lying 10 mi. W.S. W. of a harbour, when a merchant- man leaves it, steering S.E. 8 mi. an hour. If the privateer overtakes the merchantman in 2 hr., find her course and rate of sailing. 17. A fort bore E. by N. from a beacon, and was distant from it 1500 yd. From a ship at anchor the beacon bore N.N.W. and the fort N.E. by N. How far was the ship from the beacon ? QUESTIONS AND EXERCISES, 191 18. A and B are two points, 200 yd. apart, on the bank of a river, and C is a point on the opposite bank ; the angles ABG, BAG are respectively 54° 30' and 65° 30'. Find the breadth of the river. 19. (a) Two observers on the same side of a balloon, and in the same vertical plane with it, are a mile apart, and they find the angles of eleva- tion to be 22° 18' and 75° 30', respectively. What is the height ? (6) Two observers on opposite sides of a balloon, and in the same vertical plane with it, take its altitude simultaneously ; one observer finds it to be 64° 15', and the other, 48° 20'. Find the height of the balloon at the time of observation. 20. From a ship sailing along a coast a headland, C, was observed to bear N.E. by N. After the ship had sailed E. by N. 15 mi. the headland bore W.N. W. Find the distance of the headland at each observation. 21. From a certain station a fort, A, bore N., and a second fort, J5, N.E. by E. Guns are fired simultaneously from the two forts, and are heard at the station in 1.5 sec. and 2 sec. respectively. Assuming that sound travels at the rate of 1142 ft. per second, find the distance of the two forts apart. 22. From a point A in the same plane as the base of a tower, the tower bears N. 62° W., and the angle of elevation of the top of the tower is 53° 37' ; from B, 165 ft. due north of A, the tower bears west. What is the height of the tower ? 23. From a ship steering W. by S. a beacon bore N.N.W., and after the ship had sailed 12 mi. farther, the bearing of the beacon was N.E. by E. At what distance had the ship passed the beacon ? 24. From the intersection of two straight paths which are inclined to each other at an angle of 37°, two pedestrians, A and B, start at the same instant to walk along the paths, A at the rate of 5 mi. an hour, and B at a uniform rate also ; after 3 hr. they are 9^ mi. apart. Show that there are two rates at which B may walk to fulfil this condition, and find both of those rates. 25. Two straight railroads are inclined to each other at an angle of 22° 15'. At the same instant two engines, A and B, start from a station at the point of intersection, A going on one road at the rate of 20 mi. an hour, and B going uniformly on the other. After 3 hr. A and B are 25 mi. apart. Show that there are two rates at which B may go to fulfil this condition, and find those rates. 26. A tower stood at the foot of an inclined plane whose inclination to the horizon was 9° ; a line was measured straight up the incline from the foot of the tower of 100 ft. in length, and at the upper extremity of this line the tower subtended an angle of 54°. Find the height of the tower. 27. The altitude of a certain rock is observed to be 47°, and after walking toward it 1000 ft. up a slope inclined at 32° to the horizon, the observer finds that this altitude is 77°. Find the vertical height of the rock above the first point of observation. / V 102 PLANK TRIGONOMETRY, 28. If, from a point at which the elevation of the observatory on Ben Nevi8 Ib 60°, a man walks 800 ft. on a level ijlane toward the mountain, and then 800 ft. furthcir up a Klope of 30" to a point at which the elevation of the observatory is 75°, show that the height of Ben Nevis is approximately 4478 ft., the man's path being always supposed to lie in a vertical plane passing through the observatory. 29. A man walks 40 ft. in going straight down the slope of the embank- ment of a railway which runs due east and west, and then walks 20 ft. along the foot of the embankment ; he finds that he is exactly N.E. of the point from which he started at the top of the bank. Show that the inclination of the bank to the horizon is Q0°. 