Plane j rigonometry JE Edward Bright PLANE TRIGONOMETRY BY R. D. BOHANNAN PROFESSOR OF MATHEMATICS IN THE OHIO STATE UNIVERSITY ■oo^c Boston ALLYN AND BACON 1904 -31* COPYRIGHT, 1903, BY R. D. BOHANNAN. NortoooU $rfgg J. S. Cushing & Co. — Berwick & Smith Co. Norwood, Mass., U.S.A. L3*c PREFACE. Some of the features of this book are : (1) Abandonment of the " Academic Triangle." The " academic triangle " is one, certain of whose parts are known exactly and whose remaining parts are calculable free from all errors except such as are due to the " place " of the tables used. In the academic triangle one side may show one figure and another seven figures, and the angles may show any readings the caprice of the maker may suggest. (2) Replacement of the " Academic Triangle" by triangles having variety of form, but always such form as they should have if they had been actually measured in the field: (i) Showing in all measured lines of a diagram the same number of significant figures, thus indicating about the same care in each measurement. (ii) Giving in all calculated lines of a diagram not more significant figures than in its measured lines, since such extra figures are misleading and of no value. (iii) Letting the angles of a diagram show such read- ings as exhibit care in angle- reading in keeping with that indicated in line-measurement. While the student is learning to calculate, he might as well learn at the same time what form intelligent measurement should take and what reliability calculated results have as related to the data. (3) What place tables to use to suit the data and how to cut a large table to suit short data. (4) How checks should check to suit the data. (5) The fundamental principles of calculation with approxi- mate data (Chapter II.). (6) The suggestion of simple laboratory exercises in connec- tion with fundamental things. iii 797953 iv PREFACE. (7) Exercises from related topics, as Physics, Analytical Geome- try, Surveying, etc. (8) Graphs, using rectangular and polar coordinate paper. (9) Geometric treatment of DeMoivre's Theorem. (10) Order of treatment of the trigonometric functions. When many different ideas are presented simultaneously to the mind the result is confusion. When the same idea is presented again and again, with a suitable interval between presentations, its difficulties dis- appear. Most of the difficulties connected with any trigonometric func- tion are those of every other such function. I have ventured, therefore, to present the subjects in the following order : (i) The sine, — its inverse and reciprocal, (ii) The cosine, — its inverse and reciprocal, (iii) The sine family and cosine family in union, (iv) The tangent, — its inverse and reciprocal, (v) All the functions in union, (vi) The quantity ^/ — 1 in trigonometry. I am indebted to Dr. Coddington and Dr. Kuhn of the Ohio State University, and to Dr. J. W. Young of the Northwestern University, for valuable suggestions. Dr. Coddington has tested all the exercises. R. D. BOHANNAN. Columbus, Ohio, May, 1904. CONTENTS. CHAPTER PAGE I. Elementary Discussion of Logarithms — How to use Tables and what Place Tables to use 1 II. Calculation Vices and Devices 32 III. Angles and Angle-units 60 IV. Construction of Angles and of Straight-line Diagrams to Scale, and the Measurement of Angles .... 77 V. The Sine, Anti-sine, Reciprocal Sine (Cosecant), and Co- versed Sine of an Angle 82 VI. The Cosine, Inverse Cosine, Reciprocal Cosine (Secant), and Versed Sine of an Angle 192 VII. The Sine and the Cosine in Union 237 VIII. The Tangent and Reciprocal Tangent (Cotangent) of Angles 297 IX. Sines and Cosines, Tangents and Cotangents . . . 313 X. The Tangent and Cotangent in the Solution of Oblique- angled Triangles 331 XI. General Review on the Solution of Triangles. List of Formulas to memorize . . .... . . 340 XII. The Quantity V^I in Trigonometry 345 CHAPTER I. , ELEMENTARY DISCUSSION OF LOGARITHMS. — HOW TO USE TABLES AND WHAT PLACE TABLES TO USE. § 1. Definition of a Logarithm. The expressions a x = y, (1) log#/ = x, (2) V = loga" 1 ^ (3) indicate the same relation between the three quantities, a, x, y; namely, that x is the logarithm of y.to the base a. The logarithm of any number, y, to the base, a, is the exponent, x, of the power to which a must be raised to produce y. Expression (2) is read, " the logarithm of y to the base a is x" or, as is usual among engineers, "log y, base a, is x." Expression (3) is read, u y is the number whose logarithm to the base a is x" or, more briefly, "y is anti-log x, base a." Among engineers it is customary, both in writing and in speech, to cut the word logarithm to "log." Expression (1) is called the exponential form for a loga- rithm ; (2), (3) are called the logarithmic forms. 2 3 = 8, (1), is equivalent to log 2 8 = 3, (2), and to 8 = log 2 ~ 1 3, (3); and 10 2 = 100, (1), is equivalent to log 10 100 = 2, (2), and to 100 = log 10 _1 2, (3). Read these expressions. EXERCISES. 1. What is the value of a 10 ^? Of e l0 ^? Of 10 lo «io*? Of l0ga(l0g a -l*) ? Of lOga" 1 (}0g a y) ? 2. What is the base when the logarithm of 81 is 4 ? Express this in the three forms, (1), (2), (3). 3. What number has 3 for its logarithm when the base is 6 ? Express this in the three forms, (1), (2), (3). 1 2 PLANE TRIGONOMETRY. [§ 1 4. What is the logarithm of 1000 to the base 10 ? Express this in the three forms, (1), (2), (3). 5. Make up ten examples like Exs. 2, 3, 4, and solve. 6. If 10 is the base, what are the logarithms of 10, 100, 1000, 10000, 10" ? If 0.01 = — = — = 10- 2 , what is the logarithm of 0.01 ? Of 0.1, 100 10 2 » & 0.001, 0.0001, 0.00001 ? Of -A-? 10» 7. Give the usual illustration from algebra from which is drawn the interpretation that a = 1. In accordance with this interpretation, what would you say is the log of 1 for all bases ? 8. Express in the form of an identity the relation of a~ x to a x . How, then, is the logarithm of a number related to that of its reciprocal? 9. If in the expression a x = y,a is positive, what sign has y, both for positive real values of x and for negative real values of x ? Give some numerical illustrations. What would you say, then, as to the possibility of negative numbers having real logs, if the base is positive ? 10. If the base is 10, between what limits will lie the logs of all numbers between 1 and 10 ? Between 10 and 100 ? Between 100 and 1000? How many digits to the left of the decimal point have numbers which lie between 10 and 100 ? Between 1 and 10 ? Between 100 and 1000? Write three four-figured numbers which lie between 1 and 10. Between 10 and 100. Between 100 and 1000. Calling the integral part of a log its characteristic, can you make up, from your answers to the preceding questions, the rule as to the number which is the characteristic as compared with the number of digits to the left of the decimal point in the number whose log is desired ? 11. Express 0.1, 0.01, 0.001, 0.0001, 0.00001, in exponential form similar to (10) -2 = 0.01, where the base is 10 and the exponents negative. From these expressions draw conclusions as to the limits between which lie the logs of all numbers between 1 and 0.1, and between 0.1 and 0.01. Between 0.01 and 0.001. Between 0.001 and 0.0001. Write four num- bers which lie between 1 and 0.1. Between 0.1 and 0.01. Between 0.01 and 0.001. Between 0.001 and 0.0001. What sign has the log of a decimal fraction ? It is customary to make the integral part of such logs negative and the decimal part positive, taking for the integral part the log of the power of 10 next below the given decimal whose log is sought. Can you make out, from your answers to the preceding questions, the rule as to the negative number which is the characteristic as related to the number of zeros to the right of the decimal point before a figure other than zero is reached in the number whose log is sought ? §2] DISCUSSION OF LOGARITHMS. 3 § 2. Working Rules for Logarithms. (a) The logarithm of the product of two or more numbers is the sum of the logarithms of the numbers, or, log (mri) = log m + log n; log (mnp) = log m + log n + log p. Proof. If a x = m, (1), then log a m = :z, (2), and if a v = n, (3), then log a n = y, (4). Multiplying (1) by (3), a x+v = mn, or, log a (mw) = x + y = log a m + log a w, by (2), (4). Similarly for a product of more than two factors. EXERCISES. 1. If n is a positive integer, show that log (a n ) = n • log a. 2. Given log 10 2 = 0.30103 and log 10 3 = 0.47712, find, to the base 10, the logs of the following numbers : 6 ; 4 ; 9 ; 27 ; 8 ; 12 ; 36 ; 2 10 ; 3 8 ; 2 n ; 3 m ; 2 n • 3 m (n, m, integers, positive) . 3. Write down all the prime numbers from 1 to 100, and show that if the logs of these numbers are known, the logs of the other numbers below 100 can be calculated. For what other numbers could the logs be calculated ? 4. If e is the base, show that 1 + log e n = \og e (en). If 10 is the base, show that 1 + log 10 n = log 10 (10 • n). Show also 2 + log e n =5 log e (e 2 • n) ; 3 + log e n = log e (e 3 - ri) ; n + log e ra = log e (e n . n) ; 4 + log 10 n = log 10 ( 10000 • ri) ; n + log 10 n sa log 10 (10" ■ n). (5) The logarithm of an indicated quotient [ — ) is the loga- rithm of the numerator minus the logarithm of the denominator, or log ( — ) = log fn> — log n. 4 PLANE TRIGONOMETRY. [§2 6 Proof. If a x = m, (1), then log a m = x, (2), and if a y = n, (3), then log a n = y, (4). Dividing (1) by (3), *—5 (5), or, log a (^) = x-y = \og a m - \og a n, by (2), (4). EXERCISES. 1. If log 10 2 = 0.30103 and log 10 3 = 0.47712, find to the base 10, the of the following numbers: |j 1.5; 4.5; 6.75; 1|; 2f;i; ^; f£; o 9 27 o TO (-) (n, m, integers, positive or negative). 2. Express the logs of the proper fractions above with negative characteristic and positive decimal part (mantissa) to the log. For example, log f = 1.82331, where 1 alone is negative and .82391 is positive. . Show that log f - j = — log a. 3 4. Show that 1 - log e x = log fl -j 1 - log 10 n = log 10 — j x log e x- 3 = log.(^); log 10 ft - 5 = log 10 (^). 5. Show that log 10 27.34 = log 2734 - 2 ; log ]0 27.34 = log 273.4-1; log 10 3.415 = log 3415 - 3. 6. In general, show that if a and b are two numbers with the same sequence of digits, but with the decimal point differently placed, that a = 10 n • b, and that, consequently, the logs of a and b to the base 10 differ only by the integer n. What is the relation between the decimal parts of the logs of numbers with the same sequence of figures ? 7. From log 10 2, find log 10 5 ; then from log 10 3, find log 10 15. ( I one and the same scheme giving 5, c. In adding, cover the log not to be used (/) Using tables of logarithms. It is recommended that at this point the student be taught how to look up the logarithms of numbers in a four-place table and in some five-place table, both in finding the logarithms corresponding to numbers and in finding the numbers corresponding to logarithms, taking, for the present, only the cases where the logarithms and numbers can be found exactly, or to a close approximation, in the tables, without interpolation. The principles on which interpolation are based can be understood only after further study, as set forth in the next several articles. The student should carry out the operations with the tables and with- out the tables and compare results. How to use the four-place tables of this book is explained in the preface to the tables. EXERCISES. (For a four-place table.) Calculate with logarithms and without logarithms the values of the following expressions and compare results. Give results to not more than three figures and to only the nearest approximation which the 3 4] DISCUSSION OF LOGARITHMS. 13 tables will give without interpolation. In the case of fractions, calculate their values with cologs and without cologs, using negative character- istics and also characteristics positive by the aid of 10's. Find b, c, in Exs. 4, 5, 6. 1. 2x3; 3x4; 4x5; 5x6; 6x7; 7x8; 8x9; 9x10; 10 x 11 ; 11 x 12 ; 12 x 13. 2. 2x3x4; 3x4x5; 4x5x6; 5x6x7; 6x7x8; 7x8x9; 8x9xl0;2x3x9;3x4x9;4x5x9;5x6x8. 3 12 x 13 . 13 x 14 . 16 x 17 . 18 x 19 . 25 x 27 . 35 x 37 . 45 x 47 7 ' 9 ' 11 ' 13 ' 32 ' 43 ' 53 , 45 x 49 x 85 . 56 x 57 x 73 . 96 x 83 x 85 0.21 0.41 0.51 5. 7. 71 x 73 ' 81 x 91 ' 61 x 77 ' 0-31 b " c 243 x 244 . 256 x 295 . 259 x 831 . 0.031 = 0.023 = 0.033 231 ' 331 ' 923 ' 0.016 b c 341 x 441 x 551 . 349 x 837 x 624 . .317 = .231 _ .314 536 x 437 ' 555 x 989 ' .218 b c VMl • #225 #0034 • VO00l4 #316 • #719 #0^015 • #0.00042 ( _95). (-34). (-65 ), ( -32). (-41). (-51). (64 ) 27- (-86) ' (-17). (-29). (-43) 9. Find to three figures : #2; #3; #5; #6; #2; #3; #4; #5; #2; #3; #5; #6. ( 6.6) 8 . (1.02)* . ( 2.51) 6 . (3.03) 8 (2.5) 5 • (3.15) 2 ' (2.47) 8 • (1.78)4* 8 10. J( 3.11)«.(2.71)' . '/( 47)».(-85)« ' ^(2.14) 5 .(3.19) 8 ' * (23) 2 . (71) 4 ' § 4. Systems of Logarithms. While any finite positive number, other than 1, may be made the base, there are only two bases of importance. These are 10 and the number indicated by the letter e, where e = 2.7182818284-... The decimal part of e is unending and non-repeating. The number e is the limit toward which the expression H)' 14 PLANE TRIGONOMETRY. [§ 4 approaches as x becomes infinite ; that is, as will be shown presently, e is the limit toward which the infinite series approaches, as the number of terms approaches infinity. The logarithms with the base 10 are called the Briggs Sys- tem, or Common Logarithms. The logarithms with the base e are called, in honor of Baron Napier, the inventor of logarithms, the Napierian System. They are also called the Natural Logarithms.* § 5. Changing from One System to Another. Logarithms are first calculated, in the manner shown in § 10, to the base e. From the table to the base e, the table to the base 10 is then calculated, by dividing each log in the e-table by the log of 10 in the e-table. That is, the number — - used as a multiplier for the e-table will give the log e 10 10-table. The number • * = 0.4342944819 • • . is called the log e 10 Modulus of the Briggs System, or Common Logarithms, with reference to the Napierian Logarithms. In general, any table of logs can be converted into a table to another base by dividing every log in the table by the log of the new base in the given table. Proof. Let a x = m = b y , (1) so that x = \og a m, (2) and y = \ g b m. (3) * As indices had not found a place in Algebra at the time of Napier, his logarithms were very different from those now called, in his honor merely, Napierian Logarithms. The student interested in the subject may consult Cajori's "History of Mathematics" ; the section "Logarithms" in the Ency- clopaedia Britannica ; also article by J. W. Young in Am. Math. Monthly, 1903. §6] DISCUSSION OF LOGARITHMS. 15 Take the logarithm of (1) to the base a, .-. x = y\og a b i (4J) or y = -?—. (5) By (5), logarithms of the 5-system are those of the a-system divided by the logarithm of b in the a-system ; whence the general rule for changing logarithms of a given system to a new system : Divide the logarithms of the given system by the logarithm of the base of the new system taken from the old system. The number, -, is called tjie modulus of the 6-system log a & with reference to the a-system. Thus the modulus of the Common Logarithms with ref- erence to Napierian Logarithms is -JL- = 0.43429 • • -. log, 10 And the modulus of the Napierian Logarithms with refer- ence to Common Logarithms is — — = 2.303.... lo gio e Tlius, to convert a Napierian table to a Common table, multiply each logarithm by 0.43429 And to convert a Common table to a Napierian table, mul- tiply each logarithm by 2.303 § 6. log a b • log 6 a = 1. Proof. Let a x =b, (1) so that log a b — x. (2) Take the logarithm of (1) to the base 5, .-. x • log 6 a = 1, (3) .-. by (2), log a 6.1og 6 a=l, (4) 16 PLANE TRIGONOMETRY. [§ 6 Thus the modulus of the 5-system with reference to the a-system is either -, or log 6 a. In particular, — - = log 10 e. log, 10 Thus the modulus of the Common system is either the reciprocal of .the logarithm of 10 in the e-system, or it is the logarithm of e in the 10-system. EXERCISES. 1. Look up loge in tables and compare with the modulus. 2. How would you convert an ordinary table of logs to the base 10 into a Napierian table ? Ans. Divide each log by 0.4342 • • •, or multiply each by 2.303 • • •. 3. Of what number in the 10-system is the modulus, 0.4342 • • •, the log ? Of what number in the e-system is 2.303 • • • the log ? a x = m = by, . log 5 m • log c b & log c a log 10 37.31 = 1.57183 log 10 2 = 0.30103, log 2 37.31 and log 2 373.1 log 2 3731 and log 2 37310. logj e = 0.43429, log 10 2 = 0.30103, log 10 3 = 0.47712, log 10 7 = 0.84510, find to the base e, the logs of the following numbers : 2 ; 3 ; 4 ; 5 ; 8 ; 9; is f ; I; I; 2f ; xiu- 7. By the aid of the preceding logarithms, find x in the expressions : 2* = 3; 3* = 2; 5* = 12; 16* = 10; 27* = 4. (a) Using e-logs. (b) Using 10-logs. 8. Given log, 2 = 0.69315, log 6 3 = 1.09861, and log e 10 = 2.30259, find to base e the logs of following numbers : 6 ; 12 ; 4 ; 9 ; 8 ; 27 ; 5 ; 15 ; 25; 125; f ; f; |; f. 4. From the relation show that 5. If and find 1 and 6. Given §7] DISCUSSION OF LOGARITHMS. 17 9. From the values of log e 2, log e 3, given in Ex. 8, calculate log 10 2, log 10 3, and compare with values previously given. 10. If e x = V# 2 + 1 + y, then e* - e~* = 2 y ; e x + e~ x = 2Vy 2 + 1. 2y; e* + e-* I, then e x + e~ z = 2 y ; e* - c~ x = 2Vy 2 - 1. 11. If e« = y - vy § 7. The Base of the Napierian System of Logarithms. e = lim z= Jl + -J • By the binomial theorem, when y < 1, (l + y)--l + *y + *l?LrJ V + etc., 1-2-3 ■■■ {} + -) ^^-y +-TTW + 1-2-3 + etc - _i . 1 + 1 *J , \ *A */ , etc - 1 + i + 1.2 + 1-2-3 + etc- Let x approach the value infinity, liuw (l + iy = l + l + l + l + l + etc. The value of e may be calculated to five places as follows : 1.000000 1.000000 0.500000 0.166667 0.041667 0.008333 0.001388 0.000198 0.000025 0.000003 e = Sum = 2.71828 To ten places, e = 2.7182818284 .... e is an irrational number, or an unending, non-repeating decimal. 18 PLANE TRIGONOMETRY. COROLLARIES. [§7 1. 2. 3. 4. 5. 6. 7. 8. x/x= e 9. ft- AY =±. l-iy*=fl + -T=«. 10. (1 + xY =e. IV B i ( 1\na; -) =6 n . Xjx=«> x)x=*> e n 1 + i-*Y = xj l+iy =^. nx/x*=*> 11. (1+*) *=-. 12. (1 + nxY =e n n 13. (l + x)l=e*. 14. (1 — nxf = — V y x=0 e n 15. \ tf/a;=0 16. ri-;# = A_ -w § 8. The Exponential Series, or Exponential x. e*=l + y + j| + j| + ^+etc., = lim^Jl + -J , .-. ^ = lim x= ^l+^ or, by the binomial theorem, e» = li m _ of (l + ^g)+^Mlll)gJ + etc.) -S y -('-i)( y -P = lim r _^ of l + y + yy 1-2 1.2-3 + etc. §8] DISCUSSION OF LOGARITHMS. 19 V 2 V 3 V 4 or, * y=1+ ^ + ! + J3 + |I + etc * x 2 x 3 at Similarly, e x =l + x + — +.-4-.-T 4- etc. E 12 Iz - % This series is called the Exponential Series, or Exponen- tial x. It is true for all finite values of x, that is, it converges to a definite limit for all values of a?, except #=oo . No matter how large x is, if finite, the numbers in the factorial denomi- nators finally overtake it in value. From that point on, the terms decrease in value. Thus, the ratio of the general term, — to the preceding term, — — -, is , — h: r = -- This \n \n — l [w \n — 1 n ratio has the limit zero, when n = oo, for any finite x. Thus, the series is convergent. (See any College Algebra.) Note. — The teacher may find it advisable here to give the proof of the convergency of an infinite series, as covered by the statements : (a) If the ratio of the nth term to the preceding is < 1, when n = co , the series is convergent. (b) If this ratio is > 1, the series is divergent. (c) If this ratio is =1, the matter is in doubt. EXERCISES. 1. 1? l? + li 15 15 2. ^ + e ~ l) = 1 + tttt eUi ' 3. ^ttt^^ttt^ 1 ' 4. H + H + ^ 1+ ( 1+ H + ^)* 5. l + 2 + 3 A = e. 6> 2 4 + 6 1 15 15 II 2 IS 15 II e 7. ttt^ e-l 1+ tt et °- e + 1 20 PLANE TRIGONOMETRY. [§ 9 8. 1 + ^ 3 + §! + ^ + etc. = 5e. IS IS 12 +2 1+2+31+2+3+4 . _3 12 + [8 + |4 +^-"2 § 9. The Logarithmic Series. log e (l + z)=z--+----- 8eV / 2 3 4 -io ge (i-s)=s+!+!+j Proof. If ^=1+3, then iog«(i+*)=y- But .-(i4Y • \ X/ X = And \ X Jx=co \ X Jx=<*> ••• by (1), - 1/ . 1 (1) e y . .-. 1 + - = (1 + g)«, when x = cc. X 1 If 55 < 1, we may expand (1 + z)* by the binomial theorem. 1/1 .A 1/1 •-• 1 + ;- I+ ;* +i T72 Cancel the l's and multiply by x. ,e-'>, q--)(H , lrt: •■•*" J+ 1.2 + 1.2-3 + ' when # = oc. Z 2 , Z 3 z 4 . •*^ ==2 -2 + 3"4' etC -' or log e (1 + «) = i - | + | 3 - J, etc. (1) § 10J DISCUSSION OF LOGARITHMS. 21 This series has been derived under the supposition that z < 1, numerically, and is therefore subject to that limitation. Changing z to — s, log e (1 - = - (z + 1 2 + 1 3 + J + etc.). (2) § 10. Calculation of Napierian Logarithms. The series (1), (2), above, are not adapted to the calcula- tion of logarithms. They hold only when z is numerically less than 1 ; (1) holding also when z is 1. And even within the field of their convergency, they are very slowly conver- gent, that is, a large number of terms have to be taken to get an approximate value of the logarithm of a proper fraction. A rapidly convergent series, suitable for computation pur- poses, is obtained thus : Subtract (2) from (1). ••• io ^:= 2 [ z+ f + f + 7 +etc -) (3) Since z is to be less than 1, let 1 where a: is a positive integer. _ 1 + 2 + 2z + l 2z + 2 s + 1 Inen q = = — = — ~ = 1—2 - 1 zx x 2x + l Now x and x + 1 are consecutive numbers, when x is an integer. And log £±1 = log (x + 1) - log x = log 1±£ •'. log, (x + 1) 22 PLANE TRIGONOMETRY. [§ 10 As soon as x is large, a very few terms of this series will give log e (1 + x) to a close approximation, when log e x is known. Since log e 1 = 0, the Napierian table is constructed as follows : lo ge 3 = lo & 2 + 2g + i.l + ^.| 6 + etc.) log, 4 = 2 log e 2, log,5 = lo g ,4 + 2g + l.| + l.| + etc). And so on. The number of terms taken in the series is dependent upon the place of the table to be constructed. EXERCISES. 1. Calculate to four decimals log e 2; log e 3; log e 4; log e 5; log e 6; log e 7; log e 8; log e 9; log e 10. Reduce each such logarithm to the base 10, and compare with the logarithms in the table accompanying this book. f loge 2 = 0.6951 Ans.< loge 3 = 1-0986 [loge 4 = 1.3863 loge 5 = 1.6094 loge 6 = 1.7918 loge 7 = 1-9459 loge 8 = 2.0794; loge 9 = 2.1972 ; loge 10 = 2.3026. 2. What relation has log e 10 to the modulus of the e-system as com- pared with the 10-system? What relation has it to the modulus, 0.4343, of the Common system as related to the e-system ? § 11. Series for the Logarithm of Any Number whatever, without Reference to the Adjacent Number. If in + g we let z = x ~~ . , so that z < 1, if x is any posi- 1—2 x+1 1 + — .. , 1 + 2 X + 1 tive number, q = T = x. 1 — 2 -j X—l x + 1 § 12] DISCUSSION OF LOGARITHMS. 23 .-. Series (3) becomes , 1 o 8 „«.o.8«(|^ + |(|^)vi( S i; + .,4 W Series (5), (6) define what is meant by log e x and log 10 x as related to ar, in calculation form, 0.8686 being twice the modulus, approximately. EXERCISES. 1. Show that 1 - $ + $ - i + i-" = log e 2. Is this the best series from which to calculate log e 2? Is it rapidly or slowly convergent? 2. Show log.8-lo«j.2=i-|.i + |.i-|.l, etc. 3. Show that if x > 1, series (3) becomes 4. Show that log, a - lo ge ft = «^i + 1 f^zl) * + 1 («Z^ 8 + etc. a 2 \ a I 3 \ a / 5. Show that log e (1 + 3* + 2z 2 ) = 3x - ^+ ^ - i^+ ... ^ O TC if 2 a: be not > 1. 6. Show that 2 log, a: - log. (x + 1) - log, (x - 1) 7. Show 10^2=^ + ^ + ^+.. .. 8. Show log. 2 - 1 = _!_ + _1_ + _|_ + .... § 12. The Rule of Proportional Parts in Logarithms. Let x and x + e be two numbers, where e is small compared with x. Then log e (x + e) - log, rr = log e ?^p = log e f 1 + ^J . Butlo ge (l + i) = i-lgJ + lg) 3 ,etc.(§9). 24 PLANE TRIGONOMETRY. [§ 12 If - is small, (-) » (~m are s ^ smaller. For instance, if - = o.ooooi, f-) 2 = 0.0000000001. x \xj .*. log e ( 1 + - ] = - , approximately, and log 10 (l + j)« 0.4343 U* -e, nearly, where iJf is the modulus. x i 0.4343 .'. log 10 (s 4- e) - log 10 a; - — — e, nearly. TAws £A# difference of the logs of any two numbers which differ by a quantity relatively small, is proportional, approxi- mately, to the difference, e, of the numbers. This is called the Rule of Proportional Parts. In the tables of logarithms the numbers differ by unity. Thus e = 1, and M 0.43429... . ., -. . _. — = is the Tabular Difference, xx M in tables to the base 10. Thus the tabular difference is large at the beginning of the table, and decreases as the numbers increase. Also, a given difference in numbers at the beginning of the tables will make a larger difference in the logarithms of the corresponding numbers than will the same difference in numbers occurring at any other part of the table. Four-place tables are generally arranged to give directly the logarithms of three-figured numbers, and five-place tables (arranged like Gauss's) give directly the logarithms of four-figured numbers. Since the differences in logarithms at the beginning of the table are large and vary rapidly, it is advisable that a four-place table should give directly the logarithms of four-figured numbers at the beginning of the table. The table in this book follows this plan in numbers up to 1709. § 13 a] DISCUSSION OF LOGARITHMS. 25 § 13. Applications of Tabular Difference. M By § 12, log e (> -f- e) - log e a; = — e, x or log-difference = tabular difference times number-difference. (a) To find the logarithm of a number not in the table. Given log 2054 = 3.31260 and log 2055 = 3.31281. What is log 2054.2 ? The tabular difference is 21. The number-difference is 0.2. .*. the log-difference is 0.2 of 21, or 4.2, or 4. .-. log 2054.2 = log 2054 plus 0.00004 = 3.31264. Note. — The usual rule for " throwing away " is followed in the log- differences. The above difference, 4.2, is counted as 4. If it had been 4.8 or 4.5, it would have been taken as 5. When the number to the right of the point is less than 5, it is thrown away ; when it is 5 or more than 5, the number to the left of the point is increased by 1. Since the mantissa with the base 10 is independent of the position of the decimal point in the number, when the loga- rithm of a number is to be interpolated from the table- logarithms a decimal point is set in the given number after the number of digits for which the table gives the logarithm. For instance, if the table gives logarithms of numbers with four digits, and the logarithm of 2.3459 is desired, consider this number, so far as mantissa is concerned, as 2345.9, and interpolate for 0.9, as above for 0.2. Similarly, for log 456789 interpolate for 0.89 from log 4567 and 4568. Generally an interpolation for more than two figures can- not be made, since if numbers differ by 0.001 or by 0.009 beyond the table figures, their logarithms agree to the place of the table. Often the second figure is a matter of indif- ference. 4343 For, log 10 (x + e) - log 10 x = : e ; x 4343 or, log-difference = : times number-difference, and num- x ber-difference = 2.3 x times log-difference. 26 PLANE TRIGONOMETRY. [§13 a Thus the smaller x is, the larger is the log-difference for a given number-difference. If we are using a five-place table, like Gauss's, giving logarithms of four-figured numbers, the smallest value of x, so far as mantissa is concerned, is 1000, and in a five-place table the smallest appreciable log-differ- ence is 0.000005. Thus if the number-difference were 0.009 in the neighborhood of 1000, the log-difference would be 0.0004342 times 0.009, or about 0.0000039, not enough to affect the logarithm in the fifth place. Thus in no table (which gives directly logarithms for numbers of one figure less than the place of the table) is it possible to interpolate for more than two figures beyond the reading of the table ; often not for a second figure. (b) To find the number corresponding to a logarithm lying between two logarithms in the table. Given 3. 74687 = log 10 5583 and 3.74695 = log 10 5584, 1.74689 = log 10 ? The tabular difference for a unit number-difference is 8. The log-difference between the given log and that next under it in the table is 2. __ , _.„ log-difference Number-difference = - — tabular difference = f = i = .25. .-. 1.74689 = log 10 55.8325. The characteristic fixes the position of decimal point. To how many places should such divisions be carried ? This depends upon the number x. For, by the equations above, e __ log-difference x~~ 0.43429-.. For a five-place table the smallest appreciable log-difference is 0.000005 ; that is, 5 in the sixth place. •"• here I = raif - » t0 le a pp reeiabh - §13 6] DISCUSSION OF LOGARITHMS. 27 Thus in using a five-place table, the division should not be carried to a place where 1 in that place bears to the number being found a ratio smaller than 1 to 86859 (e to x). For instance, above we carried the division to two places, getting 25, where the number found was 55.8825. The ratio of 1 in the final place to the number is 1 to 558325. This is less than 1 to 86859. Thus the division was carried too far. In fact, since 5 < x 558325 86859' the final 5 would not affect the logarithms to five places. That is, the logarithm of 558325 is the same as that of 55832 to five places. The general rule for any table giving logarithms directly for numbers of one figure less than the table is : Do not carry the division to the place where 1 in that place has to the number being found a ratio less than 1 to the double modulus, considered as an integer to as many figures as the place of the table. Thus, in four-place tables not beyond 1 to 8686, in five-place tables not beyond 1 to 86859, in six-place tables not beyond 1 to 868599, and so on. It is thus apparent that the division should never be carried in such tables beyond two places ; generally not beyond one place. It is apparent from what precedes that in a five-place table, as soon as a number rises above 86859 (the double modulus) one in the fifth place of the number does not affect the logarithm in the first five figures. Thus in interpolating for numbers corresponding to logarithms in the part of the table above the double modulus, it is not possible to get the fifth figure within a unit of accuracy. In that part of the table no interpolation for the fifth figure should be made. 28 PLANE TRIGONOMETRY. [§ 13 b So in general for any place table, we should not interpolate for figures in the place of the table, when the logarithms indicate a number beyond twice the modulus. EXERCISES. 1. Look up the logarithms of 234 and 235 in the four-place table, and from them find the logarithms of 23.48, 2.341, 2.346, 23.49, 234.5, 0.2341. 2. Look up in a five-place table the logarithms of 4357 and 4358, and from them determine the logarithms of 43.572, 4.3578, 432.56, 43579. 3. "Write five logarithms at random, and find in the tables the num- bers corresponding to them. Take four-place logarithms and five-place logarithms. 4. Examine your table, and see at what part of the table the tabular differences are largest. 5. Take any two consecutive numbers greater than twice the modulus and having as many figures as the place of the table, and examine as to whether in the table they have the same logarithms to the place of the table. 6. The teacher may exercise the class at will on examples like Exs. 1, 2, 3, so selecting them that the first part of the table, the middle of the table, and the part of the table above the double modulus are all considered, until the class is thoroughly familiar with how to interpolate, when to interpolate, how far to interpolate, and when not to interpolate. § 14. What Place Table to Use. When we have before us a five-place table, we may con- sider that it has been made from a six-place table by dropping the sixth figure, following the usual rule as to the final figure as compared with 5 (see § 13, a, Note). In any particular logarithm we do not know, then, when we have a five-place table, what the sixth figure of the logarithm is. The logarithm may be in error by almost 5 in the sixth place, in either direction. We must thus allow in any given loga- rithm of five places a possible error of almost ± 0.000005. In § 13 we found e __ log-difference x modulus § 14] DISCUSSION OF LOGARITHMS. 29 If, then, there is an error of ±0.000005 in a logarithm, the corresponding value of -, or the relative error in the corre- sponding number, is ±0.000005 ±1 0.43429 . . . 86859 , nearly. Thus, if we have a single logarithm and we are certain that it is correct to five places, the corresponding number can be determined to within one 86859th part of its value. This is far beyond the degree of accuracy with which measurements in engineering work are usually carried out. Similarly, if a logarithm is known accurately to four places, the corresponding number can be determined to within one 8686th of its true value. Hence, for a great many engineer- ing operations a four-place table is sufficient. It must be stressed here that we say, If the logarithm is known to be correct to five figures, or four figures. If the given logarithm has itself resulted from a manipula- tion of values themselves uncertain in the next place beyond their given place, the resulting logarithm may itself be uncer- tain in one or more of its terminal places. The effect of such uncertainty is considered in the next chapter, so far as it relates to ordinary trigonometrical cal- culations. Later it will appear that if the sides of a diagram show only one significant figure, or only two, or only three, or only four, the angles should not show seconds, and a four-place table is sufficient. When the angles show seconds, a five- place table is called for. When the angles show tenths of seconds, a six-place table is called for ; hundredths of seconds, a seven-place table. An extension of the table by one place implies, as is clear from what precedes, a diminution of allowable error in read- ings and in the results of calculation by one-tenth, as also in the data. A diagram with its sides reading to one, two, or three significant figures and showing seconds in angles would 30 PLANE TRIGONOMETRY. [§ 14 indicate that angles had been measured more carefully than lines, and would thus be absurd. When we use a table of more places than the significant figures of the data call for, we must not take calculated re- sults to the degree of accuracy of the table, but cut them back to the pattern of the data. This will be made clear in the next chapter. EXERCISES. Calculate to four significant figures the value of the following : 1. 27.34 x 13.56. 6. 2^12. 2. 2.374 x 0.0732. 7. V3lM 71.36 x 21.27 3763 x 0.003721' 3.314.3X0.3164. 8 V0.003456 x \/56.73 4. 27.31 x 273.1 x 0.003416. 9. (26.31) 2141 x 0.3465. 316.1 21.32* 10. (2.341) ( L361 >. Calculate to five significant figures the following: 7.8321 x 0.032564 11. 73.214 x 3.2154. 14. 2.3146 x 0.056378 12. 384.62 x 2.7184. 15. V7U856 x V7\3214. 13 56.732 x 87.563 16 V0.0035678 x V0.043785 7-2134 ^0.0073214 x vU00032156 If there be more than one calculator, what is the best plan for checking such calculations as the above ? What, if there be only one calculator ? § 15. Negative Mantissa. It occasionally happens that a calculation can be carried out more readily by having the logarithms that are usually part positive and part negative all negative, as, for example, in getting the value of z = (0.00347) 00567 , and similar expressions. § 15J DISCUSSION OF LOGARITHMS. 31 The usual logarithms would be log x = 0.0567 log (0.00347) = 0.0567x3.54033. .-. log (log x) = log (0.0567) + log (3.54033). The last number is part positive (mantissa) and part nega- tive (characteristic). Getting its logarithm in this form is not possible. Instead, we use log (log x) = log (0.0567) + log (2.45967) (n) = log (0.0567) + log (2.460) (n) = 2.75358 + 0.39094 (n) = 1.14452 (*) .-. logs = -0.13948 or log x = 1.86052 .\ a; = 0.7253. In similar expressions of the form y = a"\ it is best to calculate the value of log (£> c ) = log a?, as in the preceding example, for log y = x log a, and log log y = log x + log log a. EXERCISES. Find the values of 1. (0.003468)° o 04378 to four significant figures. 2. (0.0056143)° o 0036213 to five significant figures. 3. (0.3468) (0003214) 00003614 to four significant figures. 4 - ( ^0.0346) ^0.347) °- 361 to three significant figures. CHAPTER II. CALCULATION VICES AND DEVICES. [Note. — When to stop " figuring " is what the student must know in calculations. In Trigonometry is the first place where the mature student faces this question. This chapter is intended as a preparation for judicious calculation.] § 16. When a calculation involving only multiplications and divisions is to be carried out, one may (1) Do the work as in arithmetic. (2) Shorten the arithmetic process by dropping the figures which fall, in the final result, beyond the place (decimal or integer) of possible accuracy. (See § 26 and § 40, for the shortened process of multipli- cation and division.) (3) Use logarithms. (4) Use a sliding-rule (which mechanically adds and sub- stracts logarithms and indicates the corresponding number).* (5) Use a calculating machine designed for office work. Which process to use depends upon the extent of work to be done. A single multiplication, like 83 times 72, or even like 217.3 times 31.46, can be carried out more quickly, perhaps, directly than by logarithms, since, in the latter case, one must find the table, then look up the logarithm of each number, and then look up the corresponding number. However, if more than three operations are necessary, and the numbers * Every engineer who has much calculating to do, will find a pocket sliding-rule of great convenience. The makers furnish an explanatory pamphlet. A description of the theory on which they are based will also be found in Raymond's "Surveying." 