Digitized by the Internet Archive 
 
 in 2008 with funding from 
 
 Microsoft Corporation 
 
 http://www.archive.org/details/elementsofplanetOOdu('frich 
 
THE ELEMENTS 
 
 OF 
 
 PLANE TRIGONOMETRY 
 
 BY 
 
 WILLIAM P. DURFEE 
 
 Professor of Mathematics in Hobart College 
 
 BOSTON, U.S.A. 
 GINN & COMPANY, PUBLISHERS 
 
 1901 
 
I 
 
 Copyright, 1900 
 By WILLIAM P. DURFEE 
 
 ALL RIGHTS RESERVED 
 
PEEFAOE 
 
 In preparing this book the author had two ends in view : 
 
 First, to give the student an elementary knowledge of 
 the science of Trigonometry, together with an introduction 
 to the theory of functions as illustrated by the trigono- 
 metric ratios. 
 
 Second, to give him practice in the art of computation. 
 Especial stress has been laid on this side of the subject for 
 the reason that many students have no other experience 
 with the calculation of approximate numbers. The value 
 of preliminary estimates of results and the necessity of 
 frequent checking are constantly insisted on. This feature 
 is believed to be novel. 
 
 The chapter on the Right Triangle is an informal intro- 
 duction to the subject. The use of natural functions is 
 advised here that the student may become familiar with 
 them. In the remainder of the book logarithms are used 
 in all computations. In order to meet what seems to be 
 the demand at the present time, the author has worked the 
 illustrative problems with five-place logarithms. Person- 
 ally he prefers four-place, as they are sufficiently accurate 
 for most practical purposes, and their use permits the 
 student to solve more problems in the limited time at his 
 
 1 46543 
 
iv PREFACE 
 
 disposal. By dropping the fractional part of the minute 
 
 and the fifth figure of a number, where they occur in the 
 
 data, four-place tables may be used without trouble. 
 
 The author is under great obligation to Professor E. W. 
 
 Davis, of the University of Nebraska, for his criticism and 
 
 suggestion. Much of the chapter on Computation is due 
 
 to him. 
 
 W. P. D. 
 
 Geneva, August, 1900. 
 
CONTENTS 
 
 CHAPTER I — COMPUTATION 
 
 SECTIONS 
 
 1-2. Approximate Numbers . 
 
 3. Logarithms . 
 
 4. Use of Logarithmic Tables 
 
 5. Interpolation 
 
 6. Accuracy in Computing 
 
 CHAPTER II — THE RIGHT TRIANGLE 
 
 7. 
 
 Definition of the Trigonometric Functions 
 
 
 
 8. 
 
 Applications of the Definitions 
 
 
 
 9. 
 
 Inverse Functions .... 
 
 
 
 
 10. 
 
 Functions of Complementary Angles 
 
 
 11-12. 
 
 The Fundamental Relations . 
 
 
 
 
 13. 
 
 Functions of 45°, 30°, and 60° 
 
 
 
 
 14. 
 
 Tables of Natural Functions . 
 
 
 
 
 15. 
 
 Use of Tables of Natural Functions 
 
 
 
 
 16-17. 
 
 Solution of Right Triangles . 
 
 
 
 18. 
 
 Solution of Problems .... 
 
 
 
 
 CHAPTER III — UNLIMITED ANGLES 
 
 19. 
 
 Addition and Subtraction of Lines . 
 
 20. 
 
 Angles 
 
 
 
 
 21. 
 
 Addition and Subtraction of Angles 
 
 
 
 
 22. 
 
 Measurement of Angles 
 
 
 
 
 23. 
 
 Quadrants 
 
 
 
 
 24. 
 
 Ordinate and Abscissa . 
 
 
 
 
 25. 
 
 Definition of the Functions . 
 
 
 
 
 26. 
 
 Signs of the Functions . 
 
 
 
 
 27. 
 
 Fundamental Relations 
 
 
 
 
 28. 
 
 Functions of 0°, 90°, 180°, 270° . 
 
 
 
 
 29. 
 
 Line Representatives of the Functions 
 
 
 
 
 30. 
 
 The March of the Functions . 
 
 
 
 
 81. 
 
 Sine Curve, etc 
 
 
 
 
 32. 
 
 Periodicity ..... 
 
 
 
 
VI 
 
 CONTENTS 
 
 CHAPTER IV— REDUCTION FORMULAS 
 
 SECTIONS 
 
 33. Functions of Negative Angles 
 
 34. Functions of 90° + 0, etc. . 
 
 35. Reduction Formulas 
 
 36. Applications ..... 
 
 37-38. 
 
 39. 
 
 40. 
 41-46. 
 
 47. 
 
 48. 
 
 49. 
 
 50. 
 
 51. 
 
 52. 
 
 CHAPTER V— THE ADDITION 
 
 THEOREM 
 
 Projection .... 
 Projection on Coordinate Axes 
 Addition Formula 
 Sine and Cosine of <p ± d 
 Tangent and Cotangent of <p ± 
 Functions of Double Angles . 
 Functions of Half-angles 
 Functions of Three Angles . 
 Conversion Formulas 
 Trigonometric Equations 
 
 PAGE 
 
 51 
 52 
 54 
 55 
 
 57 
 
 59 
 
 59 
 
 60 
 
 65 
 
 67 
 
 68- 
 
 70 
 
 71 
 
 73 
 
 CHAPTER VI — THE TRIANGLE 
 
 53. Nomenclature, etc. ....... 80 
 
 54. The Law of Sines 81 
 
 55. The Law of Tangents .82 
 
 56. The Law of Cosines 82 
 
 57. Functions of Half -angles in Terms of Sides . . .83 
 
 58. Circumscribed, Inscribed, and Escribed Circles . . 85 
 
 59. Area of the Triangle 87 
 
 CHAPTER VII — SOLUTION OF THE TRIANGLE 
 
 60. Formulas for Solution of Triangles 
 
 61. Use of Logarithmic Functions 
 
 62. A Side and Two Angles .... 
 
 63. Two Sides and the Included Angle 
 
 64. Two Sides and the Angle Opposite One of Them 
 
 65. Three Sides , 
 
 90 
 93 
 93 
 94 
 95 
 
 APPENDIX 
 
 Table of Important Formulas 
 
 103 
 
PLANE TRIGONOMETRY 
 
 CHAPTER I 
 ON COMPUTATION 
 
 1. Suppose I measure this table and find it 3 ft. 6 in. long. 
 Is it exactly that ? not a shade more nor less ? Obviously 
 no one can be certain of this. We find the table to be 3 ft. 
 6 in. long within some limit of accuracy beyond which we 
 do not care or are not able to go. 
 
 Should we, by the refined methods known to science, 
 attempt to get the length to within, say, a millionth of an 
 inch, we should probably find that two different measure- 
 ments would give discordant results, while neither result 
 would agree with our first rude measurement of 3 ft. 
 6 in. 
 
 This inaccuracy holds of all numbers got by measure- 
 ment ; that is, with the great bulk of numbers with which 
 we have to deal in practical computation. 
 
 Nor can we altogether avoid this approximation when 
 the numbers are ideal. For example, the square root of 2 
 is 1.4 to the nearest tenth, 1.41 to the nearest hundredth, 
 1.4142 to the nearest ten thousandth ; that is to say, the 
 square of each of these numbers is nearer 2 than the square 
 of any other with the same number of places. Similar 
 remarks apply to all surds, to nearly all logarithms, to 7r, 
 and to the various trigonometric ratios. 
 
 1 
 
2 PLANE TRIGONOMETRY 
 
 It is plain that an error of a foot in a mile is of far less 
 relative importance than an error of an inch in a yard. It 
 is indeed a wonderful triumph of measurement and calcula- 
 tion to have determined the sun's distance within some hun- 
 dred thousand miles, the whole distance being 92,900,000. 
 This being the accuracy of the sun's distance, and, further- 
 more, the earth's orbit being only approximately circular, it 
 would be impossible to determine the length of the orbit 
 with an uncertainty less than some hundred thousand miles, 
 even though we knew the value of ir to a million places. 
 
 2. The sum of a number of approximate numbers cannot 
 be accurate beyond the place where accuracy ceases in any 
 one of them. 
 
 Suppose, for example, two men had measured parts of 
 the same line, one finding his end 307.492 ft. long and the 
 other his end 602.43 ft. The length of the whole line is 
 
 307.492 
 
 602.43 
 
 909.92 
 
 Since the second man did not measure to thousandths, 
 the result cannot be accurate beyond hundredths. 
 
 A product cannot have a greater degree of accuracy than 
 that of its least accurate factor. Suppose we wish the prod- 
 uct of the approximate numbers 23.57 and 612.3. The 
 approximate number 23.57 may have any value from 23.565 
 to 23.575, while 612.3 lies between 612.25 and 612.35. 
 This product may be anything 
 
 from 23.565 X 612.25 = 14427.67125 
 
 to 23.575 X 612.35 = 14436.15125. 
 
 The mean of these results is 14432, but we cannot be 
 sure of the last figure. We do feel sure, however, that the 
 
23.57 
 
 612.3 
 
 14142. 
 
 236. * 
 
 47. ** 
 
 7. * ** 
 
 ON COMPUTATION 6 
 
 fourth figure is nearer 3 than any other digit. The product 
 to four figures is 14430. 
 
 The labor of obtaining useless figures can be avoided by 
 simply not getting those partial products that are not to 
 be retained in the final result. 
 
 A convenient arrangement is as below, keeping all deci- 
 mal points in line and multiplying from left to right. The 
 place of the right-hand figure of the product 
 of the multiplicand by any multiplier digit is 
 as many places to the left or right of the right- 
 hand figure of the multiplicand as the multi- 
 plier digit is to th& x left or right of unit's place. 
 In the example given above, 2, the first figure 
 of the partial product arising when the multi- 
 plicand is multiplied by 6, is put two places to the left of 7, 
 since 6 is two places to the left of unit's place. The stars 
 indicate places of figures not obtained and would usually 
 be omitted. We carry to the first figure retained as we 
 would carry were the work done in full ; moreover, if the 
 first figure omitted is 5 or more than 5, we carry 1 to the 
 first figure retained. This is illustrated in the second and 
 the fourth partial products found above. 
 
 The student can test the result by interchanging multi- 
 plier and multiplicand. If the multiplier or the multipli- 
 cand has only three-figure accuracy, it will readily be seen 
 that the product can have only three-figure accuracy. 
 
 In division also the accuracy of the quotient cannot 
 exceed the least accurate of the numbers which are divi- 
 dend and divisor. 
 
 The arrangement for division is given below. 
 
 To find the quotient of 14432, divided by 612.3. 
 
 Move the decimal point in both dividend and divisor as 
 many places to the right as are necessary to make the 
 
4 PLANE TRIGONOMETRY 
 
 23.57 divisor an integer. The left-hand figure of 
 
 6123. 1 144320. foe quotient is placed over the last figure 
 
 of the first product, and the decimal point 
 
 18 3 7 of the quotient is in line with the decimal 
 
 349 point of the dividend. After the last sig- 
 
 306 nificant figure of the dividend has been 
 
 used, the partial products are formed by 
 
 — omitting one, then two, then three final 
 
 figures of the divisor, remembering to carry to the first 
 
 figure retained, as in multiplication. 
 
 3. So much for the ordinary arithmetical processes. The 
 general principle underlying it all, that accuracy of results 
 is limited by accuracy of data, continues applicable when 
 tables or other labor-saving devices are used. With four- 
 place tables seven-place accuracy is not to be looked for, 
 and when our data are only four-place it is a foolish waste 
 of time and increases the liability to error to use seven-place 
 tables. 
 
 LOGARITHMS • 
 
 The labor of computation is very greatly abridged by the 
 use of logarithms. The principal facts concerning the prac- 
 tical use of logarithms are recapitulated below : 
 
 The logarithm of a number is the power to which 10 
 must be raised to produce the number. Since 
 
 10° = 1, 10 1 = 10, 10 2 = 100, 10 3 = 1000, etc., 
 we have by definition, logarithm 1, written 
 
 log 1 = 0, log 10 = 1, log 100 = 2, log 1000 = 3, etc. 
 If m and n are any two numbers, we have by definition, 
 m = 10 r , or log m = x, 
 n = 10 y , or log n = y. 
 
ON COMPUTATION 5 
 
 Multiplying these two equations, 
 
 mn = \0 x + y , or log mn = x -\- y. 
 . * . log mn — log m + log n. 
 Similarly, 
 
 log mnpq • • • = log ra + logm -f- log^> + log q H . 
 
 I. The logarithm of the product of several numbers is 
 ■equal to the sum of the logarithms of the factors. 
 
 Dividing the first of the two equations above by the 
 second, 
 
 m 10'*' m 
 
 — = --— = lO* -2 ', or log — = x — y = log m — log n. 
 
 n ■ 10 y n J . & 
 
 II. The logarithm of. the quotient of two numbers is equal 
 to the logarithm of the dividend minus the logarithm of the 
 divisor. 
 
 Raising m = 10* to the -£th power, 
 
 m k = 10**, or log m k — kx = k log m. 
 
 III. The logarithm of any power of a number is equal to 
 the logarithm of the number multiplied by the exponent of 
 the power. 
 
 Since a root is a fractional power, the logarithm of the 
 
 kth. root of a number is - times the logarithm of the number. 
 
 The following series of equations illustrates these prin- 
 
 €iples: .JSP 
 
 3 lab 2 , fab 2 \ h 1, ah 
 
 log y = log\-^- = log ( — \ = - log — by III 
 
 — i[log ab 2 — log c 4 ] by II 
 
 = -J- [log <x -j- log b 2 — log c 4 ] by I 
 
 = £[log a + 2 log & - 4 log e] by III 
 
6 PLANE TRIGONOMETRY 
 
 4. What is the logarithm of 22738 ? Since the number 
 lies between 10000 and 100000, its logarithm lies between 
 4 and 5. By calculation it has been found to be 4.35675. 
 The integral part, 4, is its characteristic ; the decimal part, 
 .35675, is the mantissa. « 
 
 log 22738 
 
 = 4.35675, 
 
 or 
 
 22738 
 
 __ ^Q4.35675 
 
 log 2273.8 
 
 = 3.35675, 
 
 or 
 
 2273.8 
 
 __ -^Q3.35675 
 
 log 227.38 
 
 = 2.35675, 
 
 or 
 
 227.38 
 
 __ 1Q2.35675 4 
 
 log 22.738 
 
 = 1.35675, 
 
 or 
 
 22.738 
 
 _ ^Ql .35675 
 
 log 2.2738 
 
 - 0.35675, 
 
 or 
 
 2.2738 
 
 __ ^QO.35675 
 
 log .22738 
 
 = 1.35675, 
 
 or 
 
 .22738 
 
 _ ^Ql.35675 
 
 log .022738 
 
 = 2.35675, 
 
 or 
 
 .022738 
 
 — ^02.^5675^ 
 
 log .002738 
 
 = 3,35675, 
 
 or 
 
 .002738 
 
 _ j[Q3i35675 
 
 Inspection of this algorithm shows us, 1st, that the char- 
 acteristic depends solely on the position of the decimal 
 point and can always be determined by inspection; 2d, 
 that the mantissa is independent of the decimal point and 
 depends on the sequence of digits constituting the number. 
 
 The student can easily make for himself a set of rules 
 for determining the characteristic. In case of a decimal, 
 say .000473, he can determine the characteristic of 473, and 
 then move the point six places to the left ; by so doing the 
 characteristic is diminished by six and is 2 — 6 = — 4, 
 written 4. The minus sign is written above the character- 
 istic because it is the characteristic alone that is negative, 
 the mantissa being always positive'. 
 
 The mantissas are found from a table. Mantissas are all 
 approximate numbers, and tables are published giving man- 
 tissas to four, five, six, and seven places. The kind of table 
 to use depends entirely on the accuracy of the numbers 
 which constitute our data. Four-place tables are accurate 
 enough for ordinary data obtained by the use of field instru- 
 
ON COMPUTATION 7 
 
 ments, five-place tables for all data except such as are 
 obtained by the use of the most delicate instruments. We 
 shall use the latter. 
 
 USE OF THE TABLE 
 
 To find the mantissa of 2273 follow down the left-hand 
 column of your table of the logarithms of numbers, passing 
 from page to page until you reach 227 : run your eye across 
 the page on this line to the column headed 3 ; the number 
 so reached, 35660, is the mantissa sought. In some tables 
 the first two figures, 35, are printed only in the column 
 headed 0. 
 
 Verify : 
 
 log 3748 =3.57380. log 165 =2.21748. 
 
 log 9741 =3.98860. log 17 =1.23045. 
 
 log 112.1 = 2.04961. log 1624 = 3.21059. 
 
 log 32.40 = 1.51055. log .0034 = 3.53148. 
 
 5. The mantissas of five-figure numbers may be obtained 
 from the table by interpolation. The mantissa of 37423 
 lies between the mantissas of 37420 and 37430. We assume 
 that it lies y\ of the way from the first mantissa to the 
 second; i.e., ^ of the way from 57310 to 57322. The 
 difference of these numbers is 12 and *fo of 12 = 3.6 = 4. 
 The mantissa of 37423 is 57310 + 4 = 57314. 
 
 We may formulate the process of finding the mantissa 
 of a five-figure number thus : Enter the table with the first 
 four figures ; subtract the corresponding mantissa from the 
 next larger mantissa to find the tabular difference ; multiply 
 the tabular difference by the fifth figure, considered as a 
 decimal, to find the correction ; add the correction to the 
 mantissa first found ; the result is the mantissa of the five- 
 figure number. 
 
8 PLANE TRIGONOMETRY 
 
 In most tables the multiplication spoken of above is per- 
 formed in the tables of proportional parts printed on the 
 margin of the page. 
 
 Verify the following logarithms : 
 
 log 127.34 = 2.10497. log 8964.3 = 3.95252. 
 log 34.876 - 1.54253. log 90002 = 4.95425. 
 log .42748 = T. 63092. log (.42748) 6 = 3.78552. 
 
 The last problem will present no difficulty if we remember 
 that the mantissa is always positive. Division of the log- 
 arithm when the characteristic is negative requires care. 
 Suppose we wish to divide 3.78552 by 6. We write it 
 6 -f 3.78552. The division is now a simple matter. If 
 we wish to divide by o, we write it 5 + 2.78552. We make 
 the negative characteristic a multiple of the divisor. 
 
 When the logarithm is given and we wish to find the 
 number, the process is the inverse of the one just considered. 
 We may formulate it thus : Find in the table the mantissa, 
 equal to or next less than the given mantissa. The corre- 
 sponding number will be the first four figures of the number 
 sought. Subtract this mantissa from the given mantissa 
 to find the correction. Divide the correction by the tabu- 
 lar difference, obtaining a one-figure quotient. Annex this 
 figure to the four already found. The result is the five- 
 figure number corresponding to the given mantissa. Place 
 the decimal point at the place indicated by the character- 
 istic. The result is the number corresponding to the given 
 logarithm. 
 
 Verify the following : 
 
 3.14216 = log 1386.3. 2.37489 = log .023708. 
 
 2.15362 = log 142.44. .96756 = log 9.2803. 
 
 1.87460 = log 74.92. 
 
ON COMPUTATION 
 
 The logarithm of — is called the cologarithm of m y 
 
 KYI 
 
 written either colog m or col m. In computing, it often 
 saves labor to add the cologarithm instead of subtracting 
 the logarithm. 
 
 If log m = 3.27463 
 
 colog m = log 1 — log m — — 3.27463 
 = 6.72537 - 10. 
 
 The subtraction is readily performed from left to right 
 by taking each digit, except the last, from 9. The — 10 
 is used to avoid negative characteristics. Some computers 
 increase all negative characteristics by 10 and take account 
 of these 10's in the final result. 
 
 6. Accuracy in computing can be attained only by prac- 
 tice and by constant care. When the computer has made 
 his interpolation, he should glance back at the table and 
 see that his result lies between the proper tabular numbers 
 and nearest the right one. This takes but an instant and 
 corrects many errors. The importance of carefully plan- 
 ning a computation before entering upon it can hardly be 
 overestimated. The plan should be written out. The com- 
 puter is then free to devote his whole attention to the 
 mechanical details of the work. Paper ruled in squares 
 conduces to accuracy. If the computation be confined to 
 one column, it can be repeated or a similar one inserted in 
 a parallel column without repeating the plan. If any given 
 number occurs repeatedly in a computation, it may be written 
 down once for all on a separate piece of paper and held 
 over any number with which it is to be combined. 
 
 The computer will avoid many errors if he accustoms 
 himself to making rough estimates of results. When the 
 
 / 
 
10 PLANE TRIGONOMETRY 
 
 nature of the subject permits, these estimates may be 
 obtained by graphic methods. 
 
 To insure accuracy the computer must continually check 
 his work. Every operation, every step in every operation 
 must be tested before going on. If two numbers are added, 
 subtract one of them from the sum. If two numbers are 
 subtracted, add the difference to the smaller. If two num- 
 bers are multiplied, interchange multiplier and multiplicand 
 and compare products. Test every step, and when the com- 
 putation is finished check the final result if the nature of 
 the problem furnishes a test ; if not, work the problem by 
 a second method and compare results. 
 