30. A man in a ship at sea sailing north observes two rocks, A and B, to bear 25° east of his course ; he then sails in a direction northwest for 4 mi., and observes A to bear east and B northeast of his new position. Find the distance from A to B. 31. To determine the distance of two forts, C, Z), at the mouth of a harbour, a boat is placed at A, with its bow toward a distant object E, and the angles CAD, DAE are observed and found to be 22° 17' and 48° 1' respectively. The boat is then rowed to B, a distance of 1000 yd., directly toward JE", and the angles CBD, DBE are observed to be 53° 16' and 75° 43' respectively. Find the distance CD. 32. A fort stands on a horizontal plane ; the angle of depression meas- ured from the top of the fort to a point i*on the plane is ^°, and to a point B, a feet beyond P, is BF'. Derive the formulas for computing h, the height of the fort, and rf, the distance from P to the bottom of the fort. 83. A person standing on a level plain at the base of a hill wishes to find the height of a tower which is in full sight on the top of the hill. Describe in detail the necessary measurements and computations. 34. A surveyor wishes to find the distance and the height of a tower which is on the same level with him, but on the opposite side of an impass- able chasm ; illustrate the problem by a lettered figure, and describe in de- tail the necessary measurements and computations. 35. What measurements must be made by an observer on the shore to find the distance between two buoys ? Give formulas necessary for solving. 36. Explain what measurements have to be made at two stations, A and i?, in order to find the distance, CD, between two inaccessible objects {A, B, C, D being in one plane) ; and state clearly the steps of the calculation by which the distance is to be found therefrom, 37. (a) From the law of sines deduce that 6 cos O + c cos ^ = a. (6) Prove this geometrically, (c) Show that a cos i; - & cos ^ = . QUESTIONS AND EXERCISES. 193 88. In any triangle ABC, prove : (a) tan B = ^ ^^^ ^ ; a — h cos C X gg + ^2 - q5 COS O _ a , .V 1 4- cos(^ - /i)cos _ a^+ 6^ , asin ^ + 6 sin J5 + c sin C 2 «in^ * 1 + co8(^ - 6')co8 B a^ + c^' CHAPTER VIII. [In what follows, 8 denotes the area of a triangle, « its semi-perimeter, R, r, Ta, Th, n, the radii of its circumscribing, inscribed, and escribed circles, respectively.] 1. Prove that any side of a triangle is equal to the second side into the f:fjsine of the angle opposite the third sine plus the third side into the cosine of the angle opposite the second side. 2. Derive expressions, in terms of the sideg of a given triangle, for the rjidii of its circumscribing circle, and of the four circles which touch the side*. 8. Derive expressions for the radii in Ex. 2, in terms of the sides and the area of the given triangle. 4L Prove B = ^, r = -, ra = — — Write similar expressions for n, r« ^S 8 8 — a 6. Prove : (a) Va + n -\- Ve — r = iR ; (6) Vr • r« • n • >*« = S. 6. Prove: (a)l + i+i = l; (6) Rr = ^^^ ra n Tc r' "• ' 4(a+6 + c)' (c)r = 458in:^sin^8in£ ^ ^ 2 2 2 7. Prove r« cot — = n cot — = r«, cot ^ = r cot — cot ~ cot — 2 2 2 2 2 2 Ann = 4 ^ cos - cos — cos -• 2 2 2 8. Prove B = — - — , r = (« - a) tan — , r« = « tan — • Write two other 2sinvl ^ ^ 2 2 similar formulas for B and r. Write similar formulas for r*, Ve. a sin— sin — 2 2 9. (a) Prove r= ^ • Write two similar formulas involving cos- 6, c. a cos— cos— 2 2 (h) Prove r^ = ^ Write similar formulas for n, r*.. cos — 2 10. Show that the length of the tangents to the inscribed circle from the angle ^ is « — a, from the angle J? is « — 6, from the angle is « — c. 11. Write and derive the formula for the area of a triangle : (a) in terms of the three sides ; (6) in terms of two sides and their included angle ; (c) in terms of one side and the two angles a^ijacent to it. 194 PLANE TRIGONOMETRY, 12. (a) Prove that /S' = s (s - c) when O = 90° ; (&) if x, y, be the lengths of the two diagonals of a parallelogram, and 6 the angle between them, show that area = \xy sin 6. 13. Find the areas of some of the triangles in Ex. 10, Chap. VII. Find the radii of their circumscribing, inscribed, and escribed circles. 14. (a) An isosceles triangle whose vertical angle is 78° contains 400 square yards ; find the lengths of the sides. (6) Find two triangles each of which has sides 63 and 55 ft. long, and an area of 874 sq. ft. (c) The angles at the base of a triangle are 22° 30' and 112° 30' respectively ; show that the area of the triangle is equal to the square of half the base. 15. (a) Show that the area of a regular polygon inscribed in a circle is a mean proportional between the areas of an inscribed and circumscribing polygon of half the number of sides. (&) The sides of a triangle are as 2:3:4; show that the radii of the escribed circles are as ^ : | : 1. 