32 § 18] CALCULATION VICES AND DEVICES. 33 are not small, there is a great saving of time in using logarithms. Speed, however, in using logarithms, as in any other labor-saving device, depends on a thorough acquaintance, from long practice, with the process. § 17. The numbers occurring in calculations may be (1) Known exactly. (2) Known only approximately. (3) Known exactly, but used only approximately. Since most of the numbers used in calculations in engineering work come from measurements which are of necessity made only ap- proximately, or else come from other calculations carried out only approximately, we may make the broad statement that an engineer is concerned chiefly, in calculations, with approximate numbers. The chief vice, then, of the engineering student (we might also include many engi- neers), as a calculator, is a more or less total disregard of the fact that the numbers used are only approximate, with the consequent super- abundant care in the extent of " figuring " done. His vice is " cipher- ing " gone mad. § 18. Significant Figures. All figures, other than zero, are significant in a number. A zero is significant unless used merely to locate the decimal point. In 207, 2.07, or 0.207, the central zero is significant. A zero used merely to locate the decimal point is not signifi- cant. Thus in 0.0003, the zeros are not significant, for an error in the next place to the right of 3 would be the same relative error in 0.0003 as in tenths' place for 3. Thus 0.0003 is not called a number of four significant figures, but of one significant figure. A final zero may be quite significant. In such cases it should always be retained. Thus if a line measured to the nearest hundredth of an inch is found nearer 29.30 than to 29.31 or to 29.29, the result should be written 29.30 and not 29.3, for the latter would mean that the measure- ment had been made only to the nearest tenth. Frequently, however, ending zeros are not significant. For example, if it is said that a certain building cost about $35,000, evidently the three zeros are not significant. When the con- 34 PLANE TRIGONOMETRY. [§18 text does not indicate the degree of inaccuracy of an approxi- mate number, there should be some accompanying statement to make the matter clear. Frequently a number is written in the form 35.246, ± 0.012, with the understanding that the error may be as great in either direction as 0.012, and that, consequently, the true value lies somewhere between 35.258 and 35.234. The same notation is used in Least Squares in an entirely different sense, namely, that it is an even chance that the error in the result is larger or smaller than the indicated error. The inaccuracy to which a result is liable may also be expressed as a per cent. Thus 217, ± 2.17, means the same as 217 with possible inaccuracy of 1 fo. A statement like 31.145693, ± 0.032, would be absurd, as the last three figures, 693, would then be meaningless. This should be written, 31.146, ± 0.032. With the exception already mentioned in the case of zero, all figures in a number are significant, that is, any change in the figures would indi- cate a different number. § 19. Rejection Error. Approximate numbers of very frequent occurrence are : logarithms ; square roots of numbers like 2, 7, etc. ; numbers like 7r = 3.14159 . . . and e = 2.71 . . . ; numbers like those treated later in this book and called sines, cosines, etc., which, with a few exceptions, belong to the large class of non-terminating, non-ending decimals. Approximate values of such numbers are obtained by stopping at any place, increasing the figure in that place by 1 when it is followed by a 5 or by a figure larger than 5, — otherwise, leaving the final figure unchanged. Thus, approximate values of it are 3.1, 3.14, 3.142, 3.1416. Such approximations are always nearer the true value than 5 in the next place to the right. Thus 3.14 is too small for tt, but not by 0.005 ; 3.142 is too large, but not by 0.0005. The errors introduced into calcu- lations by thus cutting the number of figures in a number are called rejection errors. The limit for the extreme value §20] CALCULATION VICES AND DEVICES. 35 of the rejection error in any approximate number entering directly into a calculation is 5 in the next place to the right of the ending place. This error in the elements of a calculation can effect very materially the final result of the calculation. § 20. Effect of Errors in the Terms of an Addition-subtraction Result. If a is liable to an error x v and b is liable to an error x v evidently a ± b is liable to the error ± (x x + # 2 )> for the error in a and b may both chance to lie the same way. Similarly, for any number of such operations, the largest possible error is ± (x x -f- x 2 + x z + x 4 -f- • • • + #„)• It is extremely unlikely, of course, that in a very large number of such terms, taken at random, where the error in each term is as likely to lie one»way as the other, the errors in all the terms will lie the same way. Experience has shown that in such cases there is a balancing of errors. When, however, the number of terms is small, it is not at all uncommon to find the errors all running one way. The student can give himself some acquaintance with this matter by selecting logarithms, in groups of two, of three, of four — as will be the case in most of the calculations in trigonom- etry — at random, from a four-place table, and afterwards looking up the fifth figure in a five-place table, or higher place table. The errors will be far from balancing. It is quite a common error to apply the deductions of the Method of Least Squares, based as they are on "the long run," to a very short run. A not uncommon disregard of the preceding deductions is shown in adding (subtracting) approximate numbers. For example, in the addition, 27.31 346.2159 373.5259' 36 PLANE TRIGONOMETRY. [§20 if these numbers are liable to rejection error, the final result is quite misleading, for it makes it appear that the sum is liable only to the error 0.00005, whereas it is 0.00505. In such cases the number with many decimal places should be cut back to the pattern of the other. Thus, 27.31 346.22 373.53, where the final 3 may be "off" by 1. In additions and subtractions of approximate numbers, subject to rejection error, let each term show the same number of decimal places. § 21. Applications of the Suggestions of § 20. If the number of terms of an addition-subtraction expres- sion is n, and each term is liable to rejection error beyond the rth decimal place, the extreme limit for the accumulation- effect of these rejection errors in the algebraic sum, is 5 n in the (r + l)th place. Suppose 20 terms lead to the result, 1000, the smallest sequence of four significant figures, with 20 times 5 in tenths' place as extreme accumulation rejection error, or 10. This is 1 JG of the sum. If the result had been 9999, the largest four-figured number, the accumulation rejection error would still be 10, but as a per cent, only a little more than one-tenth of 1 fo. Similarly, for 10000 and 99999, the smallest and largest five-figured numbers, the per cents are ^ of 1 fo and jfa of 1 fo. Since per cents are independent of the position of the decimal point, the following computation rule is sometimes advised : If an addition-subtraction expression is desired within the range of -^ of 1% to 1% possible inaccuracy, let the final result show four significant figures. If the permissible error is to lie between -^ of 1% and y^ of 1%, let the final result show five significant figures. This rule, deduced as just shown, takes the worst cases and the best cases, presuming in each case that the effect of §21] CALCULATION VICES AND DEVICES. 37 rejection error lies in each term all the time the same way, and with the rather unusual number of 20 terms. The " factor of safety " is thus so great that in the case of 2, 3, 4, 5, 6 terms, which are far more common than numbers of terms very much larger, it is very much better to disregard the rule and take the number of figures to suit the individual example, as illustrated in the following examples. EXERCISES. 1. Find to within 0.4 of 1 % the value of 55.6312 + 71.8371 - 4.32713. By inspection, the result is about 120, of which 0.4 of 1 % is about 0.5- Since there are only three terms, the maximum accumulation rejec- tion error is 0.15, if hundredths are dropped. This is within the limit required. Therefore as follows : 55.6 + 71.8 - 4.3 = 123.1, ± 0.15. 2. Find to within 0.6 % the value of 47.3489 + 174.32825 - 5.62147. By inspection, the result is about 200, of which 0.6 % is 1.2. Thus, hundredths may be dropped, and we have 43.3 + 174.3 - 5.6 = 216.0, ± 0.15. 3. Make up and solve some examples like the preceding, some with three terms, some with four, some with five, six, seven terms. 4. Add 273.1415 and 564.1. To what error is the sum liable ? 5. Add 32.14156, 32.718, 32.6. To what error is the sum liable ? 6. Find the value of 314.2156783 + 315.61 - 2.3124 - 87.4. 7. Find the sum of 36.732, ± 0.021, and 71.2468, ± 0.0004. The uncertainty in the first number being the greater, that decides the number of decimal places to retain in the second. Siuce 0.0004 added to the final 8 of the second number makes the 6 uncertain bjy 1, we may use 36.732 ± 0.021 71.246 + 0.001 107.978 ± 0.022 the negative limit being not quite this. 8. Add 3.607(± 0.022) and 5.84(± 0.28). 38 PLANE TRIGONOMETRY. [§22 9. Reduce to a single term 7.0845(± 0.0012)+ 7.364(± 0.001)- 3.2748(± 0.012). 10. Find the value of 53.3456(± 0.05%)+ 167.3578(± 0.01%)- 6.2315(± 0.03%). § 22. Application to Logarithmic Work. The most important practical application of the results reached in the preceding sections concerning the effect of accumulation rejection error is in the use of logarithms. The logarithm of a product is the sum of the logarithms of its factors, etc. Take the case of two logarithms added (subtracted). Consider two four-place logarithms as a special case. The result may possibly be uncertain, in the extreme case, by almost 1 in the fourth place. If ten such logarithms are added (subtracted), the result, in the extreme case, may be uncer- tain by 5 in the fourth place, so that the resulting logarithm has only the accuracy of a logarithm taken from a three-place table. These simple facts are constantly disregarded in the use of four-place tables, as, for example, when one interpo- lates to get a fifth figure in a number corresponding to a log- arithm obtained by additions (subtractions) of logarithms. It has been pointed out in § 12, page 24, that if two logarithms differ by the small quantity e, the ratio of the difference of the two corresponding numbers to the smaller number (the "relative error" in the two numbers) is T~, — or „ , * In constructing a four-place table, modulus 0.4342 ... 5 F 0.00014 counts as 1 in the fourth place, while 0.00015 counts as 2, as would also 0.00024. Thus the " relative error" corre- sponding to the accumulation rejection error 1 in the fourth place, in the case of an addition (subtraction) of two four- , , t 0.00014 ,12 3 p ace logs may be or about — or _ or _, etc. Thus if two four-place logs are added, giving a log cor- responding to a number lying between 3100 and 6200 (the §23] CALCULATION VICES AND DEVICES. 39 decimal point falling where it may), the rejection error of 1 in the fourth place will leave the corresponding number in doubt as to three consecutive numbers (and of course all intermediate numbers). If the corresponding number lies between 6200 and 9300, it will be in doubt among five con- secutive numbers and the intermediates. It will be in doubt among seven consecutive numbers and their intermediates when it lies above 9300. From this one will readily draw the conclusion that in using a four-place table one ought not, as a rule, to interpo- late for a fifth figure, and one sees that the fourth figure is in doubt as soon as the number corresponding to a given log which has arisen by adding two logs, or subtracting them, rises above 3100. This can be observed in a four-place table. EXERCISES. 1. Two logs added (subtracted) give 2.6042 ; what is the correspond- ing number? Ans. 401.9 or 402.0 or 402.1 or any intermediate number. 2. Two logs added (subtracted) give 2.8482 ; what is the correspond- ing number ? Ans. 704.8 or 704.9 or 705.0 or 705.1 or 705.2 or any number inter- mediate to these. 3. Two logs added (subtracted) give 3.9845 ; what is the correspond- ing number? Ans. 9647; 9648; 9649; 9650; 9651; 9652; 9653; or any number intermediate to these. § 23. Conclusion as to What Place Table to Use. From the preceding section it will be clear that when the number of additions (subtractions) of logs is few, and where, consequently, a balancing of errors is not to be counted on, one ought to use a table giving at least one figure beyond the number of figures desired in the calculated results. In 20 additions with a seven-place table, it may 40 PLANE TRIGONOMETRY. [§24 happen that the fifth figure is in error by 1. If the charac- teristic calls for more than four figures in using a four-place table, the figures following the fourth should be filled with non-significant zeros. Such a characteristic indicates that a four-place table ought not to be used. § 24. Application to Some of the Problems of Trigonometry. 1. It will be found later in solving right-angled triangles that two logs are added (subtracted). The effect is, there- fore, as pointed out in § 22. 2. In solving oblique-angled triangles, except when three sides are given, it will be found later that three logs are united. In using a four-place table, the fourth place becomes doubtful by 1.5. This counts as 2 in the table construction. So does 2.4. " Relative error " = — ■ = = = = 0.4342 . . . 1770 3500 5300 7000 , as approximate results. 8900 Consequently, in such work, if the resulting logarithm corresponds to a number (irrespective of decimal point) which lies between 1770 and 3500, there is doubt among three consecutives, between 3500 and 5300, there is doubt among five consecutives, between 5300 and 7000, there is doubt among seven consecutives, between 7000 and 8900, there is doubt among nine consecutives, and all intermediate numbers. EXERCISES. If three logs added (subtracted) give the following logs, state the corresponding numbers : 1. 2.2989. 2. 1.5955. 3. 2.7404. 4. 1.8633. 5. 3.9294. 3. In solving a triangle when the three sides are given, it will be found later that four logs are added (subtracted), and this result is divided by 2 on account of a square root. The resulting doubt in the fourth place is therefore 1, and this case is the same as for solving right-angled triangles. §25c] CALCULATION VICES AND DEVICES. 41 § 25. Errors in Products of Two Factors. (a) If a is liable to an error ± x v and b is liable to an error ± rr 2 , to find the error to which the product ab is liable. Let a v b v be the true values of a, b, a 1 = a ± £j, b 1 = b ± x v a x b x = ab ± ax 2 ± bx x + x x x 2 . If x v x 2 are small, x x x 2 may be neglected. ,\ error in ab is ± (ax 2 -f taj), or, a times b's error plus b times d's error. (If) If a is liable to an error of p 1 per cent, and b is liable to an error of p 2 per cent, to what per cent error is the product ab liable ? Here, . _ a Pi 1 100' and ** = m' ax 2 +bx 1 = j^(p 1 +p 2 ). .\ per cent error in the product ab is the sum of the per cents for the factors. This fact is frequently stated in a manner quite misleading, namely, "A product cannot be more accurate than its least accurate factor." Since when the errors are unknown the worst case must be allowed for, it is not how small the error may be, but how large, that is the important matter. If a is liable to 2 ^ of 1%. (2) For four significant figures, < T ^ > -^o °f !%• (3) For five significant figures, < T ^ ¥ > j-^ of 1%. (4) For six significant figures, < j-^q > -jowo" of 1%. etc., etc. § 28. Approximate Values for Per Cent Accumulation Rejection Error in a Product of Two Factors. Let the 100 in (~7 + ~#) cance ^ tne two term i na l fig- ures of A, B. Whence the rule : (1) The per cent rejection error in the product ab is given approximately by dropping the decimal points and the two terminal figures of A, B, observing the usual rule with ref- erence to the next figure being 5, more or less; invert; add ; take half §29] CALCULATION VICES AND DEVICES. 47 An approximation somewhat rougher, but in many cases close enough, is given by the following : (2) Drop the decimal points and two terminal figures; change each remaining figure, except the initial figure, to zero ; retain the initial figure, observing the usual rule with reference to the next figure being 5, or more or less than 5 ; invert ; add ; take half. EXERCISES. 1. A square has its side measured to the nearest hundredth inch, giv- ing 25.32. What is the maximum per cent accumulation rejection error in its area? Ans. About ^ z of 1 %. 2. A rectangle has its sides measured, giving 24.32 and 378.2. What is the maximum per cent accumulation rejection error in the area ? Ans. About & of 1%. 3. Apply the approximation rules to find the maximum per cent re- jection error in each product of § 26. § 29. Relation of the Two Factors of a Given Product when the Effect of Accumulation Rejection Error is Least. „ . . 100/1 , 1\ 100 (A + B\ Per cent error is -y.^ + jji or -g^-^g-J ■ Since AB is fixed by hypothesis, the per cent error is least when A + B is least. Let A=VAB + X, (1) B = VAB - Y. (2) Then A+B=2VAB + X-Y. (3) Also ( VA — VB) 2 is positive, when A is not B. ^ + j g_2VZ5>0, or A + B>2VAB. .'. Y cannot be more than X in (3). .*. the least value which A + B can have in (3) is 2 VAB, which occurs when A = B. 48 PLANE TRIGONOMETRY. [§30 EXERCISES. 1. Compare the effect of accumulation rejection error in the area of a square whose side is 12.24, and the rectangle of about the same area, its sides being 3.06 and 48.96. 2. Compare the accumulation rejection error in (31. 62) 2 , and 9999 x 0.1000, the products being each about 999.9000. § 30. To find the Per Cent Error in the Product of Three or More Factors, Each Factor being Liable to a Given Per Cent Error. If a is liable to p x per cent error, and b to p 2 per cent error, it has been shown that ab is liable to the error (P1+P2) P er cent. Evidently, then, (a6)c is liable to the per cent error CFi+A + J's)' c being liable to the error p z per cent. So, in general, the per cent error to which a product of any number of factors is liable, is the sum of the per cents of the factors. It is assumed here, of course, that the absolute error in each factor is small, for only in this case is the error in ab a times 5's error plus b times a's error. In the above deduc- tions all errors arising from product of errors have been neglected. The tendency of multiplication of factors, each subject to error, is to increase the maximum error to which the product may be liable. t This is a consequence of the preceding section, it being assumed all the while that the worst possible case may occur by the errors chancing to fall all the same way. When the number of factors is comparatively few, so that a balancing of errors is not to be relied on, the tendency of multiplication is to subject the result to an increasing possible error. From this it appears that the measured quantities which form the basis of a calculation should be measured with a greater degree of accuracy than that expected in the calculated results based on them. All measurements are subject to rejection error. §32] CALCULATION VICES AND DEVICES. 49 Later on, therefore, when the measured sides of a diagram are given to one significant figure, it will be absurd to let the calculated sides show two or more significant figures. The calculated parts of a diagram should never show more significant figures than the measured parts. It is hopeless to give calculated results seven-place accuracy when they are based on field measurements with three-place accuracy. § 31. To find Numerical Expressions for the Per Cent Accumu- lation Rejection Errors in Products of Several Factors. For two factors it is -rr-f-j + -=)« Similarly, for three factors -77- (-7 + -5 + 7^) Similarly, for four factors — — ( — + — + -x,+ — V J 2 \A B CD) where A, B, (7, D are a, 5, or about *#*. The use of any additional significant figure practically divides this result by 10. 2. Show that if the volume of a cylinder is gotten from tt = 3.142, r = 6.043, h = 12.65, the accumulation rejection error is less than -^ of 1%. 3. Use the shortened process of multiplication to find the volumes of the cylinders in Exs. 1 and 2, to within 1% and to within ^ f 1% respectively. Solution of Example 2 to within 1%. Hundredths doubtful. Tenths doubtful. Doubtful by less than 12. 6.04 36.5 12.7 6.04 3.14 115. 36.48 109.5 1270. 3.7 127. 1.4 64. 114.6 1451. The student will find it valuable exercise to calculate the error ax 2 + bx 1 for each product as formed, and notice its growth. 4. What would be the maximum effect of accumulation rejection error in Exs. 1 and 2, if we take in Ex. 1, tt = 3.1, r = 8.4, h = 9.8 ; in Ex. 2, ir = 3.1, r = 6.0, A = 13? 5. The area of a circle («• • r • r) is obtained from ir = 3.142, r = 6.043 ; in what place is the area uncertain from accumulation rejection error ? The easiest way to answer a question like this is to determine the maximum per cent rejection error, and from this get the doubtful place in the area from a rough estimate of the area. 6. Determine the area of the circle in Ex. 5, using four significant figures and the shortened process of multiplication, dropping in each product, as in the solution of Ex. 2 above, the doubtful place as new products are formed. To within what per cent will ir — 3.1, r = 6.0 give the area of the circle ? What if we take tt = 3 and r = 6 ? 7. If in the cylinder of Ex. 2, r = 6.0428, ± i of 1 % and h = 12,653, ± jJj of 1 % and 7r = 3.1416, to what per cent error is the volume liable, neglecting the error in tt? Am. About T 7 ^ of 1 %. How many significant figures should be taken in each factor of iT'T'T-h to get within this error? Ans. Four. §33] CALCULATION VICES AND DEVICES. 51 8. Consider the volume of the cone r = 6.18, h = 127, ir = 3.14, in the same manner as the preceding examples. § 33. On the Reasonableness of Retaining the Same Number of Significant Figures in the Factors of a Product. Two measurements showing the same number of significant figures indicate about the same degree of relative accuracy of measurement. Thus, in 36.27 and 4134, the former means something between 36.265 and 36.275, and the latter some- thing between 4133.5 and 4134.5, so that the largest relative error in the first is y^s? an( l ^ n the second -g-gV'S"' w l ncn are reasonably near together. The relative rejection error in the first is ^g§ y 7 and in the second ^ i|^-q, or y^V? and -g^o* The widest possible discrepancy occurs, of course, between the largest and smallest sequence of the given number of places, as, for example, in 1000 and 9999, for four figures. Thus, unless some unusual reason can be given why one factor of a product should be measured with far greater accuracy than others, all factors should show the same num- ber of significant figures, as a rule. All the measured parts of a diagram should thus show the same number of significant figures, and the calculated parts should not show more significant figures than the measured parts. They may show, generally, the same number of sig- nificant figures. EXERCISES. 1. Find 21746893 x 1.53, these representing measurements. Since the rejection error in 1.53 is 0.005, ax 2 is large and bx^ is small. Thus the larger quantity may be cut back to the number of significant figures of the smaller. Therefore find 21700000 x 1.53, as close enough. 2. What is the product of 372 and 0.0001, these numbers representing measurements ? Here the first number shows three significant figures and the second only one. Therefore take 400 x 0.0001. The rejection error, as a per cent, is in the first case 1 | ft ( 7 | T + i) an( l in the second case i%°- ( ? £ 7 + £). The difference is slight. 52 PLANE TRIGONOMETRY. [§34 3. What is the relative accuracy of measurement in each factor of the products 1.000 x 0.0001 ; 10.00 x 0.01 ; 1.000 x 0.0001 ? Find the per cent rejection error in each product. § 34. Some Extreme Cases and Computation Rules based on them. (1) Suppose there are 20 factors in a product, each factor with four significant figures, and each factor about 1000. The accumulation rejection error as a per cent is about *t* (tuW + ToV o + etc -> to twenty terms) or lf>. (2) Taking now the other extreme case of four significant figures, where each factor is about 9999 (without regard to decimal place). The per cent rejection error is now -^(9^9 + W99 + etc -> t0 twenty terms) or about -^ of lf>. Similarly, for numbers of five significant figures the results are from -^ of 1 f> to j^ of 1 fo, and so on. From these results arise the following computation rules occasionally advised : For an inaccuracy ranging from -^ of 1 fo to If in the final result, use four significant figures in each factor of the product and in each partial product as formed and in the final product. When the range of inaccuracy is to be from ^ of 1 fo to yj-g- of 1 fo, use, similarly, five significant figures, and so on. Since a product with 20 terms is uncommon, and since in the majority of cases the number of factors in the product to be found is quite few, the rough approximation rules given in § 28, for the determination of the per cent error in the final product, will, in any individual case, generally shorten the work sufficiently to justify finding the maximum error likely to arise. Then the number of significant figures to take to keep within the prescribed error can be determined. Illustrative examples have already been given in § 32. §35] CALCULATION VICES AND DEVICES. 53 The foregoing computation rules are sometimes stated incorrectly, thus : " For an inaccuracy of 1 ^ or worse, use four significant figures, etc." The error cannot be worse than 1 f> for less than 20 factors. That is the extreme case, as already shown. § 35. On what Forms of Products Accumulation Rejection Error has Least Effect. We have already shown that in a product of two factors the accumulation rejection error is least, for a given product, when the two factors are the same, decimal point being dis- regarded. (See § 29.) Similarly for three factors in a given product, the effect of such error is least when A = B = (7, and for four factors, making a given product, when A= B—C= D, and so on. EXERCISES. 1. A square and rectangle of about the same area have their sides measured and areas calculated. For the square, the side is 24.82. For the rectangle, the sides are 12.41, 49.64. Determine by the shortened process of multiplication the areas of each and the per cents of rejection error in each area. 2. Consider in the same way the cube whose edge is 12.36 and the parallelopipedon whose edges are 6.06, 4.09, 73.2. 3. Consider in the same way two cylinders of about 1450 cubic inches volume, where in one r = 6.043, h = 12.65, and in the other r = 3.022, h = 50.60. 4. What shaped cylinder can be measured most accurately to five places for volume ? Ans. r = h = ir = 3.1416. 5. What circle? Ans. r = tt. 6. Which can be measured with greater accuracy for area, a square or rectangle, both being of about the same area ? 7. For volume, cube or parallelopipedon, both being of about the same volume? 8. What shaped triangle for area, the base and altitude being measured ? 9. What shaped right-angled triangle for area, if the two legs are measured ? 10. What shaped cone ? 54 PLANE TRIGONOMETRY. [§36 8 36. Approximate Values of and when x is Small. 3 vv 1 + jrl-jr By actual division, = 1 - x + x 2 — x z -f- R v 1 + x By actual division, = l+x + x 2 + x? + iL, where B 1 — x denotes " the rest " after the terms written down. If x is less than unity, x 2 , X s , etc., are smaller than x. If x is 0.0001, x 2 is 0.00000001, and so on. When x is on the verge of the smallness taken into account in any measurement, x 2 , x 3 , etc., may be neglected. Thus, approximated, = = 1 — #, 1 + x and = 1 + x. 1 — x EXERCISES. 1. — = — - — = 1.1, nearly. 0.9 1-0.1 J 2. -i- = 1 = 1.01, nearly. 0.99 1-0.01 ' J 3. —?— = 1 = 1.001, nearly. 0.999 1 - 0.001 J 4/ — - — = 1 - 0.0001 = 0.9999. 1.0001 § 37. Approximate Values of when x is Small. l l ^ir x\_i 2 a ± x f^x\a\ a) a a EXERCISES. i. JL-l-JL. 9.1 9 810 2. -J— = -1 ?— = 0.0997, nearly. 10.03 10 10000 ' y 3. -A- = 1 = A + — 1 — = 0.10001. 9.999 10 - 0.001 10 100000 §40] CALCULATION VICES AND DEVICES. 55 § 38. If a is liable to an error x, where x is small, to what error is - liable ? a By § 37 the error is — . a 2 § 39. If a is liable to an error x x and b to an error x v to what error is the quotient j liable ? 1 o ?-•©■ This is a product, and the error in such a product is a times the error in - plus - times the error in a. o b ^2 1 = ax,, + bx x b* b l W- Thus the error in the quotient - is that in the product ab divided by b 2 . As a per cent this error is — \ —, which is the ab same as it would be in the product ab. Thus all that has been said concerning products hereto- fore holds also for quotients, so far as per cent errors is concerned. § 40. The Shortened Process of Division. Special Example : find the value of * • The first step is to determine to how many places the division may be carried accurately, assuming the terms as representing approximate numbers and thus subject to rejection error. This can be determined by calculating the error in the quotient by the formula of § 39. This is too much trouble --as much as performing the division itself. It is better to get the rejection error approximately as a per cent, and calculate roughly the quotient, and thus calculate the inaccurate place in the quotient. + fe) 56 PLANE TRIGONOMETRY. [§40 In the quotient above, the per cent error is (by § 27) 100/ 1 2 U57853 ' or, approximately, - • — = zrrj °f 1 °lo . The quotient is roughly 2, of which — — of 1% will be a figure in the fourth decimal place. The division may thus be carried accurately not beyond the third decimal place. 2.198 7182)15785.3 14364 14213 7182 7031 6464 567 574 Explanation. (1) Make the denominator an integer by multiplying both numerator and denominator by an appropriate power of 10. The quotient is set in the top line, and so that the decimal points fall the one under the other. (2) In carrying out the division, one proceeds as in arithmetic until the final significant figure of the dividend is used. This, above, is until the remainder 7031 is reached. (3) When such figure is reached, then, instead of drawing down zeros in the dividend, as in arithmetic, the divisor is cut one figure, making it 718, and 718 goes into 7031, 9 times, giving the third figure of the quotient. One carries from the rejected 2 of the divisor in multiplying 718 by 9 ; 9 times 2 is 18, and 2 is carried to the 9 times 8, making §40] CALCULATION VICES AND DEVICES. 57 74. At the next step the divisor is cut to 71, which goes into 568, 8 times. In multiplying by 8, 6 is carried from 8 times the rejected 8 of the divisor. A second example : F 673.2 23.29 6732)156780 13464 2214 2020 194 135 59 60 EXERCISES. Determine to how many places the divisions following should be carried, and then do the work as in the preceding examples : 15374 . 327.3 . 427. 4J5 624.2' 31.8384' 61.7' 7.3* The teacher may assign other examples at pleasure. In the divisions which occur in practical work, arising from measurements intelligently taken, both the divisor and divi- dend will, as already pointed out, show the same number of significant figures, and the limit of accuracy takes care of itself in the actual division by the disappearance of the divisor itself under the pruning process to which the method subjects it. A number of examples should be selected, in which numera- tor and denominator show the same number of significant figures, the number of places to which the division may be carried accurately determined by the methods given, and then the division by the shortened process should be carried out, noting that the divisor vanishes opportunely. Test results by using logarithms. 58 PLANE TRIGONOMETRY. [§41 § 41. If two numbers subject to rejection error are multi- plied together by the use of logarithms, what effect have such errors on the logarithms and on the final result as calculated by logarithms ? It has been pointed out in § 12 that if two numbers differ by e, a small number, their logarithms will differ by (0.4342. . .).£". x Also log (abc . . . V) = log a -f- log & + •••+ log I. Thus, if a, b, c'. . . Z, have the errors x v x 2 . . . # n , the error in the \og(abc . . . I) is (0.4342. . .-j/a + Sl. • .3i\ In the case where the #'s are rejection errors, this becomes, as in the cases already considered, (0.4342 . . .) /! 1 1 . . 1_\ 2 U -B C7- xy Corresponding to this error in the log of the product will be the relative error in the product itself, 2\A B B. & This is exactly the same as would be the relative error due to rejection error if the multiplications were carried out directly. (See § 25 and § 31.) Consequently the use of logarithms has no effect on the final result other than that arising from rejection error in the logs themselves, and which has been considered already in § 22. § 42. Approximate Values of Powers and Roots. When x is small compared with fa + x) n = a + nx;' (a + x) n = a n + -• -—• a ra If a: is a percentage = — — (z per cent), the above results become 2 " I I 1 - 1 z (a + a;)" = a n + rca n • j^j ; (a + a;)" = a n + - • a n • — -. Tims the per cent of error in the square of a number is, approximately, twice that in the number; in a cube, three times that in the number, and so on. The per cent of error in a square root is, approximately, one-half that in the number ; in a cube root, one-third, and so on. EXERCISES. 1. If x = 0.000008168, what is the value of Vl - a;? Vl - x = 1 - I x, nearly, = 1 - 0.000004234, = 0.9999958.' 2. If the side of a square is 1.0002, what is the area approximately V (1.0002) 2 = 1 + 2 x 0.0002 = 1.0004, nearly. 3. Find the approximate yalue of (1.0001) 8 . 4. Find the approximate value of Vl.0003. 5. Find the approximate value of V0.9999996. 6. If the mantissa is zero, what modification is made in the rule on page 11 for getting the cologarithm? What is the colog 0.01? Colog no) ioooooo 7 CHAPTER III. ANGLES AND ANGLE-UNITS. § 43. Angles in Formation and in Sign. The student beginning the study of trigonometry is already familiar with the word angle. Some extension of his ideas, as to size and sign of angles, is essential. Imagine OA, OP, to be two straight lines hinged together at ; and that OA is held fixed in position while OP turns in a plane about 0, starting from coincidence with OA. Then the amount that OP has turned, as compared with some unit-turn, is a meas- ure of the angle between OA and OP. OP may turn as do the hands of a clock (clockwise), or in the opposite way (counter-clockwise). To distinguish between the two, when this is desired, angles described by a counter-clockwise turn of OP, as M, are called positive angles ; the reverse, like N, negative angles. The line OA is called the initial line ; OP, the terminal line, or simply the terminal, of the angle. When, therefore, significance is attached to the sign of an angle, a distinction is made between its border lines. One is the initial line ; the other, the terminal line. The latter must, for positive angles, lie in the order of a counter-clockwise turn from the former ; the reverse, for negative angles. h / / V^ N 1 1 1 1 i 1 Initial \ line \ / \ / \ / s / N s Fig. 1. § 44. Adding and Subtracting Angles. Angles are added as are numbers. To add A to B, pictori- ally, first lay out A, paying attention to both magnitude and - 60 §45] ANGLES AND ANGLE-UNITS. 61 sign. Then from the terminal of A as a new initial line, lay- out an angle equal to B and of the same sign. The old initial line of A and the new terminal of B form the initial line and terminal of A + B. This applies to ± A + ( -f- i?) as well as to ± A + ( — i?) . To subtract the angle B from the angle A, lay from the terminal of A an angle equal in magnitude to B, but of opposite sign. The initial line of A and the new terminal of B form the border lines in order for A - B. This applies to ± A - ( + B) and to ± A ~(-B). In Fig. 1, OA, OB border many different angles, so that the expression, angle AOD, is indefinite. However, any angle A OB is the sum of two represented by A OP, P OB. It is also the difference of two represented by A OP and BOP. So is AOP equal to AOB + BOP, or, it is AOB-POB. EXERCISES. 1. What is the method used in geometry for constructing an angle equal to a given angle ? 