 The computer who aims at rapidity should train himself 
 to do all he safely can mentally. He should early acquire 
 the habit of remembering a number of six or seven figures 
 long enough to transcribe it. He should perform his inter- 
 polations mentally. He should add and subtract two num- 
 bers from left to right. Other devices will come to him 
 with practice. 
 
 The most important habit to be acquired is that of being 
 constantly on the watch for errors and of constantly check- 
 ing results. The computer who makes no mistakes can 
 hardly be said to exist. Such a one would be a marvel. 
 The ordinary man who forms the habit of not letting a mis- 
 take go uncorrected is more trustworthy than the marvel 
 who does not verify his work. 
 
CHAPTEK II 
 
 THE RIGHT TRIANGLE 
 
 7. The three sides x, y, r of the right triangle ABC 
 furnish six ratios : 
 
 y x y x r r 
 
 — 9 —i —9 —9 — 9 — • 
 
 r r x y x y 
 
 If a second right triangle A'B'C 
 B be constructed with angle A' equal 
 to angle A, it will be similar to the 
 first and we have : 
 
 x : y : r== x : y : ?\ 
 Therefore, 
 
 Fig. 1. 
 
 
 
 c' 
 
 Jj/S 
 
 / 
 
 V 
 
 
 /^ x' 
 
 
 
 A' 
 
 
 B' 
 
 Hi 
 
 Fig. 2. 
 
 y x' __ X y y 
 
 v r! r x' x 
 
 x' x r' r r f r 
 
 y* y x' x y ] y 
 
 Each ratio of one triangle is equal to the corresponding 
 ratio in the other. 
 
 Let us construct a third right triangle A"B"C", making 
 angle A" > A and side A"C" = AC. 
 From the construction 
 
 and 
 
 r 1! __ ^ 
 
 x' 
 
 < x, 
 
 y' 
 
 >m 
 
 y"y 
 
 "77 > ~> 
 r r 
 
 x' 
 
 X 
 
 <-> 
 
 r 
 
 x' 
 
 X 
 
 x" X 
 
 r y 
 
 x" 
 
 r 
 >-> 
 
 X 
 
 r' 
 
 y" 
 ii 
 
 r 
 <- 
 
 y 
 
 Fig. 3. 
 
12 PLANE TRIGONOMETRY 
 
 The ratios in this triangle are not equal to the corre- 
 sponding ratios in the first triangle. 
 
 The foregoing considerations lead to the conclusion that 
 these ratios depend for their values solely on the angle A ; 
 i.e., they change when A changes, they are constant when 
 A is constant. This dependence is expressed mathematic- 
 ally by saying that the ratios are functions of the angle A. 
 To distinguish them from other functions they are called 
 Trigonometric Functions. 
 
 The six trigonometric functions of A are named as 
 ' follows : 
 
 - = sine of A , 
 r 
 
 written 
 
 sin A. 
 
 - = cosine of A. 
 r 
 
 a 
 
 COS^L. 
 
 - = tangent of 'A, 
 
 a 
 
 tan A 
 
 x 
 
 - = cotangent of A 
 
 y 
 
 > 
 
 cot A. 
 
 r 
 
 - = secant of A, 
 
 X 
 
 a 
 
 sec A. 
 
 r 
 
 -^= cosecant of .4, 
 
 u 
 
 esc A. 
 
 y 
 
 8.' The foregoing equations define the trigonometric func- 
 tions. They are fundamental and should be carefully mem- 
 orized. These definitions may be put into words : 
 
 ?/ side opposite 
 
 sin A = * = -r f^ 
 
 r hypotenuse 
 
 x side adiacent 
 cos A a= - r* 
 
 r hypotenuse 
 y _ side opposite 
 Fig. 4. x side adjacent 
 
THE RIGHT TRIANGLE 13 
 
 x side adiacent 
 
 cot A = - = — J - —• 
 
 y side opposite 
 
 _ r _ hypotenuse 
 
 x side adjacent 
 
 r hypotenuse 
 esc A = - .— — ^ 
 
 t/ side opposite 
 
 EXERCISES 
 
 Find the six functions of each of the acute angles in the 
 right triangle whose sides are : 
 
 1. 5, 12, 13. 7. a, Vl - a 2 , 1. 
 
 2. 3, 4, 5. 8. a, 6, Va 2 + 6 2 . 
 
 3. 8, 15, 17. v 9. 5, 5, 5 V2. 
 
 4. 9, 12, 15. 10. a, V^ax~+ti\ a + x. 
 
 5. 5, 8, V89. 11. a + b, a-b, V2(V + 6 2 ). 
 
 6. 2, 3, Vl3. , 12. m*—n 2 , 2mn, m 2 + ^ 2 . 
 
 9. Inverse Functions. Suppose we have the expression 
 sin .4 = f; how may we describe A ? A is the angle whose 
 sine is f . It is customary. to express this by writing 
 
 A = sin-H. 
 
 This is read "A is the angle whose sine is -J." So, too, the 
 
 expressions cos -1 §, tan -1 |, sec -1 f 
 
 are read the angle whose cosine is 
 
 §, the angle whose tangent is §, 
 
 the angle whose secant is f. 
 
 These functions are called inverse 
 
 functions. They are distinguished D /<£ — ^ 
 
 from the trigonometric functions F IG# 5# 
 
 by the exponent — 1. 
 
14 PLANE TRIGONOMETRY 
 
 Let it be required to construct sin -1 !. 
 
 Construction. At B erect BC perpendicular to DB and 
 equal to 3. With C as a center, and with a radius equal 
 to 5, describe an arc cutting BD at A. Draw CA ; then is 
 A the required angle. For by definition 
 
 sin A = | , or A = sin -1 |. 
 
 EXERCISES 
 Construct the following angles : 
 
 1. sin -1 1, sin -1 f, sin _1 f. 
 
 2. cos -1 !, cos -1 1, cos"" 1 !. 
 
 3. tan -1 ^, tan -1 1, tan -1 l. 
 
 4. cot -1 f, cot -1 2, cot -1 5. 
 
 5. sec -1 1, sec -1 2, sec -1 5. 
 
 6. esc -1 1, esc -1 1, esc -1 3. 
 Show by constructing a figure that : 
 
 7. sin -1 f = cos -1 % = tan -1 §. 
 
 Show by construction that the following angles are im- 
 possible. 
 
 8. sin -1 1, cos -1 2, sec -1 f, csc -1 ^. 
 
 9. sm -1 -7? cos -1 7? a>o. 
 
 b o 
 
 10. sec -1 -? esc -1 -? b<a. 
 a a 
 
 10. Functions of Complementary Angles. The angles A and 
 C are complementary. By definition 
 
 V 
 
 sin A = - = cos C. 
 r 
 
 cot A 
 
 x 
 
 = - = tan C. 
 
 y 
 
 cos A = - = sin C. 
 r 
 
 sec A 
 
 r 
 
 = - = esc C. 
 
 X 
 
 tan A = - = cot C. 
 
 esc A 
 
 = - = sec C. 
 
THE EIGHT TRIANGLE 
 
 15 
 
 We may summarize these relations by saying that any 
 function of an angle is equal to the 
 co-function of the complementary 
 angle. In this statement we assume 
 that the co-function of the cosine is 
 the sine, etc. The cosine, cotangent, 
 cosecant are contractions for com pie- FlG - 6 - 
 
 ment's sine, complement's tangent, complement's secant. 
 
 EXERCISES 
 
 1. The functions of 30° are : 
 
 sin 30° = ±, cos 30° = i V3, tan 30° - £ V3. 
 cot 30° '= V3, sec 30° = § V3, esc 30° = 2 ; 
 write the functions of 60°. 
 
 2. sin 40° = cos 50° ; express the relations between the 
 other functions of these angles. 
 
 3. The angles 45° -f A and 45° — A are complementary ; 
 express the functions of 45° -f- A in terms of the functions 
 of 45°- A. 
 
 4. A and 90° — A are complementary ; express the func- 
 tions of 90° — A in functions of A. 
 
 5. 45° is its own complement ; show that sin 45° = cos 45°, 
 tan 45° = cot 45°,' sec 45° = esc 45°. 
 
 11. Fundamental Relations of the Trigonometric Functions. 
 
 The six functions sine, cosine, etc., 
 are connected by a number of equa- 
 tions. The more important of these! 
 are derived below. The first five\ 
 depend immediately on the defini- 
 tions, the other three on a well- 
 known property of the right tri- 
 angle. These last involve the squares of the functions. By 
 
 Fig. 7. 
 
16 PLANE TRIGONOMETRY 
 
 universal usage powers are indicated by affixing the expo- 
 nent to the functional symbol. E>g-, (sin A) 2 is written 
 sin 2 A, (cos A) z is written cos 3 A. 
 
 y v 
 Since - X -. = 1 we have sin A esc A = 1 [1] 
 
 cos A sec A = 1 [2] 
 
 tan A cot A = 1 [3] 
 
 sin A 
 cos A 
 cos A 
 
 V 
 
 
 y 
 
 
 X 
 
 r 
 
 X 
 
 r 
 
 X 
 
 1 
 
 y 
 
 X 
 
 X 
 
 X 
 
 y~ 
 
 1 
 
 y 
 
 
 X 
 
 y 
 
 r 
 
 
 r 
 
 X 
 
 X 
 
 
 y = 
 
 X 
 
 V 
 
 
 r 
 
 y 
 
 sin A 
 
 tan A [4] 
 
 = cot A. [5] 
 
 From the figure y 2 + x 2 = r 2 . 
 
 Dividing this equation by r 2 , by x 2 , and by y 2 , 
 
 S + S =1; (r)+(7-) 2=1 - ••• S inM+cos^=l. [ r,] 
 
 =/ 
 
 **?*■?/ i+ ©-w- 
 
 These eight identities constitute the fundamental rela- 
 tions of the trigonometric functions. They are very im- 
 portant and should be committed to memory. 
 
 By means of these relations, when we know one function 
 of an angle, we can find all the others. 
 
 Suppose, for example, sin A = -J. 
 
 [6] cos A = Vl - sin 2 ,4 = Vl - £ = £ V& 
 
 r a -. sin A \ 1 , . r- 
 
 [41 tan ^4 = = -~=z = --= = 4 V3. 
 
 ■ cos.l iV3 V3 
 
THE RIGHT TRIANGLE 17 
 
 sin A 
 
 The values of these functions may also be found by con- 
 structing sin -1 £ and finding the third side of the 1 triangle 
 geometrically. This side is V3. The functions can now 
 be written from the definitions (§ 8). 
 
 EXERCISES 
 
 Find all the functions of the following angles, using^each 
 of the methods illustrated above : i 
 
 1. cot" 1 2. 3. cot" 1 ! 5. cos" 1 !! 7. cos" 1 2. 
 
 2. tan -1 3. 4. sec -1 §. 6. csc~^f. 8. sin- 1 !. 
 
 12. We can express any function of A in terms of any 
 other function of A by making use of formuras [1] to [8]. 
 As an illustration let us express each of the^functions in 
 terms of the tangent. 
 
 [3] 
 
 tanvl = tan A. 
 
 cot A = -• 
 
 tan A 
 
 
 [7] 
 
 . sec A = Vl + tan 2 A. 
 
 
 
 1 i 
 
 Vl + tanM 
 
 [8] 
 [1] 
 [2] 
 
 pc , A -y/l 1 p n +2 | "\/l 1 A 
 
 * tan 2 A 
 1 tan A 
 
 tan A 
 
 esc A Vl + tan 2 A 
 
 
 sec A Vl + tan 2 A 
 
 
18 
 
 PLANE TRIGONOMETRY 
 
 EXERCISES 
 
 Express each of the functions of A in terms of 
 
 1. sin A. 2. cos A. 3. cot A. 4. sec .1. 5. coo A. 
 
 6. Tabulate the results. 
 
 Prove the following identities by means of formulas [1] 
 to [8]. 
 
 7. sin A sec A = tan A. 
 
 8. (sin A + cos A) 2 = 1 + 2 sin A cos A. 
 
 9. (sec A 4- tan A) (sec .4 — tan ^1) = 1. 
 1 — sin .4 _ cos A 
 
 cos .4 1 + sin A 
 
 11. (l + tan^) 2 4-(l -tan.4) 2 = 2 sec 2 ,!. 
 
 12. (sin A 4- cos ,4) 2 4- (sin A - cos ,4) 2 = 2. 
 
 13. Functions of 45°, 30°, and 60°. 
 
 Construct angle A == 45°, lay off yli? = 1, and complete 
 the right triangle. 
 
 Fig. 9. 
 
 Angle C = 45°. 
 From the definitions 
 
 sin 45° = cos 45' 
 
 Fig. 10. 
 BC = 1 and.4C = V2. 
 
 V2 
 tan 45° = cot 45° = 1. 
 
 sec 45° = esc 45° 
 
 V2 r 
 
THE RIGHT TR. NGLE 21 
 
 Construct the equilateral triangk _ 4452 = 26 
 Bisect angle A by AB. The triangle ii-_-jo 9 _-.o 
 with angle BAC = 30°, angle C = 60°,~ ' ~~ 
 side^=V3. -- 4470 - 
 
 By definition 
 
 sin 30° = j. sin 60° = -^ = %V3. 
 
 cos 30° = ^ = £ V3. cos 60° = j. 
 
 tan 30° = -i= = i V3. tan 60° = — - = ^3. 
 
 V3 1 
 
 cot 30° = = V3. cot 60° = — ^ = 4- V3. 
 
 1 V3 
 
 sec 30° = -4= = f V3. sec 60° = f = 2. 
 
 V3 F T 
 
 esc 30° = * = 2. esc 60° = -^ .=== # V& 
 
 V3 3 
 
 14. Trigonometric Tables. In the preceding paragraph we 
 have found the functions of 30°, 45°, and 60° by simple 
 geometrical expedients. The functions of other angles are 
 not found so easily. For purposes of computation, tables 
 of trigonometric functions are used. Such tables give the 
 values of the sine, cosine, tangent, and cotangent of all 
 angles from 0° to 90° at intervals of 10'. Examine such 
 a table. You will find in the left-hand column the angles ; 
 their sines, cosines, tangents, and cotangents are opposite in 
 appropriately headed columns. The column at the extreme 
 right also contains angles. Inspection will show you that 
 these angles are the complements of the corresponding 
 angles at the left. We learned in § 10 that the function of 
 any angle was the co-function of its complementary angle. 
 The sine of an angle at the right is the cosine of the 
 
18 PLANE 
 
 a TRIGONOMETRY 
 
 at the left. You will find this indi- 
 
 Express each of iy the word cosine at the bottom of the 
 
 1. sin A. 2. Jie - Similarly for the other functions. If 
 
 6. Tabulate / our table carefully, you will find that the 
 
 Pro^* an down the left side of the page till 45° is reached ; 
 
 they then run up the right side of the page till 90° is 
 
 reached. If the angle is less than 45°, you look for it at 
 
 the left; if more than 45°, at the right. If the angle is 
 
 at the left, the name of the required function is at the top 
 
 of the page; while if the angle is at the right the name. of the 
 
 function is at the bottom of the page. 
 
 The tables do not contain secants and cosecants. These 
 functions are reciprocals of cosine and sine, and can readily 
 be found by taking advantage of this fact. You will find 
 by experience that it is never necessary to use them in 
 computation. 
 
 Take your table and run down the column of sines. They 
 increase with the angle. So do the tangents. Examine the 
 cosines and cotangents. They decrease as the angle increases. 
 
 15. The table gives the functions of angles which are 
 multiples of 10 f . To find the functions of other angles 
 we interpolate, as explained in § 5. Care must be taken 
 to add the correction in finding sines and tangents, to sub- 
 tract it in finding cosines and cotangents. 
 
 Find the sine of 27° 34'. 
 
 sin 27° 30' -.4617. 
 The tabular difference = .4643 - .4617 == 26. 
 
 The correction = .4 of 26 = 10.4 = 10. 
 
 sin 27° 34' = .4617 + 10 = .4627. 
 
 Find the cosine of 63° 27'. 
 
 cos 63° 20' = .4488. 
 
THE RIGHT TRIANGLE 21 
 
 The tabular difference = .4488 - .4462 = 26. 
 
 The correction - .7 of 26 = 18.2 = 18. 
 
 cos 63° 27' = .4488 - 18 = .4470. 
 Find the tangent of 84° 28'. 
 
 tan 84° 20' = 10.078. 
 The tabular difference = 10.385 - 10.078 = 307. 
 
 The correction = .8 of 307 = 245.6 = 246. 
 
 tan 84° 28' = 10.078 + 246 = 10.324. 
 
 The work which is here done out in full should be per- 
 formed mentally as far as possible. In case your table has 
 a column of differences, the operation of finding the tabular 
 difference is unnecessary ; if your table is provided with a 
 table of proportional parts, the operation of finding the 
 correction is much simplified. 
 
 EXERCISES 
 Verify the following : 
 
 sin 0° 42' = .0122. sin 58° 38' = .8539. 
 
 cos 0° 42' = .9999. cos 58° 38' - .5220. 
 
 tan 0° 42' =.0122. tan 58° 38' = 1.6405. 
 
 co!; 9° 42' == 5.8505. cot 58° 38' = .6096. 
 
 sin 43° 01' = .6822. cos 28° 13' = .8812. 
 
 tan 38° 29' = .7949. cot 81° 31' = .1492. 
 
 To find sin- 1 . 4327. 
 
 The next smaller sine is .4305, the sine of 25° 30'. 
 
 The difference = .4327 - .4305 = 22. 
 
 The tabular difference = .4331 - .4305 == 26. 
 
 Correction = || = 8. 
 
 .'.sin- 1 .4327 =25° 38'. 
 
22 PLANE TRIGONOMETRY 
 
 Fin<i cos" 1 .8826. 
 
 The next larger cosine is .8829, the cosine of 28° 00'. 
 
 The difference = .8829 - .8826 = 3. 
 
 The tabular difference = .8829 - .8816 = 13. 
 
 The correction * = T 3 ^ = 2. 
 
 .*. cos- 1 .8826 = 28° 02'. 
 
 Here the next larger cosine is taken because the cosine 
 is a decreasing function. 
 
 EXERCISES 
 
 Verify the following : 
 
 tan- 1 .4329 - 23° 24'. tan" 1 3.4268 = 73° 44'. 
 
 cot" 1 .3721 = 69° 35'. cos" 1 .4268 - 64° 44'. 
 
 sin" 1 .8523 = 58° 28'. cot" 1 1.4823 = 34° 00'. 
 
 16. The Solution of the Right Triangle. To solve a right 
 triangle is to find numerical values for the unknown parts. 
 This is possible when two parts, one of which is a side, 
 are known. Two methods of solution are open to us, —the 
 Graphic and the Trigonometric. 
 
 Graphic Solution. It is desirable to solve all problems 
 by this method before proceeding to the more accurate 
 trigonometric method. It gives rough approximations and 
 enables the student to detect his grosser mistakes in the 
 application of the trigonometric formulas. The solution 
 consists in accurately constructing the figure from the data 
 given. The required or unknown parts may now be care- 
 fully measured by scale and protractor. The use of paper 
 ruled in squares facilitates this work. The only instru- 
 ments needed are a scale, a protractor, a straight-edge, 
 and a pair of dividers. Two-figure accuracy is all that 
 should be aimed at in this method of solution. 
 
THE EIGHT TRIANGLE 23 
 
 17. Trigonometric Solution. In using this method we com- 
 pute the values of the unknown parts. The first four defini- 
 tions (p. 12) furnish formulas sufficient for this purpose. 
 
 These formulas are : 
 
 (a) sin A = J. 7 \ \ ( C ) cos A = y 
 
 (b) tan A = ^. (d) cot A = -• 
 
 No matter what two parts are given, one of these four 
 formulas includes them both. This statement assumes that 
 when A is known B is known, since the two angles are 
 complementary ; and it further assumes that when A is 
 known any of its functions are known, and vice versa, 
 since the tables enable us to find the one from the other. 
 
 The student should satisfy himself of the truth of this 
 statement by selecting all possible combinations of two 
 parts as known parts. To effect the solution we proceed 
 as follows : 
 
 Select a formula containing the two known parts, and 
 substitute in it the values of these parts ; the resulting 
 equation will give a third part. Of the three parts now 
 known, one is an angle and two are sides. To find the 
 remaining side, select a formula containing it. Where pos- 
 sible, a formula should be selected which does not contain 
 the computed part. Experience will show that this is pos- 
 sible when one of the given parts is an angle. 
 
 Checks. These computations, like all others, should be 
 checked. A convenient formula for this purpose is 
 
 r 2 = x 2 -f- y 2 , 
 or y 2 = r 2 — x 2 = (r -f- x)(r — x), 
 
 or x 2 = r 2 — y 2 = (r + y)(r — y). 
 
 The problems that follow illustrate the process of solution. 
 
24 
 
 PLANE TRIGONOMETRY 
 
 1°. The hypotenuse of a right triangle is 36, and one of 
 its angles is 32° 14' ; find the other parts. 
 
 Graphic Solution. Construct the 
 
 V angle A = 32° 14', or as nearly so 
 
 as your protractor admits ; lay off 
 
 K AC = 36; drop CB _L to AB. The 
 
 triangle ABC is the required tri- 
 
 B angle. Measure AB and BC. 
 
 ~^A 
 
 Fig. 11. 
 Trigonometric Solution. C= 90° - 32° 14' == 57° 46'. 
 
 Formula (b) cos A = - • . * . x = r cos A . 
 \ / r 
 
 [table] x = 36 (.8459) = 30.45. 
 