16. Two roads form an angle of 27° 10' 25". At what distance from their intersection must a fence at right angles to one of them be placed so as to enclose an acre of land ? 17. If the altitude of an isosceles triangle is equal to its base, the radius of the circumscribing circle is f of the base. 18. An equilateral triangle and a regular hexagon have the same perim- eter. Show that the areas of their inscribed circles are as 4 : 9. 19. If the sides of a triangle are 51, 68, and 85 ft., show that the shortest side is divided by the point of contact of the inscribed circle into two seg- ments, one of which is double the other. CHAPTER IX. 1. Explain how angles are measured (1) by sexagesimal measure, (2) by radian measure. Show how to connect the radian measure of an angle with its measure in degrees. Find the number of degrees in the angle called the radian. How many degrees are there in an arc whose length is equal to the diameter ? Show that the radian measure of an angle is the ratio of the lengths of two lines. What advantage is there in using radian measure ? 2. (a) Give the number of degrees in each of the following angles : ^ tt, ItT, 27r, T^^TT, WTT, 1, -|7r, fTT, | TT, f TT, | TT, | TT, f TT, -|, - | TT, - JjW, (D^, (2|)('-), (— !)(»•). (6) Give the supplements and complements of those angles in radian measure and in degree measure. (c) Give the radian measures of 30°, 80°, 49°, 41° 30' 15", 120°, - 210°, - 175°. Give the radian measures of their supplements and complements. 3. (a) A central angle 1.25'- is subtended by a circular arc of 16 ft. ; find the radius. (&) Find the number of radians and degrees in the central angle subtended by an arc 9 in. long, iu a circle whose radius is 10 ft. QUESTIONS AND EXERCISES. 195 ^c) Find the radius of a circle in which an arc 15 in. long subtends at the centre an angle containing 71° 36' 3".6. (d) If the radius be 8 in., find the central angle, in degrees and in radians, that is subtended by an arc 15 in. long, (e) An angle of S*" is subtended by an arc of 5 in. ; find the length of the radius ; find also the number of radians, and of degrees, in an arc of 1.5 in. (/) Find the number of radians and seconds in the angle subtended at the centre of a circle whose radius is 2 mi., by an arc 11 in. long, (g) Find the length of the arc which subtends a central angle of (1) 2 radians, the radius being 10 in. ; (2) 1.5 radians, radius 2 ft. ; (3) 4.3 radians, radius 21 yd. ; (4) 1.25 radians, radius 8 in. 4. The value of the division on the outer rim of a graduated circle is 5', and the distance between the two successive divisions is . 1 of an inch. Find the radius of the circle. 5. Show that the distance in miles between two places on the equator, which differ in longitude by 3° 9', assuming the earth's equatorial radius to be 7925.6 mi., is 217.954 mi. 6. (a) The difference of two angles is 10'^, the radian measure of their sum is 2. Find the radian measure of each angle. (&) One angle of a tri- angle is TT degrees, another is tt grades. Show that the radian measure of the third angle is tt — ^ • (c) If the number of degrees in an angle be equal to the number of grades in the complement of the same angle, prove that the radian measure of the angle is — • (d) The angles of a triangle are in the ratios 1:2:3. Express their magnitudes in each of the three systems of angular measurement, (e) One angle of a triangle is 45°, another is 1.5 radians. Find the third, both in degrees and in radians. (/) Express in degrees and in radian measure the vertical angle of an isosceles triangle which is half of each of the angles at the base. 7. Prove the following statements, in which a denotes the length of a side of a regular polygon ; P, the length of its perimeter ; n, the number of its sides ; r, the radius of the inscribed circle ; i?, the radius of the circum- scribing circle : a = 2i?sin- = 2rtan-; P= 2ni?sin - = 2wrtan - ; P + r = -cot — ; n n n n 2 2n area of polygon =Vl B"^ sm — z=z nr^ tan - = — cot -. 