2. Draw with a straight-edge some angles at random. Construct an angle equal to the sum of two selected positive angles ; then to three. Add a positive and a negative angle, selected at will. Subtract a positive angle from a given larger positive angle ; from a given smaller positive angle ; a positive from a negative angle. Make the number and variety of exercises sufficient unto proficiency, but not unto weariness. Note. — The teacher using this book is advised against assigning to tiresomeness exercises that are alike. Nothing deadens intellect more than having to do too much of a tedious thing, which one can see readily how to do, but does not wish to do. * § 45. The Initial Line Par Excellence and the Quadrants. When angles are considered singly, with reference to size and sign, and with reference to certain related magnitudes, called sines, cosines, etc., which vary with the size and sign of angles, it is customary to take as initial line a right-hand horizontal line, as OA in Fig. 1 or Fig. 2. The point is called the origin; also the pole. Continuing OA Fig. 2. II I III IV 62 PLANE TRIGONOMETRY. [§45 backward through the origin, 0, and drawing through 0, at right angles to OA, another line, and extending both lines in imagination infinitely in both directions, we divide the plane of angles into four quadrants. These quadrants are numbered and named, as indicated in Fig. 2, in counter- clockwise order, the first quadrant (I), the second quadrant (II), the third quadrant (III), the fourth quadrant (IV). > ^ Angle of lei Fic , quadrant k3. Angle Fi( of 2nd j. 4. Angle of 3rd Fig. 5. / Angle of 4th Fig. 6. An angle whose terminal is in the first quadrant (I), no mat- ter what its size or sign, is said to be an angle of the first quadrant. When the terminal is in the second quadrant, the angle is said to be an angle of the second quadrant ; similarly, with reference to the third and fourth quad- rants. In all these cases the right-hand horizontal line is reckoned the initial line. / M § 46. Angles unlimited in Size. In geometry the angles considered are, for the most part, less than two right angles. In trigonometry no limit is set upon the size of angles. That is, no limit is set upon the extent of turn of OP in * Fig. 7. It may pass any posi- tion any number of times, finite or infinite, and in either direc- tion. In Fig. 7 the arrow M indicates the smallest positive Fig. 7. angle, M, corresponding to that / §47 a] ANGLES AND ANGLE-UNITS. 63 position of the terminal. If T denotes one complete turn of OP, from the position OA, back to that position again, then M± T, M± 2 T, M± 3 T, M± nT (where n is any integer), are angles whose terminals are coincident with that of M. Denoting by the arrow N, in Fig. 7, the smallest negative angle, numerically, corresponding to the given position of the terminal of M, then also, evidently, N± T, N± 2 T, N±nT, are angles whose terminals are coincident with that of M. Thus M+nT is a general formula for all angles whose terminals are coincident with that of M, n being any positive or negative integer. The smallest angle, numerically, locating any given ter- minal, is called the principal angle of the terminal. § 47. Measuring Angles. To measure an angle is to determine its relation in magni- tude to some unit-angle, that is, its number (or its ratio to the unit-angle). This is done in practice, in outdoor work, by means of graduated circles, with some means of " pointing," more or less accurate, ranging in accuracy from the u sights " on two uprights, as in the surveyor's compass, to the telescope of a transit instrument. Detailed descriptions of such appa- ratus will be found in books on surveying. Angles on paper are measured by means of a protractor. This is taken up in the following chapter. We are concerned here, for the present, with the unit-angle. (a) The Sexagesimal Angle Measure, or Degree Measure. In America, England, and Germany (and some other coun- tries) the unit-angle is one degree (1°), or ^^ of a complete turn. Fractional parts of this unit are expressed, ordinarily, in minutes and seconds ; sometimes in decimal parts of a degree. The student is assumed to be familiar, from his work in arithmetic, with the notation for degrees, minutes, and seconds, as in 24° 13' 43", and with the table: 64 PLANE TRIGONOMETRY. [§47 a 60 seconds make a minute, 60 minutes make a degree, as also with the conversions which may arise in connection with the table. From the prominence of 60 in the table, this system of angle measurement is frequently called the Sexagesimal Sys- tem ; more frequently, however, merely the Degree Measure. To the Babylonians is ascribed the credit of originating this system. It is undoubtedly of very great antiquity. The "360 degrees in a cir- cumference" dates back to that remote period when astronomy was in its infancy and men thought the year contained 360 days. Thus the unit- angle, one degree, is but what was thought then the daily step (gradus) of the sun in his apparent annual walk around the ecliptic, his step being each day about twice his angular diameter. Why " 60 minutes make a degree " and " 60 seconds make a minute," is lost in obscurity. A fairly good guess, but only a guess, is that the Babylonians had noticed that the radius of a circle, used as a chord, would go just six times around the circle, subtending 60 degrees, thus making 60 a sort of " charmed number." (5) The Centesimal Angle Measure, or the Grade System, or French System. In this system of units, the right angle is divided into 100 equal parts, instead of 90, as in degree measure, and one hundredth of a right angle is the primary unit-angle. This unit is called a grade. A minute and a second in this system are, respectively, the hundredth and ten-thousandth of the grade, or, 100 seconds make a minute, 100 minutes make a grade. In this system the angle is, therefore, always expressed decimally. The expression 19 g .3552 is 19 grades, 35 min- utes, 52 seconds, while 3 g .0407 is 3 grades, 4 minutes, 7 seconds. Also 12 g .3456 is 1234'. 56 and 123456", the unit being changed by merely moving the decimal point. The decimal system has here, as in all other cases, great advan- tages in adaptability to calculations. Can you mention any serious practical difficulty which ^sjl / §47c] ANGLES AND ANGLE-UNITS. 65 stands in the way of a change from degree measure to grade measure in America? (), curve- tracing (Graphs, §§ 102, 132), etc. Here it should be used to estimate with the eye the size in radians of degree-angles, that the student may really know what a radian is. Groat's Polar Coordinate Paper can be supplied by the H. W. Wilson Co., Minneapolis, Minn., in both radian measure and degree measure. §47c] ANGLES AND ANGLE-UNITS. 67 (4) The radian measure of an angle can also be pictured as a line (arc). Let A OP be the given angle. Draw about the unit circle : the portion of this circle-arc included between the initial line and terminal line of the angle is a picture of the radian measure of the angle ; that is, the number representing the length of the arc BD (between the arrows) is the same as that representing the angle FlG 10 BOD. For arc BD angle BOD n^n • j- — - — : = — ^— — = BOD in radian measure. radian s arc radian But the radian's arc is the radius, which is here unity. •\ arc BD— radian measure of the angle BOB. Since the scale-unit is quite arbitrary, so is the size of the adjoining picture. LABORATORY EXERCISE. Construct a pasteboard circle of one inch radius (or of one foot radius). Lay out five central angles on the circle. Cut strings equal in length to arcs of these angles. Get the lengths of the strings ou the inch rule (or in decimals of a foot, if a circle of one foot radius is used). These lengths are the circular measures of the angles. Measure the angles with the protractor. Reduce the protractor measure to circular measure (§ 48). Compare results with string lengths. (5) Similar to the preceding is the surface picture of the spherical measure of a solid angle. Draw about the vertex of the solid angle a sphere of unit radius : the portion of the surface of this sphere included within the solid angle, and bordered by arcs of great circles, is the spherical measure of the solid angle. (6) It is becoming customary, as is done in this book, to denote angles expressed in radian (circular) measure by the small letters of the Greek alphabet, while the capital letters of the English alphabet are used to denote angles expressed in degree measure. Since many students 68 PLANE TRIGONOMETRY. [§47c taking up trigonometry are not familiar with the Greek letters, we give here the Greek alphabet for reference. No attempt to memorize it is recommended. When the student meets a new Greek letter acquaint- ance (there are but few of them not met with in mathematical journals and astronomical publications), he can look it up here. GREEK ALPHABET. A B r A E Z H © I K A M a y 8 e t V 0or i K alpha N beta S gamma O delta n epsilon p zeta 5 eta T theta Y iota kappa X lambda * mu Q nu xi omikron pi rho sigma tau upsilon phi chi psi omega (7) Circular Ares in Radian Measure. given arc on v Take AB as any a circle of radius OA = r. Draw the unit circle, CBO, concentric with the given circle. Then nrc AB _ radius OA __ radius OA arc CD radius 00 1 ,\ arc AB = radius x arc OB. But arc OB is the circular (radian) measure of the angle A OB (as shown in (4), page 67) or 6. .*. arc AB = r -6. That is, the are of any circle is its radius, r, multiplied by the circular measure, 0, of the central angle which it subtends. arc AB Fig. 11. Also 6 = that is, the circular measure of an angle can be expressed as the ratio of a line to a line (the radius), making it of the same char- §48 a] ANGLES AND ANGLE-UNITS. 69 acter as the ratios which form later the chief study of this booh. Herein lies the importance of this system of measurement in theoretic mathematics. LABORATORY EXERCISE. Construct from pasteboard a circle of one inch radius. Construct five other circles of pasteboard, some larger, some smaller, than the inch circle. Lay out equal central angles on all the circles. Measure with strings the lengths of the corresponding arcs. Get the lengths of the strings (in inches) and the lengths of the radii. Show that for each circle the length of the arc is its radius times the length of the arc on the inch circle. (8) Similarly, the portion of the surface of any sphere, when the boundary lines are great circles, is r 2 times the spherical measure of the corresponding solid angle. (9) Areas of Circular Sectors in Terms of Radian Measure of their Angles. It is shown in geometry that the area of such a sector is one-half its radius times its arc. .\ area = %r • r6=\ r 2 ^, where 6 is the circular measure of the angle. (10) Similarly, the volume of a spherical sector is, J r . tty = i rty where is the spherical measure of its solid angle. § 48. Conversion of Angles from One Measure to Another. (a) Degrees to Grades and Vice Versa. One right angle is 90 degrees and is 100 grades. Let x denote the number of degrees in an angle, and y the number of grades in the same angle. .%£-¥•-*+.•!* (1) tmdx=£ j y = y- 1 \y. (2) Thus, by (1), to convert degrees and decimal parts of a degree to grades, add one-ninth of the given number to itself. 70 PLANE TRIGONOMETRY. [§ 48 a When minutes and seconds are given, they should first be changed to the decimal part of a degree. And by (2), to convert grades to degrees and the decimal part of a degree, subtract one-tenth of the given number of grades from itself. This is not a matter of very great importance. The teacher may assign a few practice examples, selected at random. (5) Degrees to Radians and Vice Versa. This is a matter of considerable importance. Denoting by it the number 3.14159 •••, as in geometry, the circumference of a circle of radius r, is 2 7rr, and any two arcs of the same circle are to each other as the central angles which they subtend. radian radius r 1 4 right angles circumference 2 irr 2tt __ 4 right angles ' 180° _ 180° ... radian- g- - — - 3.14159 '"* .\ radian = 57°. 2957 ••• degrees, or, 57°. 3, approximately, = 3437'.747- = 206264".806 - = 57° 17' 44".8. .*. The rule for converting degrees to radians : (i) When the angle is in degrees and decimals, divide by 57.2957 •••, cutting according to the accuracy desired. (ii) When the angle is expressed in minutes and decimals, divide by 3437.747 •••, cutting according to the accuracy desired. (iii) When it is given in seconds, divide by 206265. Division by the preceding numbers is the same as multi- plication by the following, respectively : (iv) For degrees, 0.017453 (v) For minutes, 0.000291 -. (vi) For seconds, 0.00000485 -. When the relation of degree measure to radians is merely §48 6] ANGLES AND ANGLE-UNITS. 71 to be indicated, the work being left unperformed, we write : r XX x' = 180 ' 10800 648000 To change radian measure to degree measure, use the pre- ceding divisors for multipliers and the multipliers for divisors. When numbers indicate radians, it is customary, in case special attention is to be called to the fact, to use r or c as an index. Thus 2 r or 2 C would mean two radians, or 114°.59. EXERCISES. 1. Express in degrees (decimally) the angles : 3 r ; 2 r .5 ; 4 r .7 ; 4 r .23. Test the results by working them backwards. 2. Express in radians the angles: 180°; 360°; 90°; 45°; 30°; 60°; 235°; 270°; 19F.37; 424°.76; 5°.39; 17° 46' 35"; 5° 43' 26"; 10'; 10"; 1°; 1'; 1". Test by working backwards. 3. If the radius of a circle is 57.3 feet, how long is the arc whose central angle is 1°? If 3437.7 feet, how long for 1'? If 206265 feet, how long for 1" ? 4. If a man 6 feet tall, leaning on the rim of a circle, covers the arc of a radian, what are the circumference and radius of the circle ? 5. If an object subtending an angle at the eye of 2£' is the smallest object that is visible, how far away is a brick building when the hori- zontal lines of plaster are just discernible? 6. Find the ratio of 13° 24' 56" to 4.57 radians. 7. Assuming the radius of the earth as 4000 miles, find the length of a radian on the equator ; of a degree ; of 1' ; of 1". 8. The circular measure of two of the angles of a triangle are £ and § ; what is the third angle in degrees and in radians ? 9. The angles of a quadrilateral are in arithmetic progression, and the greatest is double the least; express each angle in radians and the least angle in degrees. 10. The angles of a triangle are in arithmetic progression, and the number of radians in the least angle is to the number of degrees in the mean as 1 : 120 ; what are the angles in radians? 11. The diameter of the whispering gallery in St. Paul's is 108 feet; how long is the arc of a radian? Of ir radians? What is the area of the circular sector whose arc is a radian in this circle? 72 PLANE TRIGONOMETRY. [§49 12. The large hand of the Westminster clock is 11 feet long ; how- many yards does its moving extremity travel in passing over a radian? What area is described ? 13. What is the circumference of the circle on which the radian arc is 3 inches? What is the area of the sector of this circle whose arc is the radius? 14. The apparent diameter of the moon is 30' and that of the sun is 32' ; how many moons, placed tangentially to each other on the cir- cumference of a circle, would it take to cover a radian? How many suns ? Look in the almanac for the time of sunrise and sunset to-morrow, and determine how many suns, placed tangentially, would cover the day- arc of the sun's motion that day. The night-arc. 15. A railway train is going at the rate of 60 miles an hour on a circular curve whose radius is f of a mile ; how long does it take to pass over a radian ? 16. How long does it take the minute hand of a clock to pass over a radian ? 17. If 7r is taken as %?-, what common fraction expresses the number of degrees in a radian? 'What if it is taken as fff ? 18. What is the area of a circular sector whose angle is § r , the radius being 10 feet ? 19. A radian being 206265", at what distance does 1 foot (assumed as a circular arc) subtend an angle of 1" ? Of 1' ? Of 1° ? 20. The circumference of a circle is divided into 5 parts which are in arithmetic progression, and the greatest part is 6 times the least ; find in radians the angles subtended at the centre by the parts. § 49. Special Angles in Radian Measure (ir-Measure). By (5) of the preceding section, 180° in radian measure is — - times 180 radians, or ir radians. 180 Thus the ratio of 180° to the unit of radian measure is the same as the ratio of the circumference of a circle to its diameter, or 3.14159 • • •. Consequently it is the Greek letter representative (see (6), page 67) of the angle which is the equivalent of a half-turn, or 180°. It is customary to write 180°, in circular measure, simply as 7r, instead of 7r c , or 7r r . Thus: 50] ANGLES AND ANGLE-UNITS. 73 360° = 2 7T 540°= 3 7T 720° = 4 7T 90° = IT 45°=? 30 c 6 ± 180° = ± 7T ± 60° = ± IT 120° = f7T 270° = 1 7T 240°= 4 7T 135°= |tt, etc. EXERCISES. 1. Express in degrees, in radians, and in terms of w radians, the central angles subtended by the sides of each of the first twelve regular polygons. Express in terms of 7r radians the angles whose terminals are bisectors and trisectors of the quadrants. 2. Do the same for external angles of the same polygons. 3. What is the central angle of a regular polygon of n sides in terms of 7r ? In radians ? In degrees ? 4. Show that A ° expressed in 7r-measure is - — it. v 180 5. Reduce some numerical angles in degree measure to ir-measure. LABORATORY EXERCISE. Construct of pasteboard a circle of one foot radius. Lay out the angles noted in § 49. Use strings to measure the arcs, get the lengths of the strings (in feet), and test the above results numerically, in com- parison with 7r = 3.1. § 50. Special Terminals located in Radian Measnre (ir-Measure). If n is any positive or negative integer (including zero), the angles 2 nir radians (or, as it is usually writ- ten, 2mr) will have their terminals coincident with the initial line OA. Thus if a is any angle whose terminal is given, all the angles 2 mr + a will have that same terminal. The lines of the adjoining diagram indicate the quadrant and semi-quadrant lines. Thus all angles 74 PLANE TRIGONOMETRY. [§ 50 with terminal OA have the form 2»7*, as 0, 2 7r, 4-7T, etc.; with terminal Oi?, the form (2n -f l)7r, as 7r, 37r, 5 7r, etc.; with terminal 00, the form (2w + J)7r, or (4w + 1)^, aS 2'-2-'^' etC - ; with terminal OB, the form (2 w -f- |)7r, or (4 w -f- 3)^, u 37T 7 7T 11 7T aS ^"'"2"'^"' etC - ; with terminal OE, the form (2 w -f ^)7r, or (8 w 4- 1) —, 7T 9 7T 17 7T . as -, — , — , etc.; with terminal O^ 7 , the form (2 n + |)7r, or (8 n -f 3) j, 3tt 11 tt 19 tt , as T 1, "5^ ~T~' etc,; with terminal 06r, the form (2 w + |)7r, or (8 w + 5)—, 5 7T 13 7T 21 7T , as — , -^—, -j-, etc. ; with terminal OH, the form (2 n + |)tt, or (8 n -f- 7)—, 7 7T 15 7T 23 7T . as — , — , — > etc. EXERCISES. 1. Show that if n is an integer running from— go to + oo, 2 n + 1 and 2 n — 1 represent the same set of numbers. Find expressions for every alternate even number ; every alternate odd number ; every fourth even number ; every fourth odd number. 2. Consider the first twelve regular polygons as central at 0, with one vertex on the initial line, and, denoting by the symbol (VI, 4) the terminal line for the fourth vertex of the hexagon, counting that on the initial line as first and the order being counter-clockwise, write the general expression for all angles whose terminal is that of any specified vertex of any one of these regular figures. For example, for (VI, 4) it is (2 n + 4 x £)tt or (12 n + 4)§ tt.- §50] ANGLES AND ANGLE-UNITS. 75 3. Taking four different angles (two positive and two negative) whose terminal is OE in Fig. 12, write out from the general formula 2 mr+ a, where a is each of the four angles in turn, four sets of ten angles each, placing them in parallel vertical columns, with the angles obtained by the same value of n in the same horizontal line. Compare the sets and see, if n were given all integer values from positive infinity to negative infinity, whether or not the four sets of angles would be identical. Can one set be brought to coincidence with any other set by sliding ? Is the proposition true that if a is any angle of a given terminal, the formula 2 rnr + a will give all the angles corresponding to this terminal and only these ? 4. State for each of the following angles the quadrant to which it belongs : 1*5 -jp W; |*i 9tt; -|tt; -|; -|tt; -f*J n *"+?; (2n + l)7r+§7r; (2n-l)7r- 3 2 7r; 2iMr+|j 2«r-|j -§*■; -§7r; _|» 5. State for each of the following angles the quadrant to which it belongs: 50°; 120°; 240°; 396°; -50°; -60°; -101°; -380°; -1000°; -1080°; -90°; 90°; 180°; -180°; 270°; 360°; 425°; 370°; 425°; 590°; -750°; -39°. 6. Give for each of the angles of Exs. 4 and 5, two other angles which have the same terminal, expressing those for Ex. 4 in radian measure. 7. Give the general expression in degrees for the difference of all angles whose terminals are coincident ; the general expression in circular measure. 8. Give the general expression in degree measure for all angles whose terminals are symmetric to the vertical with the terminal of 30°; the general expression in circular .measure. 9. Give in degree measure the general expression for all angles whose terminals coincide with that of 45° ; — 45°. 10. Give in degree measure the general expressions for all angles which have for terminals the border-lines of the quadrants ; the general expressions in circular measure in terms of ir. 11. Give in degree measure all angles whose terminals are coincident with that of ^4°; symmetric to that of A , with the vertical; with the horizontal. 12. Give in all three systems of angle measure two angles whose ter- minals are symmetric to the initial line ; to the 90° line ; to the 180° line ; to the 270° line ; to the lines bisecting the quadrants. 76 PLANE TRIGONOMETRY. [§50 13. Give the general expressions for some selected three of the angles of Ex. 2. 14. What are the expressions for the length of a circular arc and for the area of the corresponding sector? Selecting two angles involving degrees, minutes, and seconds, calculate the lengths of the corresponding arc and areas of the corresponding sectors, on a circle of 10 inch radius. 15. Find the length of the arc subtending an angle of 2 r on a circle of 8 inch radius. Express in terms of radians the corresponding lengths and areas for 30°, 45°, 60°, 120°, 135°, 180°, 210°. 16. If the radius of a circle is 100 feet, what angle in radians does a 10 foot arc subtend? Express the same result in terms of ir radians. Make up and solve four other examples like this. 17. Find in degree measure, grade measure, and radian measure the angle between the hour and minute hands of a clock at 20 minutes of 6 ; at 2 : 30. 18. The angle of a circular sector is 22 1° and the diameter of the circle is 10 feet ; what is the area of the sector and the length of its arc? 19. The area of a sector of a circle of unit radius is 10 square feet; what is its angle in all three measures ? 20. A strip of paper a mile long is rolled tightly into a circular cylin- der. The paper is 0.001 inch thick ; what is the radius of the cylinder and the volume of the cylindrical sector whose angle is a radian ? 21. Three circles, each of 10 inch radius, are tangent to each other ; find the length of the arcs between the points of tangency, the area of the sectors and the area bounded by the circular arcs, assuming ir — -\ 2 -. 22. If a geographical mile on the earth's surface subtends 1' at the centre of the earth (radius 3960 miles), how far off is the sun if the earth's radius subtends at the centre of the sun an angle of 8.76" ? 23. The moon subtends an angle of about 30' from the centre of the earth, and is about 60 earth's radii distant ; what is its diameter, approxi- mately, and what angle does the earth subtend at the moon ? 24. The radius of a circle is 8 feet; how many radians does an arc of 13.2 feet subtend? 25. If the number of degrees in one angle of a triangle equals the number of grades in another and the number of radians in the third, what is that number ? 26. Show that mr + - will give angles whose terminals are either on the upright vertical or downright vertical, n being any positive or nega- tive integer. CHAPTER IV. CONSTRUCTION OF ANGLES AND OP STRAIGHT-LINE DIAGRAMS TO SCALE, AND THE MEASUREMENT OF ANGLES. [Note to the Teacher. — It is advised that in solutions of examples diagrams to scale be made and the ungiven parts be measured and compared with the calculated parts as a test. The diagram will serve as a check on numerical solutions. Even a rough free-hand sketch will serve often as a check on results, giving a " common sense " check, saving one from writing 139 for 1.39, and the like. Drawing to scale is valuable exercise in itself, in preparation for "graphic solutions," so useful in all engineering work.'] § 51. The Protractor (in Degree Measure and in Radian Measure). The protractor is an instrument for laying out angles of a given size on a diagram and for determining the size of the angles of a given diagram.* (a) To lay out at a Given Point on a Straight Line a Griven Angle. Place the straight edge of the protractor on the given line and the middle point of the straight edge at the angle- vertex. Then with a sharp pencil, or with a pin, make a dot opposite the given angle on the circular rim of the protractor. Connect the dot and the angle-centre by a straight line. (6) How to measure a given angle with the protractor will occur to the student without direction. * A protractor (in degree measure) and a scale of equal parts in inches and fractions of an inch should be used in connection with this book. Groat's Coordinate Paper in degree measure and in radian measure furnishes a cheap protractor, and can be used to fix in the mind the ability to estimate angles as well as measure them. Protractors can be bought made of paper, of horn, of brass, etc. See Johnson's "Surveying," or the catalogues of the instrument makers, as that of Dietzgen, New York. 77 78 PLANE TRIGONOMETRY. [§52 LABORATORY EXERCISES. 1. Make radial clippings, from Groat's Coordinate Paper, of angles of various sizes, and study them until you feel able to estimate fairly well, in degree measure, radian measure, and 7r-measure, the size of any angle selected at random. 2. Construct some angles at random. Guess at their size. Write down your guess. Measure the angles and test yourself as a guesser on angles. Repeat the process until you feel like congratulating yourself. 3. Construct angles of special sizes until you can readily estimate by eye the number of degrees and the number of radians in a given angle. 4. Construct the angles of the first ten regular polygons, beginning with a triangle. Construct a radian and a degree. 5. What angle is subtended at the end of a pencil line of average width by the width at different points along the line ? 6. What angle is subtended by the opposite edges of a chalk-spot just visible on a blackboard ? What angle is subtended by the distance between the pairs of a double star just apparently double ? * § 52. Drawing to Scale. A straight-line diagram to scale is one similar to the object represented, that is, having all its angles the corre- sponding angles of the object, and any pair of its sides bearing to each other the same ratio as do the correspond- ing lines of the object. Architects' plans of buildings, sur- veyors' plots of fields, are familiar examples. The ratio of any line of the drawing to the corresponding line of the object is the scale of the diagram. This scale should be indicated on the drawing. This may be done by noting on the map, "scale 1 to 10," or "scale 1 inch to 1 mile," * This angle is, for most very good eyes, about 2 J'. You can test your eye by seeing at which corner of the parallelogram under the bright star Vega there is a double star. The stars are about 2|' apart, and but few students can correctly report the corner. This should convince the student that when he calculates seconds, tenths of seconds, and even hundredths of seconds, in angles, as many text-books do, he is splitting hairs very fine. Visibility varies with the eye and with contrast in color of object and back- ground, running from about 30" to 2|'. Mr. F. L. O. Wadsworth has shown that much of the "fine measurement" with instruments is merely an optical delusion. §52] DIAGRAMS TO SCALE. 79 as the case may be. Look this matter up on some railway map, or map in a geography, and report as to how the scale was indicated. The scale will, of course, depend on the size of the object as compared with the size of the draw- ing desired. It may be an inch to many miles, or an inch to a few feet, according to the amount of detail desired. The drawing may be on reduced scale, as railway maps, or on enlarged scale, as in drawings of microscopic objects. EXERCISES. (Using protractor and straight edge.) 1. Draw to scale a triangle, given two sides and the included angle, the sides being 217, 250, angle 63° 15', the longest side of the diagram not to be over 5 inches. Measure the other two angles. Test. 2. Given two sides and the angle opposite one of them. Sides, 240, 224; angle, 47° 30', opposite 224. How many triangles are possible? What change would be necessary in one of the given sides to make just one triangle possible ? Can the sides be so given as to make the triangle impossible ? How many triangles are possible when the angle given is to be opposite the larger of the two sides? 3. Given two angles and the included side. Angles, 30°, 85° 30' ; side, 10 feet. Make the diagram to scale. 4. Given three sides, 5.2, 8.2, 9.3. Make the diagram to scale. 5. Given the three angles. How many triangles ? 6. In all the preceding cases, use the scale and protractor to deter- mine the ungiven parts. Do you know of any way by which you can test your results ? 7. Plot to scale a right-angled triangle of which one side is 20 miles, and one angle 57° 30'. 8. Plot to scale a right-angled triangle, one of whose angles is 76° 30', and whose hypothenuse is 500 feet. 9. Plot to scale a right-angled triangle whose two legs are 15 and 12. 10. Measure the ungiven parts in the preceding right-angled triangles, and from these measurements and the scale calculate their values. Test the same by other calculations. 11. Look up the map of your state, in some geography or wall map, and calculate from the given scale of the map the air-line distances be- tween its five largest cities. 12. Calculate to three decimal places the ratio of the sides of each of the right-angled triangles in Exs. 7, 8, 9, and the ratio of each side to the hypothenuse. 80 PLANE TRIGONOMETRY. [§53 13. Plot a four-sided field whose sides are, in feet, 500, 240, 120, 180, and whose angles are, following the order of sides, 80°, 70°, 110°, A. How large is A ? 14. Draw the diagonals in Ex. 13, and calculate from the scale their lengths. Measure in the diagram all angles not given, and test them by the three angle tests which hold, together with the corner tests. § 53. The Two Half Terminals, Three Third Terminals, etc. If the angle A is the principal angle of a given terminal, this terminal is also located by the angles, A, A ±360°, A ±720°, etc. Thus the half angles corresponding to a given terminal are |, | ±180°, | ±360°, etc. These locate two, and only two, terminals. These terminals are 180° apart. Similarly, one-third of the angle locating a given terminal will fall under the forms, |, | ±120°, | ±240°, etc. These locate three, and only three, terminals, 120° apart. Similarly, there are four terminals for one-fourth of the angle corresponding to a given terminal, and so on. EXERCISES. 1. Use the protractor to draw the terminals for the half angles cor- responding to the initial line. 2. Do the same for the terminals corresponding to 90°, 180°, 270°. 3. Show that if one vertex of an equilateral triangle is on the initial line, the other vertices are on the terminals corresponding to one-third of the angle of the initial line. 4. Show the corresponding proposition for the square, regular penta- gon, regular hexagon, etc. 5. If one vertex of an equilateral triangle is on the terminal of 180°, show that the other vertices are on the terminals of the third of the general angle corresponding to the terminal of 180°. §54] DIAGRAMS TO SCALE. 81 6. Locate the terminals for one-quarter of the general angle corre- sponding to the terminal of 180°; for one-fifth; for one-sixth. Note. — Later it will be shown that the geometric operations above correspond to the algebraic solution of the equations, x 2 = ±l, x s = ±1, z 4 = ±l, x» = ±l, etc., by De Moivre's theorem. § 54. Instruments for Measuring Angles. It is advisable that along with the study of trigonometry the student be given field-work in measuring angles with the surveyor's compass, with a sextant, and with a transit. In no other way can he learn the limits of accuracy of measurement, and the folly of calculating hun- dredths of seconds when he ought not to do it. EXERCISES. 1. Investigate and report on the following topics : (a) The accuracy of the outfit and processes of a county surveyor in your state, (b) The same for a city surveyor, (c) The same for a steam railroad surveyor. (d) The same for the surveyor of an electric railway, (c) The means of measuring small angles in engineering and astronomy and the small- est readings in different sorts of surveying. (/) Measuring base-lines in geodetic surveys, (g) Limits of accuracy in government land-surveys. 2. A city lot, about 29 feet by 290 feet, was sold at $1000 a square foot, on the measurements 29.3 feet by 293.7 feet. Show that, possibly, the owner lost about $14,600, by not having the short side measured with the same relative accuracy as the long side. Which would entail the larger loss, dropping .3 on the short side or .7 on the long side? 3. Show that J^^ = log 6 a. Show that a x = e xl0 ^°, and from the log e b series for e x (§8) deduce a series for a x . 4. Show that if a set of numbers form a G. P., their logarithms form an A. P. Show that the arithmetic mean of two logarithms is the loga- rithm of the geometric mean of the two numbers corresponding to the given logarithms. Show that the logarithm of any number can be cal- culated to any desired degree of accuracy by the continued insertion of arithmetic and geometric means. Calculate logio2 to four figures in this way. CHAPTER V. THE SINE, ANTI-SINE, RECIPROCAL SINE (COSECANT), AND COVERSED SINE OF AN ANGLE. § 55. Coordinates. (a) Rectangular Coordinates and Rectangular Coordinate Paper. Draw through a pair of mutually perpendicular lines, giving the quadrants as explained in § 45. is called the origin of coordinates, also the origin, also the pole. The border lines of the quadrants are called the axes of coordinates. And for distinction, the old initial line (right and left) is called the axis of abscissas, and the verti- cal line the axis of ordinates. From any point, P, in the termi- nal OP, of a given angle A OP, drop a perpendicular PM, on the initial line OA (produced if necessary). Then OM is called the abscissa of the point P, and is indicated by the letter x ; MP, its ordinate or y ; and OP its modulus, and also its radius vector (we shall use the word modulus and the letter r). Notice particularly the order in which these lengths are written : It is OM, not MO ; it is MP, not PM; it is OP, not P 0. This is of primary importance, for reversal of the order in which a length is written, when sign is considered, is a reversal of direction and equivalent to a change of sign. As to sign, it is now universally agreed that all lines taken upward, no matter where starting, are + ; 82 T S ^ p 4- J,- ^ X- 5* s r ? s s^ *f ^ r» J? X SO S ± ]lf . M \/ M M ^' v ^ M ^ ^^ 7 ^^ S ^-L ^ " £_ ^P f Fig. 13. § 55] THE SINE FAMILY. 83 downward, — ; to the right, -f ; to the left, — . Where no sign is stated, + is understood. A point in the plane of the axes is located by giving its coordinates in a parenthesis, as (2, 3), the abscissa being in the lead. The point (2, 3) is two scale units to the right of the vertical axis and three scale units above the horizontal axis. To reach it, go two units from along the axis of abscissas to the right; then three units vertically up from the point so reached on the abscissa axis. Similarly, to locate (—2, 3), go two units along the abscissa axis to the left from 0, and then three units vertically up from the point so reached. How are (2, - 3), and (-2,-3) located ? The teacher may select at random sufficient practice examples to make sure that the matter is understood by the class. In quadrants I and II, ordinates are -+- ; in III, IV, — . In quadrants IV, I, abscissas are -f- ; in II, III, — . The modulus OP, of the point P, is always counted plus when it lies along the terminal of the given angle. For all points in the opposite terminal (the terminal continued back- wards through the origin), it is minus. Thus, while OP might be counted plus for a certain angle, this same line OP would be minus for the angle which is 180° more than the given angle, or any odd integral multiple of 180° more than the given angle. Even when the terminal falls on the left-hand horizontal line or on the downward vertical, the modulus for any point on it is counted plus, while in the first case the ordinate is zero and the abscissa negative ; and in the second, the reverse. Any measurement along the termi- nal, away from the origin, is, by agreement, plus always, no matter whether the angle is plus or minus, and no matter what its size. Paper like that on which Fig. 13 is constructed is called rectangular coordinate paper. It can be bought in small sheets or by the yard, and of various divisions to the inch (or centimeter). That 10 x 10 to the inch is convenient for locating points in a drawing. 