 Formula (a) sin A = — • .'. y = r sin ^4. 
 
 [table] 
 Check. 
 
 y = 36 (.5334) = 19.20. 
 
 (30.45) 2 = (55.20) (16.80). 
 927.2 = 927.4. 
 This shows that our work is fairly accurate. 
 
 2°. One leg of a right triangle is 27, and the adjacent 
 angle is 67° 23' ; find the other parts. 
 
 Graphic Solution. Construct angle A = 
 67° 23', lay off AB = 27, erect the _L BC. 
 ABC is the required triangle. Measure AC 
 and BC. 
 
 Trigonometric Solution. 
 
 C = 90° - A = 22° 37'. 
 x 
 
 Formula (b) cos A = : 
 
 x 
 
 
 h 
 
 r 
 
 y 
 
 L:^ 
 
 
 Al X=2T 
 
 B 
 
 21 
 .3846 
 
 cos A 
 = 70.21. 
 
 Fig. 12. 
 
THE RIGHT TRIANGLE 
 
 25 
 
 Formula (c) tan A = : 
 
 y ■ 
 
 x tan A. 
 
 Check. 
 
 y = 27 (2.4004) = 64.81. 
 y 2 = (r + #) (7 — a;). 
 (64.81) 2 - (97.21) (43.21). 
 4200.2 = 4200.4. 
 
 3°. The hypotenuse of a right triangle is 48, and one leg 
 is 37 ; find the other parts. 
 
 Graphic Solution. Construct the 
 right angle ABC, lay off BA = 37, 
 from A as center, with radius 48, draw 
 an arc, cutting BC in C ; draw AC. 
 ABC is the required triangle. Meas- 
 ure BC and the angle BAC. 
 
 Trig ono metric Solution. 
 
 Fig. 13. 
 
 Formula (b) 
 
 Formula (a) 
 
 Check. 
 
 .7708. 
 
 A =39° 34'. 
 
 C = 90° - A = 50° 26'. 
 
 sin A ■= 
 
 y 
 
 ~7a 
 
 x=m 
 
 . y = 48 sin 39° 34'. 
 
 y = 48 (.6370) = 30.-58. ' 
 
 x 2 = (r + y)(r- y). 
 (37) 2 = (78.58) (17.42). 
 1369 = 1368.9. 
 
 Fig. 14. 
 
 4°. The two legs of a right triangle are 
 j§ 487 and 756; find the other parts, 
 ss Graphic Solution. In the right angle 
 ABC lay off BA = 487 and BC = 756; draw 
 _ AC. A B C is the required triangle. Meas- 
 ure AC and the angle A. 
 
26 
 
 PLANE TRIGONOMETRY 
 
 Trigonometric Solution. 
 Formula (d) cot A = ± f J = .6442. 
 
 A = 57° 13'. 
 
 C = 90° - A = 32° 47'. 
 
 Formula (a) sin C == 
 
 487 
 
 487 
 
 487 
 
 sinC .5415 
 
 Check. 
 
 r 
 r = 899.4. 
 
 y 2 = (r + x) (r — x). 
 (756) 2 - (1386.4) (412.4). 
 571536 = 571740. 
 Four-figure accuracy is all that we expect, and this we 
 probably have in r but not in (r -f x) (r — x). 
 
 EXERCISES 
 
 
 Exercises 1-6 refer to Fig. 15 ; 7-16 refer to Fig. 16. 
 
 1. x = 20, r = 30. 
 
 2. y = 17, r = 60. f >/ 
 j 3. x = 34, y= 45. 
 
 Fig. 15. 
 
 7 
 
 7. a = 20, .4 = 30°. 
 
 8. ft = 16, ^ = 45°. 
 
 9. c = 75, . J5 = 60°. 
 
 10. a = 12, ft = 15. 
 
 11. a = 407, c = 609. 
 
 4. A= 30° 24', r = 207. 
 
 5. c = 38° 47', r = 103.4 
 
 6. A =64° 23', x= 20.32. 
 
 Fig. 16. 
 
 12. ft = 1.306, c = 2.501. 
 
 13. A = 15° 17', = 163. 
 
 I 14. ^ = 81° 17', ft = .0143. 
 
 15. a = 137.4, ft = 101.2. 
 
 16. 5 = 65° 8', c = 3.145. 
 
/ 
 
 THE RIGHT TRIANGLE 27 
 
 ^1§! The Solution of Problems. The problems that com- 
 plete this chapter can all be solved by right triangles. 
 While different problems demand different methods of 
 solution, the following general method of procedure will 
 be found very useful : 
 
 1°. Carefully construct a diagram to some convenient 
 scale and find the graphic solution by proper measure- 
 ments. 
 
 2°. Examine the diagram for a right triangle with two 
 parts given; if this triangle contains the required part, 
 solve it ; if not, consider all the parts of this triangle as 
 known, and find another right triangle with two parts 
 known ; if this second triangle contains the required part, 
 the method of solution is obvious ; if it does not contain 
 the required part, repeat the process until a triangle is 
 found that does contain it. It may be necessary to draw 
 auxiliary lines. When you have found the several steps 
 that lead to the solution, review the work to make sure 
 that all of them are necessary. 
 
 3°. Proceed to the computation, being careful to check 
 each step. No computation should be made until the whole 
 process of solution is determined upon and written out. 
 
 Definitions. If denote an 
 observer, P an object above the 
 horizon, and POH' a vertical 
 plane intersecting the horizon 
 in HH ! , the angle POH 1 is called 
 the Elevation (or Altitude) of P. 
 If the object be below the hori- FlG * 17# 
 
 zontal plane, as at Q, the angle QOH' is called the depression 
 of Q. 
 
 The bearing of an object is its direction from the ob- 
 server. The use of the word is obvious from tfye following 
 
28 
 
 PLANE TRIGONOMETRY 
 
 illustrations. If be the observer, the bearing of P is 
 E 20° N, of Q is N 25° E, of R is W 30° iV, of T is 5 25° W. 
 
 Fig. 19. 
 
 Bearing is also often given in terms of the divisions of 
 a mariner's compass. The circle is divided into 32 eqnal 
 parts, the points of division being named as indicated* in 
 the figure. 
 
 EXERCISES 
 
 fly The center pole of a tent is 20 ft. high, and its top is 
 stayed by ropes 40 ft. long ; what is the inclination of the 
 ropes to the ground ? 
 
 \2J A man standing 140 ft. from the foot of a tower finds 
 that the elevation of its top is 28° 25'; what is the height 
 of the tower ? ^. 
 
 p /m) At what latitude is the circumference 
 
 oiuAparallel of latitude equal to two-thirds 
 ' of the circumference of the equator ? 
 I Suggestion. Let PEP'E' be a section of 
 the earth through its axis, PP' its axis, 
 ~p' EE' the equator, LL' the circle of lati- 
 
 fig. 20. tll d e . Then will EOL be the latitude. 
 
 4. The length of a degree of longitude at the equator 
 is 69.16 mi.; find a formula for the length of a degree of 
 longitude at latitude A. 
 
THE RIGHT TRIANGLE 29 
 
 /5j A ladder 40 ft. long reaches a window 33 ft. from the 
 ground. Being turned on its foot to the opposite side of 
 the street, it reaches a window 21 ft. from the ground; 
 how wide is the street ? 
 
 (ey From a window the top of a house on the opposite 
 side of a street 30 ft. wide has an elevation of 60°, while 
 the bottom of the house has a depression of 30° ; what is 
 the height of the house ? 
 
 /vN A pole stands on top of a knoll. From a point at a dis- 
 tance of 200 ft. from the foot of the knoll, the elevations 
 of the top and the bottom of the pole are 60° and 30°, respec- 
 tively ; prove that the pole is twice as high as the knoll. 
 
 8. A regular hexagon is circumscribed about a circle 
 whose radius is 20 ft. ; find the length of the side of this 
 hexagon. 
 
 9. The radius of a circle is 1. Find the side, the perim- 
 eter, the apothem, and the area of a regular inscribed poly- 
 gon of 5 sides, of 8 sides, of 9 sides, of 12 sides, of n sides. 
 
 10. A person at the top of a tower 100 ft. high observes 
 two objects on a straight road running by its foot. The 
 depression of the nearer is 45° 36', of the more remote is 
 
 ^2£ ; what is their distance apart ? x <*^~ 
 
 11. If the edge of a regular tetrahedron is 10 ft., find the 
 , fy length of a face altitude ; the length of the altitude, and 
 
 the angle between two faces. 
 
 12. The roof rises from the adjacent sides of a square 
 house at an angle of 30° ; find the angle which the corner 
 of the roof makes with the horizon. 
 
 13. At a certain port the seacoast runs X. N. E., and a 
 vessel 10 mi. out is making 12 mi. per hour S. S. W. At 
 2.30 p.m. she is due east ; what is her bearing at 2 p.m. ? 
 at 3 p.m. ? At 2 p.m. a vessel sailing 10 mi. per hour is 
 dispatched to intercept her ; what course must the latter 
 vessel take ? 
 
30 
 
 PLANE TRIGONOMETRY 
 
 Fig. 21. 
 
 Suggestions. Let P be the port and SS } 
 the course of the vessel, E its position at 
 2.30 p.m. Let PD be perpendicular to SS' ; 
 find when she will be at D. The bearings 
 at 2 p.m. and at 3 p.m. will be easily found. 
 To find course of second vessel let Q be 
 point of meeting and t the time after 2 
 p.m., when they meet. PQ and DQ can 
 now be calculated in terms of t, which 
 can be easily found. 
 
 14. A smokestack is secured by wires running from 
 points 35 ft. from its base to within 3 ft. of its top. 
 These wires are inclined at an angle of 40° to the ground. 
 What is the height of the smokestack ? the length of the 
 wires ? What is the least number of wires necessary to 
 secure the stack ? If they are symmetrically placed, how 
 far apart are their ground ends ? How far are the lines 
 joining their ground ends from the foot of the stack ? from 
 the top of the stack ? What angle do the wires make with 
 these lines ? with each other ? What angle does the plane 
 of two wires make with the ground ? What angle does 
 the perpendicular from the foot of the stack on this plane 
 make with the ground ? what is its length ? 
 
 15. On the U. S. Coast Survey an 
 observation platform 50 ft. high was 
 built. The platform was 8 ft. square. 
 The four legs spread to the corners of 
 a 12-foot square at the base. They 
 were braced together by three sets of 
 cross-pieces, as represented in the illus- 
 tration. If the cross-pieces are equidis- 
 tant and the lowest is 3 ft. from the 
 ground, find the length of each piece 
 required for the construction. 
 
 Fig. 22. 
 
THE RIGHT TRIANGLE 31 
 
 y 16. A man wishing to know the width of a river selects 
 a point, A* one bank, directly- 
 opposite a tree, TR, on the other 
 bank. He finds its elevation to 
 be 10° 30' ; going back 150 ft. 
 to B, he finds its elevation to 
 be 9°. What is the width of the 
 river? (Find A C, perpendicu- B ^° A Fiq 93 
 
 lar to BR ; then AR and A T.) 
 
 17. Upon the top of a shaft 125 ft. high stands a statue 
 - which subtends an angle of 3° at a point 200 ft. from the 
 
 shaft ; how tall is the statue ? 
 
 18. A wheel 1 ft. in diameter is driven by a belt from 
 a wheel 4 ft. in diameter. If the shafts bearing these 
 wheels are parallel and 10 ft. apart, how long will the 
 belt be (a) if crossed ? (/?) if not crossed ? 
 
 19. In a circle whose radius is 15 ft., what angle will 
 a chord of 20 ft. subtend at the center ? a chord of 25 ft. ? 
 of 10 ft. ? In a circle of radius r, what angle will a chord, 
 a, subtend ? 
 
 20. In a circle of 15 ft. radius, find the area of the seg- 
 ment cut off by a chord of 18 ft. ; the area of the segment 
 included between this chord and a chord of 25 ft. 
 
 21. The base of a quadrilateral is 60 ft., the adjacent 
 sides are 30 ft. and 40 ft., the corresponding adjacent 
 angles are 110° and 130°, respectively ; find the fourth 
 side and the other two angles. 
 
 22. The elevation of a balloon due north from A is 60° ; 
 from B, 1 mi. west of A, its elevation is 45° ; what is the 
 height of the balloon ? 
 
 23. One of the equal sides of an isosceles triangle is 
 47 ft., and one of the equal angles is 38° 24'; what is 
 the base of the triangle ? 
 
; 
 
 CHAPTER III 
 HE TRIGONOMETRIC FUNCTIONS OF UNLIMITED ANGLES 
 
 ' 19. A directed line is a straight line generated by a point 
 moving in a given direction. It possesses two qualities — 
 length and direction. 
 
 The lines AB and DC are of equal length but of opposite 
 direction or sign. We indicate the direction of a line by 
 the order of naming its extremities. For example : 
 
 A B AB = -BA, 
 
 c _ < D or AB + BA = 0. 
 
 Fig. 24. 
 
 Parallel directed lines may be added by placing the 
 initial point of the second on the terminal point of the 
 first. Their sum is the line defined by the initial point 
 of the first and the terminal point of the second. They 
 may be subtracted by placing their terminal points together. 
 The remainder is the line defined by the initial points of 
 the first and second. 
 
 D E F, G 
 
 Fig. 25. 
 AB + BC = AC. DE + EF + FG = DG. 
 
 AC + CB = AB. DG - FG + FE = DE. 
 
 AC - BC = AB. GF+ FE= GE. 
 
 AB - CB = AC. GE- FE = GF. 
 
 32 
 
UNLIMITED ANGLES 33 
 
 The difference may also be obtained by putting the initial 
 points together. The remainder is defined by the terminal 
 points of the second and the first. 
 
 AC - AB = BC. 
 
 AB - AC = CB. 
 
 By general agreement horizontal lines are positive when 
 they make to the right, negative when they make to the left. 
 Vertical lines are positive when they make upwards, nega- 
 tive when they make downwards. 
 
 Measurement. A directed line possesses two qualities — 
 length and direction. Measurement takes account of both. 
 Its length is the number of times it contains the unit line, 
 and its direction is indicated by its sign. 
 
 20. Angles. We conceive the angle L VM to be generated 
 by the revolution of L V about V till it comes into coinci- 
 dence with VM. This revolution may 
 be performed in two ways : 1st, as 
 indicated by the arrow marked a; 
 2d, as indicated by the arrow marked 
 J3. In a the motion is counter-clock- IG ' 6 * 
 
 wise (opposite to the motion of a clock hand), in f3 the 
 motion is clockwise. Mathematicians have agreed to call 
 the former motion positive. The angle denoted *by a is 
 positive, ft is negative. 
 
 Nomenclature. Capital letters denote points ; small let- 
 ters, lines ; Greek letters, angles. The angle above may be 
 named, indifferently, LVM, Im, a. 
 
 The angle ml is the angle described when m turns in 
 a positive direction to coincidence with I. The angle 
 described by m when it turns negatively into coincidence 
 with I is — Im = — a. 
 
34 PLANE TRIGONOMETRY 
 
 The turning line is the initial line of the angle ; the other 
 bounding line is the terminal line. In naming an angle, 
 the initial line is always put first. 
 
 The angle Im is generated by the revolution of I, in a 
 positive direction, until it comes into coincidence with ra. 
 If I continues to revolve, it will again come into coincidence 
 with m. The angle it has described is still called Im. It 
 differs from the former Im by a whole revolution, 360°. The 
 two angles are congruent. If I continues to revolve, it will 
 pass m repeatedly. The angles described when it passes 
 m are all denoted by Im. They differ from each other by 
 some multiple of 360°. They are congruent angles. While 
 Im denotes any one of these congruent angles, the smallest 
 is always understood. 
 
 The student may get a clearer conception of what an angle 
 is by considering the motion of the minute hand of a clock. 
 In one hour this hand describes an angle of 360°; in an 
 hour and a half, an angle of 540° ; in a half day, an angle 
 of 4320°, and so on. At 12.15, 1.15, 2.15, 3.15, etc., this 
 hand is in the same position. Counting from 12 o'clock, it 
 has described angles of 90°, 450°, 810°, and 1170°. These 
 angles are congruent. It is to be remembered that in the 
 case under consideration all the angles are negative. 
 
 21. Addition and Subtraction of Angles. Angles are added 
 and subtracted in the same way that lines are. 
 
 The sum of two angles is found by placing their vertices 
 together and bringing the initial line of the second into 
 coincidence with the terminal line of the first, preserving 
 the direction of both. The angle determined by the initial 
 line of the first angle and the terminal line of the second 
 angle is their sum. 
 
 Two angles are subtracted by bringing together their 
 terminal lines. The angle determined by the initial lines 
 
UNLIMITED ANGLES 35 
 
 of the first and second angles is their difference. The 
 same result may be obtained by placing their initial lines 
 together. Their difference is defined by the terminals of 
 the second and the first. It is to be noted that the differ- 
 ence defined above is the difference obtained by subtracting 
 the second angle from the first. 
 
 L VM + MVN = L VN 
 
 LVN+NVM = LVM. n^^ N 
 
 L VN - MVN = L VM. ^^n^^ M 
 
 L VN - L VM = MVN ?&£— L 
 
 Fig. 27. 
 LVM-LVN =NVM. 
 
 22. Measurement of Angles. The measure of an angle is 
 the number of times it contains the unit angle, and this 
 measure will be positive or negative, according as the angle 
 is positive or negative. 
 
 Definition. A Perigon is the angle generated by a 
 single, complete revolution of a line about a point in a 
 plane. Two unit angles are in common use : 
 
 The degree, which is 3^ of a perigon. This unit is too 
 familiar to require further comment. 
 
 The radian, which is the angle whose arc is equal to the 
 radius. 
 
 The radian is a definite angle. For the circumference of 
 any circle is 2 7r(7r = 3.1416) times its radius. The angle 
 whose arc is equal in length to the radius is therefore the 
 
 angle whose intercepted arc is - — th of the circumference. 
 
 Z 7T 
 
 Since angles are proportional to their arcs the radian is 
 - — th of the perigon. 
 
 Z IT 
 
 Eadians are denoted by the letter r, e.g., l r , 2 r , 6 r , ir r . 
 Generally, however, this symbol is omitted unless such 
 omission gives rise to ambiguity. 
 
36 PLANE TRIGONOMETRY 
 
 Formulas for changing from degree measure to radian 
 measure, and vice versa, are readily obtained. 
 
 2 7r r = 360°. 
 
 360° 180° 
 
 - ^_ 2tt 
 
 180° 
 To reduce degrees to radians, divide by 
 
 IT 
 
 180° 
 To reduce radians to degrees, multiply by 
 
 IT 
 
 When the angle bears a simple ratio to the perigon, its 
 radian measure is expressed as a multiple of it. 
 
 E.g., 180° -7T, 90° = |, 270°- |tt, 45° = J- 
 
 EXERCISES 
 
 1. Express the following angles in terms of it radians : 
 30°, 45°, 60°, 75°, 90°, 105°, 120°, 135°, 150°, 165°, 210°, 
 225°, 240°, 300°, 330°, 450°, 600°. 
 
 2. Express the following angles in degrees : \ 7r, § 7r, 
 
 t^ \**> tV*j A 7 ^ to^ i 77 "- 
 
 3. Express the following angles in radians : 130°, 36° 4', 
 147° 21', 200°, 340° 36', 38° 35'. 
 
 4. Express the following angles in degrees : l r , 2 r , 5'', 
 l r .4, 3 r .6, 5 r .47, 8M, 10 r , 1M. 
 
 Degree measure is used in all practical applications 
 of trigonometry, while radian measure is used in analyt- 
 ical work. In this book both systems are used indiscrimi- 
 nately. 
 
 23. Quadrants. It is customary to place the angle in 
 such a position that the initial line is horizontal and the 
 vertex of the angle toward the left. 
 
II 
 
 I 
 
 
 
 
 III 
 
 IV 
 
 UNLIMITED ANGLES 37 
 
 The initial line and a line through the vertex perpen- 
 dicular to this divide the perigon into 
 four equal parts called quadrants. 
 These quadrants are numbered I, II, 
 III, IV, as in the accompanying figure. 
 An angle is said to belong to, or to b.e 
 of, the quadrant in which its terminal 
 line lies. Thus, f ig# 2 8. 
 
 angles > 0° and < 90° are of the 1st quadrant. 
 " > 90° « < 180° " lid " 
 
 " > 180° " < 270° " Hid " 
 
 « >270° " <360° " IVth " 
 
 Angles greater than 360° may be said to belong either 
 to the quadrant of their smallest congruent angle, or to the 
 quadrant determined by counting the number of quadrants 
 passed over in the generation of the angle. 
 
 E.g., the angle 800° is congruent to 80°, since 800° = 2x 
 360° + 80°. 
 
 This angle belongs to either the 1st quadrant or to the 
 9th since 800° = 8x90° + 80°. 
 
 EXERCISES 
 
 ' To what quadrant do each of the following angles belong : 
 50°, 150°, 200°, 300°, 400°, 500°, 600°, 700°, 1000°, 2000°, 
 10000°, 100000°, - 40°, - 100°, - 200°, - 300°, - 600°, 
 
 3' ~3~' 3 ^ 7r ' 7 ^ 7r? * 7r ' <' 7r? 5 6 57r? 
 