2 n n 4 n CHAPTER X. 1. Define and illustrate the trigonometric functions. Show in tabular form the signs of these functions in each of the four quadrants. 2. (a) Construct a table showing the values, with proper signs, of the trigonometric functions of 0°, 30°, 45°, 60°, 120°, 180°, 225°, 270°, 315°, 360°. 196 PLANE TRIGONOMETRY, (6) Compare the trigonometric functions of 90° - A, 90° + A, 180° + A, 180° -A, -A, with those of A. 3. Show, from both the ratio and the line definitions of the trigonometric functions, that (1) the sine and cosine are never greater than unity, (2) the cosecant and secant are never less than unity, (3) the tangent and cotangent may have any values whatever from negative infinity to positive infinity, (4) the trigonometric functions change signs in passing through zero or infinity, and through no other values. 4. Given tan J. = — ^-^, find the values of the other trigonometric func- tions of A. 5. Find geometrically an expression for the cosine of the difference of two angles in terms of the trigonometric functions of those angles. 6. Prove that : (a) sin2 B + sin2 (A- B) + 2 sin B sin (A - B) cos A = sin2 A ; cos nA - cos (n + 2) A ^ ^^^ ^^ ^^^^ ^ ^ sin (w + 2) A - sin nA ^ ^ 7. Give the ratio definitions of the trigonometric functions, sine, cosine, tangent, and secant. These functions have also been defined as straight lines. Give these definitions, and show from them that tan 90°, sec 90° would each be infinite. Show that the two systems are consistent. 8. (a) Trace the changes, in magnitude and sign, in the values of the trigonometric functions as the angle increases from 0° to 360°. (6) Trace the changes of sign of sin as ^ increases through 360°, and show that its 6 6 equivalent 2 sin - cos - has always the same sign as sin d. 9. Trace the changes, as A increases from 0° to 180°, in the sign and value of (a) cos (tt sin^), (6) sin J. + cos^, (c) sin^ — cos A Draw the graphs of these functions. 10. Show that the radian measure of an acute angle is intermediate in value between the sine and the tangent of the angle. 11. (a) Show that the limit of ^^—. when B is indefinitely diminished is 1, i.e. sin ^ = ^, very nearly. (6) Show that the limit of ^^^ when e is indefi- nitely diminished is 1, i.e. tan ^ = 0, very nearly. 12. Find the area of a regular polygon of n sides inscribed in a circle, and show, by increasing the number of sides of the polygon without limit, how the expression for the area of the circle may be obtained. 13. (a) Find the distance at which a building 50 ft. wide will subtend an angle of 8'. (6) A church spire 45 ft. high subtends an angle of 9' at the eye. Find its distance approximately, (c) Find approximately the distance of a tower 51 ft. high which subtends at the eye an angle of 5y\'. (d) How large a mark on a target 1000 yd. off will subtend an angle of 1" at the eye? QUESTIONS AND EXEBCISES, 197 14. Show how the functions may be represented by lines connected with a circle. 15. Explain, with illustrations, how functions may be graphically repre- sented by means of a curve. Draw the graphs of the trigonometric functions. Note, — "If a function of a variable has its magnitude unaltered when the sign of the variable is changed, that function is called an even function^ but if the function has the same numerical value as before, but with opposite sign, then that function is called an odd function ; for instance, x^ is an even function of ic, x^ is an odd function of x, but x^ + x^ is neither even nor odd, since its numerical value changes when the sign of x is changed." 16. Show that the cosine, secant, and versine of an angle are even func- tions, and the sine, tangent, cotangent, and cosecant are odd functions, and the coversine is neither even nor odd. (See Art. 78.) CHAPTER XI. N.B. The problems which are purely numerical are to be solved inde- pendently of tables. The results can be verified by means of the tables. 1. (a) Deduce a general expression for all angles which have the same sine ; {h) for all which have the same cosine ; (c) for all which have the same tangent, (d) What are the general expressions for all angles which have the same secant, cosecant, cotangent, respectively ? 