84 PLANE TRIGONOMETRY. [§ 55 (b) Polar Coordinates and Polar Coordinate Paper. A point on a plane is also located by giving the length of its modulus and the angle which the modulus makes with the initial line. These are given in a parenthesis as (r, 0), the modulus being set first. By the previous agreement as to sign of the modulus, a point indicated by (2, 30°) would be two units from the origin along the terminal of 30°, while ( — 2, 30°) would be two units along the opposite terminal from the origin. Where will (2, — 30°) and (_2, -30°) fall? EXERCISES. 1. Calling the small divisions on radii units, locate on Fig. 14 the points (4, 5°), (5, 5°), (6, 5°), (4, 2(F), (4, 25°), (4, 50°), (4,90°), (-5,5°), (-5, 50°), (-5, 180°), (-5,275°), (-8, -20°), (-8, -80°), (-8, -180°), (-8, -270°), (-8, 3000°). 2. Similarly lo- cate on Fig. 15 the following points, the second figure in each parenthesis being radian meas- ure: (6,3), (7,3), (8, 3), (8, 4.3), (9, 5.5), (6, 6.3), (7, - 0.5), (4, - 0.6), (4,3.14159), (4,5, 4, Fig. 14.* 0, ■), (-4, -*■), (-4, -|), (-4, -|), (-4, 3), ( 6). 3. Construct to scale the following points in rectangular coordinates, using, preferably, rectangular coordinate paper : (2, 3), (2, — 3), ( — 2, 3), * Figure 14 is in 5° angle-spaces ; Fig. 15 is in tenths of radians. The divisions on the radius are in each case arbitrary. §55] THE SINE FAMILY. 85 (- 2, - 3), (7, 8), (15, - 3), (- 5, 4), (- 9, - 3), (3, - 3), (3, - 1), (150, 160), (900, -800), (3.3, 4.1), (-3.3, - 4.1), (5.6, - 7.3). 4. Where are (0, 0), (0, 1), (-1, 0), (1, 0), (1, 0), (0, - 1)? 5. Use scale and protractor, or Groat's coordinate paper, to construct the following points: (2, 30°), (2, -30°), (-2, 30°), (-2, -30°), (2, t), (3, 60°), (3, - 60°), (- 3, - 45°), (3, 45°), (5, - tt), (- 5, -^*j, (5, =j£), (150,2*-), (l50, ^), (2, lr), (- 2,3-), (- 2, - 1'), (2, -3'). 6. In what quadrant is a point whose abscissa and ordinate are both plus? Both minus? Abscissa plus, or- dinate minus? Or- dinate plus, abscissa minus ? 7. Use Groat's coordinate paper, taking the small divisions on the radii as units, to locate to the extent of the sheet the points (ra- dian measure for an- gles) : (0,0), (1,0.1), (2, 0.2), (3, 0.3), (4, 0.4), etc. Then join all these points by a smooth curve. This will represent a particular form of the Spiral of Archimedes, namely, r = lO0, the general curve being r = aO. Note that in each point above r is 10 times 6, unit for unit. 8. Use rectangular coordinate paper, taking the small divisions as units, to locate the points : (0, 0), (1, 1), (2, 2), (3, 3), (4, 4), (5, 5), etc., (- 1, -1), (- 2, - 2), (- 3, - 3), (- 4, - 4), etc. Join these points. What then is y = x ? 9. Use rectangular coordinate paper to locate (0, 2), (1, 5), (2, 8), (3, 11), (4, 14), where y is always 3x + 2. Join these points. What do Fig. 15. 86 PLANE TRIGONOMETRY. [§56 you get? What are some negative points on the same figure? What, then, does y = 3 x + 2 represent? What does y = mx + b represent, if m and b are fixed numbers ? 10. If a point is located by (r, A ), what variety of form may r and A take to locate one and the same point? § 56. The Sine of an Angle. The sine of an angle 1 > / / / / / t / ) / & / I 1 I - the ratio of the ordinate of any point on the terminal of the angle to the modulus of the same point, or, in Fig. 16, MP Sine of 6 = OP Fig. 16. This is known as the ratio- definition of the sine, or as Hassler's definition, to distin- guish it from the so-called line- definition. See § 67. It must be considered more as a working definition than as one of theoretic exactness. Sines are not calculated by laying out the angle with a protractor, measuring ordinate and modulus and then dividing the former by the latter. Later it will be found that connected with every angle 6, expressed in circular measure, is a certain number, the sine of the angle, which can be calculated for any given value of 6 by the following series, to any desired degree of exactness : Sin 0=6 — — +— — — ..., etc., to infinity. This series gives the correct theoretic definition of the sine of the angle 6. The sine of an angle is thus a number connected with the angle in a specific manner, as indicated by the above series, and calculable when the angle is given. These numbers have been calculated, approximately, and tabulated for certain angles, forming what is called a Table of Natural Sines. §56] THE SINE FAMILY. 87 All that Hassler's definition means is that if the ordinate were measured with exactness and the modulus with exactness, and then the division of the corresponding numbers carried out as indicated, there would be obtained a number agreeing with that obtained from the series for the same angle. Calculations of the sine from the series give the sine in the form of a decimal fraction. Since the series has an infi- nite number of terms, the sine of an angle can be calculated, as a rule, only approximately. The tables give the sines to 3, 4, 5, 6, 7, 10 places. A table of sines giving them to three places of decimals is called a three-place table; one giving them to four places of decimals, a four-place table, and so on. Ordinates and moduli and angles can be measured only approximately. Consequently, if the sine of a measured angle were obtained from Hassler's definition and the result compared with the tables, the closeness of agreement would depend upon the degree of accuracy with which the measure- ments of lines and angle had been carried out. The practical value of Hassler's definition is, therefore, this : It indicates that if the ordinate and modulus of an angle in a diagram, or in the field, are measured, and their ratio obtained and compared with the tables, the tables will show an angle (one of many) which is, approximately, the angle of the diagram, or one from which that angle can be determined, approximately. The closeness of the approxi- mation will depend upon the accuracy with which the meas- urements are made. The sine of an angle is thus a number connected with the angle and having under Hassler's definition a threefold prac- tical value : (1) When ordinate and modulus are given, it indicates the angle. (2) When the angle and modulus are given, it is the number which multiplied by the modulus will give the ordinate. (3) When the angle and ordinate are given, it is the number by which the ordinate is divided to give the modulus. 88 PLANE TRIGONOMETRY. [§56 These three uses to which the sine may be put constitute its claim to importance as a calculation device and give to Hassler's definition its value. LABORATORY EXERCISES. That the student may get firmly fixed in his mind the definition of the sine as a ratio, the teacher may here show him how to look up in the tables the sines of angles in degrees (omitting minutes and seconds). The student may then lay out to scale some five such selected angles, measure the corresponding ordinate and modulus, divide (to one or to two figures), and compare with the tables. He may also take the modulus one inch long and show that the ordinate (in decimals of an inch) is the table-sine. Do the same with a modulus ten inches long. Make ten such constructions and measure. EXERCISES. (Make diagrams to scale.) 1. If tiie modulus is 5 and the ordinate 3, what is the sine of the corresponding angle ? Calculate the sine to two decimal places and get the angle when the modulus is 34 and the ordinate 25 ; to three places with ordinate 25.3 and modulus 34.7 ; to four places with ordinate 25.37 and modulus 34.72. 2. The side of a square is V2 inches ; calculate to four decimal places the sine of the angle which the diagonal makes with the side, and com- pare with the tables. Show that the sine of the angle which the diagonal makes with the side is independent of the length of the side. 3. Assuming the side of an equilateral triangle as 2, calculate the sine of 60° to four decimal places. Do the same when the side is a. 4. By drawing an equilateral triangle whose side is 2, so that its median is horizontal and the initial line, calculate to four decimal places the sine of 30°. Compare the result with the table-value. 5. Give the sines of 30°, 45°, 60°, in the form of radicals. 6. The modulus for an angle whose sine is 0.35 is 27, what is the ordinate ? How high up on the side of a house will a ladder 50 feet long reach when tilted at an angle of 30° with the ground? 45°? 60°? How high when the sine of the angle of tilt is 0.42 ? Give all these results to only two figures. 7. The ordinate for an angle whose sine is 0.24 is 53, what is the modulus ? If the roof of a house is inclined to the horizontal at an angle 58] THE SINE FAMILY. 89 of 30° and the ridge-pole is 15 feet, how long are the rafters? How long when the angle is 45° ? 60° ? How long when the sine of the angle is 0.32 ? Give results to only two figures. 8. The student will devise seven other examples like the above, solve them and hand them in. Let five of them be practical examples (roofs, etc.) within the student's experience. § 57- The sine has many interesting and important properties, aside from its use in calculations. All these can be deduced directly from the series definition, without reference to Hassler's definition. In fact, a text-book could be based on the series definition. However, for that the times seem not yet ripe. § 58. Hassler's definition holds for all positions of the terminal, no matter in which quadrant it falls. T X X ST T x Z>U \ %S s^ S s y s * Sr- _,• s Q_) Positive or negative common fractions, (c) Unending, non-repeating decimals, positive or negative. §60] THE SINE FAMILY. 91 Ratios of the second kind, (6), are of two classes : (i) Ending decimals, like J = 0.5, ^ = 0.2, etc. (ii) Repeating decimals, like i = 0.333 •••, J = 0.1666 •••. All common fractions whose denominators are 2 or 5, or powers of 2, or of 5, or of 2 and 5, will form ending decimals. For example : 1 = 0.5; £ = 0.2; J = 0.25; £ = 0.04, etc. All other common fractions, when expressed decimally, give repeating decimals. This will become evident by con- sidering some special example, as ^f 1, and carrying out the division as follows, setting the remainders in the upper hori- zontal line : 46.451326 7 )251.000000 35.857142 The remainder, when the zeros first come into use, is 6. Since, now, in dividing by 7, there can be only six different remainders, and since the division is to be non-terminating, this special remainder, 6, appearing in connection with the use of the added zeros, must, sooner or later, reappear, as shown in the actual division. As soon as this remainder reappears, there will be a repetition of the quotient figures obtained since the 6 first appeared as remainder. What has happened here with 7 is readily seen to be general. For with any other denominator, as, say, 213, there can be only 212 different remainders for non-terminating division, and that one present when the zeros are first used must again reappear, giving a repetition of quotient figures. EXERCISES. Show that the following fractions form repeating decimals : 92 PLANE TRIGONOMETRY. [§60 Examples of numbers of the third class, (V), the unending non-repeating decimals, are : 7r = 3.14159 •••, the ratio of the circumference of a circle to its diameter. e = 2.7182818 •••, the base of the Napierian logarithms. V2 = 1.41~., and all other real numbers which do not belong to classes (a) and (5). Originally, only the roots (of rational num- bers) which could not be extracted exactly were called the irrational numbers. Since, however, there is no essen- tial difference between such roots and numbers like 7r, e, in so far as being neither ending nor repeating is con- cerned, all real numbers not of classes (V), (5), may be called irrational. If the numbers of (a), the positive and negative integers, are plotted on a straight line, as in algebra, they are repre- sented by separate points, a unit distance apart. The num- bers of (6) when plotted on the same straight line help to fill in the points between those occupied by the numbers of (a) ; but (a) and (5) do not occupy completely all the points on the straight line. Class (* 5fc S 32 S^ +, or (8*4-1) -180°. The position OP x (Fig. 24) is best reached by going forward the principal angle #, or A°, from the terminal OM 1 ; the position OP 2 , by going back- ward the same angle from the position OM 2 . .\ sin { (2 n + 1)tt - (9} = sin (9, or, sin {(2 n + 1)180° - .A°} = sin 4°. In particular, sin (jr — 0) = sin 0, and sin (180° - A ) = sin A°. (iii) Any angle of terminal OP 2 (Fig. 24) has the same sine as any angle of terminal OP v .'. sin{(2w + l>r-0} = sin(2wwr + 0), or, sin j (2 n + 1)180° - A° \ = sin (2 m- 180° + 4°), where m, w are any positive or negative integers. (iv) Thus, in general, all angles having the same sine as or A°, fall under the forms nir M'i C AU S, • - ' i~4%7 S ^T *_ x y s S *». S s^ 7 s 7 s o -J* sr S 2mr + 9 (2n + l)ir I 0r l(2w+l)180°-^°> §62] THE SINE FAMILY. 95 Since (— V) m is + 1 when m is even and — 1 when m is odd, the two preceding sets of formulas are both included in these : mir + ( - l)«*e, m . 180° + ( - 1)™ . A°, where m is any positive or negative integer. (v) The general result of (iv) is so important that a verbal statement may also be given: If any given angle in degree measure is added to an even multiple of 180°, or subtracted from an odd multiple of 180°, an angle is obtained having the same sine as the given angle. The student may give the corresponding statement for angles in radian measure. (vi) While in the preceding formulas any angle of the given terminal may be taken, it is customary to take the principal angle of the terminal. For example, while 187° and — 173° locate the same terminal, one would use as the general expression for all angles whose sine is that of this terminal, m .180-(-rpl73°, rather than m . 180 + ( - 1)™ 187°, though they both give the same set of values, when m is taken from — oo to -f- oo. EXERCISES. 1. Find twelve angles, six being plus and six minus, having the same sine as 30°. In which quadrants will these angles lie ? Give the diagram. 2. Do the same for - 60°, and for ? and ^JL 4 4 3. Select sufficient examples of the same character to fix the formula in mind. Stop when you know it thoroughly. 4. From Fig. 25, show sin (imr + (- l) m <£) = sin . 5. Show that sin (a x • b x ) = sin (rrnr + (- l)™e xl0 ?e ah ) and that sin — = sin (rrnr +(— l) m e x ° Se i). 6* v y 96 PLANE TRIGONOMETRY. [§63 6. Find the angles which satisfy the equation sin 9 6 = sin 8 6. All angles having the same sine as 8 6 are of the form 2mr + 8 0, or of the form (2 n + \)tt — 8 0. To satisfy the given equation, 9 must fall under the one or the other of these forms : .-. 90=2n7r + 80, (1) or, 90 = (2n + l)7r-80. (2) By (3), = 2n7r, and by (4), $ = (2» + 1)^. The student may make an illustrative diagram and note the positions of the terminals. 7. Solve similarly and give diagrams for (a) sin 7 6 = sin 0, (6) sin20 = sin30, (c) sin w?0 = sin r$. 8. Make up three examples like those of Ex. 7, and solve ; illustrate by diagrams. § 63. Angles with Opposite Sines. Numbers which are equal numerically but opposite in sign are said to be opposite, or oppositely equal. (i) Angles with opposite terminals have opposite sines. Here for equal moduli the ordinates are oppositely equal. .-. sin ( ± 180° + A°) = - sin A°, sin ( ± it + 0) aa — sin 6. And, in general, sin{(2 n + 1) 180° + A } = - smA°, sin {(2 n + 1) it + 0} = - sin 9, where n is any positive or negative integer. ♦ (ii) Terminals symmetric to the horizontal give for equal moduli oppositely equal ordinates. Thus all angles whose terminals are symmetric to the horizontal have opposite sines. Angles numerically equal but opposite in sign form a special case. /. sin (-0) = - sin0, sin t-A°) = - sin A°. 63] THE SINE FAMILY. 97 (iii) Any angle of a terminal has the opposite sine of any angle of the opposite terminal. sin \ (2 n + 1) ir + 0\ = - sin (2 mir + 0), or, sin j (2 n + 1) 180° + A°\ = - sin A% where m, w are any positive or negative integers. -£*- "7 y s *t- y * -i y y o i l -+ -fe «^ * r i ? jzL ^f£ Fig. 26. ^ \ .1/ iT\ Fig. 28. Fig. 29. X A ^■^^ ^^\ ''j -M ^ v ». ^2 ^"-^ ^^\^ ""• ^ ?- Fig. 27. 2j "* _5k ^ -S* *r- if - ^- — ~2 ::e. j .. -*: ■ 4- ^ ■ ' "P -T *t (iv) Any angle of any terminal has the opposite sine of any angle of the terminal symmetric to the horizontal. sin (2 nir + 0) = — sin (2 mir — 0), sin (2 n • 180 + A°) = - sin (2 m 180° - A ), where m, w. are any positive or negative integers. 98 PLANE TRIGONOMETRY. [§63 (v) Thus, while adding an angle in degree measure to an even multiple of 180° or subtracting it from any odd mul- tiple of 180° leaves the sine unchanged, the reverse changes the sign of the sine. (vi) The student is advised against attempting to memo- rize the preceding formulas as mere facts. It is better to hold in mind the effect on the terminal of any given angle, of its addition to, or subtraction from, given multiples of 180°, or of 7r radians, noting that: («) For terminals symmetric to the vertical, sines are equal. (/3) For terminals symmetric to the horizontal, sines are opposite. (7) For terminals which are opposite, the sines are opposite. EXERCISES. 1. Determine from a diagram five positive angles having the opposite sine of 30°, also five negative angles. Let one-half the number of angles be on the opposite terminal, the remainder on the terminal symmetric to the horizonal. Compare the results obtained from the diagram with the formulas. 2. Do the same for 45°; 60°; 120°. 3. Do the same for -135°; -210°; -300°. 4. Do the same for 1780°; 2117°; 3185°. 5. Do the same for |; -; - ?; - |tt. 6. What effect on the terminal of an angle has the addition or sub- traction of a double-even multiple of 90° ? What a double-odd multiple of 90° ? Find, among the first fifty numbers, those multiples of - radians which may be added to or subtracted from a given angle without chang- ing its sine. Also among the second fifty numbers, those multiples of 90° which change the sign of the sine, when added or subtracted. § 64. The Angles in a Table of Sines. From the preceding sections, it is seen that the sine of any angle can be connected, in equality or opposite equality, with the sine of an angle of the first quadrant, between 0° and 90°> inclusive. §64] THE SINE FAMILY. 99 To find such angle for any given angle, note : (a) The sine of a negative angle is the negative of that of the same angle taken positive : sin (- A ) = - sin^l . (6) All multiples of 360° may be dropped. Call the remainder R. (c) If R is less than 90°, it is the required angle. (d) If R is > 90° and < 180°, take 180° - R. 0?) If R is > 180° and < 270°, take R - 180°. (/) If R is > 270° and < 360°, take 360° - R. (#) And then if A° is the given angle, (i) sin (180 - R) = sin A°, case (<*)• (ii) sin (72 — 180°) = — sin J.°, case (e). (iii) sin (360 — R) = — sin A°, case (/). In practice it is best to sketch the terminal of the given angle. (Ji) If the terminal is in the first quadrant, its principal angle is the required angle. (i) If the terminal is in the second quadrant, draw the terminal symmetric to the vertical and take its principal angle. (/) If the terminal is in the third quadrant, draw the oppo- site terminal and take its principal angle (opposite equality). (&) If the terminal is in the fourth quadrant, draw the terminal symmetric to the horizontal and take its principal angle (opposite equality).* EXERCISE. Give the angles less than 90° and greater than 0°, which have sines equal, in numerical magnitude, to those of the following angles, stating in each case the proper sign, and using a diagram : 136°, 172°, 185°, 200°, 275°, 246° 34' 53", 301° 23', - 30° 10', 100° 45', -185° 54', -13° 13' 13", 400°, - 400°, 500°, -500°, 1200°, -1290°, 1800°, - 1800°, 2500°, - 2500°. * Some tables are so arranged that the sine of any angle less than 360° can be taken from the table directly. See Hussey's Tables. 100 PLANE TRIGONOMETRY. [§65 § 65. Supplementary Angles. Any two angles whose sum in degree measure is 180°, or, in radian measure, ir radians, are said to be supplementary. Each is called the supplement of the other. They have the forms 180° - A°, A° ; ir - 0, 0. To con- struct 180° — A% lay out from the left-hand horizontal line an angle equal to the angle A°, reversed in sign. Thus the terminal of an angle and that of its supplement are symmetrically inclined to the vertical. .-. sin (180° - A°) = sin A° ; sin (ir - 0) = sin 0, or, the sine of an angle is the sine of its supplement. EXERCISES. 1. Show from a diagram that sin (180° — A ) = sin A°. 2. If the sines of two angles are equal, are the angles necessarily supplementary ? 3. Does it follow that because the terminals of supplementary angles are symmetric to the vertical, all angles whose terminals are symmetri- cal to the vertical are supplementary ? 4. Name five pairs of supplementary angles which have the same pair of terminals. How many pairs of supplementary angles have the same pair of terminals? If A is an angle of a given terminal, give the gen- eral formulas for all supplementary pairs having this terminal. 5. Select five angles of the first quadrant and determine their sup- plements, illustrating the positions of the terminals by diagrams. Do the same for five angles in each of the other quadrants. 6. Do the same for five negative angles of each quadrant. 7. If A, B, C, are the angles of a triangle, give three formulas con- necting these angles by sines. 8. Connect by sines pairs of opposite angles of a parallelogram. 9. Connect by sines the adjacent angles, when one straight line meets another. Also the exterior angle of a triangle with the sum of the inte- rior opposite angles. 10. Select five angles in circular measure and determine their sup- plements. 11. What is the supplement of 45° - A°1 Of | + 0? THE SINE FAM'LY. J(jj REVIEW EXERCISES. 1. If 45° is a special solution of the equation sin A = — - — , what is the general solution ? + v2 2. How are the terminals of 180° - 18°, 180° + 18°, 360° - 18°, related to that of 18°? How are the sines of these angles related to that of 18°? 3. How are the terminals of all angles having the same sine as 30°, related to the terminal of 30°? 4. How are the terminals of all angles having the opposite sine to that of 60°, related to the terminal of 60°? 5. How is the terminal of the supplement of an angle related to that of the angle ? 6. What is the supplement of 120°? Of -120°? Of 360°? Of -800°? 7. What is the supplement of -? Of - -? Of 4tt? Of - 8tt? 3 3 8. If the sine of A = $, what is the sine of the supplement of A ? 9. State how the terminals of the following angles are related to the border lines of the quadrants, and give the sign of the sine : ±45°; ±60°; ±f; ±£; ±5*; ±?£; ±180°; ±135°; ±120°; o 12 o o ±240°; ±225°; ±330°; ±315°; ±2; ±^£; ± 2n7r + f ; ±(2 n + 1) 9. 2^2'^ 4 r-=S ±<2n-l)»+^. 10. State with reference to each of the following angles the quadrant to which it belongs, the sign of its sine, and the angle of the first quad- rant whose sine is numerically the same : 150°; 120°; 240°; 225°; 330°; 105°; 165°; 195°; 255°; 480°; 498° 510°; 585°; 555°; 975°; 1305°; 1590°; 1665°; ^; !^ ; ^; il^; 11^ 2mr + ^; (2n + l),r-|; (2„-l)» + JL; 2»r-|j (2n-l)»+|| (2n + l)T+-; -405°; -390°; -420°; -840°; -1200°; -1305° -1020°; -1665°; -7|tt; - 5fv; -8fir; -8|*j -6£tt; -IS}}*. 11. Express in radians 46° 30' ; £ of a right angle ; 49° 43' 45" ; 1'. 12. Express the angles in Ex. 11 in terms of it radians. Q if, 13. Express in degrees, minutes, and seconds : -^- ; 2 7r r ; .8 r ; 3.1416 r ; 14. Taking tt as V> now ^ ar °^ * s a 1-foot line when it subtends an angle of 1" ? 192 PLANE TRIGONOMETRY. [§66 15. What angle at the earth's centre does the moon subtend, supposing her diameter 4000 miles and her distance 240,000 miles, taking it = 3.1 ? 16. How far off is a bright object 350 feet tall when just visible? (2 1' is the limit of average eye-vision for a bright object on a black background.) 17. Find two angles whose difference is 1° and whose sum is l r . 18. Divide 77° into two parts such that the number of English sec- onds in one is the number of French seconds in the other. 19. Name the quadrants to which the following points belong : (2, 4) ; (-3, 4); (3, -4); (-3, -4); (2, 30°); (-2, 30°); (2, -30°); (-2, -30°). 20. Define the sine of an angle. 21. If modulus is 27.8 and sine = 0.345, what is the ordinate ? 22. If the ordinate is 3.18 and the sine is 0.432, what is the modulus? 23. If the modulus is 31.4 and the ordinate is ± 16.3, what are the sines, to three places ? 24. A tree is 50 feet high and subtends an angle whose sine is 0.37 from a point in the horizontal plane of its ground-line ; how far is it from that point to the top of the tree ? § 66. Construction of the Terminals for Given Values of the Sine. Required the terminals when sin = J . Consider 9 as modulus and 5 as ordinate. Draw a circle of radius 9 about the origin. Lay out on the axis of ordinates a distance OD equal to + 5. Then draw NDM parallel to the initial line, cut- ting the circle in iV, M. The required positions of the ter- minals are OM, ON. If is some one of the angles locat- ing the terminal OM, as, for example, the principal angle as in the diagram, then, Fig. 30. 2w7T + e (2n + l)w - 61 are the complete set of angles whose sine is f (§ 62) , or, mir + (- l) m 0, §67] THE SINE FAMILY. 103 If the sine had been given as — f, the only change in the preceding construction would have been to lay out OB 5 units downward from 0, instead of upward. In constructing terminals for table-sines, cut the table- value down to the nearest tenth or nearest hundredth. EXERCISES. 1. Construct the double terminals for each of the following values taken as sines (using coordinate paper) : - f , i, - 1, 1, -i,h- h h - h Are any of the preceding values impossible ? 2. Selecting a line to represent unity, construct a line to represent + V2 ; a line for V3 ; for V5. Can you see from the construction for V2, what angles have ±i for sine? What—? What ^? What -=-^?? V2. V2 2 2 3. Where is the terminal when the sine isl? — 1? 0? What are the general expressions for the corresponding angles ? 4. Construct terminals for five sines in the table of sines, and meas- ure the angles. Compare with table-values. § 67. Line Picture of the Sine. ordinate MP M t P t .- ., , . Since sine = n — = -^7 = ' , it the modulus is modulus OP OP x taken as unity, the sine becomes, on the same scale, the ordinate. If, therefore, 0P X (Fig. 31) is taken as a unit, M 1 P 1 is the sine, or sin 6 = ordinate M 1 P V (A). (A) is frequently spoken of as the line definition of the sine. Not many years ago all text- books on trigonometry used this definition exclusively. Two objec- tions have led to its abandonment. Fig. 31. (1) Its use gives a false impres- sion at the outset, namely, that the sine is a line, and a line of a cer- tain length, that of the diagram in the text-book studied. The fact -,£.1 ^_ . 1 7 \ jt^ ' . \ \l ^ 7 4^ n- At I +' %+ t *' l\ ? i£ .* u _£^ t r _l _,*/ .if. \ 1/ v / \ / \ / s^ / ^ s ^-' ■---:=. .-'' 104 PLANE TRIGONOMETRY. [§68 is that the sine is not a line, but a ratio, as already seen. The diagram means merely that the line MP represents the sine on the same scale (any scale) on which OP represents unity. Since the unit-representative is arbitrary in length, so is the line representing the sine. If the line representing a unit is taken half as long as that above, the sine-line is reduced one-half. (2) The second objection to the use of the line defi- nition is that those who learn trigonometry with unity as the modulus, find trouble in introducing any other modulus, whereas, with the general modulus, r, used from the outset, no difficulty is met in assigning to r the special value, 1. The ratio definition was first used in America at the University of Virginia, by Professor Bonny castle. The line definition continued to be used in American text-books to a very recent day, and is still in very general use in engineering field books. Instead of abandoning completely the line definition, as is now the drift in text-books, it appears to the author advisable to hold it for such use as is made in § 69. It presents to the eye certain properties, and thus teaches these properties with greater lucidity than does the ratio definition. § 68. The Expression "Sine of the Arc." It has been shown (§ 47, c, 4) that when the radius is taken as unity, the number representing the arc is the same as that representing the corresponding central angle in radian measure. Conse- quently, when the line defini- tion was in use, the sine of the angle, as now used, was called the sine of the arc. The notation for an angle whose sine is any special value, as |, is sin -1 !- This is generally read, "the angle whose sine is §," frequently, " the arc whose sine is f." This is a lingering influence of the line definitions, as is the designation used in German text-books, " arcns sinus." The relative magnitude of arc 6 and ordinate M X P^ (Fig. 32) expresses graphically the size of an angle com- pared with its sine. .-- •«. "2Z. 'a. 7 Sit 7 '&* r <*~X 2 s> -W t .<" rt L c. ^ i \ J s .hi - h r ~ f Ji i; \ / ^ -^ -f - S _z J 7 Fig. 32. §69c] THE SINE FAMILY. 105 § 69. Line Picture of the Sines of all Angles, and its Story. \ x J * : Fig. 33. Draw the unit-circle about 0. It cuts the terminal OP at P x (Fig. 33). Then M 1 P 1 is the sine of 6. The ordinate at any point represents the sine of all angles whose terminals pass through that point. Thus the ordinates of the unit-circle make up the sum total of all possible values of the sines of all angles. From the diagram of the line values of sines (Fig. 33) there can be learned more readily than in any other way — because seen by the eye — cer- tain very important properties of the sine. (a) When the terminal is close to the initial line, the sine is small. (6) When the terminal is on or opposite the initial line, the sine is zero. ( with the general statement, sin(4n-l)90° = -l, or, sin(4n-l)£ = -l, 8 where, as in all formulas of this kind, here as elsewhere, n is any positive or negative integer. (7) The sine of an angle is less than the corresponding arc on the unit-circle, that is, the arc of the principal angle. This is evident at once on the unit-circle, for the arc PAP X (Fig. 34) is longer than the straight line PP X ; therefore, the half of arc PAP X is longer than the half of the straight line PP V 1 Fig. '34. or, MP < arc AP, or, sin < arc AP. It has been mentioned in § 68 that when the radius is taken as unity, the number representing the arc on the unit-circle is the same as the number representing the corresponding central angle. Therefore an equivalent statement to that above is, The sine of an angle is less than its radian measure. 108 PLANE TRIGONOMETRY. [§ 69 1 The same proposition is also readily proven for any other radius than unity. For in Fig. 34, straight line GrBQ < arc Q-SQ, .: ordinate BQ < arc SQ, ordinate BQ arc SQ modulus OQ radius OQ^ or, sin 6 < radian measure, or, sin 6 <9. (m) A slight change in the angle makes a slight change in the sine. For the angle A OP 1 (Fig. 35), the sine is M X P V For the larger angle A OP v the sine is M 2 P 2 . .\ sin AOP 2 - sin A OP t Evidently, then, with diminishing difference in the angles, SP 2 di- minishes, becoming zero when the angle difference is zero. When the angle difference be- comes less than any assignable quantity, so also does the sine difference. An equivalent statement to this, and the one most commonly used is : The sine varies continuously as the angle varies continuously. This is also differently expressed by saying that if angle values form a continuum from A to i?, the sine values form a continuum from the sine of A to the sine of B. For ex- ample, as the angle passes from zero to 90°, the sine passes through all those values indicated as ordinates on the unit- circle from to 1. Similarly, as the angle changes continu- ously from 45° to 135°, the sine passes through all values V2 V2 from •— - to 1, and then from 1 to -— again. It waxes in length as does a stretching rubber string, and wanes in length as does a contracting rubber string. The illustration ex- presses fairly well what is meant by a continuous change, §69n] THE SINE FAMILY. 109 a change without sudden jumps in value, without omitting intermediate values of a continuum. One can easily, how- ever, get a false impression from the illustration, if any idea of the rate of change along the con- tinuum is allowed to enter the mind. Suppose, for instance, one considers the ordinates of the straight line AB (Fig. 36) from A to B. They form a continuum. The ordinates of the circle APB form exactly the same continuum, as do also the ordinates of any other curve, like A QB, which has no ordinate between A and B higher than OB- This is seen by drawing a line RQP parallel to the initial line. This gives three equal ordinates, so that for any ordinate of any one of the curves (line), there is an equal ordinate on the others. Thus the sets of ordinates form the same continuum, whether from the straight line or from the circle. Evi- dently the rate of rise of the ordinates of the circle is not uniform. The rise is rapid at the outset at A, and is a di- minishing rate of rise as the terminal turns uniformly from the position OA to OB. As much of the notion of continuity as we desire at pres- ent to give, is covered by the statement : If two angles in circular measure, or, what is the same thing, two arcs on the unit-circle, differ by less than any assignable magnitude, so will their sines. This follows from the diagram (Fig. 35), for the straight line SP 2 is shorter that the curve P^Py that is, the sine difference is always less than the arc difference. (n) For a given small change in the angle, the change in the sine is smaller the nearer the terminal of the angle is to the vertical ; and the larger, the nearer the terminal is to the horizontal. In particular, if two angles are very near 90°, the difference in their sines is very small compared with the difference in the angles ; whereas, if the two angles are very 110 PLANE TRIGONOMETRY. [§69o near zero, the difference in the sines is very nearly equiva- lent to the difference in the angles in radian measure. The student is expected to see this only in so far as it is clear from the diagram, that is, see it with the eye. Later the same thing will be investigated by the aid of formulas. Often we see on the diagram what a formula never makes any clearer. (0) When an angle is small, doubling the angle very nearly doubles the sine. Motion along the arc at 4$ on the cir- cle, is very nearly vertical, and the in- crease in the ordinate is very nearly the same as in the arc. (jt>) However, sin 2 a is never 2 sin a, except when a is zero or some multiple of ir. For, let APB (Fig. 38) be a semi- FlG. 37. IT circle, with a > and < — • Then, sin 2 a MP and sm a = Fig. 38. OP' In the triangle BOP, BP - ^^, or 2 sm a > — — , MP BP' 2 OP or, 2 sin a > sin 2 a, numerically. This is true for all values of a. See § 139 (1). We introduce the word " numerically " above, to correspond with the result which would be obtained if the diagram were turned over about BA as an axis. We give the statement, sin 2 a is not 2 sin a here, because the desire to treat 2 as a factor with the word " sin " mani- fests itself quite promptly among average students. EXERCISE. Deduce all the propositions from (a) to (jo), using the ratio definition for the sine, and without making use of the unit-circle. §70] THE SINE FAMILY. Ill LABORATORY EXERCISES. 1. Draw a circle of one foot radius. Divide it into five-degree spaces ; measure the ordinates in decimals of a foot, and compare with a table of sines, and show that the sines for the second, third, and fourth quadrants can be expressed in terms of those of the first quadrant. 2. On the same circle measure the sine of 10° and the sine of 5°, and see that sin 10° < 2 sin 5°. Test several other double angles. 3. Show from the diagram of Ex. 1 that sin 15° < 3 sin 5°. 4. Show from the diagram of Ex. 1 that sin 20° < 4 sin 5°. § 70. The Sine as a Function of the Angle. Whenever a quantity y is so related to another quantity x, that y can be calculated when x is given, then y is said to be a function of x. Calculations involve four operations, — addition, subtraction, multiplication, division. In fact, these are the only operations involved in calculations. When y is reached by a finite number of calculation operations on sb, y is said to be an algebraic function of x. Examples are : y = 2x, y=2x±3, f 8 a*+ 6 a* - 8 * V etc. 9s5_6z2 + 3 Here y is in each case derived from # by a finite number of calculation operations. Roots of algebraic functions, where the root-indices are integers or common fractions, are also reckoned as algebraic functions, even though it may not be possible to calculate the function exactly. Functions that are not algebraic are said to be transcen- dental, since their calculation implies an infinite number of one or more of the fundamental mathematical operations. Such functions can evidently be calculated only approximately, this being, however, as indicated above, no test of the trans- cendental character of a function. 