 24. Ordinate and Abscissa. The position of any point in 
 the plane is uniquely determined as soon as we know its 
 distance and direction from each of the two perpendicular 
 axes XX' and YY'. The distance from XX' (SP in the 
 
38 
 
 PLANE TRIGONOMETRY 
 
 X 
 
 s 
 
 figure) is called the ordinate of the point P. Its distance 
 from YY 1 (OS in the figure) is the abscissa of P. Together 
 they are the coordinates of P. The 
 abscissa is usually denoted by x, and 
 the ordinate by y. We write P = 
 (x, y), where the abscissa is always 
 put first. These coordinates are 
 directed lines, and according to the 
 convention mentioned in § 19 (p. 33) y 
 abscissas which make to the right 
 fig. 29. are positive, to the left, negative ; 
 
 ordinates which make upwards are positive, downwards, 
 negative. The initial extremity of the abscissa is on the 
 y-axis, of the ordinate on the rc-axis. The signs of the 
 coordinates in the several quadrants are therefore : 
 
 p 2 
 
 £, 
 
 Y 
 R n 
 
 
 * 
 
 , s 3 
 
 S 4 
 
 p 3 
 
 s 2 
 
 
 
 $ 
 
 T? 
 
 P A 
 
 4 
 
 Y' 
 
 X 
 
 y 
 
 i 
 
 ii 
 
 in 
 
 IV 
 
 + 
 + 
 
 + 
 
 — 
 
 4- 
 
 EXERCISES 
 
 Draw a pair of axes (preferably on coordinate or cross- 
 section paper) and fix the following points : 
 
 3, 5 ; - 3, 7 ; 3, - 7; - 6, 10 ; 4, - 8 ; - 3, - 5 ; - 1, 2 ; 
 - 2, - 6 ; 3, ; - 2, ; 0, 4 ; 0, - 5 ; 0, 0. 
 
 Note. Since RP — OS and OR = £P, the coordinates of P may 
 be taken as RP and OR instead of OS and SP. 
 
 25. The Trigonometric Functions of any Angle. We are 
 
 now in a position to define the trigonometric functions of 
 any angle. These definitions are more general than those 
 given in § 7 and include them. 
 
T ^4- 7**i /, 
 
 UNLIMITED ANGLES 
 
 39 
 
 Let Im (or XOP) be any angle. Through 0, its vertex, 
 draw YY* perpendicular to OX. Take XX ] and YY' as axes 
 of coordinates. Let P be any 
 point on m, and #, ?/ its coor- 
 dinates. Let OP = r, and let us 
 agree that r shall be positive . 
 when it lies on m and negative 
 when it lies on the backward 
 extension of m. Denote the 
 angle Im or XOP by <f>. 
 
 The functions are defined as follows : 
 
 sine of cf> = sin <j> = 1/ /r. 
 
 cosine of <£ = cos cf> = x/r. 
 tangent of <£ = tan<£ = y j x. 
 cotangent of <£ = cot cj> = x/y. 
 secant of <£ = sec<£ ~ v / x. 
 cosecant of <f> = esc <£ = r/y. 
 
 Note. These definitions do not differ from those in § 7 except in 
 generality. 
 
 These are all the possible ratios of the three lines x, y, 
 and r. 
 
 These ratios are independent of the position of P on m. 
 For if P be taken at any other point, as P\ the signs of x, 
 y, and r are unchanged, while the ratios of the lengths are 
 the same in both cases, since the triangles OSP and OS'P' 
 are similar. If the point be taken at P n on the backward 
 extension of m, the signs of x, y, and r are all changed. 
 The triangles OSP and OS"P" are similar. The ratios of 
 x, y } and r are therefore the same as before in both magni- 
 tude and sign. 
 
40 
 
 PLANE TRIGONOMETRY 
 
 26. These ratios, being independent of the position of P 
 on m } are functions of the angle <£. Their algebraic signs 
 depend upon the quadrant to which <j> belongs. 
 
 Draw an angle of each quadrant and verify the following 
 table, taking r positive. 
 
 
 X 
 
 y 
 
 sin 
 
 cos 
 
 tan 
 
 cot 
 
 sec 
 
 CSC 
 
 I 
 
 + 
 
 + 
 
 + 
 
 + 
 
 + 
 
 + 
 
 + 
 
 + 
 
 II 
 
 — 
 
 + 
 
 + 
 
 — 
 
 — 
 
 — 
 
 — 
 
 + 
 
 III 
 
 — 
 
 — 
 
 — 
 
 — 
 
 + 
 
 + 
 
 — 
 
 — 
 
 IV 
 
 + 
 
 — 
 
 — 
 
 ■f 
 
 — 
 
 — 
 
 + 
 
 — 
 
 In quadrant I all functions are positive. 
 
 " II " negative except sin, esc. 
 
 " III " " tan, cot. 
 
 " IV " " cos, sec. 
 
 EXERCISES 
 
 1. Write down the signs of the several functions of the 
 following angles : 
 
 40°, 100°, 160°, 200°, 250°, 300°, 340°, -40°, -80% 
 - 130°, - 190°, - 240°, - 300°. 
 
 2. In Fig. 31 the lines CC and BB ! are drawn equally 
 Y inclined to XX', forming the angles 
 
 ^yC <£u <ta <t>3, <£ 4 - If we take 
 
 OP 1 = OI\ = OP s = OP 4 , 
 
 the coordinates of the points P lr 
 P 2 , P 3 , and P 4 will be equal in mag- 
 nitude but not in sign. We shall 
 have 
 
 sin <£ x = sin <f> 2 = — sin <£ 3 = — sin <£ 4 . 
 Find the corresponding relations between the other 
 functions. 
 
UNLIMITED ANGLES 41 
 
 3. Show by Fig. 31 that there are always two angles less 
 than a perigon which have the same sine : 1st, when the 
 sine is positive ; 2d, when it is negative. Show the same 
 thing for each of the other functions. 
 
 4. Construct the following angles : (See § 9.) 
 sin -1 §; sec -1 3; cos -1 — i; tan -1 — 2; 
 cot -1 — 1; sec -1 — 2; sin -1 — i ; csc -1 J; 
 tan -1 2; cos -1 §; sin -1 §; esc -1 3. 
 
 Remember that in each case there are two solutions. 
 
 5. Prove the following equations by means of a diagram : 
 sin 60° = sin 120°; tan 225° = tan 45° ; 
 
 cos 30° = - cos 150° ; cos 45° == sin 135° ; 
 
 cos 120° = - sin 30° ; tan 150° = - tan 30° ; 
 
 sec 40° = - sec 140° ; cot 130° = - cot 50° ; 
 
 sin 210° = - sin 30° ; tan 135° = - tan 45°. 
 
 6. What angle has the same sine as 35°, 130°, 190°, 
 350°, 47°, - 40°, - 140°, - 230°, - 340°, cj> ? What angle 
 has the same cosine as each of the preceding angles ? the 
 same tangent ? the same cotangent ? the same secant ? the 
 same cosecant ? 
 
 7. Draw a diagram and find the functions of 120°. 
 (See §13.) 
 
 8. Find the functions of each of the following angles : 
 135°, 150°, 210°, 225°, 240°, 300°, 315°, 330°. -^ 
 
 27. Fundamental Relation of the Trigonometric Functions. 
 
 The relations [1] to [8], which were proved in § 11 for 
 acute angles, can be readily shown to hold for all angles. 
 The proof is left for the student. For convenience of 
 reference they are repeated here. 
 
42 
 
 PLANE TRIGONOMETRY 
 
 sin <£ • esc <j> = 1. 
 
 [i] 
 
 cos <£ • sec cf> = 1. 
 
 [2] 
 
 tan<£- cot 4> = 1. 
 
 [3] 
 
 , , sin <£ 
 tan <b = • 
 
 COS cf> 
 
 [4] 
 
 cos <£ 
 
 COt d> — . - 
 
 ■ sm cj> 
 
 [5] 
 
 sm 2 cf> + cos 2 <£ = 1. 
 
 . [6] 
 
 1 + tan 2 <£ == sec 2 <£. 
 
 m 
 
 1 + cot 2 cj> = CSC 2 cj>. 
 
 [8] 
 
 The following scheme may assist in remembering the 
 first three of these formulas : 
 
 sin </> 
 
 cos <f> 
 
 tan <f> — , 
 cot <f> ' 
 
 sec <£ 
 
 esc <£ 
 
 = 1. 
 
 EXERCISES 
 
 By means of the relations [1] to [8] verify the following 
 equations : 
 
 1. sin <£ = tan <f> cos <£. 4. cos <£ =Vl — sin 2 <£. 
 tan sin <j> 
 
 2. sin <j> = 
 
 sec <£ 
 
 5. tan cj> 
 
 vT^ sin 2 <£ 
 
 3. sin <£ = Vl — cos 2 $. 6. tan <£ ="Vsec 2 <£ — 1. 
 
 7. cot <f> = 
 
 Vl - sin 2 <fr 
 sin <£ 
 
 cos <£ 
 
 V 1 — cos 2 </> tan cj> 
 
 1 
 Vsec 2 <£ — 1 
 
 = Vcsc 2 <£ — 1. 
 
UNLIMITED ANGLES 
 
 43 
 
 8. 
 
 9. 
 
 10. 
 
 11. 
 12. 
 
 13. 
 
 14. 
 15. 
 16. 
 17. 
 18. 
 19 
 20. 
 
 Express each of the functions in terms of the sine. 
 cos 2 <£ — sin 2 <£ = 1 — 2 sin 2 <£ = 2 cos 2 <£ — 1. \ 
 sec 2 <£ + csc 2 <£ == sec 2 <£ csc 2 </>. ^, 
 
 sin <£ _ 1 qp cos <£ 
 1 ± cos<£ sin (j> 
 
 cos <£ _ 1 ± sin </> 
 1 qz sin <f> cos 4> 
 
 sec <£ ± 1 _ tan <£ 
 
 tan <£ sec cf> zp 1 
 
 tan <£ + cot <£ = sec <£ esc <£. 
 
 sin $ = i', find all the other functions analytically, 
 cos <£=—$; " " « " 
 
 tan<£ = §; « « " " 
 
 sec <£ = f ; " " " " 
 
 sin <£ + cos <£ == 1.2 ; find sin cf>. *-^ 
 tan 2 <£ — sin 2 <£ = tan 2 </> sin 2 <£. 
 
 2|. The functions of 0°, 90°, 180°, 
 270°, 360°. 
 
 Let P be a point on the terminal line h 
 
 sin 0° = - = 0, 
 
 Z 
 
 of <£ at a distance r from the origin. p 3 
 
 When <j> = 0, P coincides with the 
 point P 1? and its coordinates are x = r, Pj- 
 
 y = 0. Fig. 32. 
 
 cos 0° = - = 1, 
 
 tan 0° = - = 0, 
 r 
 
 cot 0° = - = oo, 
 
 sec 0° = - = 1, 
 r 
 
 esc 0° = - = oo. 
 
44 PLANE TRIGONOMETRY 
 
 When <£ = 90°, P coincides with P 2 , and its coordinates 
 
 ace x = 0, y =■ r. 
 
 sin 90° = - = 1, cos 90° = - = 0, tan 90° = £ = oo, 
 r r 
 
 cot 90° = - = 0, sec 90° ==J = oo, esc 90° = - = 1. 
 r . r 
 
 When <£ = 180°, P coincides with P 3 , and its coordinates 
 are x = — r, y = 0. 
 
 sin 180° = -- 0, cos 180° = — = - 1, 
 
 r r 
 
 tan 180° == — = 0, cot 180° = ^— = oo, 
 
 sec 180° = — - - 1, esc 180° = £ = oo. 
 
 — r 
 
 When <\> = 270°, P coincides with P 4 , and its coordinates 
 are x = 0, y = — r. 
 
 sin 270° = — =- 1, cos 270° - - = 0, 
 
 r r 
 
 tan 270° = ^ - oo, • cot 270° = — = 0, 
 
 — r 
 
 sec 270° = £ = qo, esc 270° = — = - 1. 
 
 — r 
 
 When <£ = 360°, P coincides with P lf and the functions 
 of 360° are identical with those of 0°. 
 
 It is customary to prefix a double sign to the zero and 
 infinity values of the functions, the upper sign being that 
 of the function in the preceding quadrant, the lower that 
 of the function in the following quadrant. 
 
 The results obtained are tabulated in the first table on 
 the opposite page. 
 
UNLIMITED ANGLES 
 
 (3 
 
 OS 
 cos L OP = — = length of OS, ' .' OP = unit of length. 
 
 SP LT 
 
 cot LOP = -^- = i 7 — = " NH,'.'ON = " 
 
 SP ON 
 
 OP OT 
 sec LOP = — = — = « OT,\'OL= « 
 
 OS OL 
 
 csc LOP = = = " OH'.'ON = « 
 
 SP ON 
 
 If we agree that secants and cosecants shall be positive 
 when measured on the terminal line and negative when 
 measured on the backward extension of this line, it will 
 be found on examination that these lines represent the 
 functions in sign as well as in magnitude. For example, 
 L T, the tangent, is positive in quadrants I and III, negative 
 in quadrants II and IV. 
 
 30. The March of the Functions. We will now study the 
 variation or march of each of the several functions as the 
 angle increases from 0° to 360°. As the angle increases 
 the point P travels in the positive direction around the 
 ' circumference of the circle. As P passes through the 1st, 
 2d, 3d, and 4th quadrants : 
 
 The sine, SP, increases from to 1, decreases to 0, 
 decreases to — 1, increases to 0. 
 
 The cosine, OS, decreases from 1 to 0, decreases from 
 to — 1, increases to 0, increases to 1. 
 
 The tangent, LT, increases from to go, changes sign 
 and increases from -co to 0, increases to oo, changes sign 
 and increases from — oo to 0. 
 
48 
 
 PLANE TRIGONOMETRY 
 
 The cotangent", NH, decreases from oo to 0, decreases to 
 — co, changes sign and decreases from oo to 0, decreases 
 to — oo and changes sign. 
 
 The secant, OT, increases from 1 to oo, changes sign and 
 increases from — oo to — 1, decreases from — 1 to — oo, 
 changes sign and decreases from oo to 1. 
 
 The cosecant, OH, decreases from oo to 1, increases from 
 1 to oo, changes sign and increases from — oo to — 1, 
 increases from — 1 to — oo and changes sign. 
 
 These results are tabulated below : 
 
 sin 
 
 0° 
 
 1st Quad. 
 
 90° 
 
 2tl Quad. 
 
 180° 
 
 3d Quad. 
 
 270° 
 
 4th Quad. 
 
 0° 
 
 =f° 
 
 inc. 
 
 1 
 
 dec. 
 
 ±0 
 
 dec. 
 
 -1 
 
 inc. 
 
 *o 
 
 cos 
 
 1 
 
 dec. 
 
 ±0 
 
 dec. 
 
 -1 
 
 inc. 
 
 *o 
 
 inc. 
 
 1 
 
 tan 
 
 T° 
 
 inc. 
 
 ±oo 
 
 inc. 
 
 =F° 
 
 inc. 
 
 ±oo 
 
 inc. 
 
 T° 
 
 cot 
 
 If 00 
 
 dec. 
 
 ±0 
 
 dec. 
 
 zpoo 
 
 dec. 
 
 ±Q 
 
 dec. 
 
 qpoo 
 
 sec 
 
 1 
 
 inc. 
 
 ±oc 
 
 inc. 
 
 -1 
 
 dec. 
 
 ^00 
 
 dec. 
 
 1 
 
 esc 
 
 =F°° 
 
 dec. 
 
 1 
 
 inc. 
 
 ±oo 
 
 inc. 
 
 -1 
 
 dec 
 
 ipoo 
 
 31. Graphic Representation of the Functions. The nature 
 of the variations which we have just been studying may be 
 exhibited by the following constructions. 
 
 Divide the circumference of the unit circle into any num- 
 ber of equal parts. In the figure the points of division are 
 marked 0, 1, 2, 3 ... 12. Lay off the same number of equal 
 parts on a horizontal line, and number the points of division 
 in the same way. Make the divisions of the line approxi- 
 mately equal to the divisions of the circumference. 
 
 At the points 0, 1, 2, 3 on the line erect perpendiculars 
 equal in sign and length to the sine (SP) of the correspond- 
 ing point on the circle. Join the ends of these perpendicu- 
 lars by a continuous line. The resulting curve is the curve 
 
UNLIMITED ANGLES 
 
 49 
 
 of sines. As P moves along the circle, SP changes continu- 
 ously, i.e. , it changes from one value to another by passing 
 through all intermediate values. If now we conceive S 1 as 
 
 3 
 
 i< 
 
 
 
 
 
 f 
 
 
 
 
 
 
 t 
 
 ► 
 
 
 * 
 
 
 
 
 
 \ / 
 
 
 9<r 
 
 
 \180° 
 
 
 270° 
 
 
 300 y 
 
 
 
 L 2 
 
 i 
 
 > o\ 7 
 
 8 
 
 9 
 
 10 
 
 11 
 
 /13 
 
 
 S 
 
 
 
 
 
 
 V 
 
 ^ 
 
 
 
 M 
 
 Fig. 34. 
 
 moving along LM, keeping pace with P. while S'P' is equal 
 to S'P, the point P' will trace the curve of sines. Our con- 
 struction is an attempt to realize this conception. 
 
 32. If the angle increases beyond 360°, i.e., if P makes 
 a second revolution, the values of SP would repeat them- 
 selves in the same order. If we plot these values, we shall 
 have the curve between L and N repeated beyond 2V, and 
 this curve will be repeated as many times as P makes 
 revolutions. The sine curve will take this form. (Fig. 35.) 
 
 The student should construct the curve of cosines, the 
 curve of tangents, and the curve of secants in a similar 
 manner. To find the tangents and secants, the construc- 
 tion of the preceding section should be used. The sum or 
 the difference of two functions may be plotted. To plot 
 
50 
 
 PLANE TRIGONOMETRY 
 
 sin cj> -f cos <£, erect at 0, 1, 2, 3, etc., on MN (Fig. 33), 
 perpendiculars equal to SP + OS at the several points 0, 1, 
 2, 3 on the fundamental circle in that figure. The result- 
 ing curve will represent sin <£ + cos <£. 
 
 ( 
 
 ) 
 
 T 
 
 2 
 
 7T 
 
 37T 
 
 4 
 
 IT 
 
 5 
 
 7T 
 
 6 
 
 7T 
 
 
 
 / ^\ 
 
 
 
 
 
 
 
 
 
 
 
 X 
 
 
 
 
 
 
 
 
 
 
 
 
 + \ 
 
 ( 
 
 )° li 
 
 10° 
 
 360° 
 
 540° 
 
 720° 
 
 900° 
 
 1080 
 
 Fig. 35. 
 
 Periodic Functions. Functions which repeat themselves 
 as the variable or argument increases are called periodic 
 functions. The period is the amount of change in the. 
 variable which produces the repetition in the values of 
 the function. The sine, as is evident from Fig. 35, is a 
 periodic function Avith a period of 360°, or 2 it. The tan- 
 gent has 7r for its period. 
 
 EXERCISES 
 
 Plot the following functions and determine their periods : 
 
 1. sin <£ — cos <£. 4. sin(90 o -h <£). 
 
 2. tan<£ — sin <f>. 5. sin(— <£). 
 
 3. sec <f> — tan<£. 6. cos(— <£). 
 
 7. cos <£ and sec <£ on the same axes. 
 
CHAPTEK IV 
 REDUCTION FORMULAS 
 
 33. Negative Angles. The object of this chapter is to 
 obtain a set of formulas which will enable us to express 
 any function of an angle greater than 
 90° as a function of an angle less 
 than 90°. 
 
 Let AOC be a negative angle and 
 AOC' an equal positive angle. Lay off 
 OP = OP' and draw PP'. SP' = — SP. 
 Let the coordinates of P be x, y, of 
 
 p' = x\ y. 
 
 Now x = x', y = - 
 
 -y'. 
 
 
 
 sin (— d>) = - = 
 
 
 • sin <£. 
 
 
 X 
 
 COS(— d>) = - = 
 J r 
 
 x' _ 
 r 
 
 cos<£. 
 
 
 tan (—(b) = - = 
 
 v J X 
 
 £ 
 
 x' 
 
 - tan cj>. 
 
 
 X 
 
 COt ( — d>) = - = 
 
 y 
 
 x' 
 
 - cot <£. 
 
 
 sec(— d>) = - = 
 
 v X 
 
 r 
 x' 
 
 sec </>. 
 
 
 V 
 
 csc(— d>)= - = 
 
 y 
 
 r 
 
 -y'~ 
 
 - CSC <f>. 
 
 [9] 
 
 A little reflection will show that this proof is independ- 
 ent of the magnitude of <£ and is therefore general. Its 
 
 51 
 
52 
 
 PLANE TRIGONOMETRY 
 
 results may be summed up by saying that in passing from 
 — <£ to <£ the functions do not change name but do change 
 sign except the cosine and secant. 
 
 34. Let AOB = cf> and A OC = 90° + ,<£. Lay off OP' = 
 OP. The two triangles OPS and OP'S' are congruent 
 
 (equal). Let coordinates of P 
 be x, y\ of P' f x'y'. Neglecting 
 algebraic signs, we have 
 x = y', x' = y. 
 
 No matter what the magnitude 
 of cj>, it is obvious that we shall 
 always have this relation between 
 the coordinates of P and P'. 
 