2. Define inverse trigonometric functions ; give illustrations. Define tan-ix, cos-ix. 3. (a) Explain fully the equations sin(sin-i 1)=^, sin-i(sin 0') = 0. (&) Con- struct sin-i (f), cos-i 0, tan-i oo, sec-i^sec ~\ . (c) Find tan (cos-i |). [Carefully state the limitations under whMsh the following equations are true. ] 4. Show that : (a) tan-i x+ tan-i y =tan-i ^ + ^ ; (&) tan-i x - tan'i y= _ 1 -xy tan~i — — ^ ; (c) sin-i x — sin-i y = cos-i Vl — x^ — y^ + x%2 + xy. I -\-xy 6. (a) From sin 2^=2 sin ^ cos ^, show that 2 sin-i x=sin-i(2 xVl -x^). (&) Show that for certain values of the angles, 2 cos~ix = cos~i(2x2 — 1); 2 X ^ . o 2 tan-i X = tan-i — ; 2 cot-i x = cosec" 1 - x-^ 2 X : (a') cos-ix=sin-H/- 6. Show that for certain values of the angles : (a) cos"i x=sin-i'v/ + cos-iJ^-i-^; (6) sin-i J— ^— = tan"! V^ = - cos-i ^^—^. (c) tan-im-v ^ 2 ' ^ ^ ^a + x ^a 2 a + x ^ ^ cot-im=^,or ^ 2' 2 198 PLANE TRIGONOMETRY. 7. Prove that for certain values: (1) sec-i3=2cot-iV2; (2) sec2(tan-l2) + cosec2(cot-l3)=15 ; (3) sin-if+sin-if=z90°; (4) cos-iVj-cos-i ^^"^-^ ^ 2v^3 ^; (5) cos-iff+2tan-i^=sin-if; (6) cos-ii+2 sin-i|=:120^; (7) tan-i^+ cot-i 1 + sin-i ^ = ; (8) tan-i ^ + ^ + tan-K/^ = ^• 10 V3-\/2 ^2 4 8. Prove that : (1) sin-i ^^^ + sin-i ^^ ~ ^^^ = ^ ; (2) tan-i (cot A) - to2 + 71^ m^ + n^ 2 2l__t...-ill tan-i (tan A) =nT +--2A; (3) tan-i « + tan-i -^~ = tan , (4) tan-i ^^^Lzil + tan-i ^ = wtt + Z^ ; (5) tan-i ^iLl^ + tan-i ^^^H^ z:z m 2m- 1 4 &^3 ay/3 - ; (6) tan-i ^ + tan-i n = cos"! 'i^ - mn 3 V(H-m-0(l + n'-2) 9. Prove that: (1) tan-i -^ + tan-i -^ + tan-i ^ = wtt ; (2) tan-il = 1 + a 1 — a or tan-i i + tan-i \ = tan-i ^ _j_ tan-i ^ + tan"! i ; (3) tan-i I + tan-i ^+tan-i |-+ tan-i I = 45° ; (4) cofi 1 = cot-i 3 + cofi f ; (5) cot-il^^ - cofii^ = 1 + a I + b sin-i ^ ~ ^ ; (6) 4(cot-ii + cosec-iV5) = 7r; (7) a cos f sin-i^^ = y/a^ - &2 ; (8) sin-i[ ^~^ + ^ ^^ = -cos-i^^^; (9) sincot-ia = tan cos-1a/^^-±^ ; (10) {tan (sin-i a) + cot (cos-i a)}2 = 2 a tan (2 tan-i a). 10. Find all the angles (i.e. find the general values of the angles) which satisfy the following equations : (1) sec^ A = ^; (2) 2 tan^ 6 = sec^ 6 ; (3)tan2<9-sec^=l; (4) VS tan2^+l = (l+ V3)tan0 ; (5) cos0-sin^=— ; \/2 (6) 2 sinx+2 cosecaj=5 ; (7) 2sin2y=3tan?/ ; (8) cos^+tan jB=sec jB ; (9) 3 cos2 ic + 2 V3 cos X = 5.25 ; (10) tan ^ - 2 sin = ; (11) 4 cot 2 ^ = cot2 e - tan2 6 ; (12) _cosec C + cot = V3 ; (13) cot ^ - tan ^ = 2 ; (14) cosec ic = cot cc + V3. 11. Solve cos 2 x = sin x. fSoLUTiON : cos2a; = cosf— — X ]; or, sinf — — 2a; 1= sinx. .•. ^— x = 2w7r ±2x, or --2x = WTT +(- l)'*x. 12. Solve sin 5 u4 = sin 11 A rSOLUTIONJ ll^ = W7r+(-l)«5A /. ^=0, ^, J, ....1 L lb o J 13. Find the general solutions of : (1) sin + cos = \/2 ; (2) sin 4 = ein ^ J (3) 2 C09 3 ^ - 2 sin - X = ; (4) tan2 ^4-3 cot2 e = i; (5) sin* x - _..._.___ ___ __._ _..._.._._.__ __._..__.._ __.__ J QUESTIONS AND EXERCISES. 199 cos*x = l; (6) sin"^2ic— sin2aj=.25; (7) tan J5+cot5=2, cos?/+cos2 2/+ cos3y = (Suggestion: cos t/ + cos 3 y = 2 cos y cos 2 ?/) ; (8)sin^x = cosec X — cot X ; (9) cos x + cos 7 x = cos 4 a; ; (10) cos ^ + cos ^ J. = cos I ^ ; (11) cosec z = 2sia.z ; (12) 2 tan-i cos A = tan-i 2 cosec A ; (13) tan (A - 15°) = | tan {A + 16°) ; (14) tan(45° + -B) = 1 + tan B. 14. Find x in the following equations: (1) cos-i x + cos~i (1 — x) = cos-i ( - X) ; (2) tan-i x + tan-i 2 x = tan-i ^^ ; (3) tan-i (x + 1) + 6 tan-i (X - 1) = tan-i ^ ; (4) tan-i ^-iti + tan'i ^-Ili = tan-i ( - 7) ; (5) tan-i ^^^ + tan-i ^^"^ = tan-i — • ^ ^ X + 1 2 X + 1 36 15. A flagstaff a feet high is on a tower 3 a feet high ; prove that, if the observer's eye is on a level with the top of the staff, and the staff and tower subtend equal angles, the observer is at a distance a V2 from the top of the flagstaff. 16. In any triangle ABC, if tan - = -, and tan - = — , find tan C with- 2 6 2 37 out tables. Verify the result by means of the tables. Show that in such a triangle, a + c = 2 6. CHAPTER XII. 1. Explain the advantages of measuring angles by the sexagesimal, cen- tesimal, and radian methods, respectively. 2. Show that, if x be the radian measure of a positive angle less than -, then (a) cos x is less than 1 but greater than 1 — ^ x^ ; (&) sin x is less than x but greater than x — ^ x^. By means of (b) show how the sine of 10" may be calculated approximately. 