112 PLANE TRIGONOMETRY. [§ 70 That the sine of an angle is a function of the angle, in the sense of the foregoing definition of a function, is evident at once from the definition of a sine ( ). For one can Vmodulus/ draw the ordinate, measure it and the corresponding modulus, divide, measure the angle. It is evident, however, that while such a procedure establishes the fact that the sine is a function of the angle, the value of the sines of all angles could not be ob- tained with any great degree of accuracy from this procedure without great care in the measurements. The ratio definition of the sine is not given with a view to its use in calculating the tabulated sines of angles, but rather with the object of calculating the ordinate when the angle and modulus are given, or calculating the modulus when the angle and ordi- nate are given. That is, the definition is not used to calculate a table of sines of angles. Frequently, however, from meas- ured ordinates and moduli, sines are calculated and the corresponding angle determined by comparison of the result- ing calculated sine with a table of sines. It will be shown in § 156 that if 6 is the radian measure of an angle, sin 6 = 6 — — + — — — -f — ... etc., without end. [3 [6. (I [9 From this relation the sine of any angle can be calculated to any desired degree of accuracy, but never when 6 is other than zero can it be determined completely from this series, since an infinite number of calculation operations is indicated. The sine is thus a transcendental function of the angle, and in most cases can be known only approximately. While the formula above for the sine would be used to calculate the sine of any given angle, if that were unknown, it is not the formula from which tables of sines of angles have been calculated. It will be shown later that when the sine of any angle is known, that of an angle differing from it slightly can be determined more readily than by using this formula. The chief point we wish to make here is, that the §72] THE SINE FAMILY. 113 sine is a function of the angle ; that is, that when the angle is given, the sine can be calculated. These calculations have been made with great care for all angles between 0° and 90°, at intervals of every " second of arc " for small angles ; for every 10" for larger, but still small, angles ; then for every 1', or for every 10', the angle -interval depending upon the place of the table. Such tables of sines are frequently called tables of natural sines, to distinguish them from the tables of logarithms of the same numbers, called tables of logarithmic sines. § 71. Many- valued Functions. When to each value of the quantity x corresponds a single value of the calculated quantity y, y is said to be a single- valued function, or unique function, of the quantity x. Examples are the algebraic functions, free from radicals, previously given. If y = V# 2 -f- 3x — 7, to each value of x will correspond two values of y. In this case, and all similar ones, y is called a two-valued function of x. While, as already seen, the sine is a single-valued function of the angle, the angle is a many-valued function of the sine, since all angles with the same terminal, or with terminals symmetric to the vertical, have the same sine (§ 62). § 72. The Sine as a Periodic Function. When it is desired to indicate that y is a function of x without stating the character of the relation specifically, we write y = F(x). This is read, "y equals a function of #." If two or more functions are to be used at the same time, some other letter may take the place of F. Thus we might have y = F(x), y = 0(x), y = (x), y=G(x), etc. In such cases, if the functions are read in succession, dis- tinction is made among them by reading the function letter, as "y equals the F-i unction of as," u y equals the theta- function of #," etc. 114 PLANE TRIGONOMETRY. [§72 If y = F(x), then y t = F^x^ means that y x is the special value which y takes when x is given the special value x v For example, if y = z 2 -f- 3 z - 7 = F(x), F(x 1 ) = x 1 2 + Sx 1 -1, and F(2) = 2 2 + 3-2-7 = 3, etc. Similarly, if y = FQc), then y x = F(x x + K) means that y 1 is the special value which y takes when x takes the special value x 1 + h. With these agreements, if y = F(x), then F(xJ = F(x x + K) signifies that y takes the same value for x = x 1 as for x = a?j + h, and F(x{) = JX^ 4- K) = .FOi + 2 K) = F(x x + 3 K) = F(x x -f 4 ft), etc. = #<>! - A) = #0^ - 2 A) = .F(>i - 3 A), etc., signifies that all values of x separated from x 1 by multiples of the value ^, give the same value to the function y. When we write F(x) = F(x + nh), where n is any integer, positive or negative, and x is free from subscripts, it signifies that for any value whatever of x, values of x separated from this value by multiples of h give all the same value to the function as did the initial value. Such a function is called a periodic function, and h is called the period. We have seen that for the sine, sin (0) = sin (6 + n • 2 tt), for all integral values of n, and for all values of (§ 62), and that this is the only angle relation which leaves the sine unchanged both in magnitude and sign. Thus the sine is a periodic function whose period is 2 7r. §73] THE SINE FAMILY. 115 EXERCISE. Point out the periodicity of the sine on the unit-circle, taking the arc, its number being the same as that of the corresponding angle, to repre- sent x, while the sine, or ordinate, is y. § 73. Inverse Functions. Notation for Angles having a Given Sine. When y is expressed directly in terms of S, y is said to be an explicit, or direct, function of #, as in y = x 2 + 2 x — 1 or y = sin x. When x is given here, y can be calculated directly. Generally, if y is a function of x, x is also a func- tion of y ; that is, if y is given, x can be calculated. In the first of the relations above, if x is 2, y is 7. The inverse process of finding what x is when y is given as 7, is quite different. For when z 2 + 2z-l = 7, by the usual process for solving quadratics, -2±V4 + 32 x = ^— — ! = 2 or — 4. A Here, while y was single -valued, x is two-valued. Similarly, if y is a cubic in x, as y=2x*-5x 2 + ±x-S, y will be single-valued, and x, as shown in algebra, three- valued, with corresponding relations for quartics, quintics, etc. From the foregoing it may readily appear that if one quantity, «/, is expressed directly in terms of another, x, it may be quite difficult, or impossible, to express x in terms of y, or to calculate x when y is given. For example, if y is a quintic in #, as y = x 5 + ax* + bz? -h ex 2 + dx + e, # cannot be expressed in terms of y, and the quantities a, 5, i The infinite degree of the second member would carry with it, to one familiar with algebra, an infinite number of values of x for any value of y. We have already seen that for any given value of the sine there are an infinite number of angles, differing by multiples of 2 7r, or 180°. The direct expression of x in terms of y from (3) we do not take up here. Read " Reversion of Series " in any college algebra. Relation (2) is read in several different ways : (i) x is an angle whose sine is y ; (ii) x is an arc whose sine is y ; (iii) x is anti-sine y ; (iv) x is the inverse sine of y ; § 74] THE SINE FAMILY. 117 (iii) is preferred, on account of its brevity. It is, perhaps, best to use (i) until the significance of the symbol is well in mind. The statements j sin • sin -1 ?/ (3) ) , J sine anti-sine y) \ sin -1 • sin y (4) ) ' { anti-sine sine y ) " (3) is also read : " The sine of the angle whose sine is y," and thus makes immediately the mental impression that its value is none other than y itself. After some practice, and after the notation is familiar, " sine anti-sine " will make the same impression. Expression (3) is thus single- valued. On the contrary, (4) may be read " any angle whose sine is that of y." It is thus many-valued, being any one of the an S leS IMT+(-l)-0, where 6 is any particular angle of the set (§ 62). § 74. Origin of the Notation, sin -1 . a x is x. That is, a • a' 1 • x is x. a Therefore a and a~\ considered as operators on #, annihilate the effect of each other. . Corresponding to this is the effect of the expressions " sine of " and " whose sine is," as in the expression " the sine of an angle whose sine is g, is a?." Thus, while ( Bin • sin" 1 x = x) ^ correspond to each other, ( a • a -1 • x = x ) • (sin-i-sinz) (2) donot . ( a -1 • a - x ) for while the last is x, the next to the last is, as already pointed out, any angle whose sine is that of x. However, the origin of the notation, sin -1 a;, is in the similarity here noted in (1) ; so also log e _1 x. 118 PLANE TRIGONOMETRY. [§74 EXERCISES. 1. Read in all possible ways each of the following expressions : sin-?; **», rin-l(-l); dn-i(-f); sin--L; *,->(_ f)j sfc-.f; sin-^; ^(-1). 2. Read the following : Bin-»i + sln-i(-|)i sin-^- sin-- i; sin->| + sin-i(-|); sin- ^ + sin -i ( _ ^ . sin -.| ± sin-(-|). 3. If 0, A are the radian and degree measure, respectively, for the principal angle of any expression of Ex. 1, what is the general expression ? Write the general expressions for the angles of Ex. 2, in both radian and degree measure. 4. Read the following expressions in all possible ways : sin" 1 - + 3 sin- 1 - ; 2 sin" 1 — - 5 sin-U ; 2 sin" 1 ! - 3 sin — ; 3 4 _2 _ 2 2' sin" 1 ^ 2 8^-4; 4sin^5-2sin- 1 (-^ > ); s iW-§) -sin" 1 ?; 3 2 2 \ 2 / V 3/ 3 and write the general expressions in circular measure and in degree measure for the corresponding angles, using the tables when necessary. 5. Construct the terminals corresponding to each expression in Exs. 1, 2, and 4. GENERAL EXERCISES. 1. Construct the terminals when sin A has the following values: z 4 1. i * -n. V§. VI. -V2 . -V3 1 1 |, 3, l, l, u, — , — , -— -, -£- ; -; --. 2. Give, with diagrams, general solutions of the equations : sin = 1 ; sin 6 = — 1 ; sin 6 = ; sin 2 $ = 1 ; sin 3 6 = - 1 ; sin 4 = ; sin md = 1 ; sin ( — m0) = — 1 ; sin 2 mO = 0. 3. Name two angles not in the same quadrant whose sines are the same as that of 17°, with diagram for terminals. 4. Name two angles not in the same quadrant whose sines are the opposite of that of 17°, with diagram for terminals. 5. What is the best plan for determining for any given angle what angle of the first quadrant has numerically the same sine ? 6. What are the general values of when 6 — sin -1 1, $ = sin- 1 (-1), 6 = sin- 1 (0)? Of A when SA = sin- 1 (+ 1), mA = sin- 1 (-1), 7 A = - sin- 1 (0) ? §75] THE SINE FAMILY. 119 7. Write the general form of a periodic function of x whose period is K. 8. Write the sine equation which indicates that the sine is a periodic function. 9. If sin 3 6 = 15, may we divide by 3 and write sin 6 = 5? 10. Give the general formula for all angles whose terminals are either on the upright or downright vertical. 11. Find the number of seconds in the angle 0.3 r ; the number of minutes in 0.4 r ; the number of degrees, miuutes, and seconds in 0.5 r and in A**. 12. Express 11" in radian measure. 13. The three angles of a triangle have the same numerical measure, one being in degree measure, another in grade measure, the third in radians; find the third angle in terms of 7r. 14. Construct the terminals of 0, when 2 = sin -1 § ; when 3 = sin -1 ( — £); when 4 = sin -1 (1); giving, in each case, all possible solutions. 15. What angle of the first quadrant has the same numerical sine as 2317°? 16. How are the sines of ir + 6, ir — 6, 2 ir — 6 related to sin 0? 17. Express 3^.12 in radian measure in terms of ir. 18. When the modulus is 25.4 and the sine — 0.312, what is the ordinate ? 19. When the ordinate is ± 37.2 and the sine is ± 0.341, what is the modulus ? 20. When the ordinate is ± 21.3 and the modulus is 32.4, what are the sines to three figures ? § 75. Angles whose Sines can be determined readily from a Diagram. We have already stated (§ 70) that to calculate the sine of any angle given in radian measure and selected at random, one must use the relation : sin0 = 0-^ + ^-etc. [3 [5 There are, however, a few angles whose sines can be deter- mined quite readily without the use of this formula. Prominent among these are the .ingles whose terminals pass through the vertices of the regular polygons of 3, 4, 6, 120 PLANE TRIGONOMETRY. [§75 8, 12, sides, the polygons being central at the origin, with one vertex on the initial line. Such terminals will form the bisectors and trisectors of the quadrantal angles, together with the border lines of the quadrants, in all sixteen termi- nals, as in the above diagram, in which the terminals are indicated by numerals and also in degree measure : (a) For terminals 1, 5, 9, 13, the sines have already been given (§ 69). The student may restate the results, giving the general formulas. Try it without looking at the refer- ence, and then compare results with the reference. (5) For terminals 3, 7, 11, 15, use the diagram adjoining, in which a right-angled tri- angle whose hypothenuse, co- incident with terminal 3, is 2 units of length, and taken with one vertex at the origin and one leg (length V2) on the initial line. §75c] (i) Then, sin 45° = THE SINE FAMILY. MP 121 OP V2 . 7T 2- = sm T (ii) Terminals 3, 7, are symmetric to the vertical, and therefore belong to angles of the same sine (§ 62), .-. sin 135° = sin 45° = ±^? = sin ? = sin — . 2 4 4 (iii) Terminals 3, 11, are opposite and therefore belong to angles of opposite sine (§ 63), .-. sin 225° = - sin 45° = -^ = _ s i n Z = s in — • 2 4 4 (iv) Terminals 3, 15, are symmetric to the horizontal, and therefore belong to angles of opposite sine (§ 63), .-. sin 315° = - sin 45° = V2 . IT 7 7T sin — = sin— — 4 4 EXERCISES. 1. Find the sines of the negative angles numerically equal to those above. 2. Give in degree measure and in radian measure the general formulas for all angles having the same sine as 45°. Also the general formulas for all angles having the opposite sine of 45°. 3. Find from the diagram (Fig. 40) six positive angles having the same sine as 45°. Which lines may be their terminals? What values of m in the general formula give the angles which you have found? 150° 30° 4. Determine similarly six negative angles. 0) For terminals 2, 8, 10, 16, use the diagram adjoining, in which an equilateral triangle whose side is of length 2, is set with one vertex at the origin and a bisector on the initial line. The side OP (Fig. 41) will fall on terminal 2, MP will be 1, and OM will be VS. FlG . 41 . 210 122 PLANE TRIGONOMETRY. [§ 75 c Then, exactly as in the preceding case, (i) sin 30°=l = sin|; (ii) sin 150° = sin 30° = J = sin - = sin^ ; (iii) sin 210° = - sin 30° = - £ = - sin % = sin- v j 2 6 6 (iv) sin 330° = - sin 30° = - £ = - sin ^ = sin 7tt 6 11 7T EXERCISES. Take those on page 121, replacing 45° by 30 c (c?) For terminals 4, 6, 12, 14, use Fig. 42, an equilateral triangle, whose side is of length 2, one side coinciding with the initial line and another with terminal 4. /-•\ • r»AO +v3 . 7T (l) sin 60° = ^— = sin-; (ii) sin 120° = sin 60° = + V3 . 7T • 2 7T = sin — = sin -— — ; 3 3 (iii) sin 240° = - sin 60 -V3 2 • 7T . 4 7T -sin-=sin_; -V3 (iv) sin 300° = - sin 60' . IT . 5 7T = sm-=sm_. EXERCISES. Those on page 121, replacing 45° by 60°. Note. — The student is advised against attempting to memorize the foregoing results as mere facts. Let the diagrams be called up when needed, and read from the mental diagram the value sought : §75] THE SINE FAMILY. 123 (a) For 45° and the related angles, use the diagram shown in Fig. 43. (b) For 30° and the related angles, use the diagram shown in Fig. 44. (c) For 60° and the re- lated angles, use the dia- gram shown in Fig. 45. These diagrams deter- mine the sine in magni- tude. The sign is then determined by the loca- tion of the terminal. Fig. 44. Fig. 45. These results may be tabulated as follows : f degrees ] 0, 30, 45, 60, 90, 120, 135, 150, 180, I radians j 0, -, -, -, -, — , "jr. -J-,'* , n .\ +V2 +V3 - +V3 +V2 , A whose sines are 0, J, , — — , 1, — — ,— — , J, 0. EXERCISES. 1. What are the sines of the corresponding negative angles? 2. Read from mental diagrams the sines of 0°, 45°, 30°, 60°, 90°. 3. Considering the terminals of the angles in Ex. 2 as the funda- mental terminals, state the relative position of any other one of the sixteen terminals of Fig. 39 with reference to some one of these, and thereby determine at once the sine of the corresponding angle. The class may be drilled several days in succession at this, until the sine of any one of the sixteen terminals can be read off at once from the proper mental diagram. 4. Using the principal angle corresponding to each of the following sines, determine the general angles for the following expressions : sin sin - 2 un-M --J+shr 1 (-«. 3 sin" 1 5 sin" 1 !; (-•#♦«**$ (-*. ~ x — - + 2 sin «-& ^..(-^^(-a 124 PLANE TRIGONOMETRY. [§75 sin-i(-i) + sin(-^?); sin-i(-l) + 3sin-i(-i)-2sin-i^; sin- 1 + sin- 1 ! ; 2 sin" 1 ; 3sin- 1 l + 2sin- 1 (l)-4sin- 1 (-^);6sin- 1 (^)-5sin- 1 (^^); and so on, until the sines of these fundamental angles are known. 5. Verify the following statements (sign not considered) : sin 2.45° < 2 sin 45°; sin 2.30° < 2 sin 30° ; sin 2.90° < 2 sin 90° ; sin 2.60° < 2 sin 60° ; sin 2.120° < 2 sin 120° ; sin 2.135° < 2 sin 135° ; sin 2.150° < 2 sin 150° ; sin 2.180° = 2 sin 180°. 6. Consider similarly the remaining positive angles less than 360° corresponding to the sixteen terminals of Fig. 39. 7. Compare results when the angles of Ex. 5 are taken as negative. 8. Give three special solutions and the general solution of : sin 2 a; = ; sin 2 a: = 1 ; sin 2 a; = \ ; sin 2 a; = f ; sin 2 x = i; sin3a; = i; sin 2 — = ?; V2sin5a?=l. 4' 2 2 4 9. Using the general formula for angles of a given sine, determine the general values of the angles represented by the expressions in Ex. 4. 10. Taking the general form of the quadratic equation as ax 2 + bx + c = 0, show that the general values of x are _ _ b ± Vb 2 - 4 ac 2 a Sufficient special numerical examples may here be assigned to make sure that the class can write at once the solution of any quadratic equa- tion without going through the usual process of completing the square. 11. Find general expressions for the angles satisfying the following equations : 4 sin 2 x - 2 (1 + V2) sin x + V2 = 0. 4 sin 2 x + 2 (1 + V2) sin x + V2 = 0. 4 sin 2 a; + 2 (1 - V2) sin x - V2 = 0. 4 sin 2 x + 2 (V2 - 1) sin x - V2 = 0. 4 sin 2 a; - 2 (V3 + 1) sin x + V3 = 0. 4sin 2 a: + 2(\/3 + 1) sin* + V3 = 0. 4 sin 2 a: + 2 (V3 - 1) sin x - V3 = 0. §76] THE SINE FAMILY. 125 4 sin 2 a; - 2 ( V3 - 1) sin x - V3 = 0. 4 sin 2 a; - 2 (\/2 + V3) sin x + V6 = 0. 4 sin 2 x + 2 ( V2 + a/3) sin x + V6 = 0. 2 V2 sin 2 a: - V2 (V2 - V3) sinx - V3 = 0. 2 V2 sin 2 a; - V2 (V3 - V2) sina; - V3 = 0. 2 sin 2 x - ( V2 + 2) sin a: + V2 = 0. 2sin 2 x+(\/2 + 2)sina;+\/2 = 0. 2sin 2 a:+(>/2 - 2) sin a: - V2 = 0. 2 sin 2 a: - (V2 - 2) sin x - V2 = 0. 12. Find what values of x will satisfy the following equations, giving x in degrees and in terms of it in each case : sin" 1 (a; 2 - x) = 30° ; sin" 1 (x 2 - x) = ± 180° sin- 1 (a: 2 - 5 x) = - 30° ; sin" 1 (a; 2 - x) = ± 210° sin"^- 4a: 2 + 7x)= - 45°; sin" 1 (a; 2 - 2ar) = ± 90° sin" 1 (a; 2 + 6 x) = 60° ; sin" 1 (4 x 2 - x) = ± 270° ; sin" 1 (3 x 2 - 8 x) = - 60° ; sin" 1 (7 x 2 - x) = ± 225° ; sin" 1 (4 x 2 - 4 x) = 120° ; sin" 1 (2 x 2 - 3 x) = ± 240° ; sin" 1 (3 x 2 + 7 x) = - 120° ; sin" 1 (3 x 3 + 5 x) = ± 360°; sin" 1 (2 x 2 - 4 x) = ± 135° ; sin- 1 (x 2 - x) = ± 300°. sin _1 (x 2 + x) = ± 150°; 13. Name two angles for each of the examples of Ex. 12, which might replace the second member of the equation without altering the final result. 14. Show that, knowing the sines of 30°, 45°, 60°, one can, by con- structing a diagram, find the sines of the following angles: ±405°; ±390°; ±420°; ±840°; ±660°; ±690°; ±1050°; ±960°; ±1305°; ±1200°; ±1020°; ±945°; ±855°; ±870°; ±810°; ±900°; ±780°; ±570°. § 76. Calculations using Sines, without use of Tables except as Check. Since the three quantities, sine, ordinate, modulus, are connected by the relation of definition, ordinate sine = — T~r~ ' modulus if any two are given, the third can be calculated as follows : 126 PLANE TRIGONOMETRY. [§76 (i) When ordinate and modulus are given, _ ordinate modulus (ii) When angle and modulus are given, ordinate = modulus times sine. (iii) When angle and ordinate are given, ordinate modulus = sine These three formulas are to be memorized each as a primary relation, so there will be no hesitation in writing any one of them, due to halting to deduce it from some other. The author's experience is that where students learn trigonometry from the ratio definitions, the tendency is to hold in mind as a primary relation that of (i), there being a halt always in seeing (ii), (iii), due to their deduction from (i). The class should be drilled on exercises like the following until the three relations are known equally well. EXERCISES. N.B. Never carry the number of figures in calculated results beyond the number in the given quantities. All measured sides should show the same number of significant figures. All lengths and angles given must be assumed to represent measurements and are thus approximate. 1. Draw diagrams to scale, illustrating the following data, and calcu- late the corresponding ordinates. Test each result by calculating back- ward the given modulus from the calculated ordinate. Also draw the ordinate in the diagram, measure it, and compare the result with the calculated value. Measure the angles in the diagram and check by com- paring the given (or calculated) sine with the table-sine. Modulus. Sin A. Ordinate ? Modulus. A°. Oedinate ? 250 ±f 316 ±30 350 ±| 250 ±45 32 ±\ 216 ±60 16 ±\ 215 ±90 85 ±\ 16 ±120 90 ±\ 80 ±135 100 ±| 100 ±150 25 ±J 120 ±210 17 ±| 150 ±225 76] THE SINE FAMILY. 127 Modulus. Six A. Ordinate? Modulus. A°. Ordinate ? 50 ±i 30 ±240 8 ±1 70 ±300 10 ±1 80 ±315 25 ±1 90 ±330 80 ±f 100 ±360 20 ±1 200 ±390 60 ±* 8 ±405 75 ±1 10 ±420 80 dbA 12 ±450 100 ±A 14 ±270 200 ±1 18 ±0 2. Interchange the words " ordinate " and " modulus " in the above table, construct to scale the corresponding diagram, calculate the corre- sponding moduli, test as in Ex. 1 ; measure ; compare. Note. — Construction to scale will be found quite helpful in fixing in mind the sine as a ratio. The actual lines of the diagram will be seen to bear to each other the ratio indicated by the sine. 3. Construct to scale diagrams illustrating the following data, and calculate the corresponding sines to as many decimal places as the data. Sin A. Modulus. Ordinate. Sin A . ODULUS. Ordinate. 5 ± 3 8 ± 4 13 ±12 13 ± 8.2 17 ±15 17 ±17 17 ± 7.1 25 ±24 29 ±20 2.9 ± 2.1 37 ±12 37 ±35 41 ± 9.2 41 ±19 53 ±28 5.3 ± 4.5 Modulus. Ordinate. 61.2 ±60.1 61 ±11 6.53 ± 3.32 65 ±56 65,24 ± 16.32 6.5 ± 6.3 73 ±48 7.31 ±48.2 0.85 ± 0.36 85 ±77 36 ± 9.2 85.3 ±84.2 89 ±80 9.721 ± 6.532 61 ±16 43 ±34 4. On the diagrams representing the data above, measure any other ordinate and its modulus; divide, and compare the resulting sine with that gotten from the given numbers. Also measure the angle and com- pare calculated sine with the table sine. 128 PLANE TRIGONOMETRY. [§77 § 77. Selecting and Cutting a Table of Sines. As already indicated, the number of angles for which the sine can be determined directly from a diagram is quite limited, and, in fact, such angles here, as in other text-books, are given far more prominence than they deserve. It is customary to use the tables for all angles except multiples of 90°. What place table to use depends upon the accuracy with which the measurements have been made and upon the accuracy to which the calculated result is desired. As pointed out in §§ 20, 22, 24, the tendency of calculation manipulations is to make the calculated results less accu- rate than the measurements on which they are based. The resulting inaccuracy depends on the number of additions, multiplications, etc., made. It must be assumed in calculations with triangles that the lines whose lengths are given have been actually measured, as also the angles given. All the given parts of a diagram should indicate about the same degree of accuracy in meas- urement. In calculating triangles, the number of operations being always few, the following rules may be laid down : (1) All measured lines must show the same number of significant figures. (See § 33.) (2) Calculated lines, sines, and logarithmic sines must not show more significant figures than the measured lines. They may generally show the same number. (See § 30.) (3) When the measured lines of a diagram show only one significant figure, measured angles and calculated angles may read to the nearest five degrees; with only two significant figures in lines, angles may read to the nearest half-degree, as with a surveyor's compass. (4) When the measured lines of a diagram show only three significant figures, angles may read to the nearest five minutes ; while with only four significant figures, the meas- ured angles and the calculated angles may read to the near- §79] THE SINE FAMILY. 129 est minute when a four-place table is used, and to the nearest ten seconds when a five-place table is used. (5) In cases (3), (4), a four-place table is sufficiently accurate, generally. At times four significant figures in measured results call for the use of five-place tables. (See § 23.) Whatever table is used should be cut back to one place more than the data. (6) When the measured lines show five significant figures, a five-place table or a six -place table should be used. The measured angles and calculated angles may then read to the nearest second with a six-place table, and to the nearest five seconds with a five-place table. (7) In astronomical calculations, when the number of terms makes a balancing of table-errors likely, a five-place table is used when the given angles and calculated angles are given to the nearest second ; a six-place table, when tenths of seconds appear; a seven-place table, when hun- dredths of seconds appear. (8) Do not use a seven-place table when a four-place table is sufficient, nor a four-place table when a seven-place table is necessary. One place beyond the data is all that is ever needed. The smaller the suitable table, the greater is the saving of time in manipulation. Proof of the Rules. Sin # = 0.1 means that sin # lies between 0.05 and 0.15. Taking from the tables the angles opposite the sines 0.05 and 0.15, which stand closest to the sine 0.1, we get 3° 26' and 8° 38', whose mean may be taken as x when sin#= 0.1, and half whose difference, or 2° 36', will be the possible error in x. Similarly, sin x = 0.2 has as extremity angles, 9° 15' and 14° 29', whose mean may be taken as #, with the possible error of 2° 37'. Treating 0.3, 0.4, •••, 0.9, the same way, the average error for the whole table is about 3° 30' in determining an angle from one figure in the sine. Treating 0.10, 0.20, etc., in the same way, the average is about 20'. 130 PLANE TRIGONOMETRY. [§78 Safe working limits then are 5° and 30', respectively, as in rule (3). An increase of one figure in the line-data, when the table has at least one place more than the data, divides the error by 10. This and § 23 give the remaining rules. § 78. A Four-place Table. In the tables accompanying this book is a four-place table of sines and of logarithmic sines. It is not to be used : (1) When the measured lines of the diagram show more than four significant figures. (2) When the measured angles of a diagram show seconds. (3) For small angles (less than about 5°) when the sines (angles) can- not be obtained directly from the table without interpolation. Here the sines change too rapidly to use their differences. (See § 186.) How to use these tables is explained in the preface of the tables. In practical work in electrical and mechanical engineering the data are largely three-figured. A four-place table meets most of the re- quirements. § 79. Some Calculations using Sines of the Tables and Logarithmic Sines. (Four-place Tables.) From the relations (i) Sme = JT-' modulus (ii) ordinate = modulus times sine, (iii) -, i ordinate modulus = : , sine follow the relations (iv) log sin = log ord — log mod, (v) log ord = log mod + log sin, (vi) log mod «s log ord — log sin. A. Examples on Calculating the Ordinate (case ii). (i) -,. I modulus = 8 ^^ * lnc \. GlV6n I angle = 25° - ordlnate Fig. 46. §79] THE SINE FAMILY. 131 Ordinate = modulus times sine, log ord = log mod + log sin. Modulus being given to only one significant figure, the ordinate should be calculated to only one significant figure. By natural sines, to as many figures as the data, is the proper method. We give logs here and in (2) as a mere check, or rather as a mere sample process in cutting logs to one figure beyond the data. SOLUTION BY NATURAL SINES. SOLUTION BY LOGARITHMS. sin 25° = 0.4 log 8 = 0.90 mod = 8_ log sin 25° = 1.63 .-. ord = 3 log ord = 0.53 Possible error = ± 0.6 .*. ord = 3 (2) Given, modulus = 27, angle = 22°. HON BY NATURAL SINES. SOLUTIC a 22°= 0.38 1 mod = 27 log sin 22° = 1.574 7.6 log ord = 1.005 2.6 ... ord =10 SOLUTION BY NATURAL SINES. SOLUTION BY LOGARITHMS. sin 22°= 0.38 log 27 = 1.431 10 ± 0.3 The solution by natural sines is to be carried out by the shortened process of multiplication. (See § 26.) The num- bers 0.6 and 0.3 after the ordinates show the possible error in these results if tenths were written down, and are obtained by § 25. The student is advised to follow some scheme like that above. Place the data on one side of a diagram. Place the things to be found on the other side. Draw the given parts of the diagram in heavy lines ; the parts to be found in dots. Make the diagram to scale on coordinate paper ; measure as a check. Place the formulas to be used under the diagram. Check by combining the given and calculated quantities in some new relation, directly or by logarithms. 132 PLANE TRIGONOMETRY. [§79 Note. — The student may solve the following examples both by natu- ral sines and by logarithmic sines, and compare results. He may also work the examples backward, as a test of accuracy. Since every calcula- tion ought to be tested in some way by the person making it, no answers are given. With one-figured and two-figured data, practical engineers use the natural functions. EXERCISES. Make all diagrams to scale on coordinate paper. Measure the ungiven part as a check on calculation. Keep in mind, in making the diagrams, that with data like Ex. 1, a measured modulus 9 means not 9.0, but something between 8.5 and 9.5, and that the angle 35°, given as going with such a modulus, means that 35° is the angle to the nearest 5°, and that the angle lies between 30° and 40°, but nearer 35 c than to 30° or to 40°. If diagrams are made repre- senting the extreme cases of measurement signified by the data, it will become very apparent why calculated lines should not be carried to more places than the data. Make such diagrams for the following examples. In examples like 5, 6, 7, 8, modify the angle readings to suit the table used, according to the rules on page 129. 1. Modulus 9, angle ± 35° ; modulus 7, angle ± 55° ; modulus 6, angle 85°. 2. Modulus 9, angle ± 105° ; modulus 8, angle ± 200° ; modulus 7, angle ± 305°. 3. Modulus 23, angle ± 34° ; modulus 5.6, angle ± 123° ; modulus 6.8, angle ± 196° 30' ; modulus 84, angle ± 302° 30' ; modulus 23, angle ±4325°. 4. Modulus 45.3, angle ± 34° 5' ; modulus 4.57, angle ± 174° 10' ; modulus 34.8, angle ± 267° 20' ; modulus 41.3, angle ± 27942° 15'. 5. Modulus 43.41, angle ± 23° 14' ; modulus 4.587, angle ± 174° 23' ; modulus 0.4752, angle 234° 43' ; modulus 678.4, angle ± 4545° 2'. 6. Modulus 345.61, angle 34° 25' 43" ; modulus 23.598, angle 172° 34' 56" ; modulus 4567.8, angle 256° 31' 36" ; modulus 0.58923, angle 295° 17' 13" ; modulus 7.8519, angle 862° 44' 44" (at least five-place table). 7. Modulus 234.567, angle 45° 47' 53".2 (at least six-place table). 8. Modulus 2564.321, angle 82° 13' 27".34 (at least seven-place table). 9. What data would suit an uncut ten-place table ? 10. What data would suit a three-place table ? 79] THE SINE FAMILY. 133 B. Examples on Calculating the Modulus (case iii). (i) Given ( ordinate = 8 1 angle 17° ^^yn* A Fig. 47. Modulus ordinate sine Find modulus = 30 .*. log mod = log ord — log sin. SOLUTION BY SINES. SOLUTION BY LOGARITHMS. sin .3)8 26 which must be taken as 30, since ordinate has only one significant figure. Are the data consistent ? (2) ns„„„ (ordinate = 13. 27 GlVen (angle =42°17' A log 8 = 0.90 log sin 17° = 1.47 log mod = 1.43 .-. mod = 27 .-. mod = 30 A 4>y v42 17 g; Find 9 modulus = 19.73 Modulus Fig. 48. ordinate sine log mod = log ord log sin. SOLUTION BY NATURAL SINES. (Shortened division process.) sin 19.73 mod 6728 )132700 673 654 605 49 47 2 SOLUTION BY LOGARITHMS. log ord =1.1229 log sin = 1.8279 log mod = 1.2950 mod = 19.73 134 PLANE TRIGONOMETRY. [§79 The solution by natural sines is made by the shortened process of division. (See § 40.) To determine to how many places the division may be carried, use may be made of § 27 and § 28, from which it follows that the per cent rejection error in the modulus is 1§^- (g yV ¥ + TFT2") P er cen ^? or about KeV + iV)> or about KeV + tV)> or about A of 1 P er cent ' or about -fa of 1 per cent. The modulus being about 20, the error is about 0.01. The division should not be carried, therefore, beyond hundredths. By the shortened process of division, the operation is self-limited, when the data have the same number of significant figures. It ceases at the limit of possible inaccuracy. The results by natural sines and by logs might have differed by 1 in the last place. The work may also be tested by the backward calculation, ordinate = sine times modulus. 0.6728 sin 19.73 mod 6.728 ' 6.055 .470 .020 13.27 ord EXERCISES. Interchange the words "ordinate" and "modulus" in the exercises of (A), page 132, and solve. Determine in each case the maximum error in the result due to rejection error in the ordinate and sine. Make dia- grams to scale on coordinate paper, and measure the ungiven parts as a check on calculation. C. Examples on Calculating the Sine and Angle from the Ordinate and Modulus (case i). f ordinate = 36.16 4$^ S Find 1VGn I modulus =67.32 S^ t * angle = 31° 29' O M Fig. 49. 79] THE SINE FAMILY. 135 _ ordinate modulus log sin = log ord — log mod. SOLUTION BY LOGARITHMS. log ord = 1.5582 log mod = 1.8281 log sin = 1.7301 Table angle, 32° 29'. General angle, m .180 + (-rr32°29'. SOLUTION BY NATURAL SINES. (Shortened division.) 0.5371 sin 6732 )3616. 3366 • 250 202 48 47 1 Table angle, 32° 29'. General angle, m . 180° + (-1)™ 32° 29'. ADDITIONAL TEST OF ACCURACY. ordinate = modulus times sine. 67.32 mod .5371 sin 33.660 2.020 .471 .007 36.16 =ord To determine the number of places to which the sine may- be determined from the data, assuming rejection error, use may be made of § 27 and § 28. Per cent rejection error in sine is 4* CnW + t&f> or about i (eV + A)i or about \ (y 1 ^ + ^), or about if ^ of 1 c jo. Now, ^ of 1% of the sine, 0.5371, is about 0.0001. So the division may not be carried beyond four figures. The results here and in the previous cases agree with the general statement made in rule (2) of § 77, that calculated 136 PLANE TRIGONOMETRY. [§79 results must not be carried beyond the number of significant figures in the data. It will be observed in any large table of sines, where seconds are taken into account, that over a large part of the table, when the angles differ by only a second, the sines agree in the first four figures. Thus, in four-figured data, calculated angles cannot, as a rule, be determined to the nearest second. It is sufficient to give only minutes. EXERCISES. Determine the sine and angle in the following cases : Ordinate. Modulus. Ordinate. Modulus. Ordinate. Modulus. 72.16 95.73 71.6 94.3 2.31 8.56 1.831 3.924 54.3 83.9 0.5673 1.906 21 67 5 7 5.43 9.56 Take the ordinates as both plus and minus. § 80. Using a Five-place Table. Engineering students should have some good five-place tables, as Gauss's or Hussey's. The teacher may here explain the use of such a table, if it is to be used. He may also make five-figured data, corresponding to the cases above, and have the class make the corresponding calculations. Given angles will read now to seconds ; also calculated angles will be given to seconds, or five seconds. See § 77, (6). Similarly, some practice may be given on examples where a six-place or seven-place table is called for. § 81. Solution of Right-angled Triangles by Sines. The selection of the position of the initial line is altogether arbitrary, as is also the direction for positive or negative turn. If in Fig. 50 AH is taken as initial line and AB as modulus, with the counter-clockwise turn as positive, then HB is ordinate, .:smA = M. AB Similarly, counting BR as initial line, BA as modulus, and the clockwise turn as positive, then HA is ordinate, §81] THE SINE FAMILY. 