 Fig. 37. «-./..,. -, ^ 
 
 By definition we have, neglect- 
 ing algebraic signs, 
 
 ^ 
 
 
 
 Y 
 
 
 \P 
 
 
 A / 
 
 
 
 
 S 
 
 <x'v 
 
 
 
 
 y 
 
 
 
 
 I 
 
 p' 
 
 Y l 
 
 sin (90° + $) = ~ = ~ = cos <£. 
 cos (90° + <£) = - = ^ = sin <£. 
 
 tan (90° + <£) = ^ = -, = cot <£. 
 x y 
 
 cot(90° + 4,) = ^ : 
 
 tan <f>. 
 
 v v 
 
 sec (90° + <£>)= — = - = esc <f>. 
 v J x' y 
 
 esc (90° + <£) = -, = - = sec <f>. 
 y x 
 
 By studying this table (p. 53) of the signs of the functions 
 in the several quadrants, it appears that sin (90° -f- <£) and 
 cos <£ always have the same algebraic sign. For if (90° + <£) 
 
REDUCTION FORMULAS 53 
 
 falls in quadrant IV, <£ falls in quadrant III, n i 
 
 and sin (90° 4- <j>) and cos <£ are both neg- sin + 
 
 ative. If (90° + <£) falls in III, <f> falls in II, cos - 
 
 and both are negative, etc. We conclude : tan ' 
 
 SID + 
 
 cos + 
 tan + 
 cot + 
 
 the cosine of an angle in the preceding quad- cgc ■ ! csc , 
 rant have the same algebraic sign. 
 
 The sine of an angle in any quadrant and 
 
 sin — 
 cos + 
 tan — 
 cot — 
 sec + 
 
 CSC — 
 
 sin — 
 Cos (90 + <£) and sin <j> have different signs. cos ._ 
 
 For the sign of the cosine in any quadrant tan + 
 
 in the table is different from the sign of the cot + 
 
 sine in the preceding quadrant. sec ~~ 
 
 By employing the same method of reason- ~" 
 ing we can show that 
 
 tan (90° 4- </>) and cot <£ have different signs 
 
 cot (90° 4- <t>) " ■ tan <£ " " " 
 
 sec (90° 4- <£) " csc <j> " " " 
 
 csc (90° 4- </>) " sec <t> " the same " 
 
 The preceding formulas (p. 52), written with the proper 
 signs, are: sin(90 o + ( ^ )= cos ^ 
 
 ^ cos (90° 4- 4>) = — sin ^. 
 
 tan (90° 4- <£) = - cot <£. [10] 
 
 cot (90° +<£) = - tan <£. 
 
 sec (90° 4- <£) = - csc </>. 
 
 csc (90° 4- </>) = sec <£. 
 Since 180° 4- <£ = 90° 4- 90° 4- <k 
 
 sin (180° 4- 4>) = cos (90° 4- <£) = - sin <£. 
 
 cos (180° 4- <f>) = - sin (90° 4- <t>) = - cos <£. 
 
 tan (180° 4- <f>) = - cot (90° + <A) = tan<£. [11] 
 
 cot (180° 4- <£) = - tan (90° 4- <£) = cot </>. 
 
 sec (180° + <£) = — csc (90° 4- <£) = - sec <£. 
 
 csc (180° + <#>)= sec (90° 4- </>) - - csc <£. 
 
54 PLANE TRIGONOMETRY 
 
 Functions of 270° + <£ are found by putting 270° + £ 
 = 180° + 90° + <£• The results are tabulated below. The 
 student is advised to verify these results by drawing dia- 
 
 grams. 
 
 
 
 A 
 
 
 
 
 
 90° -0 
 
 90° + 
 
 180° - 
 
 180° + 
 
 270° - 
 
 270° + 
 
 360° - 
 
 sin 
 
 COS <£ 
 
 COS </> 
 
 sin <f> 
 
 — sin <j> 
 
 — cos <£ 
 
 — cos_<£ 
 
 — sin <f> 
 
 cos 
 
 sin <£ 
 
 — sin cj> 
 
 — cos <f> 
 
 — COS <f> 
 
 — sin <£ 
 
 sin </> 
 
 cos <£ 
 
 tan 
 
 cot <j> 
 
 — cot <£ 
 
 — tan</> 
 
 tan <j> 
 
 cot <£ 
 
 — cot cj> 
 
 — tan<£ 
 
 cot 
 
 tan<£ 
 
 — tan </> 
 
 — cot <f> 
 
 cot <£ 
 
 tan<£ 
 
 — tan<£ 
 
 — cot <j) 
 
 sec 
 
 CSC <£ 
 
 /sec <j> 
 
 — sec cf> 
 
 — sec <f> 
 
 — CSC <j) 
 
 CSC <£ 
 
 sec <j> 
 
 CSC 
 
 sec <£ 
 
 CSC <j> 
 
 — esc <£ — sec <j> 
 
 — sec <f> 
 
 — CSC <f> 
 
 This table includes, beside the cases we have already 
 discussed, the functions of 90° - <j>, 180° - <£, 270° - <£, 
 and 360° — <£. These are reduced as follows : 
 
 sin (90°-<£)=sin [90°+(-<^)]= cos(-^)= cos<£,by[9] 
 sin(180 o -^)=sin[180°+(-<#))]=-sin(-^)= sin <£, " 
 sin(270 o -^)=sin[270°+(-^)] = --cos(-^) = -cos <£, « 
 sin(360°-<£)=sin (-^)=-sin <£. 
 
 The other functions of these angles are derived in a 
 similar manner. 
 
 35. If we inspect the table carefully, we find that it can 
 be summed up in the two rules that follow : 
 
 1°. If 90° or 270° is involved, the function changes name 
 (from sine to cosine, from tangent to cotangent, from secant 
 to cosecant, and vice versa), while if 180° or 360° is involved 
 the function does not change name. 
 
 The second rule has to do with the algebraic sign. 
 
 When we write 
 
 cos (90° + <£) = — sin <£, 
 tan (180° -£)==— tan <£, 
 
REDUCTION FORMULAS 55 
 
 iDoth terms must have the same sign. If <£ is less than 90°, 
 sin <£ is positive and cos (90° -4- <£) is negative. The equality- 
 is secured by putting the minus sign before sin <f>. Since 
 these formulas are general, the signs are the same, no matter 
 what the value of cf>. Our rule is then : 
 
 2°. Assume that <£ is less than 90° and make the signs 
 of both terms alike. ^ 
 
 36. Applications. Any angle greater than 90° can be 
 expressed in two of the following forms : 
 90°+<fc 180°-<£, 180°+ <£, 270° - cf>, 270°+4>, 360° -<fc 
 where <f> is less than 90°. 
 
 E.g., 200° = 180° + 20°, or 270° - 70°. 
 
 300° = 270° + 30°, or 360° - 60°. 
 135° = 90° + 45°, or 180° - 45°. 
 
 The functions of 200° are : 
 
 sin 200° = sin (180° + 20°) = - sin 20°. 
 cos 2.00° = cos (180° + 20°) = - cos 20°. 
 tan 200° = tan (180° + 20°) = tan 20°. 
 cot 200° = cot (180° + 20°) = cot 20°. ' 
 sec 200° = sec (180° + 20°) = - sec 20°. 
 esc 200° = esc (180° + 20°) = - esc 20°. 
 
 They may also be written : 
 
 sin 200° = sin (270° - 70°) = - cos 70°. 
 cos 200° = cos (270° - 70°) = - sin 70°. 
 tan 200° = tan (270° - 70°) = cot 70°. 
 cot 200° = cot (270° - 70°) = tan 70°. 
 sec 200° - sec (270° - 70°) = - esc 70°. 
 esc 200° - esc (270° - 70°) = - sec 70°. 
 
56 PLANE TRIGONOMETRY 
 
 EXERCISES 
 
 1. Express the following functions as functions of angles 
 less than 90°: tan 130°, sin 160°, cos 100°, cot 215°, sec260°, 
 esc 280°, sin 310°, cos 310°. 
 
 • 2. Express each of the preceding functions as the func- 
 tion of an angle less than 45°. 
 
 3. Express each of the following functions as the func- 
 tion of an angle less than 45° = — • 
 
 . 2 7T 3 7T 5lT 47T , 5 77 11 7T 
 
 sm— ? tan——? cos— — > sec -=-> cot — ^- > esc . • 
 3 4 6 3 3 6 
 
 4. By using formulas [9] express the following functions 
 as functions of positive angles less than 90° : 
 
 sin (-160°), cos (-30°), tan (-300°), sec (-140°), 
 cot (-240°), esc (-100°), sin (-300°). 
 
 5. The angle — <£ is obviously congruent to 360° — <j>, 
 and their functions are identical. Reduce the functions in 
 problem 4 by making use of this identity.. 
 
 6. Express the following functions as functions of angles 
 less than 45° : 
 
 cos 117° 17', sin 143° 21' 16", tan 317° 29' 31", 
 cot 90° 46' 12", sec (- 135° 14' 11"), cos (- 71° 23'). 
 
CHAPTER V 
 THE ADDITION FORMULA 
 
 37. Projection. The projection of & point on a line is the 
 foot of the perpendicular from the point to the line. 
 
 The projection of a line-segment on a given line in the 
 same plane is the portion of the second line bounded by 
 the projections of the ends of the first line. 
 
 Fig. 38. 
 
 The projection of AB is CD in I, II, and III, and AD 
 in IV. 
 
 B D % 
 
 / 
 
 /N 
 
 , Z£ 
 
 >> 
 
 E 
 
 
 D 
 
 ^^ 
 
 V 
 
 B 
 
 a{- 
 
 . r 
 
 F 
 
 
 
 a( 
 
 /J r 
 
 V 
 
 / 
 
 
 A' 
 
 B' 
 
 0' D f 
 
 E' 
 
 A 1 
 
 D' 
 
 a' 
 
 B' 
 
 
 
 Fig. 39. 
 
 
 
 
 
 Fig. 40. 
 
 
 
 The projection of a broken line is the sum of the projec- 
 tions of its parts. 
 
 57 
 
58 
 
 PLANE TRIGONOMETRY 
 
 The projection of ABODE (Fig. 39) is A'B' -f B'C + CD' 
 4- D'E'=A'E'; and the projection of ABCD (Fig. 40) is 
 A'B' + B'C + CD' = A 'D'. (See § 19.) 
 
 It is obvious that the projection of a broken line is equal 
 to the projection of the straight line connecting the ends 
 of the broken line. It is to be noted that here we take 
 the direction of the lines into account. The projection of 
 ABCD (Fig. 40) is equal to the projection of AD, and is 
 the negative of the projection of DA. 
 
 38.( The projection of a line-segment on any line in its 
 plane is equal, in length and direction, to the length of the 
 segment multiplied by the cosine of the angle which the 
 segment makes with the line.) In the figure the line-seg- 
 ment is AC, the line of projection is LM, and the angle, 
 measured according to § 20, is </>. 
 
 The projection of AC is A'C = AB. 
 
 Now cos <f> = - 
 
 AB 
 AC 
 
 \ AB = A'C = AC cos<£. 
 
 The projection is positive when </> is an angle of the 1st 
 or 4th quadrant ; it is negative when <f> is an angle of the 
 2d or 3d quadrant. 
 
THE ADDITION FORMULA 
 
 59 
 
 39. Projection on Coordinate Axes. The proj ections of A C on 
 XX' and YY' may be called the ^-projection of AC and the 
 ^/-projection of AC. Let <j> be 
 the angle which A C makes with 
 XX'. Draw OB parallel to AC. 
 
 <£ = XOD, .'. YOB = cf>- 90°. 
 
 If <£ is less than 90°, as in the 
 case of A'C, the angle YOB' is 
 negative and equal to 90° — </>. 
 
 But - (90° -<£) = <£- 90°. 
 
 In any case the angle which 
 AC makes with YY' is 90° less than the angle it makes 
 with XX'. 
 
 cc-projection of AC = AC cos <£. 
 y-projection of AC = AC cos(<£ — 90°) . 
 
 = AC cos(90°-4>) by [9] 
 
 — AC sin <£. 
 
 40. The Addition Formulas. These formulas enable us to 
 express the functions of the sum or the difference of two 
 angles in terms of the functions of the constituent angles. 
 Without examining the matter, the student might make the 
 mistake of writing : 
 
 sin (<f> -f- 0) = sin <£ + sin 0. 
 tan (<£ + 0) = tan cf> + tan 0, etc. 
 
 In the accompanying figure the 
 points F and C, on the terminal lines 
 of <f> and 0, respectively, are taken 
 on the circumference of the unit 
 circle. We have, therefore, in line 
 
 Fig. 43. 
 
 representatives (see § 29) : 
 
60 
 
 PLANE TRIGONOMETRY 
 
 sin <j> = AF, sin 6 = EC, sin (<f> + 0)= BC. 
 tsii<t> = LT, tnnO=FD, tan (<£ + 0) = Z Q. 
 
 It is evident that AF + EC> BC, 
 or sin <j> -f- sin > sin (<£ + 0), 
 
 and LT + FD< LQ, 
 
 or tan <£ + tan < tan (<£ + 0). 
 
 Since the formulas fail in this particular case, they are 
 obviously untrue. 
 
 41. The Sine and Cosine of <|> + 8. 
 
 vA T 
 
 Y 
 
 
 <2\\ 
 
 
 /M 
 
 
 /X 
 
 A 
 
 
 
 ii 
 
 N 
 
 Fig. 44. 
 
 Let LOM= <j> and MON = 0, then LON = <f> + 0. 
 
 In I, <£ + (9 < 90°; in II, <£ + (9 > 90°; in both, <£ < 90°, 
 < 90°. 
 
 Through P, any point in OM, draw TtPQ perpendicular 
 to OM. 
 
 Angle LRQ — 90 + <t>, being the exterior angle of the 
 triangle OPtf. 
 
 Since OQ is a line connecting the extremities of the 
 broken line OPQ, we have, by § 37 : 
 
 ^/-projection of OQ 
 
 = ^-projection of OP -f- ?/-proj ection of PQ, [12] 
 
 x-proj ection of OQ 
 
 = x-proj ection of OP -f cc-proj ection of PQ. [13] 
 
U -f 4 ) c e^~ A^ X - - — * 
 
 v O y THE ADDITION FORMULA ^ 61 
 
 Applying § 39, these equations become: 
 
 OQ sin (<£ + 0) = OP sin <j> + PQ sin (90° -f <£), [14] 
 6>Q cos (<£ + 0) = OP cos <t> + PQ, cos (90° + <£). [15] 
 
 But OP = OQ cos 0, PQ = OQ sin (9. 
 
 sin (90° + <£) = cos <£, cos (90° + <£) = - sin <£. 
 
 Substituting these values in [14] and [15], we have 
 
 OQ sin (4> + 0) = OQ sin <£ cos + OQ cos <£ sin (9, [16] 
 
 OQ cos (<£ -f- 0) = OQ cos <£ cos - OQ sin <£ sin 0. [17] 
 
 .*. sin (<j> + 0) = sin <£ cos + cos <j> sin 0, [18] 
 
 cos (<£ -f 0) = cos <£ cos — sin <j> sin 0, [19] 
 
 fir 
 
 where <£ and are both less than 90° ( — 
 
 42. To establish the truth of these formulas where <f> and 
 are unlimited we proceed as follows : 
 
 Let <t> = 90° + P, where < 90°. 
 sin (cj> 4- 0) = sin (90° 4-0 4-0)= cos (0 4- 0), [20] 
 cos (<£ 4- 0) = cos (90° 4-04-0) = - sin (0 4- 0). [21] 
 
 Since and are each less than 90°, 
 sin (<£ 4- 0) = cos (04-0)= cos cos - sin sin 0, [22] 
 cos (<£ 4- 0) = - sin (04-0) = - sin cos - cos sin 0. [23] 
 
 But sin = sin (<£ - 90°) = - sin (90° -<£)=- cos <£, 
 cos = cos (<£ - 90°) = cos (90° - <£) = sin <£. 
 
 Substituting these values in [22] and [23], 
 
 sin (<£ + 0) = sin <f> cos + cos <£ sin 0, [18] 
 
 cos (<£ 4- 0) = cos $ cos — sin <£ sin 0. [19] 
 
 Here <£ is an angle of the 2d quadrant, an angle of the 
 1st quadrant. By a repetition of this process we can show 
 
62 PLANE TRIGONOMETRY 
 
 that [18] and [19] hold when <£ is of the 3d quadrant, etc. 
 Treating similarly, we prove these formulas true for all 
 positive values of <j> and 0. 
 
 43. They also hold when one or both the angles are 
 negative. 
 
 Let <f> = p - 90° where p < 90°. 
 sin (<£ + 0) = sin (fi - 90° + 0) 
 
 = - sin (90° - f3 — 0) = - cos (p + 0), [24] 
 
 cos (<£ + 0) = cos (p - 90° + 0) 
 
 = cos (90° - p - 0) = sin (£4-0). [_2^ 
 
 Since, p and are positive, 
 sin (<£ + 0) = - cos (p + (9) = - cos cos (9 + sin £ sin (9, [26] 
 cos (<£ + 0) = sin (0 + 0) = sin cos 4- cos sin 0. [27] 
 But sin p = sin (<£ + 90°) = + cos <£, 
 
 cos p = cos (<£ + 90°) = - sin </>. 
 Substituting these values in [26] and [27], 
 
 sin (<£ -|- 0) = sin </> cos 4- cos <£ sin 0, [18] 
 
 cos (<£ 4- 0) = cos <£ cos — sin <£ sin 0. [19] 
 
 A similar process of reasoning would show that these for- 
 mulas remain unchanged when both <f> and are negative. 
 They are true for all positive and negative values of </> and 0. 
 
 These formulas are so important that they should be 
 carefully memorized. They may be translated into words 
 as follows : 
 
 I. The sine of the sum of two angles is equal to the sine 
 of the first into the cosine of the second, plus the cosine of 
 the first into the sine of the second. 
 
 II. The cosine of the sum of two angles is equal to the 
 product of their cosines minus the product of their sines. 
 
THE ADDITION FORMULA 
 
 63 
 
 44. The sine and the cosine of <£ — 0. 
 Putting - for in [18] and [19], we have 
 
 sin (<£ — 0) — sin <£ cos (— 0) + cos <£ sin (— 0), [28] 
 cos ((f) — 0)= cos <f> cos (— 0) — sin <j> sin (— 0). [29] 
 
 But sin (— <f) = — sin <£, cos (— <£) = cos <$>. 
 
 Substituting these values in [28] and [29], we have 
 
 sin (<f> — 0) = sin <f> cos — cos <£ sin 0, [30] 
 
 cos (<£ — 0)= cos <£ cos + sin <£ sin 0. [31] 
 
 Formulas [18] and [30], and [19] and [31], may be com- 
 bined as follows : 
 
 sin (<f> ±0)= sin <£ cos ± cos <£ sin 0, [32] 
 
 cos (4>± 6)= cos <£ cos 9 qp sin <£ sin 0. [33] 
 
 It should be noted that in [32] the double sign in the 
 second member is like the double sign in the first member, 
 while in [33] it is unlike. 
 
 9. 
 
 45. Formulas [18] and [19] are so important that other 
 geometrical proofs are added. 
 
 
 
 
 M 
 
 zt 
 
 X E 
 
 (P 
 
 
 J^\ 
 
 
 
 
 Let 
 then 
 
 BAG 
 
 I 
 
 Fig. 45. 
 
 LOM= <j> and MON = MON' = 0, 
 
 LON = <£ + and LON 1 = <f> - Q. 
 
64 PLANE TRIGONOMETRY 
 
 Through P, any point in OM, draw QPR perpendicular to 
 OM. Draw PA, QB, and RC perpendicular to OL. Draw 
 PD and RE parallel to OL. 
 
 Z.DQP =AEPR = <l> } since their sides are perpendicular 
 to LO and OM. 
 
 AP DP ER OA QD EP 
 
 sm </> — = = ? cos d> = = = 
 
 " OP QP PR * OP QP PR 
 
 A PQ PR A OP OP 
 
 sin v = — = 9 cos 6 = — = — • 
 
 OQ OR OQ OR 
 
 sin(* + «) - — -- _- 
 
 " OQ 0Q~ OP' OQ QP' OQ 
 
 = sin <£ cos + cos <£ sin 0. [18] 
 
 cos (</> + 0) = 
 
 [19] 
 
 cos(<£ — 0) 
 
 OQ OQ 
 
 
 
 OA DP _ OA 
 
 OQ OQ ~ OP 
 
 OP DP 
 OQ QP' 
 
 QP 
 OQ 
 
 cos <£ cos — sin 
 
 <£ sin 0. 
 
 
 RC AP - EP 
 
 
 
 OR ~ OR 
 
 
 
 AP EP _ AP 
 or" OR" OP 
 
 OP EP 
 OR PR 
 
 PR 
 OR 
 
 sin <£ cos $ — cos 
 
 <j> sin 0. 
 
 
 <9C OA + J577? 
 
 
 
 0i2 ~ OR 
 
 
 
 OA ER __ 0-4 
 OR OR" OP 
 
 OP ER 
 OR PR 
 
 PR 
 OR 
 
 [30] 
 
 cos <£ cos + sin <f> sin 0. [31] 
 
THE ADDITION FORMULA 
 
 65 
 
 46. Still another proof of [18] and [19] is given below. 
 
 Construction. Lay off OA = unity. Draw AB and AQ 
 perpendicular to OM and ON, respectively. Draw BC per- 
 pendicular to ON. Draw AD perpendicular to BC. 
 