3. (a) Express in terms of functions of A, each of the following : sin 2 A, cos 2 A (three different forms) , tan 2 A. (b) Find cos 3 A in terms of cos A. (c) Find sin 3 ^ in terms of sin A. (d) Find tan 3 ^ in terms of tan A, and from the formula determine the numerical value of tan A if 3 ^ = 90°. (e) Investigate a formula for expressing the cosine of half an angle in terms of the sine of the whole angle ; and if the angle lies between 270° and 360°, show which signs of the roots must be taken. 4. Show that sin {A + B - C) + ^m{A + G - B) -{■ %m{B + C - A) - sin (^ + -B + O) = 4 sin ^ sin B sin C. 5. If A + B-\-C= 180° (i.e. if A, B, C be the three angles of a tri- angle), show that: (a) sin^ + sin ^ + sinC = 4 cos |^ cos^lf cos | O; (b) tan^+tan^+tanC'=tan^tan5tanO; (c) cos^+cos i?+cos C-1 ^ A Ti r sm^+sm^+smO tan ^ tan :^ tan ^. 2 2 2 200 PLANE TRIGONOkETET, 6. If sin u4 = f , sin 5 = ^|, and sin C = ^5, where ^, B, C are positive angles and less than 90°, find sin {A-\- B + C). 7. Assuming the equation cos 3 a; = 4 cos^ a; — 3 cos x, find sin 18°. [Solution : 64° + 36° = 90°. .-. cos 54° = sin 36° ; i.e. cos 3 . 18° = sin 2 . 18°. Hence, 4 cos^ 18° - 3 cos 18° = 2 sin 18° cos 18°. .-. 4 cos'^ 18° - 3 = 2 sin 18°. On putting 1 — sin2 18° for cos^ 18°, and solving for sin 18°, there is obtained the result, sin 18° = ^"^ .3 8. Assuming the result in Ex. 7, find the other trigonometric functions of 18° and the functions of 72°. 9. Assuming the result in Ex. 7, show that cos 36° z=— — ^i^- = sin54°. 4 Hence, deduce the other trigonometric functions of 36° and 54°. Also, deduce the trigonometric functions of 9° and 81°. (The results in Exs. 8, 9, can be verified by means of the tables.) 10. Prove the formulas : sin (36° + A)- sin (36° -A)- sin (72° + A) + sin (72° -A) = sin A, cos (36° + A.)+ cos(36° -A)- cos (72° + A)- cos (72° -A) = cos A, ^nd explain their use. (See Art. 97.) 11. (a) Show that sin2 30° = sin 18° sin 54°. (&) Solve x^ cot 108° = 128 sin 72° cos 18°, without tables, (c) Find the trigonometric functions of 48°. [Hint : 48° = 30° + 18°.] 12. Two parallel chords of a circle lying on the same side of the centre of a circle subtend angles of 72° and 144° at the centre. Show that the dis- tance between the chords is equal to half the radius of the circle, (a) using tables, (6) not using tables. 13. (a) Solve : (i.) cos ^ = ; (ii.) sin jc + cos jc = 1 ; (iii.) tan y + tan 4 2/ + tan 7 y = tan y tan 4 y tan 7 y. (b) If tan tan 3 ^ = - .4, find tan 0, tan 3 6. 14. (a) If in triangle ABC, A = B, show that sinJ5 = -'vl' 36 -g A 2^ 6 ' (&) Given cosJ[ = .28, find tan—. Explain the reason of the ambiguity that presents itself in the result. (c) If sin^= — ^— -, find tan—. (d) (riven tan lx = 2 — v3, find sin x. 15 Prove the following : (i.) tan A — tan IA = tan ^yl sec A. (ii.) 1 + tan6 A = sec* A (sec2 A-S sin2 A). (iii.) sin 4 + sin 3 J. + sin 5 ^ + sin 7 yl = 16 sin A cos2 A cos2 2 A. (iv.) cos 6 4 = 16 (cos6 A - sin^ A) - 15 cos 2 A. (v.) sin 3 ic + sin 5 aj = 8 sin x cos2 x cos 2 x. QUESTIONS AND EXERCISES. 201 (vi.) 4 cos8 ^ sin 3 ^ + 4 sin^ ^ cos 3 J. = 3 sin 4 A. (vii. ) cos 20° cos 40° cos 80° = \. (viii.) sin3 A + sin^ (120° + ^) + sin^ (240° + ^) = - | sin 3 A (ix.) 1 + cos 2 (^ - B) cos2B = cos2 A + cos2 (^ - 2 B). (x.) 2 cosec 4 ^ + 2 cot 4 ^4 = cot w4 — tan A. 16. Show that cos (36° + A) cos (36° -A)+ cos (54° + A) cos (64*' - ^) = cos 2 -4 ; sin 3 ^ = 4 sin A sin (60° + A) sin (60° - -4). 17. Show that 2 cos ^ = ± Vl + sin^ ± VI - sin^, 2sin^ = ±Vl + sinvl T VI -sin4. rSuGGESTiON : cos2 — -f sin^ — = 1 ; 2 sin — cos — = sin A, 1 L 2 2 ' 2 2 J 18. Prove that the following equations are true for certain values of the angles : (i.) 3 sin-i X = sin-i (3 a; - 4 x^). (ii.) 3 cos-ix = cos-i (4a:3 - 3x). (iii.) tan-i x + tan-i y + tan-i z = tan-i ^ -^ V + ^ - ^V^ . 1 — xy — yz — zx (iv.) tan-i X + tan-i y + tan-i '^ - x - y - xy ^ tt^ 1+x + y -xy 4 (v.) Given tan « = |, tan /3 = |, tan 7 = |, find tan (a + /3 + 7). (vi.) tan-i ^ + tan-i | = | cos-i f . (vii.) tan-i - = 1 tan-i f ■=^'\ = - tan"! f^^V ^ ^ 3 2 V 7 y 3 \ 117 j (viii.) sin-i^ + sin-iA + sin-ii^ = ^. ^ ^ 5 13 65 2 (ix.) sin-i If = 3 sin-i |. (X.) tan-i 1 + tan-i 1 = 2: = sin-i— + cot-i 3. ^ ^ 2^ 3 4 V5 (xi.) sin-i -^ + cos-i -^^ + sin-i 1 = -^ ^ ^ V73 V146 2 12 (xii.) tan-i I = 1 tan-i J/. 