137 ••Bin B = M. BA In considering right-angled triangles, it is not customary to pay any attention to the question of direction, either of turns or of lines, but to use the general statements _ side opposite the angle hypothenuse The angles being acute, their sines are always positive. Using A, a-, B, b, to denote angles and their opposite sides, or pairs of opposites, h being the hypothenuse, we have : (1) sinA = j; sin B=-, or, h h the sine of either angle is the side opposite divided by the hypothenuse. (2) a = smA-h; b = sinB'h,or, either side is the sine of the opposite angle times the hypothe- nuse. (3) h = ^~7 =: - — 5' or ' sin A sin B the hypothenuse is either side divided by the sine of its opposite angle. (4) h=Va ir +¥. The relation (4) can be used advantageously to determine h only when a, b are small. After h is found, either angle can be calculated from its sine. When a, b are large, the angles are best found by a method which will be given later. (See § 169.) These relations, (1), (2), (3), are easily memorized, and should be memorized thoroughly, each as a primary fact. Note that in (1), (2), opposites, as A, a, etc., are on opposite sides of the relations. Thus, as in (1), when sin A is written, write at once on the other side of the equation its opposite, a, and then under a write h, since a sine is a ratio. Like- wise, when a is placed on one side of the equation, write the 138 PLANE TRIGONOMETRY. [§81 sine of its opposite, sin A, on the other, and then h, to make the relation homogeneous, — a line equal to a line. Examples in right-angled triangles can be of only four types, corresponding to the four sets of relations above : (a) Given the hypothenuse and a side, to find the two angles and the remaining side. For example, given a, h, then sin^l = ^; B=90-A; b=smB-h. h (5) Given either angle and the hypothenuse, to find the other angle and the two sides. For example, given A, h, B = 90 — A ; a = sin A • h ; b = sin B • h. (tf) Given a side and an angle, to find the remaining parts. For example, given a, A, h = -A-r; B=90-A; b^sinB'h. sin A (c?) Given the two sides, to find the remaining parts. h = ^aT+¥', sin^.= ^; B = 90-A. h See note above in connection with (4), page 137. § 82. Testing the Solution. With A + B =90°, h = -JL- = -^—. sin A sin B ,\ a • sin B = b • sin A. h 2 -a 2 = P. Thus the log-checks : log a + log sin B = log b -f log sin A. log (h-a)+ log (h + a) = 2 log b. In applying log-checks to "answers," the logarithms of such answers must be looked up in the tables, instead of checking with the logarithms of the calculation scheme. The latter may be correct and yet the " answers " be incorrect, an §83] THE SINE FAMILY. 139 error having been made in looking up the numbers corre- sponding to logarithms. Logarithmic checks are not to be held to a closer agreement than the data call for. It is always sufficient, as pointed out in Chapter II., if they agree within 2 in the last place, such last place not being necessarily the last place of the table used, but the place to which the results are called upon to be correct, in accordance with the data. For example, if the data are three-figured, the check is met if the logs differ by not more than 2 in the third place. (See the example, p. 140.) In extended astronomical calculations, where the number of terms is sufficient to allow for balancing of errors, it is also reckoned sufficient if the log-checks agree to within 2 in the last place. EXERCISES. Test the following triangles for consistency, using log-checks : 1. a = 42, b = 56, h = 70 ; a = 231, b = 95.7, h = 250. 2. a = 231, A = 67° 30', b = 95.7, B = 22° 30'. 3. a = 229, A = 37° 20', b = 300, B = 52° 40'. § 83. Solution Scheme for solving Right-angled Triangles when Logarithms are used. Construct to scale a diagram on which the parts given are indicated by heavy lines, and the parts to be found, by dotted lines. Write on the given parts their numerical values. Do the same with the dotted parts, after they have been found. Set the parts given in one column, as in the following example. Set off a similar column for the parts to be found, in the order in which they will be found. Fill in with the parts found after the calculations have been carried out. Give under the diagram the formulas to be used. The following calculation scheme is suitable for all cases except when the two sides are given. Merely the order of filling in the scheme will change with change of data, the scheme itself remaining unchanged. 140 PLANE TRIGONOMETRY. [§83 Model Example. Given !* = 27 - 3 I B = 35° 15' Solution formulas f .4 = 54° 45' Find J ft = 15.8 I a = 22.3 Fig. 51. A = 90 - B b = sin B • A. . a = sin -4 • h. . LOGARITHMS. (4) 6 = 1.1975 (2) sinB = 1.7613 (1) h = 1.4362 (3) sin A = 1.9120 (5) a = 1.3482 log b = log sin B -f log ^ log a — log sin J. + log h Solution Scheme. test. a sin B = b sin A LOGARITHMS. 1.1987 1.9120 1.1107 1.3483 1.7613 1.1096 Here, in applying the checks, we have taken the logs of 15.8 and 22.3, the nearest three-figured approximations cor- responding to the logs of b, a, and not the logs of 5, a in the scheme. We are cheeking our answers. The data being three- figured, the answers are three-figured. The log-check checks to within 2 in the third place, which is all that is required in the case of three-figured data. (See § 82.) The order of filling in the solution scheme is indicated by the numerals, (1), (2), (3), (4), (5). First look up logarithms of (1), (2), (3). Then add (1), (2), and set the result after (4). Next add (1), (3), and set the result after (5). » Then look up the numbers correspond- ing to logarithms (4), (5), to as many places as the data, and enter the results after " Find " in the calculation scheme. It is generally best not to enter numbers on the solution scheme, but letters, taking their logarithms to correspond to their values in the data. § 84] THE SINE FAMILY. 141 If the given parts are a, A, the same solution scheme is used, but the order of rilling in would be as indicated by the numerals following : LOGARITHMS. (5) b = (3) sin B = (4) h = (2) sin A = (1) Enter the logs (1), (2), (3). Next subtract (2) from (1) for (4). Then add (3), (4), for (5). EXERCISES. 1. Indicate the order of filling in for any other case which may arise. 2. Make schemes into which cologs enter where subtractions occur above. Which do you prefer ? NUMERICAL EXAMPLES. (Make diagrams to scale, for check.) Determine both by natural sines and by logarithms, using the scheme given above, the ungiven parts of the following triangles, testing the final results as indicated above, finding also to how many significant figures the results should be taken if the data are assumed subject to rejection error : 1. Given £ = 234.5, a = 136.8; £=234, a = 156; £ = 23, a = 15. 2. Given £ = 46.7, 6 = 23.8; £ = 46, a = 25; £ = 4, a = 2. 3. Given £ = 9.68, a = 5.76; £ = 95, a = 57; £ = 9, a = 5. 4. Given £ = 468.2, ,4=46° 34'; £ = 465, ,4 = 45° 35' ; £ = 46,4=45°. 5. Given £ = 34.43, B = 38° 23' ; £ = 54.4, 4 = 56° 15' ; £ = 5, 4 = 40°. 6. Given £ = 543.2, A =68° 43'; £ = 543, A =66° 45'; £ = 54,4 = 80°. 7. Given a =45.6, 4 =76° 35'; a = 46, 4 =45° 30' ; £=4,4=45°. 8. Given 6=213.4, £ = 64°46'; 6 = 21.3, A =54° 5' ; £=21, 4=54°. 9. Given a = 96.2, B = 49° 45'; a = 96, 5=49° 30'; £ = 9,5=40°. § 84. Solution with Five-place Table. The teacher may construct and assign corresponding data calling for the use of a five-place table. 142 PLANE TRIGONOMETRY. [§85 § 85. Calculations with Sines, with Practice Examples in the Use of Logarithms and the Shortened Processes of Multipli- cation and Division on Exact Data. Let AHB (Fig. 52) be a right-angled triangle. Drop the perpendicular HO on AB. Then BO is called the pro- jection of a on AB, and will be denoted by a r Similarly A is &'s projection, or b p . The angle AHQ is the same as B, while BHQ is the same as A. a„= a • sin CUB = a • sin A -p h h h = ft. sin OHA = :b. sin B -I- b 2 = h' h = a • sin CBH, or b- sin CAH = ab ~~ h CH (a) Oral exercises in connection with the preceding triangle : Taking a, 5, h from the table of triangles, page 145, as in the adjoin- ing diagram for Ex. 1 of that table, excellent practice on the sine may be had, as follows : a) (2) (3) a=3 TEACHER (QUESTION). sin^.? sin B? sin CHB ? sin CHA ? CH? AC? BC? STUDENT (ANSWER). I i = sin A = | = sin B = f = sin B times BH or, sin A times AH | of 3 = # fof4 = -^ sin OKI times AET = f of 4 = -^_ sin CHB times ## = f of 3 The exercise may be varied by tilting the triangle in various attitudes, until the student is thoroughly familiar with : §85 6] THE SINE FAMILY. 143 (i) The sine is the side opposite divided by the hypothe- nuse. (ii) Any side is the sine of the opposite angle times the hypothenuse. (iii) The hypothenuse is any side divided by the sine of opposite angle. (5) Numerical exercises in connection with the triangle, Fig. 52, and the table, page 145 : In addition to (1), (2), (3), for practice in the use of logarithms and in the shortened processes of division and multiplication, one may calculate the area, A, of the triangle AHB, and the radii of the circles tangential to the sides of this triangle. The area of AHB is AH ' BH or, ab A (4) The radius of any tangential circle is found by joining its centre to the vertices A, H, B (Fig. 54). In the case of the internal circle the sum of the three new triangles formed is the original triangle AHB. In the case of an external circle (an escribed circle, it is generally called), the original triangle is the sum of the two new tri- angles which have for sides the two sides of the original triangle touched on the border of the orig- inal triangle, minus that new triangle whose side is that side of the original triangle touched on the border of the domain outside the original triangle. Fig. 54. 144 PLANE TRIGONOMETRY. [§85 J For the internal circle, whose centre is 0, aCAH+aOHB + AOBA m AAHB, or, r % (a + b + h) = a • 5, 1 a+b+h K J For the escribed circle touching AB and whose centre is C, AC'AH + AC KB - AC'BA = AAHB, or, r A (a + 5 — Ji) = a • 5, or, r h = — — -. (6) a + b — h EXERCISES. 1. Deduce expressions corresponding to (6) for r a , r b . 2. Show that if A is the area of any triangle, right-angled or oblique, and s its semiperimeter, A A A A r b = s — a s — b s — c s where ft is the radius of the inscribed circle. 3. Show that -+- + — = -, for all triangles. r« r h r c r t 4. If r Q is the radius of the circle circumscribing any triangle, _ a b _ c _ a bc ~ 2 sin ,4 ~ 2 sini>' ~ 2 sin C ~ 4~A 5. Show that ry 4 = — 2(a + b + c) 6. Show that the lengths of tangents from the vertices A, B, C to inscribed circle are s — a, s — b, s — c, respectively ; and for the escribed circles s, s — b, s — c ; s, s — c, s — a ; s, s — a, s — b. (c) Table of triangles (taken from Dr. Conradt's " Trigo- nometry") for use in connection with exercises (#), (6) preceding. The table on page 145 gives a large number of integers for which a 2 + b 2 = h 2 , and from them a great variety of exercises on sines (and later cosines and tangents) will suggest themselves to the teacher. The numbers here being exact, and not representing measurements, calculated parts can be carried to any number of decimals. §85c] THE SINE FAMILY. 145 H lO 00 iO Tt( CO t* CO r-J t^H IO CO rH r-i CO t* CO 71 C7 17 rH X X CO IO o 00 o "7 — CM "HH CD _ CD IO rt< © • t— r-K N H N iO OJ ^ CO t—{ CO OS IO »H os CO H H CO CO CO t-( io rH CO HH lO O OS W H CO rH CO HN s 1- 71 I- lb •Hi o iO CO T* "* 1- r-\ iO — rH X rH OS . x o os :* • CM m CO CO rH CM ^ f-i -HH ,-1 CD CO rH s& X O H CO CO iO CO s CO iO iO "^ ^ ^7 CO 00 CD CM rtl ,-1 H- 71 71 — OS CM o OS lO CM 0J CO HH 71 — — N CO N .- : ^ 9 °5 H rH CO ri CO N H CM rH rH : 1-i \ 00 CD CM IO rH o X O O H rH B0 CM CM CO co © os © OS rH »— t iO O M DO io CO o OS H io © 00 oo 1- CO iO N CO _ "^ -* * £ CO N O CO CD-CM CD 0O IO ^ IO &» CO r-\ O IO 00 io oo ^i CM CO io b- y-i OS N ?] N co cd hh (N W H O o rH 00 o CO io O 01 rH -<* — iO 71 rH oo h a . ^ ^ ^ N X h a io to h •H^ O rH CM rH CO CO CM r^ X rH Qj CO > 3c> H HH 00 CM 00 !>. © OS X OS rH CO o X rH "* >7 — OS 1-H o 0> H CM .7 CO 1—1 1—{ N rH OS X O OS o ^ £ rH ^ ^ CM O Ol '"H CM 00 (0 iO CM CO rl CO K5 H 05 CO H 00 rH CO CO rH CO o rH CM IO 71 — o 00 H r-i OS o r— Q S CO O O OS O CO rH d ib co fc ^ lO N CO ^ CO O CO CO IO in rH O CM CM OS CI 00 rH OS i; ci s HH lO CO i-i r-t r-{ O 71 CO io HH CM o CO H OS OS p 71 rH — rH O HH CO _ . OS OS os r~ co os T-\ OS 10 O CO d '-' X * 3^ >7 71 CM HH CO NO OJ CO CO rH CM -*- CO CO CO 71 O CO <0 H CM 17 CO I- iO HH CO o CM HH iO hh O CO CO CO 25 ^ H^ IO HH CV| U7- i>. lO CO £<*> I- rH CD CO CO qoi n N CO CO y-{ © 99 © 71 CD lO o 1— 1 CO in H O 00 00 OS r. CO rH OS O O OS O CO lO !C rA N S ^ CO N H H N S « co rH CM 00 CM CO © OS tH y-i rA CO 71 lO CO CM o co H 00 co —1 00 IO rH OS -^ 00 os «£ „ • H^ CD CM CO rH lO rH X CM rH - 1 OS X H ^ CO iO HH CM CO CM CO rH w PH CO CO o CO CO H ^7 >7 00 rH IO IO CD CO N O] 00 ■* cm' os 10 2 rH CO CO H H osfe 05 HH CO >c MS * ^ ft < V. j? ^ e ^C: M( r£ #^< K e^ 146 PLANE TRIGONOMETRY. [§86 Jh § 86. Vertical Projections. If perpendiculars AA V BB X (Fig. 55), are drawn from the y points A, B, of the straight line AB, to the vertical straight line MN, A X B X is B called the vertical projection of AB ' on MN. A X B X = A 2 B = AB -sin 6. .*. the length of the vertical projection of a given length, I, is I times the sine of the projecting angle, this angle being meas- ured from the right-hand horizontal line as initial line, the starting-point for the length being taken as the pole, A for AB, B for BA Similarly, the projection of ABO (Fig. 56) is l x • sin $ 1 + l 2 • sin 6 2 , or, the projection of a sum is the sum of the projections. Counting projec- tions upward as plus and downward as minus, the projection of any closed outline is evidently zero. That of Fig. 57 is • * FlG - 56 - A X B X + B x O x + Ci©, + A^i + -#1*1 + F X A V or zero, since there is a return to the starting-point. In surveying this is expressed by saying that the sum of the jy* northings is the same as the sum of the southings, in a closed diagram D \ representing a plotted survey. The lengths of the sides being l v l 2 , l 3 , . . . l n , the sum of the projections is l x • sin 0j -|- l 2 ■ sin 6 2 + l 3 • sin 3 + . . . +Z w .sin0 n =O. What is here termed vertical projec- tion is identical with difference of latitude in surveying, or, as com- monly called, latitude. § 86] THE SINE FAMILY. 147 .-. Difference of latitude of two points is their distance apart multiplied by the sine of the bearing of their connecting line (or " course ") from the east and west line. LABORATORY EXERCISE. Draw a line and its projection on a vertical line. Measure the lines and the angle. Compare the measured length of the projection with its calculated length. Taking the projection as given, calculate the length of the projected line and compare with the measurement. Carry out the preceding exercise in the field instead of on paper. Make the projection of a closed broken line carefully in a drawing, measure each projection, and add algebraically. See how near the result is to zero. Do the same in the field. EXERCISES. Find the latitude of each of the following courses, both by natural sines and by logarithms, and test the results. The lengths of the courses are given in chains, and the bearing, or direction from the north and south line, is indicated by the corresponding letters. Thus, N 20° E, 10 chains, signifies that the course is 10 chains long, and is directed 20° toward the east from the north line. So is S 20° E, 20° east from the line running south from the beginning of the course. 1. N 20° 13' W, 18.34 chains. 2. N 34° E, 17.35 chains. 3. S 54° 34' E, 85.45 chains. 4. S 47° 47' W, 24.17 chains. 5. N 56° 56' E, 34.34 chains. 6. Calling north latitude plus and south latitude minus, what is the bearing of a course whose latitude is -f 17.21 and length 56.13 chains? What the bearing when the latitude is — 19.23 and length 75.34? How many different courses satisfy these conditions? 7. The notes of a survey give the following courses; determine whether the sum of the northings is numerically the same as the sum of the southings. Solve by natural sines and by logarithms and compare results : N 69° E, 437.0 feet. S 19° E, 236.0 feet. S 27° W, 244.0 feet. N 71° W, 324.0 feet. N 19° W, 183.0 feet. 8. In an example like 7, with the data on sides reading to four signifi- cant figures, are the angles given sufficiently exact? Should they not read to minutes? Would dropping the final zeros leave the indicated accuracy unchanged ? 148 PLANE TRIGONOMETRY. [§87 9. Make up an example calling for the use of a five-place table. 10. One calling for a six-place table. 11. One calling for a seven-place table. 12. If a velocity along a line inclined at an angle A to the hori- zontal is represented by a line of given length, what line will represent the vertical component of such velocity ? 13. What is the initial vertical velocity of a projectile which comes from a gun at an angle 6 to the vertical with a velocity of V ft. per second? What is such velocity when V = 26.3 and = 21°? 14. If a weight of w pounds rests (tied) on a smooth plane inclined at an angle 6 to the vertical, what is the pressure on the plane? What is the pressure when $ = 23°, and the weight 10 pounds? What is the pull on a string held parallel to the plane and holding such a weight ? 15. A lever a feet long and inclined at an angle A° to the vertical has a weight of p pounds at one end, what is the moment of this weight about the other end of the lever? What is it when p = 10 pounds, a = 10 feet, ^° = 10°? § 87. Horizontal Projections. To use the sine in horizontal projections, the upright vertical is taken as the initial line and the clockwise turn as positive. Then projections running £ —* toward the right are plus, and those toward the left, minus. What has been said with reference to sums of vertical projections holds A - also for sums of horizontal projec- tions. In surveying, the horizontal projection of a course is called its departure, or the differ- ence in longitude of its extremities. The algebraic sum of the departures of a closed survey is zero. EXERCISES. 1. Determine the departures of the courses given in Exs. 1-5 on page 147. 2. In Ex. 6, page 147, change the word " latitude " to departure, the word " north " to east, the word " south " to west, and solve. 3. Determine whether the sum of the eastings is the same as the sum of the westings in the survey of Ex. 7, page 147. THE SINE FAMILY. 149 EXERCISES TO BE SOLVED WITHOUT THE USE OF TABLES. Express the results in terms of the sines given in § 75, and in general terms. 1. A regular polygon of n sides is inscribed in a circle of radius r ; find the general expression for the length of its sides in terms of the radius and half the central angle subtended by a side. To what inscribed polygons may the expression be applied without using the tables? Determine the sides of such polygons when the radius is 250 feet. 2. Determine the general expression for the perpendicular from the centre upon the sides of the polygons of Ex. 1, in terms of the radius and the half of the vertex angle. To what polygons may this be applied with- out using the tables ? Make the calculations when the radius is 100 feet. 3. If a building is 50 ft. wide, calculate the length of rafters when the pitch of the roof is 30°, 45°, 60°, respectively, the roof being an isosceles tri- angle. Also the rise of the roof in each case. If A is the pitch and I the length of rafter, what is the rise ? What the width ? If b is the breadth and A the pitch, what is the length of rafter? What the rise? If p is the rise and A the pitch, what is the length of rafter? What the width? 4. A ladder 50 feet long is tilted at angle of 30° to the horizontal and leans against a building ; find the distance of the top of the ladder from the ground, and of the foot from the building ? Answer the same ques- tions when the tilt is 45° ; 60°. 5. A kite is flying 50 feet high ; what is the length of string when the angle between the string and ground is 30°, assuming the string straight? When the angle is 45°? 60°? 6. The upper part of a tree broken over by the wind makes an angle of 30° with the ground, and the top lies 50 feet from the root ; how tall was the tree ? 7. The angle of elevation of a hill (its rise from a horizontal line) is 30°; what distance would one travel in walking up the hill to an elevation of 100 feet? What if the angle were 45°? 60°? What if the angle were A and the height h ? 8. Find x, y, in the adjoining diagram. If 30°, 60° are changed to A, B, and 10 yards to a yards, find expressions for x, y, in terms of sines. (Drop perpendicular from B on AC.) 9. A regular polygon of n sides is circum- A .10 yds. b y scribed about a circle of radius r; find an expres- Fig. 59. sion for the length of the diameter of the polygon in terms of the sine of half the central angle ; find also an expression for 150 PLANE TRIGONOMETRY. the length of the side of the polygon. Taking the radius as 50 feet, find the lines just mentioned for all regular polygons for which they may be found without using the tables. 10. A person on the top of a tower 80 feet high observes two objects on a straight road in line with the tower. The angle of depression of one object below a horizontal line passing through the top of the tower is 30°, and that of the other is 45°. Find the distance apart of the two objects. Solve the same problem when the angles are A, B, and the height h. 11. A man wishing to know the width of a river, selects a tree on the opposite bank, and finds its angle of elevation to be 45°. Going back 150 feet in a straight line with the tree and first point of observation, the elevation is 30° ; how wide is the river ? Solve the same problem when the angles are 60° and 30°; 60° and 45°; also when they are A, B, with the distance 150 changed to d. 12. How far from the polar axis of the earth is a point whose latitude is 30° (the radius of the earth being taken as 4000 miles)? 45°? 60°? Solve the same problem when the latitude is L and radius r. 13. Find the area of an equilateral triangle whose side is 100 feet. Solve the same problem when the side is s. Find also the distance of the centre of gravity of such a triangle from the vertex and from a side. 14. The hypothenuse of a right-angled triangle is 15 yards ; what is its area when one of its angles is 30° ? What when one of its angles has its sine equal to § ? 15. The base of a right-angled triangle is 24 feet; what is its area when one of its angles is 30°? What when one of its angles has its sine equal to f ? EXERCISES USING THE TABLES. 1. Make up from each of the preceding fifteen examples an example for whose solution it is necessary to use the tables, by taking the angles other than 30°, 45°, 60°, etc. Solve the same, and apply in each case a suitable test. Solve each example by Natural Sines and by Logarithmic Sines, and compare results. Make in each case an example calling for the use of a four-place table, and one calling for the use of a five-place table. Suggest data suitable for a six-place table ; for a seven-place table. 2. A smooth lever, turning in a vertical plane, has a 1-pound weight attached to its end ; what connection with a table of sines (of the angles which the lever makes with the vertical) have the pressures at right angles to the lever? What if the weight is 10 pounds? 88] THE SINE FAMILY. 151 § 88. Solving by Sines Triangles Other than Right-angled Triangles. To solve an oblique-angled triangle there must be given not less than three parts, including a side. Only three dif- ferent cases, so far as method of solution is concerned, can arise : % (a) A pair of opposites (an angle and its opposite side) is among the parts given. ($) The three sides are given. (7) Two sides and the included angle are given. Case (a) is always solved by sines. Case (/3) may be solved by sines, though this is not the best solution when all three angles are desired (see § 185). Case (7) cannot be solved by sines conveniently when the sides are long. It can be solved by dividing the triangle into right-angled triangles. A pair of opposites are given. Drop a perpendicular, p, from on AB (Fig. 60), or AB pro- duced (Fig. 61). Then from the two triangles formed we have p = a • sin J?, p = b • sin A, .\ a • sin B = b • sin A. Case (a). sin A sin B By symmetry, each one of these expressions is the same as . c . This may be proven, if thought necessary, by drop- ping the perpendicular from some other vertex than C. How- ever, it is advisable to note early the principle of symmetry, that where certain parts of a diagram are involved symmet- rically in an expression, the remaining parts are similarly involved. 152 PLANE TRIGONOMETRY. [§88 It is evidently a matter of indifference whether the per- pendicular falls within the triangle or without, since the c sine of the exterior angle of a triangle is the same as the sine of the interior adjacent supplementary angle. Thus, in all cases, a b _ c sin A sin B sin C Each of these expressions is, in fact, the diameter of the circle circumscribing the triangle. The central angle BOO (Fig. 63) is twice the angle A on the circumference. Taking OD perpen- dicular to BO, angle BOD is, therefore, A. .-. DB = OB -sin BOB; r • sin A. or, 2 r, or the diameter of the circle. Fig. 63. sin .4. Show how to construct the diagram to bring in the angle B or 0. The same relation holds also, of course, for right- angled triangles. Then is the middle point of AB, and a b _ c _ c c sin A sin B 1 sin 90° sin In using the relations above in solving an oblique triangle, one should proceed as in right-angled triangles, writing on the first side of an equality the part desired, then immedi- ately on the other side the part opposite, followed by the remaining given parts, in the order, sine/side or side/sine, according as the first member is sine or side, as in ■, when J., B, b are given. sin A a = sin A sin B c sin A = a sin A = a sin sini? sin C , when A, C, c are given. , when a, 5, B are given. , when a, 90° (i) In Case (2) the given angle is | < g()< / \/ In Case (i) each of the remaining angles must be less than 90°, and there is no ambiguity. In Case (ii) there may arise four different cases : (8) The side of the given pair of oppo- sites is longer than the other given side. For example, given a, A, b, with a> b. Since, by hypothesis, A is acute, so is B, for if a > b, so is A > B. (e) The side of the given pair of oppo- sites is less than the other given side. Under this there may fall three sub- cases, as in the following diagrams : (e x ) In Fig. 65, the side opposite the given angle is not long enough to reach to the border of the angle. Here a < b and also < b • sin A, and there is no triangle. (e 2 ) In Fig. 66, a is just long enough to reach AB. Here a b - sin A. This is the only case, when a pair of opposites are given, in which both possible angles determined by the sine are to be taken. It is known as the ambiguous case. In solving problems where a pair of opposites and another side are given, one should note, therefore : (f ) If the given angle is greater than 90°, the calculated angle is to be taken less than 90°. (77) If the given angle is less than 90°, with the opposite side greater than the other given side, the calculated angle is to be taken less than 90°. (/c) If the given angle is less than 90°, with its opposite side less than the other given side, there may be no triangle, one triangle, two triangles. Which of the three cases of Case (/e) is present, may be settled by making a preliminary sketch to scale of the data, with the results already indicated in (ej), (e 2 ), (e 3 ). Lay out the given angle with the protractor. From its vertex, lay out along one of its terminals the side bordering the angle. From the other extremity of this side as a centre, describe a circle having for its radius the side opposite the given angle. // x N -^"~ "'^°V There will be either no triangle, or one triangle, or two triangles (Fig. 68). A preliminary sketch is, however, not essential. The calculation of the unknown parts will itself show which case is present. For example, if a, A, 5, are given, with A < 90°, . t> 1 sin A sm B — §88] THE SINE FAMILY. 155 • (X) If a < b - sin A, sin B will be greater than 1, showing there is no triangle. Using logarithms, log sin B would appear greater than zero. (/*) If a—b • sin J., sin jB=1, and there is but one triangle, with B = 90°. Using logarithms, log sin B will be zero. (v) If a > b • sin A, sin B < 1, and there are two solu- tions. Using logarithms, log sin B will appear less than zero. Model Examples. c 1. Given ' a = 65 4 = 73° .6 = 97 sin B = b Find < C No triangle. C =180 -(A + B) c = sin The order here is that in which the quan- tities will be found, as that also after "Find" above. sin A log sin B = log b + log sin A — log a log b = log a + log sin A log c = log sin + log a — log sin A. SOLUTION BY LOGARITHMS. log 5 = 1.9868 log a = 1.8129 log b -log a = 0.1739 log sin A = 1.9806 SOLUTION BY SINES. sin A = 0.9563 5= 97. 86.0 6.7 65 [93 sin > 1 no triangle. log sin B= 0.1545 This is > .\ no triangle. 156 J LANE TRIGONOMETRY. [§88 f a = 1072 2. Given] A=12°W { 6 = 4849 * sin B = b - N 90 Fig. 70. r£ = 90° Find] C = 77°14' I c=4731 sin A a (7=180- Oi + £) c = sin C sin ^4 log sin B = log b — log a + log sin A log e a log sin C + log a — log sin J.. SOLUTION BY NATURAL SINES. 5 = 4849. sin .4 = 0.2210 969.8 97.0 4.8 1072, about sin B =1, about. SOLUTION BY LOGARITHMS. log b = 3.6857 loga= 3.0302 log b — log • a = 0.6555 log sin A = 1.3444 log sin #=1.9999 log sin B = 0, about. Thus the triangle is seemingly a right-angled triangle. In determining 90°, with a > 5, there is one and but one solu- tion. The student may make the calculations, using both natural sines and logs, determining also how far the signifi- cant figures of the calculated results should be taken. 5. The data as in Ex. 4, with the values of a, b, interchanged. The solution is impossible, since if a < 6, then is A < B. As A is obtuse, this would require B to be obtuse. A tri- angle cannot have two obtuse angles. 6. Given 1 Find a = 20.24 \ \ C = 57° 3' A = 103° 36' \103 O 36' \ 19'21?^ b = 6.901 B = 19° 21' A Fig. 72. B c = 17.48 C= 180° -(A + tf) ,_sini?. a C - sin C - a sin A sin A Order of logs { log b = log a ~ log sin A + log sin B I log c = log a — log sin ^4 -f- log sin O Since A > 90°, with a > J, there is only one solution. this example is worked out completely as a "model example." The calcu- lated sides are carried to seven significant figures, and the calculated angles to hundredths of a second. A procedure like that is justified only in the case of data known exactly, which is never the case with any triangle met "on land or sea." When one is called upon to measure a line and finds himself satisfied on knowing the length to within one five-hundredth of its value, the spirit of calculation in, him seems somewhat insatiable if nothing short of pushing the apparent error in results to within one ten-millionth of the calculated line meets his wishes. This refinement of calculation is all a sham. The figures beyond the place of the data are altogether worthless. 160 PLANE TRIGONOMETRY. [§88 SOLUTION BY NATURAL SINES. sin B = 0.3314 a — 20.24 6.628 .066 sin A .013 b 0.9720 6.707 6.900 5.832 ' .875 .875 sin (7 = 0.8392 a = 20.24 16.784 .168 sin A .033 c 0.9720 16.985 9.720 17.48 7.265 6.804 .461 .389 SOLUTION BY LOGARITHMS. log a = 1.3062 log sin A= 1.9876 diff. =1.3186 log sin # = 1.5203 log sin 0= 1.9238 log b = 0.8389 log o = 1.2424 (The student may supply a log-check.) .072 § 89. Schemes for Solution. The preceding examples have been worked out to serve as models for subsequent solutions of the same sort. The stu- dent will note the forms followed in the calculations. He will state in tabular form the parts given, followed by a tabu- lar statement of the parts to be found, in the order in which they will be found, accompanied by a diagram to scale on which the given parts are noted in full lines, with the lines to be calculated indicated by dotted lines. As the parts cal- culated are determined, they may be entered in the table for such parts. The formulas to be used in calculating the parts are to follow the data as in the examples above, and when 89] THE SINE FAMILY. 161 logs are to be used, the order in which they are to follow each other in the scheme of solution may be indicated as above, while the student is still in the novitiate stage. The order in which the logs follow each other in the solu- tion scheme should be such as to make the manipulations easy. Some regular scheme ought always to be followed. That adopted in the preceding examples seems as economical as any. Having settled on a scheme, enter on the scheme all logs which are called for before manipulating these logs at all. (A) The complete scheme, using logs (and not cologs) when a pair of opposites and another side are given (assum- ing only one solution), is for a, A, b as follows : c ml Am FindJ(7 = I b = l c Fig. 73. Formulas sin B = b • sin A C=1$Q°-(A + B) c = sin C' Log order sin A log sin B = log b + log f^A\ log c = log sin C + log ( ) Vsin A) Log Solution Scheme. (i) (2) (3) LOGARITHMS. sin B = (6) b = (3) sin A a (4) sin A = (1) a = (2) a (5) sin A sin = (7) c = (8) 162 PLANE TRIGONOMETRY. [§89 In working examples, make the scheme first. Then fill it in. First with (1) ; then (2), (3) ; subtract (1) from (2) for (4) ; then (2) from (1) for (5) ; add (3), (4), for (6). Then find B. Then 0. Next fill in (7) ; add (5), (7), for (8). Then look up c. TESTING RESULTS. The student should test his own results. Combining the logs in some new relation is an appropriate test here,, as in b . sin = e • sin B. logc = log b = log sin C = log sin B Make sure to look up all logs (of "Ans.") in the tables. The log-check should check by 2 or less in the final figure, the final figure being the figure in the place to-which the data go. For two-figured data, the second figure of mantissa is the final place, etc. Test also by making diagrams to scale. (i?) Solution scheme for case (J.) when cologarithms are used: log sin B = (6) log 6 = (3) colog a = (4) log sin A = (1) colog sin A = (5) loga= (2) log sin 0= (7) lo g<; = (8) The numerals indicate the ordering of performing the work: (1), (4), (3) added give (6). Then find B. Then C. Then (7). Then add (5), (2), (7), for (8). Then find c. EXERCISES. Arrange calculation schemes for all similar cases, and the logarithmic tests. Arrange similar schemes when cologarithms are used. 89] THE SINE FAMILY. 163 Solve the following examples both by logarithms and by natural sines. Compare results. Give also the logarithmic tests. Make diagrams to scale. Measure the ungiven parts, and compare with the results of cal- culation. Make sure the results are given appropriately as to significant figures and angle readings. 1. a8,A3o°,bS. 2. a 05, A 47°, b 35. 3. a 5.4, A 67° 30', b 6.7. 4. a 71.5, A 65°, b 63.2. 6. a 67.34, ^ 45° 43', 6 87.34. 7. a 6715, ,4 63° 13', 6 7123. 8. a 23.579, A 34° 52' 23", b 25.431. 9. a 234.671, A 34° 45' 67".3, b 234.678. 5. a 7.45, ^47° 25', 6 9.36. 10. a 2345.76, ,4 56° 57' 58".45, b 2531.89. 11. Construct similar examples for a, A, c, taking data to one signifi- cant figure in lines; two significant figures ; three; four; five; six; seven. See that the angles read appropriately. Solve and test them. 12. Do the same for 6, B, c, as given. 13. Do the same for C, c, b, as given. 14. Construct some examples which have no solution. Test. 15. Construct some examples with two solutions. Solve. 16. Construct some examples with only one solution. Solve. ((7) The complete scheme when a pair of opposites and another angle are given, is similar to the following, assuming a, A, B given : A b = Given a = A = B = A B C Fig. 74. r (7= 180° — (A + B) Find • b = sin B • - — - sin A c = sm • - — - sin A J • c = Log order log b = log a — log sin A + log sin B log c = log a — log sin A + log sin 164 PLANE TRIGONOMETRY. [§89 LOG SOLUTION SCHEME. log b = log sin B = \sm A) log sin A = log a — log sin = logo = (7) (3) (5) (2) (1) (6) (4) (8) SOLUTION TEST. b • sin Q — c - sin i?, or log b + log sin C = log , BDC as 0, BCD as /?; prove, AB = CD sin< l> sin l 3 ~ sin (/? + <£) 166 PLANE TRIGONOMETRY. 7. At each end of a horizontal base of length 2 a it is found that the angular height of a certain peak is and that at the middle point of the base it is . Prove that the vertical height of the peak is a sin $ sin — sin 6) (sin + sin 6) 8. A flagstaff, NP, stands on level ground. A base, A B, is measured at right angles to AN, the points A, B, N being in the same horizontal plane. The angles PAN, PBN are 6, respectively. Prove that the height of the flagstaff is a t> sin 6 sin (sin 6 — sin ) (sin 6 + sin <£) 9. Solve Ex. 7 for 0, 24° 23'; <£, 34° 56'; a, 317.4. 10. Solve Ex. 8 for 0, 67° 45'; , 34° 24'; AB, 213.3. 11. A man standing due south of a tower on a horizontal plane observes the elevation of the top of the tower to be 50° 45'. Going 100 yards due east, he finds the elevation to be 46° 25'. Find the height of the tower. 12. A man in a balloon observes the angle of depression of a ship at anchor due north of him to be 40°. After the balloon has drifted 3 miles due west at the same elevation, the angle of depression of the ship is 30°. Find the height of the balloon. 13. From the extremities of a horizontal base, AB, whose length is b, the angles BAF, ABF are measured, F being the foot of a tower and in the same horizontal plane as A, B. At A the angle of elevation of the tower is also measured. Call this a; BAF, 0; ABF, . Show that the height of the tower is 7 sin • sin a > sin (90° -«) sin (# + <£)' 14. A vertical object stands on a hill whose angle of slope is A°. At a distance a from the foot of the object, this distance being measured along the slope of the hill and down the hill, the object subtends an angle B°. Show that the height of the object is sin B sin (90° - A - B) 15. A house stands on a hill whose slope is 15°, and at a point 80 feet from the house down the hill the house subtends an angle of 33°. Find the height of the house, and the distance from the point of observation to the top of the house. THE SINE FAMILY. 