 Z ABB = ; their sides are perpendicular. 
 Let sin cj>, cos <£ = s u c x , 
 
 sin 0, cos = s 2 , c 2 . 
 
 Fig. 46. 
 
 The lines in the figure evidently have the lengths indi- 
 cated. For example, BC = c x s 2 , etc. 
 
 AQ 
 
 sin (<f> + 0) =-— = AQ = CD = BD + CB = s x c 2 + c t s 2 
 UA 
 
 sin cf> cos -f cos <f> sin 0. 
 
 47. Tangent and Cotangent of <j> + 9 and ((> — 0. 
 
 sin (<fr + 0) _ sin <£ cos -f cos <£ sin $ 
 cos (<£ + 0) cos <£ cos — sin <f> sin 
 
 Cf. [4], [18], [19] 
 
 tan(<£ + 0) = 
 
66 PLANE TRIGONOMETRY 
 
 Dividing both numerator and denominator by cos <f> cos 0, 
 we have 
 
 tan d> -f tan __ . _ 
 
 **« + 9 >= i-L*» r6 [34] 
 
 Other forms for tan (<£ + 0) may be found by dividing 
 by sin <j> sin <£, sin <f> cos 0, cos <j> sin 0, instead of cos <f> cos 0. 
 Find them. Why is [34] preferred ? In like manner we 
 find 
 
 fato/j, m sin(<ft-fl) _ tan<fr-tan0 
 
 tan (*-*)- CO s(</>-0) ~ lTta^tanT [3 ° ] 
 
 Note. [35] might be obtained from [34] by putting — </> for <p. 
 Verify this statement. 
 
 Similarly 
 
 - cos (^ ~^" ^) _ cos ^ cos ^ Z s ^ n ^ E2 ^ 
 ' sin (<£ -f- 0) sin <£ cos + cos </> sin 
 
 Dividing numerator and denominator by sin </> sin 0, we 
 have 
 
 . ', , , *v cot</> cot — 1 
 
 cot(<£ + 0) = - — — — • [36] 
 
 J cot -f- cot </> 
 
 In like manner 
 
 ^ vv ^ cot - cot </> 
 
 Find other forms for [36] and [37] by dividing by cos <£ 
 cos 0, by cos <£ sin 0, by sin <j> cos 0, instead of by sin <j> sin 0. 
 
 EXERCISES 
 
 1. Deduce [36] and [37] from [34] and [35] by using [3]. 
 
 2. Deduce [36] and [37] from [34] and [35] by substi- 
 tuting (90 + <£) for cf> in the latter. 
 
 3. Prove 
 
 sin (<£ + 0) sin (<£ — #) = sin 2 <£ — sin 2 = cos 2 — cos 2 <£. 
 
THE ADDITION FORMULA 67 
 
 4. Prove 
 
 cos (<£ -j- 0) cos (<£ — 0)= cos 2 <£ — sin 2 = cos 2 — sin 2 <£. 
 
 5. Find formulas for sec (<f> + 0), sec (</> — 0), esc (<£ + 0), 
 esc (<£ — 0) in terms of the secants and cosecants of <f> and 0. 
 
 6. Find the sine of 75°. 
 
 sin 75° = sin (45° + 30°) = sin 45° cos 30° + cos 45° sin 30° 
 = iV2.£V3_+iV2.i (p. 18) 
 = i(V6 + V2). 
 
 7. Find the other functions of 75°. 
 
 8. Find all the functions of 15°. (15° = 45° - 30°.) 
 
 9. Find all the functions of 180°. (180° = 90° + 90°.) 
 
 10. Find all the functions of 135°. (135° = 90° + 45°.) 
 
 11. sin <£ = ■£-, sin = -J- ; find the functions of <j> + and 
 <j>-6. 
 
 12. Prove sin _1 # + sin _1 ?/= sin -1 (x Vl — y 2 \-y Vl — x 2 ). 
 
 13. Prove ^_ 
 
 cos -1 # + cos -1 ?/ = cos -1 (xy — V(l — x 2 ) (1 — y 2 )). 
 
 14. Prove tan -1 # + tan -1 ?/ = tan" 
 
 , x 4- v 
 
 -1^/ — fo-n — 1 1£— 
 
 1-xy 
 
 48. Functions of Double Angles. If we put = <fy in for- 
 mulas [24], [25], [34], and [36], we shall have 
 
 sin (<£ + <£)= sin <£ cos <£ -+- cos </> sin <£. 
 
 . ' . sin 2 <£ = 2 sin <£ cos <£. [38] 
 
 cos (<£ + <£) = cos <£ cos <£ — sin <j> sin <£. 
 cos 2 <£ = cos 2 <£ — sin 2 <£ I ^| 
 
 = 2 cos 2 <£ - 1, since sin 2 <£ = 1 - cos 2 <£ II [ [39] 
 = 1 - 2 sin 2 <£, " cos 2 <£ = 1 - sin 2 <£. Ill . 
 
68 PLANE TRIGONOMETRY 
 
 
 EXERCISES 
 
 1. Given the functions of 30°, find those of 60°, of 120°, 
 of 240°. 
 
 2. Given the functions of 45°, find those of 90°, of 180°, 
 of 360°. 
 
 Prove the following : 
 
 2 — sec 2 ^c 
 
 3. = = cos 2 x. 4. tan x -f- cot x = 2 esc 2 x. 
 
 sec 2 # 
 
 5. (sin x ± cos a?) 2 = 1 ± sin 2 #. 
 
 2 tan x . A ^^...^ 1 + tan 2 # 
 
 6. — — — — = sm2a;. Cf. [401. 7. -- L — — — = sec 2 x. 
 1 + tan 2 ic J L J 1 - tan 2 x 
 
 8. Find formulas for sec 2 and esc 2 0. 
 
 9. 2 sin (45° -f </>) sin (45° - <£) = «**$£ C^3L ^ $> 
 
 49. Functions of Half-angles. If in III and II of [39] 
 we substitute \ <f> for <$>, 
 
 cos <£ = 1 — 2 sin 2 -^ <£. 
 cos <£ = 2 cos 2 ^ <£ — 1. 
 
 
 
 • i <*/l-~ cos <£ 
 
 [42] 
 
 , ' , . /l + cos <f> 
 cos£<£ = ±\ r - 
 
 [43] 
 
THE ADDITION FORMULA 69 
 
 By formula [4] 
 
 t , 1 — COS <f> 
 
 * 1 + cos <f> 
 
 _ 1 — cos <£ 
 sin <j> 
 sin <£ 
 
 II 
 
 III 
 
 [44] 
 
 1 + cos <j> 
 
 II is derived from I by multiplying numerator and 
 denominator by 1 — cos <£ ; while 
 
 III is derived from I by using 1 + cos <f> as multiplier. 
 By formula [3] 
 
 4. 1 A ^ l 1 + C0S <t> 
 
 cot^-<ft =\ 
 
 COS <j> 
 
 sin <£ 
 
 1 — cos (j> 
 
 1 -f cos <f> 
 
 sin <£ 
 
 II 
 III 
 
 [45] 
 
 EXERCISES 
 
 1. Given the functions of 60°, find those of 30°, of 15°. 
 
 2. Given the functions of 45°, find those of 22° 30'. 
 
 3. Given sin <£ = £, find the functions of -t^- 
 
 4. cos <£ = a* ; find the functions of ^* 
 Verify the following : 
 
 1 + sec <£ <£ 
 
 5. -— - = 2cos 2 tt- 
 
 sec <£ 2 
 
 6. cos 2 |( 1 + tan| J = 1 + sin <£. 
 
 7. esc as — cot x = tan -• 
 
 8. sin 2 - ( cot - — 1 ] = 1 — sin x. 
 
 / 
 
y 
 
 70 PLANE TRIGONOMETRY 
 
 50. Functions of Three Angles. 
 
 sin(K+^V§=sin(^+f+y) u ^ 
 
 = sin (a + ft) cos y + cos (a + /3) sin y 
 
 = (sin a cos ft + cos a sin ft) cos 
 
 + (cos ^ cos 'ft — sin a sin 2) 
 
 = sin a cos /? cos y + cos er sin 5 cos y 
 
 + cos a cos /3 sin y — sin a sin ft sin y. [46] 
 
 Similarly " 
 
 cos (a + ft -{- y) = cos a cos ft cos y — sin a sin /3 cos y 
 
 — sin a cos ft sin y — cos a sin /3 sin y. [47] 
 
 tan a -h tan /3 + taiiy — tanatantftany _ JA _ 
 
 tan (a + + y) = =— — — -^ -— *— — ^- — £• [48] 
 
 \ ^ ' 7 1— tan /Stan y— tan y tan a— tanatan/2 L J 
 
 Formula [48] may also be obtained by dividing [46] by 
 [47], and then dividing numerator and denominator by cos a 
 cos ft cos y. 
 
 If now in [46], [47], and [48] we put ft = y = a, we 
 shall have 
 
 sin 3 a = 3 cos 2 a sin a — sin 3 a 
 
 = 3 sin a — 4 sin 8 a. [49] 
 
 cos 3 a = cos 3 a — 3 sin 2 a cos a 
 
 = 4 cos 3 a — 3 cos a. [50] 
 
 3 tan a — tan 3 a _ M - 
 
 tan3a= ^g^ • [«t] 
 
 EXERCISES 
 
 1. Put the last six formulas into words. 
 
 2. Find the sine, the cosine, and the tangent of a + ft — y, 
 of a — ft — y. 
 
 3. Find the cotangent of a -\- ft + y in terms of the 
 cotangents of the constituent angles. 
 
THE ADDITION FORMULA 71 
 
 4; Find sin 3 <£ by developing sin (2 <£ -f <£) by formula 
 [18] and simplifying the result. 
 
 5. Find sin 4 <£ by putting 2 <£ f or <£ in [38]. 
 
 6. Deduce formula [47] from [46] by putting 90 + a 
 for a. 
 
 7. Find cos 4 <£. 
 
 8. Find sin 5 <£ and cos 5 <£. 
 
 9. Given the functions of 30°, find those of 90°. / 
 
 51. Conversion Formulas. By adding and subtracting [18] 
 and [30], and [19] and [31], we have 
 
 sin (<f> + 0) + sin (<£ — 0) = 2 sin <£ cos 0. 
 
 sin (<£ -f- 0) — sin (<£ — 0) — 2 cos <£ sin 0. 
 
 cos (<£ -f- 0) + cos (<£ — 0) = 2 cos <£ cos 0. 
 
 cos (<£ + 0) — cos ($ — 0) = — 2 sin <£ sin 0. 
 Putting (<£ + 0) = a, (<£ - 0) = p, 
 whence <£ = ^ (a + /?), = £ (a — ■/?), we have 
 
 sin a + sin f3 = 2 sin £ (a + /?) cos £ (a — /?), [52] 
 sin a — sin /J = 2 cos £ (a + /J) sin -J- (a — /?), [53] 
 cos a + cos /? = 2 cos £ (a + /?) cos -J- (a — /?), [54] 
 cos a — cos /3 = — 2 sin -J- (a -f- /3) sin £ (a — /?). [55^\ 
 
 , These formulas enable us to express the sum or the 
 difference of two sines or two cosines as a product. 
 
 EXERCISES 
 
 1. Express the last four formulas in words. 
 Verify these formulas when 
 
 2. a = 60°, p = 30°. 4. a = 180°, £ = 90°. 
 
 3. a = 90°, /? = 60°. 5. a = 270°, (3 = 180°. 
 
72 PLANE TRIGONOMETRY 
 
 Verify the following identities : 
 
 sin a + sin ft _ tan %(a-\- ft) 
 ..- ' sin a — sin ft tan %(a — f3) 
 
 sin a + sin ft 
 
 7. ■ ^ = tan £ (a 4-/3). 
 
 ^ cos a + cos $ " v ^ y 
 
 cos a + cos 5 . , 
 
 J' cosa-cosg = - COt ^ a + ^ COt ^ a -^ 
 
 9. sin 60° + sin 30° = 2 sin 45° cos 15°. 
 
 10. sin 40° - sin 10° = 2 cos 25° sin 15°. 
 
 11. cos 75° + cos 15° = 2 cos 45° cos 30°. 
 
 12. sin 5 x + sin 3 x = 2 sin 4 cc cos x. 
 
 sin 3 as 4- sin 2 # x 
 
 13. 5 — = cot-- 
 
 cos 2x — cos 3x 2 
 
 14. cos (60° + x) + cos (60° - a) = cos a;. 
 
 15. tan 50° + cot 50° = 2 sec 10°. 
 
 16. sin 2 cos _1 # = 2a; Vl — x 2 . 
 
 17. cos 2 sin -1 # = 1 — 2x 2 . 
 
 18. cos 2 cos -1 # = 2 x 2 — 1. 
 
 19. tan 2 tan _1 # = 
 
 1 — x' 
 
 20. tan _1 ir 4- tan -1 ?/ = tan -1 — • (Take the tangent 
 
 of both members.) 
 
 IT 
 
 21. sin -1 # 4- cos _1 # = — • 
 
 22. sin -1 a 4- cos -1 ?/ = sin -1 (xy 4- V(l — x 2 ) (1 — y 2 )). 
 
 23. tan -1 ^ 4- tan -1 -J- = ->or — • Cf. example 20. 
 . 24. tan" 1 ! = tan- 1 ! 4- tan" 1 !. 
 
 7T 
 
 .'. y = tan -1 ! 4- tan -1 ! 4- tan -1 J. 
 
THE ADDITION FORMULA 73 
 
 52. Trigonometric Equations. Trigonometric equations 
 are generally best solved by expressing all the functions 
 involved in terms of some one function and solving the 
 resulting equation. 
 
 Illustrations. 1. sin cf> = tan cj>. 
 
 sin <£ . - A 1 \ 
 
 sin d> = 7 • . . sin d> 1 = 0. 
 
 cos (j> \ COS $J 
 
 .'. sin <£ = ; and cos <£ = 1. 
 
 <£ = and 7r, <j> — 0. 
 
 The solutions are therefore <f> = 0, 7r. 
 
 2. tan <£ = esc <£. 
 sin <£ 1 
 
 sin 2 <£ a= cos <£, 1 — cos 2 <£ = cos <f>, 
 
 cos <£ sm <j> 
 
 cos 2 <j) + cos <f> = l, 
 
 cos <£ = £(— 1±V5), 
 
 ^cos'^-liVo"); 
 but since £(— 1 -Vo) is numerically greater than unity, 
 this solution is impossible ; and 
 
 ^ = cos- 1 i(-l+V5). 
 
 3. sin + cos = 1. 
 
 sin +Vl - sin 2 (9 = 1. 
 
 l-sin 2 (9=(l-sin0) 2 . 
 (1 - sin Of - (1 - sin 2 0) = 0. 
 (1 - sin 0) (1 - sin $ — • 1 - sin (9) == 0. 
 (1 - sin (9) (- 2 sin 0) = 0. 
 .'. sin = 1 and 0. 
 
 . 6 = |, 0, 7T. 
 
 The solution q= 7r does not satisfy the original equation. 
 
74 PLANE TRIGONOMETRY 
 
 EXERCISES 
 
 Find the values of <f> that satisfy each of the following 
 equations : 
 
 1. cos 2 <£ -f- cos <j> = 0. 6. tan <f> -f cot <j> = 2£. 
 
 2. tan <£ = n cot <j>. 7. cot = 2 cos 0. 
 
 3. sec </> — tan <£ == cos <£. 8. tan <£ -f- cot <£ = m. 
 
 4. sin -f- cos <f> = tan <£. 9. tan cj> + sec <f> = a. 
 
 5. 3sin0 + 4 cos = 5. 
 
 10. If sin + cos (9 = a, then sin 2 = a 2 - 1. 
 
 11. Z cos + m sin = 0, find tan -• 
 
 MISCELLANEOUS EXERCISES 
 
 1. From sin 30° = cos 60° = \ % cos 30° = sin 60° = \ V3, 
 sin 45° = cos 45° = \ V2, find all the functions of 15°, of 
 75°, of 105°. 
 
 2. From sin a = f , sin /? = f , find all the functions of • 
 a + fi and a — f3. 
 
 Prove the following : 
 
 n sin (a + b) + sin (a — 6) 
 
 3. ) — — ^- ) -f = tan<x. 
 
 cos (a + o)-\- cos (& — 6) 
 
 sin (a dob) . 
 
 4. * '- = tan a ± tan #. 
 
 cos a cos 6 
 
 c cos (a qp ft) . , , 7 
 
 5. -t— ^ — =! —t = cot a ± tan 6. 
 sin a cos & 
 
 tan # + tan ?/ _ sin (x + ?/) 
 ~-~-^*~" tan x — tan ?/ sin (x — y) 
 
 7. 
 
 1 — tan x tan ?/ _ cos (x 4- ?/) 
 1 4- tan x tan y cos (x — y) 
 
 _. cot ?/ + cot a? , s , . . 
 
 8. — r 2 ; — = esc (x — y) sin (x + y). 
 
 coty — cotx \ ■ */ \ ■ &/ 
 
&ts 
 
 THE ADDITION FORMULA 
 
 75 
 
 10. 
 11. 
 
 12. 
 13. 
 14. 
 15. 
 
 16. 
 
 17. 
 
 18. 
 19. 
 20. 
 21. 
 22. 
 
 23. 
 24. 
 
 25. 
 26. 
 
 27. 
 
 tan x cot y 4- 1 _ sin (# + y) 
 tan x cot y — 1 sin (as — y) 
 
 sin (as + y) sin (as — y) 
 
 cos 2 as cos 2 y 
 
 tan 2 as — tan 2 ?/. 
 
 tanas. 
 
 tan 2 as — tan 2 ?/ ■ , . , . 
 
 — — - — if- = tan (as + y) tan (a: — y). 
 
 1 — tan 2 cc tan 2 y v VJ K VJ 
 
 V 2 sin (a ± 45°) = sin a ± cos a. 
 
 sin (as + 3/) cos x — cos (as + y) sin as = sin y. 
 
 sin (as — y) cos y 4- cos (as — y) sin ?/ = sin as. 
 
 cos (as + y) cos as + sin (as + y) sin as = cos y. 
 
 tan (a? — y) 4- tan y _ tan (x + y)— tan ?/ 
 
 1 — tan (as — y) tan y 1 + tan (as + y) tan y 
 
 2 sin (45° + a) cos (45° - ft) = cos (a - ft) 4- sin (a 4- ft). 
 (7/*. exercise 12. 
 
 2 sin (45° - a) cos (45° + b) = cos (a - ft) - sin (a 4- ft). 
 
 2 sin (45° 4- a) cos (45° 4- b) = cos (a 4- ft) 4- sin (a - ft). 
 
 2 sin (45° — a) cos (45° — ft) = cos (a 4- ft) — sin (a — ft). 
 
 tan as = £, tan y = i ; find tan (as 4- y) and tan (x — y). 
 
 tan as = 3, tan y = -J-; find tan (as -f ?/) and tan (as — ?/). 
 
 tan as = k, tan y = t} find, cot (as 4- y) and cot (as — y). 
 
 rC 
 
 i, , ^on cot as — 1 ^/l — sin2as 1 — sin2as 
 cot (as 4-45 ) = -— —7 = \/t— — r-3- = — 
 
 nn t. * 4-1 M 4" Sin 2 05 
 
 cot as 4- 1 14- sin 2as 
 
 cot (a; - 45°) = cotx + 1 = tan ^ + 1 . 
 v } \ — cot as tan as — 1 
 
 tan (as ^ 45°) + cot (a ± 45°) = 0. 
 
 cos 2 as 
 
 tanas = 
 
 m 4- 1 
 
 cot y = 2 m + 1 ; 
 
 find tan (as 4- y) and cot (as — y). 
 
76 PLANE TRIGONOMETRY 
 
 28. If x + y + z = 90°, then tans = 1 ~ t ^ xt ^y 
 
 tan x -f- tan 3/ 
 
 29. sin 7 # — sin 5 a; = 2 sin # cos 6 x. 
 
 30. cos 5 # + cos 9 x = 2 cos 7 # cos 2 #. 
 
 31. cos # — cos 2x = 2 sin | x sin £ x. 
 
 sin 2 03 — sin 03 
 
 32. — = cot I x. 
 
 cos ^ — cos 2x J 
 
 m sin 3 x — sin 2x , 5x 
 
 33. — -=*COt — 
 
 cos 2 03 — cos 3 03 2 
 
 34. 
 
 sin a? + sin y _ cos # + cos y 
 cos x — cos y sin 2/ — sin x 
 
 35. cos ( - — x j — cos ( — + cc j = sin x. 
 
 36. cos f — + x j -f- cos f — — x ) = cos 03.V2. 
 
 Express each of the following prodncts as the sum or 
 difference of two trigonometric functions : 
 
 37. 2 sin x cos y. 40. 2 sin 3 03 cos 5 x. 
 
 38. 2 cos x cos 2/. 41. 2 cos (x + 3/) cos (x — y). 
 
 39. 2 sin 2 a; cos 3 y. 42. 2 cos f x cos -J- as. 
 