19. The hypotenuse and shortest side of a right-angled triangle are 5 ft. and 3 ft,, respectively. Find the length of the perpendicular from the right angle upon the hypotenuse, and show that it is inclined at sin-i ^\ to the straight line drawn from the right angle to the middle point of the hypotenuse. 20. If a triangle ABC is to be solved from given parts A, a, 6, show that the solution is sometimes ambiguous ; and that in such a case the difference of the two values of C is 2 cos-i 5il^. a 202 PLANE TRIGONOMETRY. 21. The tangent of an angle is 2.4. Find the cosecant of the angle, the cosecant of half the angle, and the cosecant of the supplement of double the 22. The angle of elevation of a tower at a distance of 20 yd. from its foot is three times as great as the angle of elevation 100 yd. from the same point ; show that the height of the tower is 300 : \/7 ft. 23. DE is a tower on a horizontal plane. ABCD is a straight line in the plane. The tower subtends an angle ^ at -4, 2 ^ at JB, and 3 at C. If AB = 50 ft., and BO = 20 ft., find the height of the tower and the distance CD. 24. A ship sailing at a uniform rate was observed to bear N. 30° 57' 30" E. After 20 minutes she bore N. 35° 32' 15" E., and after 10 minutes more, N. 37° 52' 30" E. Find the direction in which she was sailing. [Ans. S. 44° 38' E.] 25. A spectator observes the explosion of a meteor, due south of him, at an elevation of 28° 45'. To another spectator, 11 mi. S.S.W. of the former, it appears at the same instant to have an altitude of 42° 15' 30". Show that there are two possible heights above the earth's surface at which it may have exploded, and find these heights. lAns. 4.33 mi. or 13.21 mi.] ANSWERS TO THE EXAMPLES. 3>0?C N.B. Not all of the answers to the exercises are given. In various ways, the student should test, or check, every result that he obtains in working the problems. CHAPTER I. Art 2. 1. logs 27 = 3, log4 256 = 4, logu 121 = 2, ..., log^p = 6, 2. 23 = 8, 54 = 625, ..., n« = P. 3. 0, 1, 2, 3, 4, 5, 6, 7, 8. 6. 1, 4, 16, 64, 256, 1024. 7. 0, 1 ; 1, 2 ; 2, 3 ; ^, 4 ; 3, 4 ; 0, -1 ; -1, -2 ; -2, -3. Art. 6. 6(a). 1.4007. 6((?). .09856. 7(6). 7.2767. 8(a). 7.937. 10. 9.214. 12. .6443. 13(a). 3.236. 13(c). 1.5563. CHAPTER II. Art. 8. 1. 8, 24, 42, 54, 720, 36 a, 12 6, c. 2. •.., a, | |-. 3. ..., 3 a, b,±. 4. 12 a, 4 6, |. 5. 440, ^j. ^ Art. 10. 5. 1 : 1200, 1 : 253440, 1 : 1920, .... 7. 2.446 mi. Art. 11. (In these answers, h, p, b represent hypotenuse, perpendicular, and base, respectively.) 2. 35°, 7i = 34.86, p =28.56, -=.574, - = 1.743, ^=.819, ^=1.22,^=1.428, h b h p b - = .7. 3. 65°, b = 27.19, p = 12.68, - = .906, - = 2.37, ^ = .423, - = 1.1, p h p h b |=.466, -=2.14. 4. 56° 19', 33° 41' (nearly), ^ = 54.08, -=.67, |=1.6, ^=1.8, |=.555, -=1.2, |=.833. 5. 41° 25', 48° 35' (nearly), i)=39.7, |=.75, - = 1.33, f=.66, -=1.51, -?=.88, -=1.13. 6. 50°, p=59.6, A=77.8, h ' b h p ' b p | = .643, ^=1.56, ■?=.766, -=1.31, |=1.19, -=.839. P Art. 14. 11. (1) tan gent is a : Vb-^ - a ^ ; (2) tangent is V ft^ _ ^2 . ^ ; (3) sine is a : Va^ + b^ ; (4) sine is b : Va^ + 52 . sine is y/a'^ -b'^ :a; tangent is b : Va'-^ - b'^. 12. 41° 24' 35". 13. 19° 28 16". 203 204 PLANE TRIGONOMETRY, Art. 15. 1. 2.28025. 2. 2.3333. 3. 5.846. 9. 2.75. 10. -.708. Art. 18. 15. 90°, 36° 52' 12". 16.45°. 17. 45°, 71° 34'. 18. 63° 7' 48". 19. 30°, 48° 35' 25". 20. 36° 52' 12", 16° 15' 36". CHAPTER III. Art. 21. 5. ^=65° 14', 6=7.834. 6. ^=50° 12' 24". 9. ^=30° 12' 12". 11. 6 = 215.6. 12. a = '312.23. CHAPTER IV. Art. 28. 1. 24.948, 12.71. 2. 58.78. Art. 29. 4. 398.19 ft. 5. 228.4,258 ft. 6. 63.88 ft. 7. 276.95 ft. 10. 86.6, 50. 12. 219.45 ft. Art. 30. 1. 26.172, 52.345 mi., second ship bears E. 19°42'.l N. from first. 2. LB = 14.197 mi. Art. 31. 2, 3. 2392.18 sq. ft. 5. 22.5 sq. ft. 6. 435.7 sq. ft. Art. 32. 2. Base = 187.9 ft.; height = 350.63 ft.; area = 32,943 sq, ft. 3. Base = 358.21 ft.; height = 161.26 ft.; area = 28,881 sq. ft. Art. 33. 1. 14.54 ft., 16.13 ft., 48.45 sq. ft., 105.2 sq. ft. 2. 16.516 ft., 20.415 ft., 133.94 sq. ft., 318.4 sq. ft. Art. 34 b. 2. 10.954 mi. 3. 96 ft. 4. 14.454 mi. 5. 67.08 mi., Dip. = 57' 39". 6. 140.7 mi., Dip. = 2° V 53". Art. 34 c. 4. 2.852 acres. 5. 12 acres 3 roods 6.45 poles. CHAPTER Y. Art. 44. 18. 