167 16. A vertical object stands on a hill whose angle of slope to the horizon is A°. At two points, whose distance apart measured along the slope of the hill is a, the object subtends angles B°, C°. Show that the height of the object is sin B sin C sin (C - B) sin (90° + A)' 17. Solve Ex. 16 when A = 20°, B = 13°, C = 19°, a = 553 feet. 18. A person standing on the slope of a hill whose angle of slope to the horizon is A, finds that the line running up hill to the top of a build- ing from the point where he stands makes an angle B with the face of the hill. The observer is a feet from the building, this distance being measured along the slope of the hill ; show that the height of the build- ing is sin£ tt sin (90°- A-B)' 19. Solve Ex. 18 when a is 75, A is 18°, and B is 35°. 20. By means of the relation that the sines of the angles of triangle are to each other as the opposite sides, prove that if a line is drawn bisecting one angle of a triangle, it divides the opposite side into seg- ments proportional to the sides about the bisected angle. 21. A regular pyramid on a square base has an edge 150 feet long, and the length of the side of the base is 200 feet. Find the sine of the inclination of the face of the pyramid to the base, and the corresponding angle. 22. A pyramid has a square base, the length of whose side is a. The vertex of the pyramid is over the centre of the base, and at a distance h from this centre. Show that the angle between the two lateral faces of the pyramid is given by the equation a 2 + 4 h* Find also the sine of the angle between a lateral face and the base. 23. A flagstaff 100 feet high stands in the centre of an equilateral triangle. From the top of the flagstaff each side of the triangle subtends an angle of 60°. The triangle is horizontal and the flagstaff vertical. Show that the length of the side of the triangle is 50 V6 feet. 24. A rectangular target faces south, being vertical and standing on a horizontal plane. If the sun's angular altitude is A° while the sun is 5° from the south, show that the area of the shadow of the target on the horizontal plane is found by multiplying the area of the target by sin (90° -4) sin (90° -B) sin^. 168 PLANE TRIGONOMETRY. 25. The angles of elevation of the top of a tower, standing on a hori- zontal plane, from two points distant a, b, from the foot of the tower and in a straight line with the foot of the tower, are together equal to 90°. Prove that the height of the tower is a mean proportional between a and b. Prove also that the sine of the angle subtended at the top of the tower by the line joining the two points is a ~ . a + b 26. A cloud observed simultaneously from two points on a north and south line and distant a miles apart, appears in the south from one point and at an elevation of A, and in the north from the other point and at an elevation of B ; find the distance of the cloud from each point, and its vertical height. 27. A man ascends a mountain by a direct course, the inclination of his path to the horizon being at first a and afterward changing suddenly to (3, which continues to the top. If the height of the mountain is a feet, and the angle of depression of the starting-point as observed from the top is y, show that the length of the ascent is j sin (/J - y) + sin ( ? -«)}-—! sin y t j sin (/? — a) 28. At noon a person on a cliff h feet above sea level observes the altitude of a cloud in the plain of the meridian to be a, and the angle of depression of its shadow to be /?. If y is the angular elevation of the sun at the time of observation, show that the height of the cloud above sea level is . . nx ^ sin y sin (a + /?) sin/3 sin (y±«) ' the plus sign preceding a when, if the cloud is in the south, the sun is in the north ; the minus holding when both sun and cloud are on the same side of the observer. 29. Two circles of radii a, b, touch each other externally ; find ex- pressions for the sines of the half angles between all possible pairs of common tangents. 30. A ship's observer notices that two objects are in a straight line whose bearing from the north is A° toward the east. The ship sails a miles in a course B° west from north, and then the bearing of the more distant object is D° east by north from the direction of the ship's course, and that of the other object is C° east by north from the direction of the ship's course ; show that the distance between the objects is sin (A + B) sin (C - D) sin (C -B - I) sin (D - A - B)' § 90] THE SINE FAMILY. 169 31. Show that if, in the preceding problem, the bearing of the line of objects is 15° to the east of north, and the ship's course northwest, and the bearings of the objects east and northeast, the distance between the objects is sin 60° sin 45° a sin 75° sin 30° 32. A ship observed another ship A° from the north, sailing in a direc- tion parallel to its own. Inp hours its bearing was B° from the north, and in q hours afterward its bearing was C° from the north. Show that if D° is the bearing from the north of the ships' courses, the following equation holds : sin (D - C) _ p . sin (B - C) sin (D-B)~ q- sin (A - B)' § 90. Use of the Sine in Calculating the Areas of Triangles. Drop a perpendicular from any vertex, as (7, on the oppo- site side. This perpendicular is a • sin B, or b • sin A, in all cases. The double area of the triangle is base times altitude. .\ 2'A = ac-sin£, (1) = be • sin A, (2) = ab - sin (7, by symmetry. (3) The area of a triangle is one-half the product of any pair of its sides and the sine of their included angle. We have previously shown (§ 88) that a b c sin A sin B sin (f abc abc abc abc (4) be sin A ae sin B ab sin C 2 A Thus each expression in (4) is — - 7 ,\ diameter of circumscribing circle = — (§ 88). ... 2A= ahc diameter By means of (4), the area expressions (1), (2), (3), may be changed so as to involve only one side and the three angles. 170 PLANE TRIGONOMETRY. [§90 For example, by (4), c = sin - - — -• Setting this in (1), 2 Double area = - — - • sin B • sin O smA b 2 sini? sin sin C • sin A, by symmetry sin A • sin i?, by symmetry. The student may express these relations in a single verbal statement. EXERCISES. 1. Find the areas, the three altitudes, and diameters of the circum- scribing circles for the following triangles, without using the tables : a = 14, b = 10, C = 30° 6 = 15, c = 12, A = 120° a = 15, c = 17, B = 135° a = 3, b = 6, C = 60° a = 6, c = 5, 5 = 150° a = 3, 6= 5, C = 180° a = 21, e = 12, 5= 45°; a = 3, A = 30°, J5 = 30° ft = 21, ^ = B = C. 2. Calculate by natural sines and by logs the areas of the following triangles and the diameters of circles circumscribing them. Test. a = 23.4, b = 34.5, C = 34° 25' ; b = 3423, c = 2145, A = 34° 23' a = 3.45, c = 5.43, B = 45° 45'; a= 231, &= 432, C = 23°45' a = 23.4, ^ = 45° 25', B = 67° 35' ; a = 23.1, A = 32° 25', C = 43° 35' c = 6.45, ^ = 57° 13', B = 67° 53'. 3. Show that the double area of any quadrilateral is the product of its two diagonals into the sine of the angle between them. 4. Show that the perimeter of any triangle is the diameter of the circumscribing circle into the sum of the sines of its angles, and the area of any triangle is the product of the radii of the inscribed and circum- scribed circles into the sum of the sines of its angles. 5. Show that if triangles of equal area are circumscribed about the same circle, their perimeters are the same. What is the sum of the sines of their angles ? 6. The diagonals of a quadrilateral are 150, 131, and their angle 45° 15' ; what is its area? §91] THE SINE FAMILY. 171 7. The diagonals of a parallelogram are 123, 321 ; find its sides, angles, and area, when the inclination of the diagonals is 45° 25'. 8. The sides of a parallelogram are 150, 130, and one of its angles is 37° 15' ; find its area. 9. Construct examples similar to Ex. 2 for one-figured data; two- figured ; four-figured ; five-figured. Solve. Test. §91. To Determine the Sine of Any Angle of a Triangle, and the Area of the Triangle, when the Three Sides are Given. Let ABC he the given triangle, with CD perpendicular to AB, and lengths as in the diagram. P *=b 2 -X>, (1) ( ■» p = a 2 - (e - xy. (2) / ... a 2 -0 - x) 2 =h 2 -x 2 . (3) y P \a .\ a 2 - (10) may be used to find the area of a triangle when the three sides are given. (7), (8), (9) may be used to find the sines of the angles. The angles themselves are not uniquely determined, since two angles less than 180° have the same sine. For this reason any angle is found from the sine of its half rather than from the foregoing expressions, since the half angle is acute. § 92. The Sine of the Half Angle of a Triangle in Terms of the Sides. Inscribe in the given triangle a circle (Fig. 77). AD = AF, BD = BE, OF=OE. .-. 2AD + 2BO = perimeter of triangle. J, .-. AD = s-a. (1) The three triangles A OB, BOO, COA make up the tri- angle ABO, r(a + b + - g ) * 8 . A DO r 2 AO AO' (2) (3) 93] THE SINE FAMILY. 173 But ZO 2 = r 2 + A& (s — a)(s — 6)(s — (?) + (*-«)* by (2), (1) — a {(t _*)(•- 0>+ •(•--•)! C!_^ {(a _( 6 . c)Xa + ( ft_ c)) + (ft + c + aXft+c _ a)j • 5c, when multiplied out. v^* )(«-ft)(«-<0 . sin#=-^ = 8 = J<« -»)(«-«). (4) sin | = V (8 ~ a ^ 8 ~ C) > (5), by symmetry. sin %=^ {8 ~ a) ^~ b \ (6), by symmetry. § 93. Calculation of the Half Angles of a Triangle when the Three Sides are Given. It is best, when using logs, to set the sines in the forms n \„ A - ~ K* - a)o - &x* - g ) a :~^- J(« -«)(«- 6)(«-<0 b sin =V i£ ^ £ abc s — b Z i n c -J (°- *)(» - b-)C - o) . _J_. 2 * afo « — c The scheme of solution will then be, the numerals giving the order of filling in, all logs being looked up before manipu- lation begins : 174 PLANE TRIGONOMETRY. [§93 (1) b= (2) c= (3) 2«= (4) • - (5) a — a = (6) .-J = (7) s-e= (8) CAecft.- (6) + (7) + (8) = (5) log («-«)= (9) log(»-6)= (10) log (•-«)= (11) log (s — a)(s — 6)(s — c) (15), by adding (9), (10), (11) \oga= (12) log 6= (13) log*= (14) log «6(? = (16), by adding (12), (13), (14) log O-a)O-ft)<>-<0 abc = (17), by subtracting (16) from (15) log— — = (18), by subtracting (9) from (12) 8 — a log -A- = (19), by subtracting (10) from (13) 8 — log— £_= (20), by subtracting (11) from (14) » — ^ ^J^if \ ° MJ N sf B N \M J Fig. 79. Fig. 80. Fig. 81. Fig. 82. In the unit-circle (Figs. 79, 80, 81, 82) MP is the sine, and, consequently, OS the cosecant. ^ MP ITS MP For op = W or T" os -. OS MP Since MP is the sine, OS is the cosecant. The cosecant is obtained by drawing the tangent at the 90° point of the unit circle, and prolonging it to meet the terminal of the angle. In the first and second quadrants the direct terminal is met, and the cosecant is positive. In the third and fourth quadrants it is the opposite terminal which is met, and the cosecant is negative. 97] THE SINE FAMILY. 179 § 97. Line Picture of the Cosecants of All Angles. Fig. 83. For the position, OA, of the terminal the cosecant is positive infinity (+ oo). It diminishes as the terminal turns, counter- clockwise, taking the values (where OB = 1) + oo - 0S± .. 0S 3 - 0S 2 .• 0S 1 - OB. Then the values OB -. 0S_ X .- 0S_ 2 -. 0S_ 3 - + oo, as the terminal goes from 90° to 180°. As the terminal passes through 180°, there is the discon- tinuity from + oo to — oo, since the cosecant passes to the opposite terminal, and the values indicated above, with signs changed, are passed over, as the terminal moves from 180° to 360°. As the terminal passes 360° there is again a discontinuity, from — oo to 4- oo, since the cosecant passes from the opposite terminal to the direct terminal. As the terminal continues to turn there is a repetition of what has just taken place. EXAMPLES. 1. Solve the equation sin x + cosec x = f . Hint : Let cosec x = Solve the resulting quadratic for sin x. sin a; Recall the corresponding angles. 180 PLANE TRIGONOMETRY. [§98 2. Can there be an angle whose cosecant is £? What is the least value of the cosecant ? When the sine increases, does the cosecant increase or decrease ? 3. Construct an angle whose cosecant is f ; — f . 4. From the corresponding matter in connection with sines, read the following expressions: cosec -1 f ; cosec -1 ( — f ) ; 2 cosec -1 1 ; 3 cosec -1 ( — f ) . Construct the terminals for these expressions. How many terminals for any given value of the cosecant ? 5. Give the general value of all angles which have the same cosecant as any given angle A, a. 6. What is the relation of the terminals of angles which have the same cosecant ? Opposite cosecants ? 7. Determine and tabulate the cosecants corresponding to the sixteen terminals of § 75. 8. Solve the equation sin x + cosec a; = — § (general value). 3 V2 9. Solve the equation sin x + cosec x = ± (general value). 10. Solve with tables the equation sin x + 3 cosec x = 3.1432. 11. Solve the equation 4 sin = cosec 6 (general value) . 12. Solve the equation 4 sin = 3 cosec 6 (general value). 13. Solve the equation sin + 3 cosec 6 = \ (general value). 14. Solve the equation 3 sin 0—8 cosec = ■ (general value). 15. Solve the equation sin a; + cosec x = 2.5618. § 98. The Graph of a Function in Rectangular Coordinates. Suppose y is a single-valued function of a?, and that for any arbitrary value, a, of #, y has been calculated, giving b. This pair of corresponding values may be taken as the co- ordinates of a point in a plane, the x as abscissa, and the y as ordinate. Let this point be constructed. Suppose it is A in Fig. 84, with the abscissa 0M= a and the ordinate MA = b. Suppose a second point, B, obtained in the same way, with x — ON '= c, y = NB = d. If we imagine now that x takes all values of the continuum from x = a to x = c, and 98] THE SINE FAMILY. 181 ~r ~r x > ,--«,. i -/ *£ A-. Z s ^ I - if M Fig. 84. that the corresponding values of y are calculated and the corresponding points plotted, as were A, B, we shall have on the paper a set of points which viewed as an assemblage will con- stitute a line from A to B, straight or zigzag or crinkly or curved, or, as commonly called, a curve (this term includ- ing also straight lines as a special case). Such a curve is called the graph of the function, y, from A to B. In practice we cannot construct an infinite number of points. A sufficiently large number of points to give us a more or less close approximation to the true curve, depending on the use that is to be made of the result, are plotted and joined by a smooth curve. (In practice a so-called " French Curve " is used.) The graph shows to the eye the relation between the function, y, and the variable, x, both as to corresponding magnitudes and as to rate of change of the function as the variable changes. The graph of a function is frequently called the locus of the equation connecting x, y ; that is, it is the places of all the points whose coordinates give pairs of values of x, y, which satisfy the equation. A graph is also called a locus of a point, moving in a specified manner. In the case of a two-valued function, each value of x will give two values of y. So two points will have the same abscissa. The corresponding curve, or graph, will have two branches. Similarly, an n-valued function will have a graph of n branches. A graph can also be used to show pictorially the connec- tion between quantities whose relation to each other is not expressible in the form of an equation. For example, we may take, from the census reports, the population of a city for every ten years for a number of years. Laying out the years, to any scale, as abscissas, and the population as corre- 182 PLANE TRIGONOMETRY. [§98 sponding ordinates, to any convenient scale (500 inhabitants to an inch, or any other convenient number), then connect- ing the plotted points by a smooth curve, we have the popu- lation graph. The ordinate for any selected year will give the population, approximately, for that year. The graph will thus enable one to calculate the population fairly well for years other than the census years. Such graphs are used extensively in engineering work, also in economic reports for the exhibition of statistics. They tell the whole story pictorially, and enlighten where the same report given in figures would confuse or make no impression at all. ILLUSTRATIVE EXAMPLES. 1. y = x. This is evidently satisfied by any point on a straight line passing through the origin and bisecting the first and third quadrants. This line is the graph. 2. y = -x. This is satisfied by any point on the bisector of the second and fourth quadrants. 3. y = x + 1. This is evidently satisfied by any point on a line passing through the point (0, 1) on the ordinate axis and making an angle of 45° with abscissa axis. 4. y s x — 1. This is a straight line passing through (0, — 1) on the ordinate axis and making 45° with the abscissa axis. 5. x = ±2. This is a pair of straight lines parallel to the ordinate axis and at distances 2 to the right and left of the origin. 6. y = ±2. This is a pair of lines horizontal and at distances 2 above and below the abscissa axis. f /> X 1 f, {) Fig. 85. 98] THE SINE FAMILY. 183 7. x 2 + y* = 25. This is satisfied by any point on a circle of radius 5, with its centre at the origin. This is an example of a two-valued function, since for any x there are two y's, V- y =±V2o-x 2 . Thus, there are two points equidistant from the abscissa axis which satisfy the given equation, both points having the same abscissa, or x. Similarly for x in terms of y. 8. The preceding examples are those of functions whose graphs can be seen readily without calculation. To illustrate the calculation process, we may take y 2 = 4 x. Fig. 86. Giving to x the values following, one can calculate the corresponding y's, or get them from a table of squares : \ X ±y .1 .6 .2 .9 .3 1.1 .4 1.3 .5 1.4 .6 1.5 .7 1.7 .8 1.8 .9 1.9 X ±y 1.0 2.0 1.1 2.1 1.2 2.2 1.3 2.3 1.4 2.4 1.5 2.5 1.6 2.5 1.7 2.6 1.8 2.7 1.9 2.8 X ±y 2.0 2.8 2.2 3.0 2.4 3.1 2.6 3.2 2.8 3.3 3.0 3.5 3.2 3.6 3.4 3.7 3.6 3.8 3.8 3.9 X ±1 4.3 4.1 4.6 4.3 4.9 4.4 5.2 4.6 5.5 4.7 5.8 4.8 6.1 4.9 - These values plotted to scale (1 inch = 1, say ), taking y to the nearest 10th, give the curve. The student may plot this curve. Use coordinate paper. 9. Construct the graphs of the following : # 2 = -4x, x 2 = 4y, x 2 = -iy, y = 2x+3, y = 3x-4, y = -2x + 3, y=-2x-l, 3x + 2y = 6, 3x-2y = 6, -3x+2y = 6, -3x-2y = 6. 10. Look up in the census reports the population of the three most important cities in your state, and construct corresponding graphs to the same scale and compare. Calculate from the graphs the population for some year not a census year. 11. On rectangular coordinate paper make graphs of y = log 10 a: and V = 1°&« x -> to the same scale. What is the relation between correspond- ing #'s for the same #? (See § 5.) 184 PLANE TRIGONOMETRY. [§99 a § 99. Graph of the Sine, or Locus of the Equation/ = sin Jr. (a) Imagine the unit circle made of wire, with each ordi- nate a wire hinged at its extremity to the unit circle. Clip the circle wire at a, and stretch it out straight, with the positive ordinates pointing up and the negative ordinates down. The ends of the wire ordinates give the graph of the sine for all values of x from zero to 2-7T radians. (5) An easy practical process for making the graph of the sine. Draw a circle of any radius on a piece of pasteboard. Draw on it a large number of ordi- nates. Cut the circle out with a sharp knife, so as to give a smooth edge. Roll the circle in a vertical position on a straight line of a horizontal sheet of paper. Dot the line with a pencil at the places where the ends of ordinates touch it, laying out at such dots the corresponding ordinates, in length and sign. Join the ends by a smooth curve. This is the sine graph corre- sponding to the radius of the selected circle as unit. T-£-\ £"g c -"■^l 3^^ ■^g • / V*. ~3Z Sb J \ r i 1 c [_ ] k d t i 5 t v \ 7 s z s / S* ^ ^.~* > x: Fig. 87. Fig. 88. — Sine graph, x = to x = 2 ir. ( \ / I 1 t \ -oo — CO 1 1 i — OO — OO 1 1 190 PLANE TRIGONOMETRY. [§ 102 What is r = a, where a is a constant ? What is r* — 5r + 6 — ? What is F(r) = 0, where the function is many valued ? What is = ? What is = a, where a is a constant ? What is £2-50 + 6 = 0? What is F(&) = 0, where I 7 is many valued ?, EXERCISES. 1. Use Groat's radian polar coordinate paper to make the graph of r = 10 0, r = 20 0, r = aO. (Plan : give values to 0, as 0.1 radian, 0.2 radian, etc., and calculate corresponding values of r. Then plot the corresponding points and join them by a smooth curve. The points will be for the first curve: (0, 0), (1, 0.1), (2, 0.2), etc. These curves are spirals of Archimedes.) 2. Use Groat's radian polar coordinate paper to make a graph of 1 2 n r = -, r = -, r = -- (Hyperbolic spirals.) (Such a curve with r about if an inch, or inch and a half, when = ^ > will be found almost as useful, when made in hard wood, for drawing purposes as a " French Curve." Such a curve was found most useful in plotting Professor Orton's Re- ports on Ohio Clays.) Plot r 2 = a. (Lituus.) 3. Show that a circle is the graph for r = a sin 0, a being the diameter (vertical), being measured from a horizontal tangent. 4. Use Groat's degree-measure polar coordinate paper to plot r = 10 sin by points, taking as radial unit one of the small divisions on the paper. See if your curve looks like a circle. 5. Show that r sin = a is a straight line parallel to the z-axis at a distance a. 6. Use Groat's degree-measure polar coordinate paper to make a graph for r sin = 10. 7. Use Groat's degree-measure polar coordinate paper to make graphs of r sin 2 = a ; r = a sin 2 0; r sin 3 = a ; r = a sin 3 0. Give a some convenient numerical value. See if the number of loops in the graph of r = a sin nO depends upon n being even or odd. 8. Draw a circle. Take a horizontal tangent as initial line and its point of tangency as pole. Prolong each radial line (vector) from this point a distance equal to a diameter beyond the circle. Connect their ends by a smooth curve. This is the graph then of r = a (sin + 1). It has the shape of a heart, and is called a cardioid. § 103] THE SINE FAMILY. 191 9. Use Groat's polar coordinate paper (degree measure) to plot r = a (1 + sin 0) ; r = a (1 - sin 0) ; r (1 + sin 0) = a ; r (1 - sin 0) = 0, for selected values of a. 10. Draw a horizontal line. Select a point above it at a distance a as pole. Sketch points that are as far from the pole as from the line. The curve is a parabola. Show that any point on it satisfies the equa- tion, r (1 — sin 6) = a. What is r (1 + sin 6) = a? 11. Make the graphs for 6 = log e r ; 6 = log 10 r ; r = e fl ; r = 10*. § 103. The Coversed Sine. When the line definition was in use, with the correspond- ing diagram for the sine to the radius 1, the balance of the radius beyond the sine was given the name, the Coversed Sine. And still, coversin A = 1 — sin A. The coversed sine is used in engineering field books. EXERCISES. 1. Can the coversed sine be negative ? 2. "What are its limits of value ? 3. Construct terminals when coversin A = \ ; coversin A = f . 4. What are the coversed sines for angles whose terminals border the quadrants? 5. What are the coversed sines for angles whose terminals are the bisectors and trisectors of the quadrants? 6. How could a table of coversed sines be made for angles from 0° to 90°? 7. Make the graph for the coversed sine, y = coversin x. 8. Use the following method to construct a sine-curve: draw a circle of one inch radius, and a horizontal line through its centre, A, AO being a horizontal radius to the right. Divide the circumference into 9° spaces from O. Lay out from O, right and left, on the horizontal line, to the extent of the paper used, equal distances, each 0.16 inch (9° in radians, inches). Prolong the abscissas of the extremities of the 9° divisions to meet the ordinates at the ends of the horizontal divisions. Intersections of corresponding lines will be points on the sine curve. Use a similar process to make a cosecant curve. CHAPTER VI. j > y / < ' \ ) ( ) s 1 f THE COSINE, INVERSE COSINE, RECIPROCAL COSINE (SECANT), AND VERSED SINE (1 - COSINE) OF AN ANGLE. § 104. The Cosine of an Angle is the ratio of the abscissa of any point on the terminal of the angle to the modulus of the same point, or QM This is the ratio-definition, or Hassler's definition. The numerical relation of an angle in circular measure to its cosine is given by the following series, which may be considered also as a definition of the cosine : Similar remarks to those made with reference to the defini- tions of the sine hold with reference to those of the cosine. As with the sine, the cosine has a threefold use in cal- culations : (1) When the abscissa and modulus are given, their ratio, expressed as a decimal fraction, and compared with a table of cosines, will indicate the corresponding angle, or angles. (2) When the modulus and angle are given, the cosine of the angle is the number which, used as a multiplier with the modulus, will give the abscissa. (3) When the abscissa and angle are given, the cosine of the angle is the number by which to divide the abscissa to obtain the modulus. 192 Fig. 94. § 104] THE COSINE FAMILY. 193 LABORATORY EXERCISES. Construct with the protractor an angle of 26°. Lay out on it five different moduli and corresponding abscissas, including the modulus of unity (one inch or one foot). Measure moduli and abscissas. Divide each abscissa by its modulus to one or two figures. Compare results with one another and with the table. Do the same for five different angles, each with a single modulus and its abscissa. Compare results with the tables. Lay out an angle with the protractor. Measure the modulus. Mul- tiply it by the table-cosine (to one or two figures). See if the result agrees with the measured abscissa. Test for another angle the measured modulus with the value obtained by dividing the measured abscissa by the table-cosine. EXERCISES. 1. Take the exercises under § 76, changing the word a sine n to " co- sine," and " ordinate " to " abscissa." 2. The student may make and solve five simple practical examples which can be solved by the cosine ; by the sine. 3. If the initial velocity of a projectile at an angle A° to the hori- zontal is v, what are the vertical and horizontal components ? Solve a numerical example. 4. Assuming that the velocity due to gravity is 32 1 (t in seconds), how long will the projectile in Ex. 3 rise before gravity destroys the initial vertical velocity? How far will the projectile drift horizontally in that time (range) ? 5. A smooth plane, inclined A ° to the vertical, has a weight of \ pound tied on by a string running from the upper end. What is the pressure normal to the plane ? What is the pull on the string ? Solve a numerical example. 6. How far from the centre of the earth is the centre of the 40° par- allel of latitude? What is the radius of this parallel? Give general formulas for sphere of radius r. 7. A particle is moving with uniform angular velocity a in a circle. What are the component velocities along horizontal and vertical diameters at the time t (seconds), counting time from the moment when the particle is at the right-hand end of the horizontal diameter ? Solve a numerical example. 8. A particle is acted on by a force, F, in the direction A° to the horizontal line. What are the horizontal and vertical components of the force? Solve a numerical example. 194 PLANE TRIGONOMETRY. [§104 9. A particle has a velocity v in the direction A to the horizontal line. What are the horizontal and vertical components? Solve a numeri- cal example. 10. Compare numerical solutions of the foregoing problems with their graphical solutions. Hassler's definition holds for all positions 'of the terminal, no matter in which quadrant it may fall. p I 5s X- -**" ■ S S i s ^L* M _c Fig. 95. -M O - 2 JL _/ 7 7 z / 7* \ n "S \^ ^ S^ s^ ^ ]\ V. ^ M Fig. 96. ^ -.*- M ^^ v. n n. ^v i ^£ £ ^ Fig. 97. Fig. 98. Representing by any one of the angles corresponding to the terminal in any one of the above diagrams, £ n abscissa OM x n cosine of o == — — = - = cos v. modulus OP r 106] THE COSINE FAMILY. 195 § 105. The Sign of the Cosine. The sign of the cosine is deter- mined by that of the abscissa, as that of the sine by the ordinate. Quadrant I E K K Cosine -h — — +■ Sine + + — — Fig. 99. EXERCISES. 1. By means of the protractor, locate the terminals of the following angles (in degree measure). Determine the signs of their sines and cosines: 30; -30; 60; -60; 110; -110; 130; -130; 271; -271; 375; -375; 456; -456; 548; -548; 638; -638. Do you notice any connection in size and sign between the cosine of positive and negative angles numerically equal? Can the sign of the sine and cosine be determined without knowing accurately the position of the terminal ? 2. Give the signs of the sines and cosines of the following angles (radian measure), taking each angle both positively and negatively: 1; 2; 3; 4; 5; 7; 9; 10; 12; ?; ?; £; £2 6 o 4 o T' T' T' 7tT 9tT. 13 7T 3tt 2 ' § 106. Angles with the Same Cosine. The cosine, like the sine, depends only on the position of the terminal. (i) All angles with the same terminal have the same cosine. .\ cos (2 mr -f- 0) = cos 0, cos (2 n • 180° + A°) = cos A°, where n is any positive or negative integer. Fig. 100. (ii) All angles with their terminals symmetric to the horizontal have the same cosine, since points with equal moduli on such terminals have the same abscissa. 196 PLANE TRIGONOMETRY. [§106 Fig. 101. .*. as a special case, cos 6 = cos (— #), in radian measure, cos A° = cos (—J. ), in degree measure, or, positive and negative angles, numeri- cally equal have the same cosine. And generally, cos (2 rnr + 0) = cos (2 m • ir — 0), cos(2n-180 o + A°) = cos(2m.l80 o -A°), where m, n are any positive or negative integers. (iii) Thus, in general, any angle having the same cosine as 0, A° are of the forms 2 nir ± 0, in radian measure, 2 n • 180° ± A°, in degree measure, where n is any positive or negative integer. The student may give a verbal statement for each of these formulas. (iv) In the preceding formulas any angle of the given terminal may be taken as 0, A . It is customary, however, to take the principal angle. For example, 185° and — 175° locate the same terminal. One would use 2rc- 180° ±175°, rather than 2 n • 180° ± 185°, for the set of angles having the cosine of this terminal. Fig. 102. EXERCISES. 1. Find five positive angles and five negative angles having the same cosine as 30°, and also give in degree measure and in radian measure the general formulas for all such angles. 2 . Do the same, replacing the word " cosine " by the word " sine." Give a diagram for Exs. 1 and 2. § 106] THE COSINE FAMILY. 197 3. Consider in the same way as in Exs. 1 and 2, the angles 100°, 193°, 275°, using the principal angle of the terminals. 4. For terminals in the first quadrant, where are the terminals for all angles having the same cosine ? For terminals in the second quadrant ? In the third ? In the fourth ? 5. Give the general formulas, both in radian measure and in4§Are0^ measure, for all angles having the same cosine as -; — • — . 2JF. UL. r 8 8 4' 3' 3 ' 4 ' 4 ' -; — • For the same angles give the general formulas, in both meas- 6 6 ures, for all angles having the same sine. Give illustrating diagrams. 6. Find the two angles numerically less than 180° which have the same cosine as ± 1085°. Do the same for ± 365°; ± 800° ; ± 1100°; ± 3000°; ± 185° ; ± 275°. Give in radian measure and in degree measure, for each of these angles, the general formulas for all angles having the same cosines. 7. Consider Ex. 6, replacing the word " cosine " by the word "sine." 8. Using tt radians as the unit angle, give the general formulas for Exs. 5 and 6. 9. Solve the equations : cos 9 6 = cos 8 ; cos 7 = cos 6 ; — cos 3 = cos 5 $ ; cos 2 = cos ; — cos 3 = cos ; cos n$ = cos ; cos md = cos nO ; sin 2 = sin 6 ; — sin 8 $ = sin 7 6 ; sin md = sin n9. Illustrate each solution by a diagram. Solution of the first equation in Ex. 9. cos 9 6 = cos 8 6. Here 9 6 is not necessarily 8 6, but 9 6 may be any one of the angles whose cosine is that of 8 0. But cos 8 6 = cos (2 mr ± 8 0). .-. cos 9 6 = cos (2 nir ± 8 0). .-. 9 6 = 2 tot + 8 0, or 9 m 2 mr - 8 0. .-. = 2 nir ; (1) or, 17 = 2 nir ; (2) Termtnal or, = n -— . (3) FlG - 103 ' (•«*«*•> 17 198 PLANE TRIGONOMETRY. [§106 80 Jfermfi jertft Fig. 104. ('-*■) For (1) the terminal line is the initial line. Turning that line nine times or eight times around leaves it unchanged. The diagram is Fig. 103. J9_ For (3), 9 (9 is 9 n . t 2 y tt, or n(l- T 8 y )2 tt; 8 is 8 n • & w, or n(l - T 9 y )2 tt. Thus the terminal for 9 6 is symmetric to the horizontal with that of 8(9. The dia- gram for n = 1 is Fig. 104. The student may give similar solutions for the remaining examples, and state what the general process is. § 107. Angles with Opposite Cosines. (a) Any pair of terminals symmetric to the vertical give angles with opposite cosines, since for equal moduli the ab- scissas are oppositely equal. .\ (i) Supplementary angles have opposite cosines, or, cos (180° - A°) = - cos A , cos (tt — 0) = — cos 6. (ii) Any angle of a terminal has the opposite cosine of any angle of the ter- minal symmetric to the vertical. Fig. 105. cos { (2 n + 1) tt - 6 } = - cos (2 m • tt + 0), cos S (2 n + 1) 180° - A \ = - cos (2 m . 180° + A°). (iii) Thus all angles of the form (2 n + 1) tt - (9, (2rc + l)180°-^°, have the opposite cosine of 0, A°. In words, subtracting an angle from an odd multiple of 180° (or tt radians} Fig. 106. gives an angle of opposite cosine. M. V M l 107] THE COSINE FAMILY. 199 jr. (#) Angles of opposite terminals have opposite cosines, since for equal moduli the abscissas are oppositely- equal. .\ cos(180° + ^L o ) = -cosJ.°, cos (tt -f- 0) = — COS #, or, in general, Fia. 107. cos ( (2 n + 1) 180° + A ) = - cos (2 m • 180° + A°), cos((2tt+l)?r + 0)= -cos (2 wtt + 0). In general, angles having the oppo- site cosine of 0, A° are of the form (2n + l)7r±!9, (2 7i + 1)180° ±A°. Thus, while adding an angle to an even multiple of 180° or subtracting it from such an angle leaves the cosine Fio 108 unchanged (§ 106), an odd multiple of 180°, similarly treated, gives the opposite cosine. EXERCISES. In solving examples, it is advised that the formulas be not used, but rather the location of the terminal, as with the sine. 1. Write five positive angles and five negative angles having the oppo- site cosine of 30°; of ± 60°; of ± 45°. Let one-half the angles found in each case have the opposite terminal, the remaining half the terminal symmetric to the vertical. Give the general formulas in each case. 2. Do the same for -2tt. 3tt 3' 6' 3 ; 2"; 3"; 4-. 3. If an angle has its terminal in the first quadrant, in which quadrants will the angles of opposite cosines have their terminals ? Consider each quadrant similarly. 4. Change the word " cosines " in Ex. 3 to " sines," and solve. 200 PLANE TRIGONOMETRY. [§ 108 § 108. On the Effect on a Diagram of a Quarter-turn, Half- turn, Three-quarter-turn, Counter-clockwise. By a "quarter-turn counter-clockwise" is meant a tilt through 90° counter-clockwise. Evidently after such a tilt: A line now vertically up was originally right-flat. A line now right-flat was originally vertical-down. Or, a line now a positive ordinate was a positive abscissa. A line now a positive abscissa was a negative ordinate. To the word " ordinate " we may consider the word " sine " attached ; to positive ordinate, positive sine ; to negative ordinate, negative sine. The word " cosine " we attach simi- larly to the word "abscissa." Thus, " a positive ordinate was a positive abscissa," is the same as " a positive sine was a positive cosine," or, " a sine was a cosine"; while "a positive abscissa was a negative ordinate" means "a cosine was a negative sine." .-. sin (90° + ^4°) = cos A° ; cos (90° + AT) »— sin A°. Similarly, in considering any question of this kind, it is necessary only to determine where the ordinate, upright after the tilt, was before the tilt, and where the abscissa, flat to the right after the tilt, was before the tilt. What effect has a half-turn counter-clockwise ? A line up, was down; right-flat was left-flat. Thus, a positive ordinate was a negative ordinate, and a positive abscissa was a negative abscissa. .-. sin (180° + A°) m - sin A° ; cos (180° + A ) = - cos A°. What effect has a three-quarter counter-clockwise turn? A positive ordinate was a negative abscissa, and a positive abscissa was a positive ordinate. .-. sin(270°-h^ o ) = -cos^L°'; cos (270° + A ) = sin A . § 108] THE COSINE FAMILY. 201 What effect has a complete turn ? The diagram is unaffected. .\ sin (360° + .4°) = sin ^°; cos (360° + 4°) = cosui°. What effect on a diagram has any number of quarter-turns, counter-clockwise t From all such tilts all complete turns may be dropped, as be- ing without effect. There will remain a quarter-turn, a half- turn, or a three-quarter-turn. The corresponding angles are : 4 rc.90°, (4ti + 1)90°, (4rc + 2)90°, (4ti + 3)90°. f sin (4 n.90° + ^°) = sin ^1°. | { cos(47i.90° + 7l°) = cosJ.°. J | sin ( (4 7i + 1)90° + ^4°) = cos ^1°. \ (cos((47i + l)90° + ^°) = -sinA°.J f sin ( (4 7i + 2) 90° + A°) = - sin^4°. | { cos( (4 n + 2)90° + 4°) = - cos^4°. J f sin((47i + 3)90° + A°) = -cos^4°.