 43. 2 sin 50° cos 10°. 
 
 aa O 7T . 7T 
 
 44. 2 cos — sm— • 
 
 4 12 
 
 Simplify : 
 
 45. 2 cos 3 x cos 03 — 2 sin 4 x sin 2 #. 
 cos x — cos 5 as 
 
 46. 
 
 47. 
 48. 
 
 sin x -f- sin 5 03 
 
 sin 3 x — sin 03 sin 3 x — sin x 
 cos 3 03 + cos 03 cos 3 x — cos 03 
 (sin 4 03 — sin 2 03) (cos x — cos 3 x) 
 (cos 4 03 + cos 2 03) (sin 03 + sin 3 x) 
 
49 
 
 THE ADDITION FOKMULA 77 
 
 Verify : 
 
 cos x + cos 3 x __ cos 2 x 
 cos 3 x + cos 5 a; cos 4 a; 
 
 # + ?/ x — y 2 sin y 
 
 50. tan n J — tan —7-^ = — - 
 
 2 2 cos a? + cos y 
 
 51. 2 sin 2 x cos a; + 2 cos 4 a; sin cc = sin 5x + sin a;. 
 
 Lob X _ 
 
 52. — r - = sec2a;. 
 
 csc 2 a; — 2 
 
 53. cos 2 a; (1 — tan 2 x) — cos 2 a;. 
 
 54. cot a; — tan x = 2 cot 2 a;. 
 cos 2 x 1 — tan a; 
 
 55. 
 
 1 4- sin 2 a; 1 + tan x 
 
 56. cos 2 a; + cos 2 ( ■« +* ) + cos 2 (7r-f-#)-bcos 2 ( —+x 1=2. 
 
 57. sin x + sin 3 a; -f sin 5 a? + sin 7 a; = 4 sin 4 a; cos 2 a; cos a;. 
 
 58. cos x 4- cos (120 + x) 4- cos (120° — a?) =0. 
 
 sin 3 x cos 3 a; 
 
 59. — ; = 2. 
 
 sin x 
 
 
 COS X 
 
 cos 3 a; 
 
 4- 
 
 sin 3 a? 
 
 sin x 
 
 COS X 
 
 sin 5 a; 
 
 
 cos 5 a; 
 
 60. — : 1 = 2 cot 2 x. 
 
 61. — = 4 cos 2 x. 
 
 sm a; cos x 
 
 62. tan a; = I, tan ?/ = T 2 T ; ■ find tan (2 x 4- y). 
 
 63. sin (y -\- z — x)-\- sin (z 4- x — y) 4- sin (a; + y — z) — 
 
 sin (a; + y 4- 2) = 4 sin a; sin y sin z. 
 
 64. cos (?/ 4- z — x) 4- cos (z + x — y)-\- cos (a; 4- y — z) 4- 
 
 cos (as + ?/ + «) = 4 cos a; cos ?/ cos z. 
 
 65. sin a; sin (y — z) 4- sini/sin(^ — a;) -{-sin 2 sin (a; — y) = 0» 
 
 66. cos a; s in (?/ — 2) + cos 2/ sin (2 — x) 4- cos 2 sin (x —y) = 0. 
 
78 PLANE TRIGONOMETRY 
 
 €7. cos a? cos (^/ — z)— sin 7/ sin (^ — x) — cos£COs(# — y) = 0. 
 68. sin x cos (y — z) + cosy sin (2 — #) — sin^cos(x — y) = 0. 
 
 ' X * - y* 4 j N 
 
 69. cot - * (aj — y) — cot~ 1 (a? + y) = cot - * ^ 
 
 70. 
 
 tan -1 — tan -1 = tan -1 ~-^- 
 
 a — 1 a la 1 
 
 71. 
 
 • 1 i ^Wl-a 2 
 
 2 sin x a = tan -1 — — - — 
 
 1 — 2 a 2 
 
 72. tan- 1 *, + 2 tan- 1 ft = tan" 1 " (1 7 P + 2 * . 
 
 1 — ft 2 — 2 aft 
 
 ,1 . , a 4-Va 2 - 1 
 
 73. tan _1 a 4- cos -1 -= sin -1 . 
 
 a a Va 2 4 1 
 
 74. cos 4 # — sin 4 x = cos 2 #. 
 
 75. sin 2 (x + y) — sin 2 (x — y)= sin 2 ^ sin 2 y. 
 
 ,76. (sin x — sin y) 2 + (cos x — cos y) 2 = 4 sin 2 — r-^- 
 
 1 4- sin .r — cos x 
 
 77. — ; = tant^. 
 
 1 4- sm x 4 cos x 
 
 tan x ± tan 7 . , 
 
 78. ; - = ± tan x tan y. 
 
 cot x ± cot y 
 
 79. (sin <£ 4 sin 0) (sin <£ — sin 0) = sin (<£ + 0) sin(<£ — 0). 
 
 80. (Vl 4 sin a — Vl - sin a) 2 = 4 sin 2 |a. 
 
 81. (Vl 4 sina+Vl — sin a ) 2 = 4 cos 2 J a. 
 
 2 
 
 82. sec 2 a + tan 2 a + 1 = 
 
 83. sin A = 
 , 84. cot (x + y) = 
 
 1 — tan a 
 sin (30° + A)- sin (30° - A) 
 V3 
 1 1 
 
 tan x 4- tan y cot # 4 cot y 
 
CHAPTER VI 
 THE TRIANGLE 
 
 53. The object of this chapter is to study the relations 
 between the sides of a triangle and the trigonometric func- 
 tions of its angles. Other properties of the triangle are 
 also considered. 
 
 NOTATION 
 
 A, B, C = the vertices of the triangle. 
 a,b,c = the sides opposite A, B, C, respectively, 
 a, ft, y = the interior angles at A, B, C, respectively. 
 s = % (a + h + c), the semi-perimeter. 
 B, r = the radii of the circumscribed and inscribed 
 circles. 
 r a> r hi r c s the radii of the escribed circles opposite A, B, C, 
 respectively. 
 jp a , p b , p c = the altitudes from A, B, C to a, b, c. 
 K = area of the triangle. 
 
 MEMORANDA 
 
 a + p + y = 180° = tt. 
 
 a,p,y = 7r-(p + y),7r-(y + a),7r-(a + f3). 
 
 sin a, sin /?, sin y = sin (/J + y), sin (y + a), sin (a + /?). 
 
 cos a, cos /?, cos y = — cos (/? + y), — cos (y + a), — cos (a + /?). 
 
 79 
 
80 
 
 PLANE TRIGONOMETRY 
 
 tan a = — tan (/? + y), etc. 
 cot a = — cot (f3 -+- y), etc. 
 sin £ a = cos %(fi + y), etc. 
 cos i a = + sin %({$ + y), etc. 
 tan -J- a = + cot £ (/? + y), etc. 
 cot i a = 4- tan £ (/? -f y), etc. 
 2 JT = ap a = bp h = cp c = 2 rs. 
 b -f- c — a = 2 (s — a), 
 c + a -b = 2(s —b). 
 a 4- b — c = 2 (s — c). 
 
 54. The Law of Sines. 
 
 In either figure, let AE = q, then, § 19, EB = AB — AE. 
 
 Fig. 47. 
 
 sin a = =^7 sin/3 = — • 
 b a 
 
 sin a _ff c £c_« 
 
 sin/3 
 This may be written 
 
 Similarly 
 
 sin a sin /? 
 
 b c 
 
 sin/3 siny 
 
 a b c 
 
 sin a sin/3 siny 
 
 [56] 
 
THE TRIANGLE 81 
 
 This is the law of sines. It may be stated in words as 
 follows : The ratio of the side of any triangle to the sine 
 of its opposite angle is constant. 
 
 Let ns denote this constant by M. The formula becomes 
 
 JL. = -J- = -J^ = M. [57] 
 
 sin a sm ft sin y L J 
 
 55. The Law of Tangents. From [57], 
 
 aba sin a 
 
 t or 
 
 sin a sin ft b sin ft 
 
 by composition and division 
 a + b _ sin a + sin ft 
 a — b sin a — sin ft 
 
 2sm|(tt+£)cosi(tt-£) [52]and[53] 
 2 cos £ (a + ft) sin £ (a — ft) J L J L J 
 
 ^ tanj(a+ft) 
 tan i (a — ft) 
 § a + 5 = tan fr (a + ft) = cotjy' ^ 
 ''a- b tallica- ft) tan£(a-ft)' L J 
 
 If b is greater than a we can avoid negative signs by 
 writing b — a and /3 — a instead of a — b and a — ft. 
 
 Similar formulas may be derived involving b and c, and 
 c and a. 
 
 This is the law of tangents. In words it is : The ratio 
 of the sum of any two sides of a triangle to their differ- 
 ence is equal to the ratio of the tangent of one-half the 
 sum of the opposite angles to the tangent of one-half 
 their difference. 
 
 56. The Law of Cosines. From Fig. 47, 
 
 a 2 = ( c — 9.Y + V?, P c * = b 2 — q 2 . 
 .-. a 2 = (c - q) 2 -\-b 2 -q 2 = b 2 -{-C 2 -2 cq. 
 
82 PLANE TRIGONOMETRY 
 
 But q is the projection of b and, therefore, 
 
 q = b cos a. 
 Substituting in the preceding equation, 
 a 2 = b 2 + c 2 — 2 be cos a. ' 
 
 Similarly b 2 = c 2 -f a 2 — 2 ca cos /3. i- 
 
 [59] 
 
 c 2 = a 2 + b 2 - 2 ab cosy. J 
 
 Solving for cos a, etc., v 
 
 6 2 + c 2 -a 2 • 
 cos a = — ■ 
 
 2&c 
 
 fa 
 
 [60] 
 
 This is the Zaw 0/ cosines. 
 
 
 57. Functions of the Half -angles in Terms of the Sides. 
 Substituting % a for <£ in [39] III, 
 
 cos a = 1 — 2 sin 2 \ a. 
 2 sin 2 \ a == 1 — cos a 
 
 by [60] 
 
 1- 
 
 b 2 + c 2 - 
 
 a 2 
 
 
 2 be 
 
 
 a 2 . 
 
 -{b 2 
 
 ■Ye 2 - 
 
 -2 be) 
 
 
 
 2 be 
 
 
 a 2 - 
 
 -(b 
 
 -e) 2 
 
 
 2 be 
 (a + b — c) (a — b -f c) 
 2be 
 
 -s A — J_i 1. (See Memoranda.) 
 
 Similarly 
 
 2 be 
 
 171 
 
 * . sin ^ a 
 
 sin£/2 
 sin | y 
 
 =V ( - 
 
 be 
 
 [61] 
 
 -# 
 -# 
 
 - c) (s — a) 
 
 ca 
 -a)(s- b) 
 
 ab 
 
 [61] 
 [61] 
 
THE TRIANGLE 
 
 83 
 
 Substituting \ a f or <£ in [39] II, 
 
 cos a = 2 cos 2 \ a — 1. 
 2 cos 2 |a = l + cos a 
 
 + 2fo 
 
 _ 2 5c + 6 2 + c 2 - a 1 
 
 " 26c 
 
 _ g + c )2 - ^2 
 
 2fo 
 
 "" 2 6c 
 
 = b -• (See Memoranda.) 
 
 2 be v ' 
 
 /S (5 — ») 
 
 Is (s - ^ 
 008 i") 
 
 ca 
 
 *V-^- 
 
 [62] 
 
 Dividing [61] by [62], 
 
 smja 
 
 = tan \ a 
 cot -J- a 
 
 \ (s 
 
 -b)(s-c) 
 
 cos^-a 
 
 — \ 
 
 s(s — a) 
 
 
 =v. 
 
 s(s — a) 
 
 -b)(s- c y] 
 
 l[63] 
 
 Since % a, £ /?, -J- y < 90°, the functions of these angles 
 are positive and the radicals in [61], [62], [63] are also 
 positive. 
 
84 PLANE TRIGONOMETRY 
 
 EXERCISES 
 
 Verify the following relations : 
 
 s _ cos \ /3 cos ^ y s — a _ cos ^ a sin -J- y 
 
 a sin \a b cos ^ /3 
 
 s — a ^ sin j- ft sin j- y 
 a sin ^- a 
 
 4. cos a + cos /? cos y = sin ft sin y. 
 
 5. a cos ft + b cos a == c. 
 c cos a + & cos y = b. 
 c cos ft + & cos y = a. 
 
 „ , a 2 - 6 2 
 
 6. a cos B — b cos a = 
 
 c 
 
 7. a cos ft cos y + £ cos y cos a -|- c cos a cos ft 
 
 = \ \a cos a + 6 cos ft -f- c cos y] 
 
 = a sin ft sin y = b sin y sin a = c sin a sin ft. 
 
 8. a sin (ft — y) -f- b sin (y — a) + c sin (a — ft) = 0. 
 
 58. Circumscribed and Inscribed Circles. 
 
 Circumscribe the circle about the triangle ABC. Draw 
 CD, a diameter. Angle a = angle D. (Fig. 48.) 
 
 sina = sinD = ^ = ^. 
 
 .'. 2 i? = -A- = -A- = -A- = M. [64] 
 sin a sm ft sm y 
 
 Inscribe the circle in the triangle ABC. By geometry 
 
 &i = C* h = a 2 > c i = K ( F ig- 49.) 
 .-. s = %(a + b + c)= a x + b l + c x = a x + a 2 +.«, = a + c^ 
 .*.^4JF = (Si = 5 — a. 
 
THE TRIANGLE 
 
 85 
 
 Now 
 
 Similarly 
 
 Fig. 48. 
 Combining [63] and [65^\, 
 
 c, F c 2 £\ 
 
 Fig. 49. 
 
 
 ~ ft) (« ~ <0 
 
 s(s — a) 
 
 .. ,. J(* -<*)(*-•*)(*- c ) 
 
 [66] 
 
 o 
 
 From [65] we have 
 r = (s — a) tan £ a = (5 — b) tan k P = (s — c) tan J y. [67] 
 
 Let be escribed to the 
 triangle ABC opposite A. 
 We have by geometry 
 
 BD == J3F, CD = C% 
 0£ = BF + C#. 
 .'. 2s = AE + AF, 
 s = AE. 
 
 tania = — -~ = — 
 2 ^4£ 5 
 
 [68] 
 
 Similarly 
 
 tan -J /?, tan |- y = 
 
86 PLANE TRIGONOMETRY 
 
 Combining [63] and [68], 
 
 r a = J ( s ~ b )( s ~ c ) 
 s * s (s — a) 
 
 .:r a ,r h> r c = ^( S - b ^ S - C ) ,et, [69] 
 s a 
 
 Comparing [65] and [68], 
 
 rs = r a (s — a) = r b (s — b) = r c (s — c). [70] 
 
 EXERCISES 
 
 Verify the following identities : ^ 
 
 , , o ar a + hr b + CT c 
 
 1- r a +r b + r c — 3r = - 
 
 11 1,1 
 
 2. - = - + - + -• 
 r r a r b r c 
 
 r — v 
 
 3. tan i a = — 
 
 2 a 
 
 4. OO x = (r a — r) esc % a. 
 
 5. OO x = a sec ^ a = b sec -J- ft = c sec \ y. 
 
 v 
 
 6. tan-J-a tan-J-jStan^y = -• 
 
 7. sina + sin/2 + sin y = — • 
 
 R 
 
 / 5£. Area of the Triangle. 
 
 / • The area of a triangle may be expressed in different 
 ways, depending upon the parts known. We have 
 geometry (Tig. 51) 
 
 2K=ap a = bp b = cp c , 
 p a = c sin (3 = b sin y. 
 .\ 2 K = ac sin f3 = ab sin y = be sin a. 
 
THE TRIANGLE 
 
 87 
 
 
 a smy 
 
 c = — : ' - 
 
 sin a 
 
 From [57], 
 
 Substituting this value in [72], 
 
 2K 
 
 _ a 2 sin ft sin y _ b 2 sin y sin a 
 
 sin a 
 
 c 2 sin a sin ft 
 
 siny 
 
 sin ft 
 
 Fig. 51. 
 
 We have from geometry 
 
 K = rs. 
 Combining [74] and [66~], 
 
 K = -\/s(s-a)(s- b) (s - c). 
 
 [73] 
 
 [74] 
 [75] 
 
 EXERCISES 
 
 Find the areas of the following triangles : 
 
 1. a » 13, b = 10, c = 17. 
 
 2. a = 143, ft = 100, y = 74°16'. 
 
 3. b = 200, a = 47° 24', y = 63° 25'. 
 
 4. The sides of a triangle are 175, 120, 215; find its 
 area and the radii of its inscribed and escribed circles. 
 
 abc 
 
 5. Prove K = 
 
 ±R 
 
 ^A2 Jw^ A? £C~ CL 
 
 UxCuZ^. 
 
 •2. 
 
 *—*£• 
 
 * frVvc 
 
 wQ^« _ 
 
PLANE TRIGONOMETRY 
 
 Verify the following identities : 
 
 6. cos i a cos i /3 cos £ y = -^ . (U se [62].) 
 
 7. cot*acot*£cot£ y = ~ (Use [63].) 
 
 2 
 
 . 8. cot£a + cot£0 + cot£y == -• 
 
CHAPTER VII 
 THE SOLUTION OF THE TRIANGLE 
 
 60. We have learned in geometry that a triangle can be 
 constructed when we are given three of its parts, of which 
 one, at least, is a side. The formulas of the preceding 
 chapter enable us to compute the values of the unknown 
 parts when we know the measures of the given parts. 
 The three given parts may be : 
 I. One side and two angles. 
 II. Two sides and the included angle. 
 
 III. Two sides and the angle opposite one of them. 
 
 IV. Three sides. 
 
 Formulas [57], [58], and [60] are sufficient to solve all 
 four cases. In the computations in Chapter II we used 
 natural functions ; here we propose to use logarithms, and 
 formula [60] is not adapted to logarithmic calculation. 
 In its place we shall use formulas [65^ and [66^, which 
 are derived from it. 
 
 The necessary formulas are : 
 
 = M, [57] 
 
 tan i (a + p), [58] 
 
 [65] 
 where r ^('-»H«-») Eg. [66 ] 
 
 a b 
 
 c 
 
 sin a sin f3 
 
 sin y 
 
 tan % (a — (2) = 
 
 a — b 
 a + b 
 
 tan ia = 
 
 r 
 
 
PLANE TRIGONOMETRY 
 
 ho 
 o 
 
 CD 
 
 ■-a 
 
 CD 
 
 r- *» 
 
 02. 
 
 II 
 
 « e 
 
 ^1^ 9 
 
 00. 
 
 d 
 
 5 =8 
 
 >"i CO 
 
 d d 
 
 CO. 
 + 
 
 T 
 
 o 
 
 O 
 
 Q5. 
 
 + 
 
 + 
 
 00. 
 + 
 
 + 
 
 CO. 
 
 o «« » a 
 
 <m ,_, n ~r I 
 
 
 1 
 
 <2 J> 
 
 c6 
 © 
 
 CD 
 
 I 
 
 oo. 
 
 + 
 
 CO. 
 
 + 
 
 e I 8 
 
 00. 
 
 I 
 g 
 
 oo. 
 
 I 
 
 00. 
 
 + 
 
 00. 
 
 M 3 
 .9 Ls 
 
 *co I "53 
 
 8 
 
 d 
 
 d 
 
 8 
 
 O 
 
 xc2 or 
 
 " » M 
 
 o£ 
 
 + 
 
 S I « 
 
 i e Ls 
 
 o I cc 
 
 § II 
 
 II ^ 
 
 00. 
 
 d 
 
 d 
 
 *? 
 
 ? 
 
 
 + 
 
 1 
 
 <*- 
 
 CO 
 
 00 
 
 d 
 
 
 
 OG 
 
 H" 
 
 H» 
 
 00 
 
 
 a 
 
 d 
 
 cr; 
 
 a 
 
 +2 
 
 +j 
 
 DO 
 
 + I 
 
 o 
 
THE SOLUTION OF THE TRIANGLE 
 
 91 
 
 GO 
 
 O 
 
 o 
 
 CQ 
 
 O 
 
 t— i 
 
 « 
 
 H 
 
 M 
 
 P3 
 
 O 
 O 
 
 3 
 
 1— 1 
 H 
 
 + . U 
 
 -* ^ 111 
 
 * 5 s oa. "3 <**• M 
 ^ + 1 o •§ n 
 
 . + ^ ^ 1 op 
 
 1 X g>^ *• , * 
 
 . S i i I'ssf g 
 
 *° ^ e ^ to ^ & 
 . II >S + o © o 
 
 + 71 ^ e « £ 
 
 S _ II g 5 S « 
 
 1 ^ II o° g> 
 g be 
 
 ■M r— 
 
 be 
 
 > 
 
 N 
 
 r oT l 
 
 bo 
 
 o 
 
 I e S" ^ 
 
 ~JL i i i 
 
 ' be be be o 
 
 ^ no O O O 00 
 
 to be 1 i i .. 
 
 w ~* ' ^ S- ** s* ^ 
 
 ^ + + be be be 
 
 ^ f U 7\ li <3a - 
 
 g 5 a c a 
 
 tZll +J +3 +J 
 
 H« bo bo bo 
 
 £ r2 r2 
 
 be 
 c 
 
 i— i 
 
 + 
 
 s 
 
 tf » +; + ^ 
 
 ^- o e ^ ^ , h*, 
 
 II II II II ^toto 
 
 w be he | | 
 bo o o ' ' 
 
 o ^ ^ tr 
 + 
 
 be 
 o 
 
 M 
 
 1— 1 
 
 2 ^ 
 
 + + 
 
 * w ^ .2 " « be^ 
 
 bO bO a- ^ gJ W) O T 
 
 ^ 1 1 -3 £ + + " S) 
 
 ^ S ^fe^bc^ 1 1 
 
 H I' % + || U bS 
 
 ^^ z 
 
 ii To 
 
 09 
 
 O 
 
 Data 
 Solution 
 
 Check 
 
 CO 
 
 O 
 
 Data 
 Solution 
 
 Check 
 
92 PLANE TRIGONOMETRY 
 
 61. Logarithmic Functions. Tables of logarithmic func- 
 tions are arranged like tables of natural, functions. They 
 consist of the logarithms of the natural functions. When, 
 however, the characteristic is negative, 10 is added. For 
 this reason the characteristics of all sines and cosines, 
 of tangents of angles less than 45°, and of cotangents of 
 angles greater than 45°, are 10 too large. This fact must 
 be kept in mind when computing. A little experience will 
 correct any liability to error from this source. Sines and 
 tangents of very small angles, cosines and cotangents of 
 angles near 90°, cannot be accurately obtained by interpo- 
 lation. Supplementary tables are generally furnished for 
 this purpose. 
 