45°, 135°, - 225°, - 315° ; 45°, 135°, 405°, 495°. 19. 60°, 240°, - 120°, - 300° ; 60°, 240°, 420°, 600°. 20. 135°, 225°, - 135°, - 225° ; 135°, 225°, 495°, 585°. 21. 150°, 330°, - 30°, - 210° ; 150°, 330°, 510°, 690°. CHAPTER VI. Art. 46. 8. cos (x + y) = .7874, sin (x + y) = .6164. (Verify by tables.) Art. 47. 3. sin (x-y)=- .1582, cos {x - y) = .9874. (Verify by tables. ) Art. 50. 7. cps6a; = cos'-^Sa; - sin23x = 1 - 2sin23x = 2cos23a; - 1, sin6x=2sin3a:cos3cc. 9. sin f a;=2sin f a;cos| ic, cos|cc=cos2|a;— sin^fx = 1-2 sin-2 |x = 2 cos'2 |x - 1. 10. cos 6 x = \/| V 1 + cos 1 2 x, sin 6x = V^Vl — cosl2x. 12. sin|x = \/}Vl — cosf X, cos|x = V^Vl + cosfx. ANSWERS. 205 CHAPTER VII. Art. 55. 2. 6 = 70.8, a = 56.1. 4. 6 = 185, c = 192. 5. 6 = 8.237, ; = 5.464. Art. 56. 2. 5 = 36° 18.4' or 143° 41.6', c = 52.71 or 5.98. b. A = t8°25' or 131° 35'. Art. 57. 3. ^ = 80° 46.44', C = 63° 48.56'. 4. i? = 33° 3.33', 9 =100° 56.67', 6 = 39.56. Art. 58. 2. 16° 47.3', 58° 46.07'. 3. 48° 11.4', 58° 24.7', 73° 23.9'. Art. 60. 2. 6 = 698..3, c = 845. 3. 600,240. 6. 6 = 749.1. t. B = 46° 52' 10", C = 111° 53' 25", c = 1767.3, or ^ = 133° 7' 50", G = J5°37'45", c = 823.8. Art. 61. 2. c = 374.04. 4. ^ = 109° 15' 30", c = 440.46. Art. 62. 2. A = 53° 7.8', B = 59° 29.4'. 4. P = 44° 48.25', B = 32° 15.8'. Art. 63. 3. 444.72 yd. 4. 1112.8 yd. 6. 179.28 ft. 7. 87.88 ft. i. 104.08 ft. 9. 479.8 ft. CHAPTER VIII. Art. 67. 1. 11977.8 sq. ft. ; 46° 13.8', 133° 46.2'; 111.3 ft., 149.1 ft. 4. 73° 30.7', 106° 29.3' ; area = 587637.5 sq. metres (approximately). CHAPTER IX; _ TT Sir lir 5_ 11 _ 3_ 6_ 6_ 7 3 11 5 Art. 73. 2. (a) ^, ^, ^, ^, ^tt, llir, ^tt, -tt, .., „, ^ ^ 4' 3 4 ' 6 ' 3 ' 6 ' 2 ' 4 ' 12 ' 20 ' 20 11 TT, -^TT ; (6) .786, 1.048, 2.357, 3.667, etc. 4. 90°, 60°, 45°, 30°, 36°, 60 6 etc. 5. -135°, -900°, -240°, -165°, -4500°. 7. 28° 38' 52.4', 229° 10' 59.2", 171° 53' 14.4", 19° 5' 54.9", etc. 9. Sine, cosine, tangent, cotangent, secant, cosecant, respectively, are : -, -, — -, — ^, "n/3, — ^, 2 ; 6 2 2 V3 V3 f ' v? ^ '' '' ^' ^ ' '' "' -'' "• -"' -'' "'-!"• f' I' ^' ^s' 2 2, — -. 10 a. 3.75. 11. The interior angles of the polygons are respec- tively, 3 ^, I ^, 5 ^, 3 ^, 4 ^, I ^, ^3 ^. 12. (1) 20^^ 163° 42' 8.3" ; (2) ^, 4° 5' 33.2" ; (3) ^, 26". 13. 10, 5, 2.5, |, 1.25, |, |, i, 20, 30, 25, 26| in. 15. 10, 40, 70, 80, 120, 5, 3.75, 1.25 in. 206 PLANE TBIGONOMETRY, CHAPTER X. , Art. 83. 4. 238,890 mi. (approximately), 347.5. ,5. About 57' 2"; about 1 : 13.5. 6. About 93,757,000 mi. 7. 206,265 times the distance of the earth from the sun, 3.26 yr. 8. 9 ft. 2.6 in. 9. 76 ft. 9.5 in. 10. 4' 35". 11. 15.708 yd. 12. 13' 1.3". CHAPTER XI. Art. 85. 3. ^=W7r+(-l)"-,60°, 120°, 420°, 480°. 4. n7r+(-l)"+i-, 3 . 6 210°, 330°, 570°, 690°. 5. n • 180°+ (-1)« 72° 30', 107° 30', 432° 30', 467° 30'. Art. 86. 2. 2 WTT ± ^, w . 360 ± 30°, 30°, 330°-, 390°, 690°. 3. w • 360 ± 7° 40', 7° 40', 352° 20', 367° 40', 712° 20'. 5. n - 360° ± 136° 35', 136° 35', 223° 25', 496° .35', 583° 25'. Art. 87. 2. WTT + - , w . 180° + 60°, 60°, 240°, 420°, 600°. 3. n • 180° + o 20° 10', 20° 10', 200° 10', 380° 10', 660° 10'. 6. n • 180° + 138°, 138°, 318°, 498°, 678°. Art. 89. 11 (first part), (a) .97302 ; (&) ±.97302, .±.12117. 12. 43° 6.5'. 13. 45°. Art. 90. 5. W7r±J, W7r±J. 6. W7r + -, W7r + f7r [i.e. ^ + -V D o 8 \ 2 8 / 7. w. 180°+ 36° 62.2'. 8. n • 180° + 63° 26', n • 180°- 71° 33.9'. 9. ^-180°+ 63° 26', (4 n- 1)45°. 10- ^, !^ + (_l)n-E-. n. w7r+(-l)n^, 4 3 18 2 ,1.360° -46° 23.85'. 12. (2n + l)90°, {4n+(-l)"}15°. 13. {6n+(-l)«}30°, (10w + (-l)«}18°, {10n-3(-l)-}18°. 14. wtt ± ^, wtt ±^. 15. wir±J, 4 3 6 W7r±J. 16. 0, ±\. CHAPTER XII. Art 94 9 4 tan A — i tan^ A 6 tan ^ — 10 tan^ A + tan^ A * 1 - 6 tan2 A + tan* A' 1-10 tan2 ^ + 6 tan* A ' 6 tan ^ - 20 tan^ ^ + 6 tan^ A 7 tan ^ - 35 tan^ ^ + 21 tan^ A - tan' ^ 1-15 tan2 A + \b tan* A - tan^ A' 1-21 tan2 ^ + 35 tan* A-1 tan^ A UNIVERSITY OF CALIFORNIA LIBRARY This book is DUE on th e l a st date stamped beiow!^ centson first da^ overdue j '^' ^^|5<)'cierifs^on fourth day overdue ^« JOn^floUar on seventh day orerdtte. . 0ct6'48RF 7 iutv^b^^^* APR! 9 19561. LD 21-100m-12,'46(A2012sl6)4120 4 REC'D LD ECD LD SEP 26 1957 Dp 20 1957 28Apr'58Fr Rt=:C'P LD APR 141958 SlanSSKK 27Jan'59Lc| RE:C'D LD Ja|| 14 1959 i^"-: --i^- UNIVERSITY OF CALIFORNIA LIBRARY ■^' ■ -^ ,^