 62. The actual work of computation in each case will 
 now be illustrated by the solution of specific problems. 
 The first step in the solution of every problem is the 
 careful construction of the figure and the graphic solution 
 by measurement. The results so obtained serve as a rough 
 estimate of what is to be more accurately determined by 
 computation. 
 
 In the following illustrative problems the work is ar- 
 ranged in convenient form, and this form should be fol- 
 lowed by the student. 
 
 , Case I. Two Angles and a Side. 
 
 Given a = 571, a = 57° 21 '.3, (3 = 43° 16 '.8, find the 
 other parts. 
 
 f a = 571. 
 Data J a =57° 21 '.3. 
 
 [£=43°16'.8. Check 
 
 y =79°21'.9. e + b = 1131.41. 
 
 log a = 2.75664. c-b = 201.59. 
 
 log sin a = 9.92532 - 10. i (y + P) = 61° 19'.35. 
 
THE SOLUTION OF THE TRIANGLE 
 
 93 
 
 log M =2.83132. 
 log sin p = 9.83605 - 10. 
 log sin y - 9.99248 - 10. 
 log b = 2.66737. 
 log c = 2.82380. 
 b = 464.91. 
 c = 666.5. 
 
 ±( y -0)=18 o 2'.55. 
 log (c + $)= 3.05362. 
 log (c - b) = 2.30447. 
 
 log quotient = .74915. 
 log tan £ ( y + p) = .26204. 
 log tan | (y - £) = 9.51288 - 10. 
 
 log quotient = .74916. 
 
 1. a =137.43, 
 
 2. a = 437.18, 
 
 3. 5 = 943.49, 
 
 5. = 637.23, 
 
 6. a = 63.72, 
 
 7. 6 = 6.372, 
 
 8. b = .0641, 
 
 9. = .0037, 
 10. a = 4.003, 
 
 EXERCISES 
 
 a =43° 21 '.3, 
 p = 83° 25'. 7, 
 a = 12° 17'.6, 
 P = 102° 35'.3, 
 a = 46° 46', 
 a = l°20', 
 a = 88° 14'.5, 
 a = 36° 17'.1, 
 = 36° 17', 
 a = 36° 17', 
 
 £=65°23'.5. 
 y = 73° 32'.8. 
 y = 121° 07'.2. 
 y = 80° 12 M. 
 p=56° 56'. 
 p=75° 40'. 
 y = 88° 14 '.2. 
 y = 53° 43'.6. 
 y = 72° 34'. 
 P=10S° 5V. 
 
 63. Case II. Two Sides and the Included Angle. 
 Given a = 1371, 6 = 1746, y = 46° 30', find the other 
 parts. 
 
 r a = 1371. 
 Data -I 6 = 1746. 
 (. y = 46° 30'. 
 ft + a = 3117. 
 b - a = 375. 
 £(£ + a) =66° 45'. 
 log (b - a) = 2.57403. 
 
 Check 
 log a = 3.13704 
 log sin a = 9.89116 — 10 
 
 3.24588 
 
94 
 
 PLANE TRIGONOMETRY 
 
 log tan£(0 + a) = 
 
 ■ .36690. 
 
 
 log b = 3.24204 
 
 
 colog (b + a) = 
 
 = 6.50626.- )• 
 
 log 
 
 sin p = 9.99616 - 
 
 10 
 
 log tan -|- (/J — a) = 
 
 : 9.44719 - 10. 
 
 
 3.24588 
 
 
 £(/?-«) = 
 
 : 15° 38'.6. 
 
 
 
 
 a = 
 
 : 51° 6'.4. 
 
 
 log c = 3.10644 
 
 
 P = 
 
 = 82° 23'.6. 
 
 log 
 
 sin y = 9.86056 - 
 
 10 
 
 loga = 
 
 = 3.13704. 
 
 
 3.24588 
 
 
 colog sin a = 
 
 : 0.10884. 
 
 
 
 
 log sin y = 
 
 : 9.86056. 
 
 
 
 
 logc = 
 
 3.10644. 
 
 
 
 
 c = 
 
 1276.7. 
 
 
 
 
 
 EXERCISES 
 
 
 
 1. a = 127, 
 
 b = 145, 
 
 
 y = 24° 37'.2. 
 
 
 2. a =127, 
 
 b = 145, 
 
 
 y = 84° 13'.6. 
 
 
 3. a = 127, 
 
 b = 145, 
 
 
 y = 173° 28'.5. 
 
 
 4. 5 = 231, 
 
 5. a =231, 
 
 c = 31, 
 
 
 a = 74° 15'.2. 
 
 
 b =221, 
 
 
 y = 100° 14'.5. 
 
 
 6. c = 347, 
 
 a = 34, 
 
 
 /3=10°46'.3. 
 
 
 7. 6 = 12.473, 
 
 c = 34.257, 
 
 a = 146° 24 '.1. 
 
 
 8. a = 100, 
 
 b = 200, 
 
 
 y = 100°. 
 
 
 9. a =100, 
 
 b = 200, 
 
 
 y = 10°. 
 
 
 10. The line AB is divided at D into two segments, 
 AD = 200, DB = 100 ; from C each of these segments 
 subtends an angle of 35°. Find the angles CAB and CBA. 
 
 sy 64. Case III. Two Sides and an Angle Opposite One of 
 Them. This case sometimes admits of two solutions. Let 
 the given parts be a, b, a. Construct the angle a. On 
 one side lay off A C = b ; from C as center with radius a, 
 describe an arc, cutting the other side AM at J5 X and B 2 . 
 
THE SOLUTION OF THE TRIANGLE 
 
 95 
 
 The triaogles AB X C and AB 2 C both satisfy the conditions, 
 and both are therefore solutions. Study of the diagram 
 will show that we shall have 
 two solutions when, and only 
 when, 
 
 a<90°, b>a>:p. 
 
 In any particular case the FlG 52 
 
 graphic solution will deter- 
 mine whether there is one or two solutions. 
 
 The angles at B 1 and B 2 are obviously supplementary. 
 In the computation we find sin /3. Now we learned in § 26 
 that there were two angles less than 180° with the same 
 sine, the one the supplement of the other ; when we find 
 /3 from sin /?, we must therefore take not only the value 
 given in the table but also the supplement of this value. 
 If there is but one solution, later steps in the computation 
 will compel the rejection of the second of these values. 
 
 Given, 1. a = 44.243, b = 30.347, a = 34° 23'.2. 
 2. a = 44.243, 5 = 60.347, a = 34° 23'.2. 
 
 
 1. 
 
 
 2. 
 
 Data ^ b 
 
 44.243, 
 
 
 44.243. 
 
 30.347, 
 
 
 60.347. 
 
 [a 
 
 34° 23'.2, 
 
 
 34° 23'.2. 
 
 log a 
 
 1.64585, 
 
 
 1.64585. 
 
 log sin a 
 
 9.75188 - 
 
 10, 
 
 9.75188 - 10. 
 
 log M 
 
 1.89397, 
 
 
 1.89397. * 
 
 log b 
 
 1.48212, 
 
 
 1.78066. 
 
 log sin ft 
 
 9.58815 - 
 
 10, 
 
 9.88669 - 10. 
 
 P 
 
 22° 47'. 5, 
 
 
 50° 23U> 
 
 2'. 
 
 129° 36'.9. 
 
96 
 
 PLANE TRIGONOMETRY 
 
 y 122° 49'.3, 95° 13'. 7, 15°59'.9. 
 
 log sin r 9.92447 - 10, 9.99819 - 10, 9.44030 - 10. 
 
 log c 1.81844, 1.89216, 1.33427. 
 
 c 65.833, 78.012, 21.591. 
 
 
 
 Check 
 
 
 c + b 
 
 96.180, 
 
 138.36, 
 
 81.938. 
 
 c-b 
 
 35.486, 
 
 17.665, 
 
 38.756. 
 
 t(f + P) 
 
 72° 48'. 4, 
 
 72° 48'. 4, 
 
 72° 48'.4 
 
 i(y-» 
 
 50° 00'.9, 
 
 22° 25'. 3, 
 
 56° 48 '.5. 
 
 log(c + ft)* 
 
 1.98308, 
 
 2.14101, 
 
 1.91349. 
 
 log (p - b) 
 
 1.55006, 
 
 1.24712, 
 
 1.58833. 
 
 log quotient 
 
 .43302, 
 
 .89389, 
 
 .32516. 
 
 log tan c 
 
 .50945, 
 
 .50945, 
 
 .50945. 
 
 log tan y 2 P 
 
 .07641, 
 
 9.61555 - 
 
 -10, .18431. 
 
 log quotient 
 
 .43304, 
 
 .89390, 
 
 .32514. 
 
 1. a = 145, 
 
 2. a = 2.37, 
 
 3. ft = 147.3, 
 
 4. a = 32.14, 
 
 5. ft = 13.47, 
 
 6. ft = .149, 
 
 7. a = 1.243; 
 
 8. a = 432.1, 
 
 9. o = .0027, 
 10. a = 124, 
 
 EXERCISES 
 ft =160, 
 
 c =3.14, 
 a = 124.2, 
 
 ft' = 270, 
 e =18.75, 
 c =.137, 
 ft =2.345, 
 ft =321.4, 
 a = .0031, 
 ft =83, 
 
 a = 47° 38'. 
 y = 65° 23'. 
 P = 142° 17'. 
 £ = 75° 48'.3. 
 P = 110° 43'. 
 y = 38° 47'. 
 a = 10° 57 '.5. 
 = 28° 47'. 
 a = 84° 21'.6. 
 p = 68° 43'. 
 
THE SOLUTION OF THE TRIANGLE 97 
 
 11. I = 241, m = 214, fi = 43° 27'. 
 
 12. p =13.17, ? =17.13, Q = 71°31'. 
 
 13. a = 187.5, b = 201.1, a = 67° 47'.4. 
 
 14. a = 5872, ft = 7857, £ = 78° 5'. 
 
 15. a - 1, b = 2, a = 23° 32'. 
 
 16. a = .0003, ft = .0004, a = 50° 5'. 
 
 17. a = 3000, ft = 4000, a = 5° 50'. 
 
 18. a = 1241, ft = 2114, a = 63° 36!. 
 
 19. ^ = 1899, ft =2004, a=73°l'. 
 
 20. ft = 173, a = 74° 12'; find the limits of a for two 
 solutions. 
 
 21. <x = 127, ft = 143 ; find the limits of a for two 
 solutions. 
 
 65. Case IV. Three Sides. Given a = 1573, ft = 2044, 
 c = 2736. 
 
 r a = 1573, colog 5 = 6.49805. 
 
 Data \ ft = 2044, log (5 - a) = 3.20507. 
 
 L c = 2736, log (s - ft) = 3.05404. 
 
 2 5 = 6353, log (s - c) = 2.64395. 
 
 * = 3176.5, log r 2 = 5.40111. X,^ 
 
 5 _ a _ 1603.5, log r = 2.70056. 
 
 s - ft = 1132.5, log tan £ a = 9.49549 - 10. 
 
 5 _ c = 440.5, log tan £ £ = 9.64652 - 10. 
 
 ia = ir22'.7, ^g tan i y = . 05661. 
 
 i fi = 23° 53'.9, 
 
 1- y = 48° 43'. 4, Check 
 
 a = 34° 45'.4, a + £ + y = 180° OO'.O. 
 
 = 47° 47'.8. 
 
 y = 97° 26'.8. 
 
98 PLANE TRIGONOMETRY 
 
 
 
 EXERCISES 
 
 
 - 1. 
 
 a = 51, 
 
 6 = 65, 
 
 c = 60. 
 
 2. 
 
 a s= 51, 
 
 b = 65, 
 
 c = 20. 
 
 3. 
 
 a = 431, 
 
 b = 440, 
 
 c = 25. 
 
 <~ 4 - 
 
 a = 78.43, 
 
 b = 101.67, 
 
 c = 29.82 
 
 5. 
 
 a = 111.1, 
 
 b = 120, 
 
 c = 130. 
 
 A 
 
 a = .003, 
 
 6 = .007, 
 
 c = .011. 
 
 8. 
 
 a = .431, 
 
 b = .34, 
 
 c = .7. 
 
 « == 6, 
 
 b = 6, 
 
 c = 2. 
 
 9. 
 
 a = 6, 
 
 b = 6, 
 
 c = ll. 
 
 10. 
 
 a = 12, 
 
 b = U, 
 
 c-16. 
 
 11. 
 
 a = 4, 
 
 b = 6, 
 
 c = 9. 
 
 12. 
 
 a = 4, 
 
 b = 6, 
 
 (5 =±=8. 
 
 13. 
 
 a = 4, 
 
 b = 6, 
 
 = 11. 
 
 EXERCISES 
 
 1. One side of a triangular lot is 1427 ft. ; the adjacent 
 angles are 48° 15' and 75° 35'; find the perimeter and the 
 area. 
 
 2. Prove that the area of a quadrilateral is one-half the 
 product of its diagonals into the sine of the angle between 
 them. 
 
 3. The diagonals of a parallelogram are 17 ft. and 30 ft., 
 and the angle between them is 64° 27' ; find the sides of 
 the parallelogram and its area. 
 
 4. A balloon is directly over a straight road. From two 
 points 3 mi. apart and on opposite sides its elevation 
 was found to be 30° 28' and 47° 22' ; what was its height ? 
 If the two points of observation had been on the same side 
 of the balloon, what would its height have been ? 
 
THE SOLUTION OF THE TRIANGLE 99 
 
 5. What is the angle between two faces of a regular 
 tetrahedron ? of a regular octahedron ? 
 
 6. Three circles whose radii are 12, 17, and 19 are tan- 
 gent, two and two externally ; find the area of the surface 
 enclosed by them. 
 
 7. From two successive mile posts on a straight and 
 level road the elevation of the top of a hill in line with 
 them is 8° and 10° ; find the distance and height of the hill. 
 
 8. The longer sides of a parallelogram are 18, the 
 shorter sides 10, and one diagonal is 12 ; find the other 
 diagonal and the angles. 
 
 9. The parallel sides of a trapezoid are 34 and 50, the 
 non-parallel sides 20 and 25 ; find the angles and the 
 diagonals. 
 
 10. Two sides of a triangle are 20 and 30, and the 
 median from their intersection is 16 ; find the base and 
 the angles of the triangle. 
 
 11. A field is 500 ft. square ; a post stands 350 ft. from 
 one corner and 400 ft. from an adjacent corner ; what are 
 its distances from the other two corners, 1°, when it is within 
 the field ; 2°, when it is outside ? If the second corner were 
 opposite the first instead of adjacent to it, what would the 
 distances be? 
 
 ^12. Wishing to find the height of a mountain, I measure 
 a line of 600 yds. in the same vertical plane with the top 
 of the mountain. The upper end of this line is 40 ft. 
 higher than the lower end, and the elevation of the moun- 
 tain top at the former is 6° 23', at the latter 3° 23' ; 
 what is the height of the mountain above the lower end 
 of the base line ? If the lower end of the base line were 
 next to the mountain, what would its height be ? 
 
/. 
 
 100 - PLANE TRIGONOMETRY 
 
 13. A straight and level road runs along a seacoast. 
 From two points on this road, 2 mi. apart, the top of 
 a lofty mountain is visible ; what measurements must I 
 make to find its height without leaving the road? 
 
 14. The parallel sides of a trapezoid are 42 and 32, one 
 oblique side is 20, and it makes an angle of 65° with the 
 longer parallel side ; find the other side, the diagonals, and 
 the angles ; find the same parts if the oblique side makes an 
 angle of 65° with the shorter parallel side. 
 
 15. A tower 50 ft. high has a mark 20 ft. from the 
 ground. At what distance from its foot do the two parts 
 of the tower subtend equal angles ? at what distance does 
 the lower part subtend twice the angle that the upper does ? 
 
 16. The altitude of a certain rock is observed to be 47°, 
 and after walking 1000 ft. towards it, up a slope of 22°, the 
 observer finds its altitude to be 77° ; find the height of the 
 rock above the first point of observation. 
 
 17. From two points A and 
 B, 5000 ft. apart, two inac- 
 cessible points P and Q are 
 visible. I find the angles 
 PAB = 107° 37', 
 PBA = 34° 23', 
 QAB = ±3° 46', 
 
 Fig. 53. 
 
 QBA =81° 11'; 
 
 what is the distance from P to Q, 1°, when both are on the 
 same side of AB\ 2°, when they are on opposite sides ? 
 
 18. Two flag-poles are 203 ft. apart. From the middle 
 point of the line joining them the elevation of the taller is 
 double that of the shorter ; but on going 43^- ft. nearer the 
 shorter, their elevations are equal. What is the height of 
 each ? 
 
THE SOLUTION OF THE TRIANGLE 101 
 
 19. From the top of a hill the depressions of the top 
 and bottom of a flagstaff 25 ft. high, standing at the foot 
 of the hill, are 45° 13' and 47° 12', respectively. What is 
 the height of the hill above the foot of the flagstaff: ? 
 
 20. A column on a pedestal 20 ft. high subtends an 
 angle of 30° ; on approaching 20 ft. nearer, it again sub- 
 tends an angle of 30°. What is the height of the column ? 
 
 21. From the middle point of the longest side of the 
 triangle, whose sides are 10, 14, 17, a circle is described 
 with radius 12 ; where will it cut the other sides ? 
 
 / 22. Two towers stand near each other in a plane. Their 
 altitudes, each measured from the base of the other, are 
 46° 6' and 33° 45', respectively, and the distance between 
 their summits is 87 ft. What is the height of each, and 
 what is their distance apart ? 
 
 23. Three circles with radii 16, 7, 5 touch each other 
 externally ; what is the area of the curvilinear triangle 
 so formed ? If the two smaller circles are within the larger, 
 what is the area of the curvilinear triangle ? 
 
 24. The sides of a triangle are 20, 30, 40 ; find the 
 lengths of, 1°, the three altitudes ; 2°, the three medians ; 
 3°, the bisectors of the three interior angles ; 4°, the bisec- 
 tors of the three exterior angles ; 5°, the radii of the cir- 
 cumscribed circle, the inscribed circle, the escribed circles. 
 
 25. Near the foot of a flagstaff, 150 ft. high, are two 
 posts, A 70 ft. north, B 100 ft. east. What is the shortest 
 distance from T, the top of the staff, to the line A B? 
 what angle does this line make with the ground? 
 
 26. Three sides of a convex quadrilateral inscribed in a 
 circle 30 ft. in diameter are I = 14 ft., m = 18 ft., n = 12 ft. ; 
 find the fourth side and the angles when, 1°, I is the middle 
 one of the three given sides ; 2°, when m is the middle one ; 
 3°, when n is the middle one. 
 
102 
 
 PLANE TRIGONOMETRY 
 
 27. Standing on a headland 250 ft. high, I observe a 
 ship. At first it bears N.N.W., and its angle of depression 
 is 16° 8', ten minutes later it bears E. by S. and its depres- 
 sion is 32° 18' ; find what direc- 
 tion the ship is sailing, its speed, 
 and how near its course lies to the 
 headland. 
 
 28. A, B, and C are three buoys; 
 AB = 320 yds., BC = 435 yds., CA 
 = 600 yds. A ship S finds that 
 AB subtends an angle of 8° and BC 
 an angle of 26°. How far is the 
 ship from each of the buoys ? 
 Suggestion. Draw a circle through A, C, and S, cutting 
 SB produced in D. Draw AD and CD. 
 
 Fig. 54. 
 
14 DAY USE 
 
 RETURN TO DeIk FROM WHICH BORROWED 
 
 LOAN DEPT. 
 
 Renewed books are sub je« to immed^erecaU. 
 
 WAR 3 1 W L 
 
 *&ZM- 
 
 _JE3-Z£-\k 
 REC'D LD 
 
 FiB_i6j?^L_ 
 
 &9 
 
 LD 2lA-50m-8.'61 
 ^O17Q5r10>476B 
 
 General Library 
 University of : California 
 
 OCT 1 1936 
 
;i) 
 
 UNIVERSITY OF CALIFORNIA LIBRARY