| {cos ((4 7i + 3)90° + ^l°) = sin ^4°. J If B, <7, D, are three angles in degree measure free from multiples of 360°, and of the second, third, fourth quadrant, respectively, then £-90°, (7-180°, B - 270°, are three angles, each less than 90°, whose terminals, by a quarter- turn, half-turn, three-quarter-turn, respectively, come into coincidence with those of B, C, D, respectively. Thus, «) JsinB= cos(J5-90°). } |cosB = -sin(B-90°). J (sinC = -sin(C-180°).K... { cos C = - cos (C - 180°). ) w f sin D = - cos (Z> - 270°). ) ( . . .. [cosI> = sin(i> - 270°). ) K J Formulas (i), (ii), (Hi), are of great practical importance in using the tables, obviating subtractions on minutes and seconds. The teacher may assign practice exercises. 202 PLANE TRIGONOMETRY. [§ 109 EXERCISES. 1. Cut out a right-triangle from paper, carry out the tilts indicated in § 108, and thus prove the formulas. 2. Determine what the formulas above become when n is a negative integer. 3. Change A to — A in the formulas of this section, and from the facts sin A = — sin (— A), cos A = cos (— A), deduce the values of the following expressions in terms of A: sin (90° — A); cos (90° — A) sin (180° - A) ; cos (180° - A) ; sin (270° - A ) ; cos (270° - A) sin (360° - A) ; cos (360°- A) ; sin (4n- 90° - A) ; cos (4n • 90°- A) sin ((4 n + 1)90° - A) ; cos ((4 n + 1)90° - A) ; sin ((4 n + 2)90° - A) cos ((4 n + 2)90° - A) ; sin ( (4 n + 3)90° - A) ; cos ((4 n + 3)90° -4). 4. Solve Ex. 3 by diagrams. 5. Determine the values of the expressions in Ex. 3 when — 90° is written for 90°. 6. Construct diagrams for the expressions 90° ± A, etc., of this section and the preceding examples, and show the results directly from the diagrams. § 109. The following generalization may be made from the preceding results : If the subtracted (added) 90°'s are even, the word " sine " remains the word " sme," and the word " cosine " remains the word "cosine." When, however, their number is odd, the word "sine" changes to "cosine" and the word "cosine" to "sine." The proper sign is then to be attached, according to the relative position of the original terminal and the new terminal. EXERCISE. Determine by the method of tilts and by the preceding suggestion, the effect on the sine and cosine of adding (subtracting) the first fifty multi- ples of 90° to an angle A, taking A in each quadrant. For example, what is the effect of adding twenty-seven 90°'s to the angle 123° ? Dividing 27 by 4, the remainder is 3. The effect, then, is the same as that of a three-quarter tilt, counter-clockwise. § 110] THE COSINE FAMILY. 203 The positive ordinate was the negative abscissa. The positive abscissa was the positive ordinate. .-. sine was a negative cosine, and cosine was a sine. .-. sin (27 x 90° + 123°) = - cos 123°, and cos (27 x 90° + 123°) = sin 123°. Using the second method : Since 27 is an odd number, the words interchange, " sine " becoming "cosine" and "cosine," "sine." Omitting from 27 all multiples of 4, the remainder is 3. The angle 123° is of the second quadrant. Adding three 90°'s will bring the new terminal to the first quadrant. A sine in the first quadrant is opposite in sign to a cosine of the second. .\ sin (27 x 90° + 123°) = - cos 123°. But a cosine of the first quadrant is of the same sign as a sine of the second. cos (27 x 90° + 123°) = sin 123°. I prefer the method of tilts. The student may take his choice, or devise a better method. § 110. Cosines of all Angles are First Quadrant Cosines. It is evident that if the cosines of all angles of the first quadrant, between 0° and 90°, are known, the cosines of all other angles are known in magnitude. For a terminal of quadrant II, use the symmetric terminal of quadrant I, with change of sign of the cosine. For a terminal of quadrant III, use the opposite terminal of quadrant I, with change of sign of the cosine. For terminal of quadrant IV, use the symmetric terminal of quadrant I, without change of sign of the cosine. 204 PLANE TRIGONOMETRY. [§110 Take in each case the principal angle of the corresponding terminal in quadrant I. Without constructing the terminal one may proceed thus : (a) Disregard the sign of the given angle, since cos ( — A) = cos A. (5) Drop all multiples of 360°, as being without effect. 0) For remainder, R, if R > 90° and < 180°, take - cos (180° - R), or - sin (R - 90°) (§ 108). (d) For remainder, R, if R > 180° and < 270°, take -cos (5- 180°), O) For remainder, R, if R > 270° and < 360°, take cos (360° - E), or cos(i2 - 270°) (§ 108). The student will observe, on comparing these statements with terminal diagrams, that they are nothing but : Terminals symmetric to the vertical, give opposite cosines. Terminals opposite, give opposite cosines. Terminals symmetric to the horizontal, give the same cosines. EXERCISES. 1. Give the angles between 0° and 90° which have the same cosine in numerical magnitude as the following degree-measure angles, and give the sign in each case, this being obtained from the quadrant of the given angle : 140°; 175°; 187°; 200°; 280°; 305° 23'; -30° 23'; -100° 32'; - 185° 43' ; - 5° 43' ; - 150° 25' ; ± 432° ; ± 403° ; ± 506° ; ± 603° ; ±1850°; ±2317°. 2. Solve the preceding when for the word " cosine " there is read the word " sine." 3. Express two of the preceding angles in radian measure, (i) in terms of a radian, (ii) in terms of it radians. 4. Solve Ex. 1 by constructing terminals. §111] THE COSINE FAMILY. 205 § 111. Construction of the Terminals of Angles having a Given Cosine. (i) Let cos A = § . Draw a circle with the radius 3. Lay out on the abscissa axis the distance OM= 2 ; then draw through M the verti- cal line, cutting the circle in P x P 4 . OP v OP± are the required terminals. For every given value of the cosine, other than ± 1, there are two and only two positions of the terminal. These two posi- tions are always symmetric to the horizontal, as were terminals for a given sine symmetric to the vertical (§ 66). fig. 109. The angles of the two preceding terminals are 2w.l80°±JL°(§106, iii). (ii) Let cos A = — J. The construction is the same as that above, except that OM is laid out to the left, or OM = - 2. EXERCISES. 1. Construct terminals for the cosines £, |, |, I, and give the general value of the angles, using tables. Test with protractor. 2. Show that taking some line for unity, one can construct lengths representing the square roots of 2, 3, 5, 6, 7, 8, or any other integer. Use the process to construct the terminals for the angles whose cosines are+ V 2 ,V3 V5 are ±_ , ±— , ±— 3. Can you determine from the constructions, without measurement or tables, what angles have the cosines ± — , ± — , ±1, ±1, 0? 4. Change the word "cosines" to "sines" in Exs. 1, 2, and 3, and solve. 5. Construct terminals for some cosines in the tables. Measure the angle with the protractor and compare with the table. 206 PLANE TRIGONOMETRY. [§112 § 112. Line Picture of the Cosine. ::zj::-:-:-::::_::: ~k^f ^ D tit ...... ^ ~$ ~ _^z s *T- 7 v 5 - y X h •-■■■■ Z Ij! 2 O - -ifef t H 4 A A \ -*i V 5 ^Z ^ Z \ »_ ^ --J — - Fig. 110. As on the unit-circle the ordi- nates represent the sines for the corresponding terminals, so do the abscissas represent cosines. For the terminal OP, the cosine is This is the line definition of the cosine. Compare § 67. LABORATORY EXERCISE. Divide a circumference whose radius is one foot into 5° spaces ; measure the abscissas ; compare with the table-cosines. § 113. Line Pictures of the Cosines of All Angles. As all the ordinates of the unit-circle represent, as a continuum, the sines of all angles (see § 69), so do the abscissas give, pictorially, a cor- rect impression of the relative magnitude of the cosines of all angles, with their manner and ex- tent of variation as the terminal passes from the initial position OA, counter-clockwise, through a perigon, and continues its mo- tion, with a repetition of the cosine values. From the diagram (Fig. Ill) one can readily draw the follow- ing conclusions : (a) When the terminal is horizontal to the right, the cosine is +1. Fig. ill. §113d] THE COSINE FAMILY. 207 .-. cos0°=l = cos(±360 o )=cos(±720°), etc., or, in general, cos (2 n • 180°) = 1 = cos (2 n • tt), where n is any ± integer. That is, the cosine of all even multiples of 180°, or of ir radians, is +1. (5) When the terminal is horizontal to the left, the cosine is — 1. .\ cos ( ± 180°) = - 1 = cos ( ± 540°), etc., or, in general, cos(2 n + 1)180° = - 1 = cos(2 n + 1)tt. What is the cos ( ± ir) ? The cosine of all odd multiples of 180°, or of ir radians, is -1. (c) When the terminal is vertical, up or down, the cosine is zero. .-. cos ( ± 90°) = = cos ( ± 270°) = cos ( ± 540°), etc. cos( ± ^j= 0= cos( ± -£-)= cos( ±-^j, etc. In general, cos(2w + l)90° = = cos(2w + l)^- The cosine of all odd multiples of 90°, or of — radians, is zero. ((i) Thus, for terminals bordering the quadrants : Terminal Right Up Left Down Right Cosine + 1 -1 + 1 Sine +1 -1 Fia. 112. EXERCISE. Name five positive angles and five negative angles in degree measure and in radian measure for each terminal corresponding to Fig. 112. 208 PLANE TRIGONOMETRY. [§ 113 e (e) As the terminal passes from 0° to 360°, the cosine, starting with the value + 1, diminishes continuously from 1 to 0, which is reached at 90°; then diminishes continu- ously to — 1, which is reached at 180° ; then increases con- tinuously to 0, which is reached at 270°; then increases continuously to -f- 1, which is reached at 360°. As the ter- minal passes again around the circuit, the same set of values is repeated, and so on. Thus, as the angles form a continuum from 0° to 360°, the cosines form a double continuum from + 1 to - 1, and - 1 to + 1. 0° 90° 180° 270° 360 + 1 The Cosine. The Sine. -l + i -l ^ ^ ST ^V^^ Fig. 113. 90° 180° 270° 360' ^ar "^ Sk Repeating, forward and backward, without end. Repeating, forward and backward, without end. Fig. 114. EXERCISE. What effect has a reversal of the order of angle description on the order of description of the double cosine-continuum ? On the sine-continuum ? (/) For any position on the terminal, a slight change in the ter- minal position makes a slight change in the corresponding cosine. In Fig. 115, OM v OM 2 are the cosines of two angles nearly equal. The cosine-difference is M 1 M 2 . As P 2 is brought nearer and nearer to P v the smaller becomes M 1 M i . When the angle- difference is less than any as- signable quantity, so is the cosine-difference, for M^M 2 is less Fig. 115. than P X P 2 , this arc-difference on the unit-circle being the same, numerically, as the angle- difference in radian measure. See (4), page 67. Thus the cosine is a continuous function of the angle. § 113 i] THE COSINE FAMILY. 209 (#) For all angles whose terminals are in the first or second quadrant, increasing the angle diminishes the cosine. Therefore all angles with positive sines have for an in- crease in the angle a decrease in the cosine. Similarly, considering the third and fourth quadrants, all angles with negative sines have, for an increase in the angle, an increase in the cosine. (Ji) The points at which a function changes from an in- creasing function to a decreasing function, or vice versa, are called the turning-points of the function. EXERCISES. 1. In what position is the terminal when the cosine is at a turning- point ? 2. When the sine is at a turning-point? 3. Give in circular measure and in degree measure general expres- sions for all angles whose cosines are at a turning-point. Do the same for the sine. (i) When the terminal is nearly flat, a slight change hi its position does not affect the cosine very materially. When, how- ever, the terminal is very nearly vertical, slight changes in its position are accompanied by marked changes in the cosine. From this it follows that small angles cannot be deter- mined with great accuracy from the cosine, just as angles near 90° cannot be determined accurately from the sine. In practical work, it is well to avoid, when possible, observations which lead to the determination of an angle from its sine, when the angle is within 5 or 6 degrees of 90°, or from its cosine, when the angle is within 5 or 6 degrees of zero. Information in detail will be given on this point later. See §§ 185, 186. EXERCISE. Examine four-place tables and five-place tables for the changes made in the sine and cosine for a change of 1' in the angle at different parts of the tables. Examine also the corresponding changes in log sin, log cos. 210 PLANE TRIGONOMETRY. [§ 114 § 114. The Cosine as a Function of the Angle. The relation between the cosine and the corresponding angle in radian measure is, as will be shown later, cos0 = l-- + --- + ..,etc. Compare this with the sine series, or sine function, as given in § 70. The cosine, like the sine, is thus a transcendental function of the angle. While the sine involves only odd powers of the angle, the cosine is expressed wholly in terms of the even powers. This is essential, since sin(<9)=-sin(-0), and cos(0)= cos(— 0). The cosine is thus an even function of the angle, while the sine it an odd function of the angle. EXERCISES. 1. Show from the function expressions for sine and cosine that sin (0) = — sin ( — 0) and cos ($) = cos ( — $). 2. Show in general that if any function is expressed in powers of x and if F(x) = F (— x), there can be only even powers of x, while if F (x) = — F (— x), there can be only odd powers of x. 3. Use the sine and cosine function expressions to calculate the sines and cosines, to two significant figures, of the following angles : 0, 1", 1', 1°, ± 2°, ± 3°, ± 5°, 1', 2T t Ti-. § 115. The Cosine as a Periodic Function. Since cos 6 = cos (6 -f- n • 2 7r), the cosine is, like the sine, periodic, with the period, 2 7r. Reread § 72. 116] THE COSINE FAMILY. 211 116. The Anti-cosine, or Inverse-cosine. When y = cos x, then x = cos -1 y . Reread § 73. EXERCISES. 1. Read in all possible ways, as in the corresponding exercise on the sine, the expressions : cos -1 1; cos -1 1; cos" 1 — ; cos -1 -—; sin -1 ^; sin -1 ( ); sin (cos -1 £); cos (sin -1 0); cos (cos -1 x) ; sin (cos -1 0) ; 3 sin -1 \ + 2 cos -1 — ; 3 sin (cos- 1 \) - 2 cos (sin- 1 f ). 2 2. Construct the terminals for such of the expressions in Ex. 1 as represent angles. 3. How many values have the following expressions? sin (sin -1 x) ; sin (cos -1 a;) ; cos (cos -1 x) ; cos (sin -1 x). Read these expressions in all possible ways. 4. How many values have the following expressions ? cos -1 (cos x) ; sin -1 (sin x) ; sin -1 (cos x) ; cos -1 (sin x). Read these expressions in all possible ways. 5. What are the values of the following expressions ? sin (cos -1 0) ; sin (cos -1 1) ; sin (cos -1 ( — 1)) ; sin ( — cos -1 1) ; sin (— cos -1 (— 1)) ; sin (— cos -1 0). 6. Give the values of the expressions in Ex. 5 when the words " sine " and "cosine" are interchanged. When both words are "sine." When both words are "cosine." 7. Give two special values and the general values of each of the following expressions: sin- 1 (cos 0°); sin" 1 (cos (±90°)); shr 1 (cos (±180°)); sin" 1 (cos (±270°)); sin" 1 cos (±360°)). 8. Interchange the words " sine " and " cosine " in Ex. 7 and solve. 9. Also answer Ex. 7 when the word " cosine " is preceded by the nega- tive sign. Then interchange the word " sine " and " cosine," and answer. 212 PLANE TRIGONOMETRY. [§117 § 117. Some Angles whose Cosines can be determined readily from a Diagram. These are the same angles for which the sines were deter- mined in § 75. The diagrams used for sines give also the cosines. These diagrams are: For 45° and all For 60° and all For 30° and all angles of the form angles of the form angles of the form 2. T nir ±- 7 4 2 n • 180° ± 45 c 2 nir ± 3 2 n ■ 180° ± 60 c 2nir±l 6 2 n • 180° ± 30 c Fig. 116. Fig. 117. Fig. 118. The cosines of all angles noted in § 75 are readily obtained from these diagrams by observing whether, as related to the diagram terminal, the new terminal is 0) opposite, (ii) symmetric to the horizontal, (iii) symmetric to the vertical, remembering: at -.lt_ -. • t t f opposite sines, Angles with opposite terminals have i ., & ri I opposite cosines. Angles with terminals symmetric to the [opposite sines, horizontal have I same cosine. Angles with terminals symmetric to the rsame sine, vertical have [opposite cosines. §117] THE COSINE FAMILY. 213 EXERCISES. 1. Name five angles having their terminals opposite to that of 45°, and give their sines and cosines. 2. Name five angles having their terminals symmetric to the horizon- tal with reference to the terminal of 45°, and give their sines and cosines. 3. In Ex. 2, replace the word " horizontal " by " vertical," and answer. 4. In Exs. 1, 2, and 3, replace 45° by 30°, and answer. 5. In Exs. 1, 2, and 3, replace 45° by 60°, and answer. 6. Verify the following tables : Angle 0° 30° 45° 60° 90° 120° 135° 150° 180° Sine i \/2 2 2 1 2 2 4 Cosine 1 V3 2 V2 2 i -* -V2 2 -V3 1" —1 Angle 0° -30° -45° -60° -90° -120° -135 -150° -180° Sine -4 -V2 2 -S3 2 -1 -at* 2 2 ~\ Cosine 1 V3 2 V2 2 h -i -V2 2 -V5 2 -1 And fill in the following tables : > Angle 0° 210° 225° 240° 270° 300° 315° 330° 360° Sine Cosine Angle 0° -210° -225° -240° -270° -300° -315° -330° -360° Sine Cosine Note. — The student is advised against attempting to memorize the results of the preceding tables as independent facts. It is better to hold in mind the diagrams, and read from these mental diagrams the numeri- cal values of the sines and cosines, attaching the proper sign according to the location of the terminal. 214 PLANE TRIGONOMETRY. [§ 117 7. Determine from the diagrams the smallest angles (numerically) satisfying the following expressions, together with some two other values ; also the general values : sin- 1 — ± sin" 1 — ; ± sin^O ± 2 sin^; 3 sin" 1 ± 4 cos" 1 ; ± 2 sin- 1 ( - ^?) ± 3 cos- 1 ^? ; ± sin" 1 (1) ± 5 cos" 1 (1) ; ± sin" 1 (J) ± 2 cos- 1 (|) ; 3 sin" 1 ( - 1) - 2 cos" 1 ( - |) ; 2 sin(- 2^) ± 3 «*"*(- ^?); sin" 1 (0) + 2 sin" 1 ^? - 3 cos^? ; sin- 1 + sin" 1 — + sin" 1 — + sin- 1 - + cos- 1 + cos" 1 2j= + cos" 1 ^?, ti i £ 2, mm and this last when the sines and cosines are negative. 8. Solve the exercises at the end of § 75, changing the word " sine " to " cosine " where possible. 9. Verify the following results (using diagrams and not the tables) : (a) sin 30° cos 60° + cos 30° sin 60° = sin 90° = cos 0° = - sin 270°. (b) sin 45° cos 60° ± cos 45° sin 60° = ^ (1 ± V3)- (c) sin 60° cos 30° + cos 60° sin 30° = sin 90°. (d) cos 60° cos 30° + sin 60° sin 30° = cos 30°. (e) cos 60° cos 30° - sin 60° sin 30° = cos 90°. (/) sin 45° cos 0° ± cos 45° sin 0° = cos 45°. (g) cos 2 30° + sin 2 30° = 1. (h) cos 2 45° + sin 2 45° = 1. (i) cos 2 60° + sin 2 60° = 1. (J) cos 2 120° + sin 2 120° = 1. (k) cos 2 30° - sin 2 30° = cos 60°. (I) cos 2 45° - sin 2 45° = cos 90°. (m) cos 2 60° - sin 2 60° = cos 120°. / \ ™ a *r\o A .11 — cos A m A M A . /l + cos /I (n) If A = 60°, sin— = +*V 5 , and cos— =? +yj— ■ L — (0) Can you show that sin 2 A + cos 2 A = 1, where A is any angle? 10. Determine from the diagrams the value of sin A + cos B, and sin A cos B ± cos A sin B, when A, B are any pair of angles appearing in the tables on page 213. Do not use the table for values, but get the values direct from a diagram. §118] THE COSINE FAMILY. 215 11. Verify from diagrams the following results, when A = 30°, 45°, 60°, or any angle of tables, page 213 : (a) sin 2 J. = 2sinvl cos 4. (b) sin 3 A = 3 sin A — 4 sin 8 A. (c) cos 2 A = cos 2 A — sin 2 A. (d) cos 2^4 = 1 -2 sin 2 A. (e) cos 2 A = 2 cos 2 A - 1. (/) cos 3 A =4 cos 8 A — 3 cos A. § 118. Complementary Angles. Any pair of angles whose sum in degree measure is 90°, or radian measure, — , are called a complementary pair. Each is said to be the complement of the other. Their form is A°, 90° -A°; 0,^-6. To construct the terminal of 90° — A°, lay out from the upright vertical an angle, in the reversed direction, equal to A. The terminals of an angle and its complement are, therefore, inclined to the right-hand horizontal line and the upright vertical in opposite equality. Thus, if the upright vertical were made the initial line and abscissa axis, and counter-clockwise motion turned into clockwise motion, and vice versa, angles would become their complementary angles by equal turns. This is equivalent to changing ordinates into abscissas, and abscissas into ordinates, or sines into cosines and cosines into sines. Thus, the sine of an angle is the cosine of its complement^ and the cosine of an angle is the sine of its complement. EXERCISES. 1. Show on a diagram that if equal moduli are taken on the terminals of an angle and its complement, the ordinate of one modulus is the abscissa of the other, and vice versa. Use each quadrant. 216 PLANE TRIGONOMETRY. [§119 2. Show that the sine (cosine) of any angle can always be expressed in terms of the sine or cosine of an angle less than 45°. Illustrate by diagrams, with the terminal in each quadrant. Make up and solve five numerical examples. This is made use of in trigonometric tables ; columns headed " sine " at the top are marked " cosine " at the bottom. The teacher may point this out in whatever five-place table is used, showing that such tables need go only to 45° direct and that the sine (cosine) of any angle less than 360° can be taken directly from the table, by the aid of formulas (i), (ii), (iii), page 201. See special arrangement in Hussey's Tables. § 119. The Complementary Arc. In the early treatment of trigonometry, with the unit circle, and with arcs used instead of angles, their numbers, or numerical measures, being the same, arcs like AP and BP (Fig. 119) were taken as complementary arcs. NP, the sine of the arc BP, is equal to the cosine, OM, of the arc AP. So also the sine, MP, of the arc AP is the same in length as ON, the cosine of the arc BP. The com- plementary arc, BP, behaved toward the upright vertical just as the arc itself to the horizontal. The eo- of cosine thus indicated the sine of the complementary arc. Fig. 119. EXERCISES. 1. Select a terminal in each quadrant, construct its complementary terminal, and show from the diagram that sin (90° - A) - cos A ; cos (90° - A) = sin A, using both the ratio definitions and the line definitions. 2. Select at random two angles (involving minutes and seconds) in each quadrant, and find their complements. 3. Test by diagrams the relations sin (90° - A) = cos A, cos (90°- A) = sin ^4, for each of the following angles in degree measure, taking each both positive and negative : 0; 30; 45; 60; 90; 120; 135; 150; 180; 210; 225; 240; 270; 300; 315; 330; 360. §119] THE COSINE FAMILY. 217 4. Show that if sin A = cos B, the terminals of A, B are equally inclined to the vertical and horizontal respectively, but that A, B are not necessarily complementary. Find the general formula connecting A, B, when sin A = cos B, and when cos A = sin J3, as in Ex. 5. 5. Solve the equation sin 2 = cos 3 by sines. Solution. sin 2 = cos 3 = sin (* - 3 Y Thus ^ — 3 6 must be some one of the angles which have the same sine as 2 0. 2 ... £-30 = 2n7r + 20; (1) or !-30=(2n + l)7r-20. (2) Remembering n is any ± integer, its sign, in transposing, is not con- sidered. .-. by (1), 50=(2n+i)7r; (3) by (2), 0=(2n-\)^ (4) by (3), 0=(4n + l)^; (5) or by (4), 0=(4n-l)|. (6) 6. Locate the terminals for a few of the angles determined by (5), (6), of the preceding solution and show that the terminals of 2 0, 3 6 are equally tilted toward the upright vertical and right-hand horizontal, so that sin 2 = cos 3 0. 7. Show from (5) and (6), of Ex. 5, that 2 0, 3 as determined by (5) are complementary only when n = 0, giving = 18°, and that as determined from (6) they are never complementary. 8. Solve Ex. 5 by cosines instead of by sines. 9. Give the general solution of sin A = cos B by sines, and show that A + B = (4 n + 1)90° or A - B = (4 n + 1)90°. For what value of n do we get complementary angles? For what value would we get A = 90° + B ? 10. Show from the results of Ex. 9 that sin ((4n + 1)90° +.4) =cos A ; sin (90° + A) = cos A ; cos ((4 n + 3)90° + A) = sin A ; cos (270° + A) = sin A . Compare with the formulas on page 201. 11. Give the general solution of sin A = cos B by cosines, and deduce the results in Ex. 10. 218 PLANE TRIGONOMETRY. [§ 120 12. Solve by sines or by cosines the equations : — sin A = cos B ; sin A = sin B ; sin A = — sin 5 ; cos A = cos 5 ; cos ^4 = — cos -S ; and deduce the results of § 108, as in Ex. 10. 13. Solve by sines or by cosines the following equations : sin 5 A = cos 6 A ; sin 7 A = — cos 9 A ; sin (sA) = cos(L4) ; sin (sA) =s — cos (tA). 14. State the general method of solving equations like those in Exs. 12 and 13. It is given in Ex. 5. § 120. Calculations using the Cosine, without the Use of Tables. From the relation of definition connecting the three quan- tities, cosine, abscissa, modulus, when any two of the quanti- ties are given, the third can be calculated: (i) When modulus and cosine are given, abscissa = modulus times cosine. abscissa modulus abscissa (ii) When modulus and abscissa are given, cosine = (iii) When abscissa and cosine are given, modulus m cosine EXERCISES. Change the words " ordinate " and " sine " in the examples of § 76, to " abscissa " and " cosine," respectively, and solve. § 121. Calculations with Cosines, using the Tables. These calculations may be made either with the tables of natural cosines or with logarithms. As the student is now calculating for calculation's sake, merely to learn, he may use both processes with each example, the one as a check on the results of the other. 121] THE COSINE FAMILY. 219 For natural cosines abscissa = modulus times cosine, abscissa cosine = — — , modulus modulus = ^m. cosine For log cosines : log abscissa = log modulus + log cosine, log cosine = log abscissa — log modulus, log modulus = log abscissa — log cosine. EXERCISES. After the manner of the corresponding exercises on the sine, § 79, carry out the following calculations, with tests of accuracy. Let calculated parts show the same number of significant figures as the data. In one-figured data on lines, read angles to the nearest five degrees ; in two- figured data on lines, read angles to the nearest half-degree ; in three-figured data on lines, read angles to the nearest five minutes; in four-figured data, to the nearest minute ; in five-figured data, to the nearest second, etc. (See § 77.) A. Find the general value of the angle in degrees and in circular measure in terms of ir: 1. Modulus 9, abscissa ± 7 ; modulus 8, abscissa ± 3. 2. Modulus 34, abscissa ± 23 ; modulus 6.7, abscissa ± 3.4. 3. Modulus 23.4, abscissa ± 15.4 ; modulus 4.56, abscissa ± 2.34. 4. Modulus 23.78, abscissa ± 17.34 ; modulus 2.674, abscissa ± 1.789. 5. Modulus 234.98, abscissa ± 189.90 (five-place table). 6. Modulus 453.764, abscissa ± 389.031 (six-place table). 7. Construct a set of examples similar to the six preceding and con- sistent as representing measurements. B. Calculate the abscissa, given : 8. Modulus 5, angle ± 25° ; modulus 9, angle ± 40°. 9. Modulus 34, angle ± 55°; modulus 7.8, angle ± 67°. 10. Modulus 35.7, angle ± 43°; modulus 3.42, angle ± 67° 25'. 11. Modulus 23.45, angle ± 67° 23' ; modulus 271.8, angle ± 45° 32'. 12. Modulus 234.67, angle ± 34° 45' 56" (five-place table). 13. Modulus 4578.67, angle ± 23° 45' 56.7" (six-place table). 220 PLANE TRIGONOMETRY. [§122 14. Construct a set of examples similar to Exs. 8-13 preceding and consistent as representing measurements. 15. Express the angles of Exs. 8-13 in terms of 7r; also in radian measure. C. Calculate the modulus, given : 16. Abscissa 7, angle 15° ; abscissa 9, angle 45°. 17. Abscissa 34, angle 23°; abscissa 7.9, angle 70° 30'. 18. Abscissa 45.7, angle 34° 50' ; abscissa 3.89, angle 59° 5'. 19. Abscissa 34.91, angle 73° 23'; abscissa 6.789, angle 45° 21'. 20. Abscissa 45.831, angle 56° 56' 56" (five-place table). 21. Abscissa 2345.67, angle 34° 45' 56".7 (six-place table). 22. Abscissa 45.69879, angle 67°23'34".51 (seven-place table). 23. Construct a similar set of consistent examples representing meas- urements. § 122. Calculations with Right-angled Triangles, using the Sine and Cosine. Right-angled triangles may be solved by using the sine B alone, as already shown in § 81, or by using the cosine alone, since the acute angles are complementary. c . side opposite the angle oine = , t= > hypothenuse p . _ side adjacent to the angle hypothenuse We have thus for the solution of right-angled triangles : sin A = sin B — USING SINES. a I b h a = sin A • h b = sin B • h h = sin A sin B USING COSINES. cos A = - h cos£=y h a = cos B • A b = cos A • h h = cos B cos A §123] THE COSINE FAMILY. 221 It is customary to use a combination of sines and cosines, the sine in connecting the hypothenuse and the side opposite an angle and the cosine in connecting the hypothenuse and the side adjacent to the angle. We may think of opposite the angle as in the angle, and adjacent to the angle as on the angle. As in is in sine and the o of on is in cosine, you will remember that the side in the angle involves the sine and hypothe- nuse, while the side on the angle brings in the cosine and hypothenuse. This may be helpful when the triangles are tilted away from a hori- zontal (vertical) position. ORAL EXERCISE. Practise with the cosine and sine in connection with the Conradt col- lection of right-angled triangles on page 145. § 123. Calculation Scheme for Right-angled Triangles, using Log Sines and Log Cosines. Given Model Examples. A f B = 35° 12' h = 27.34 A = 90° - B. b = sin B - h. a = cos B • h. Fig. 121. Find A =54° 48' b = 15.76 a = 22.34 ■. log b = log sin B -f- log h, \ log a = log cos B + log h. Solution Scheme, logarithms. log-check. a- sin B = b'SinA. LOGARITHMS. (4) b 1.1975 (2) sin B 1.7607 (1) h 1.4368 (3) cos£ 1.9123 (5) a 1.3491 1.3491 1.7607 1.1975 1.9123 1.1098 1.1098 222 PLANE TRIGONOMETRY. [§ 123 Order of work is indicated by the numerals. First enter the logs (1), (2), (3). Then add (1), (2), for (4), and (1), (3), for (5). Then look up the numbers corresponding to logs (4), (5), and enter the results in the column marked "Find." If A takes the place of B in the scheme, a, b are inter- changed. The order of filling in the scheme will change with the data, but not the scheme itself. EXERCISES. Arrange the data and results to be found, diagram, formulas, log- formulas, and calculation scheme for the following examples. Give scheme for other possible cases, omitting the case when two sides are given. 1. b 8, B 65°; b 6, B 35°; a 5, B 55°; a7,A 25°. 2. b 12, B 46°; b 3.4, B 76° 30'; a 2.7, B 61°; a 26, A 26° 30'. 3. b 12 A, B 45°; b 34.5, B 56° 50'; a 678, A 45° 15'. 4. b 23.45, A 34° 45'; b 345.7, A 45° 53'. 5. b 234.56, B 56° 34' 31". 6. h 8, B 45°; h 9, A 15°. 7. h 23, B 34°; h 4.5, A 67° 30'. 8. h 23.8, B 23°; h 765, ^ 34° 25'. 9. ft 236.9, 5 56° 21' ; I 4.798, A 34° 51'. 10. h 234.59, 5 45° 51' 34"; h 34.761, ^L 48° 48' 48". 11. h 234.786, B 45° 43' 21".3. SINE AND COSINE PROBLEMS IN PHYSICS. (Such of these as do not lie outside the students' training in physics may be used.) 1. A force F acting at an angle A° to & horizontal plane has what vertical and horizontal components ? Solve a numerical example, getting the sums of the horizontal and vertical components when three forces act at a point in a plane in different directions in the plane. Select the directions so that some of the resolved forces are negative. § 123] THE COSINE FAMILY. 223 2. A right-angled triangle in a vertical plane, with its hypothenuse horizontal and right angle above, has a uniform chain covering the other two sides. Find the pressure of the chain on each side, and also the force acting down each side, and see if a continuous chain on such in- clines would be in " perpetual motion." 3. A string inclined at an angle A° to a smooth plane inclined B° to the horizontal sustains a weight, W, on the plane. Find the pull on the string and the pressure on the plane, and also in what position the string is most effective in preventing the weight sliding down the plane. 4. A mule hitched to a canal-boat is effective only 75%. What is the angle of the pulling rope with the tow-path? 5. Three horses of equal strength, and two mules, the one as strong as the other and the two as strong as the three horses, pull on five ropes tied to another rope which passes over a pulley and is attached to a weight, W. The angles of the five ropes with the single rope are in order for the horses 17°, 19°, 21° on the same side, and for the mules, on the other side, 18°, 20°, — all in the same plane. How many times the pull of a horse is the lifted weight, and will the single rope maintain its direction as the animals pull? 6. The resultant of two forces is 8. One of them is 5, and the direc- tion of the other 35° from the resultant. What is the angle between the two forces ? 7. Two velocities represented by 100, 125, give a resultant velocity 153 directly north. What equations connect the angles that the com- ponents make with the north and south line? 8. The force 816 makes angles 25°, 35° with its components. Calcu- late the components. 9. A movable pulley rests on a rope with one end fixed, the other end of the rope passing over a smooth peg in a horizontal line with the fixed end of the rope. A weight is attached to the movable pulley. Find the equation connecting the pull on the rope and the weight. Does the pull increase or decrease as the weight rises ? 10. A sphere of weight W and radius a rests in the angle of two smooth planes inclined at angles A°, B° to the horizon. Find the press- ure on each plane. 11. A man weighing 200 lbs. stands at the apex of an ordinary tri- angular roof (equal rafters). What part of his weight tends to spread the walls ? 12. When a roof like that in Ex. 11 is covered uniformly with snow, what part of the weight of the snow tends to spread the walls ? 224 PLANE TRIGONOMETRY. [§ 124 § 124. Application of Sines and Cosines to a Problem in Field- surveying (a Projection Problem). What is meant by the bearing of a course of a survey has already been explained (§ 86). While the latitude of a course is its length multiplied by the cosine of its bearing, its longitude is its length multiplied by the sine of its bear- ing, — bearings being taken from the north and south line. In the diagram, the lati- tude AL is AP cos B, and the longitude, or departure, AM is AP sin B, B being the bearing. North latitudes are counted plus, and south latitudes minus ; east departures, plus ; west, minus. In a closed survey, carried out with absolute accuracy, the algebraic sum of the latitudes is zero, as is the algebraic sum of the departures. In practice this is never reached. The difference between the sum of the northings and the sum of the southings is called the error in latitude; the difference between the sum of the eastings and sum of the westings, the error in departure. The square root of the sum of the squares of these two errors is called the error in closing. If the survey is plotted on a diagram, the error in closing is the distance between the beginning and end of the plot when expressed in the unit of length in which the lengths of the courses are given. The error of closing is a measure of the accuracy of the field work. When it lies within the limit allowable for the kind of land surveyed, it is customary to "balance the survey." This consists in distributing the errors of latitude and depar- ture among the courses in proportion to their length as com- pared with the length of the border of the survey. The following calculation scheme applies to each course. The scheme for the entire survey is made by placing such schemes one under another in the order of the survey. At the bottom of such a scheme will appear the errors in lati- tude and departure, with the error in closing. 124] THE COSINE FAMILY. 225 (4) D (2) sin B (1) l (3) cos B (5) L Logarithms Latitude Departure Balauoe Errors Balanced N + S — £ + w- Lat. Dep. Lat. Dep. i l 1 1 i I l I . Fig. 123. Under Balance Errors and Balanced Latitude and Depar- ture, the proper sign is given, along with the magnitudes. The numerals indicate, as heretofore, the order of entry. Logs (1) and (2), added, give (4) ; (1) and (3), added, give (5). Then the numbers corresponding to (4) and (5) are looked up and entered under Departures and Latitudes, in the proper columns. LABORATORY EXERCISE. If apparatus is obtainable, make a survey of a few fields, and carry out the calculations as indicated above. Get also the area of each survey. The best laboratory for calculation-trigonometry is the field. A real stump and a real river are much more difficult to manage than paper stumps and paper rivers. EXERCISES. Fill out the scheme for the following survey, the numbers 1, 2, 3, etc., indicating the courses in order. Calculate the latitudes and departures of the courses, and the errors of latitude, departure, and closing. Bal- ance the survey. Make the plot. 1. N. 69° E., 437 ft. 3. S. 27° W., 244 ft. 2. S. 19 E., 236 ft. 4. N. 71° W., 324 ft. 5. N. 19° W., 184 ft. Do the same for the following survey : 1. S. 89° 55' E.,. 25.42 ch. 4. N. 86° 50' W., 25.58 ch. 2. N. 27°40'E., 34.68 ch. 5. S. 47° 30' W., 1.50 ch. 3. N. 19° 10' W., 7.40 ch. 6. 8. 77° 45' W., 13.60 ch. 7. S. 89° W., 3.53 ch. 226 PLANE TRIGONOMETRY. [§125 § 125. Use of the Cosine as a Check on Solutions of Triangles. In the triangle ABO, if OD is perpendicular to AB, AD is called the projection of AC on AB, or the projection of b on c. DB is the projection of a on c. AD— b -cos A. DB = a • cos B. But AD + DB = AB, for all positions of A, B, D, when signs are taken into account. ,\ c = a • cos B + b • cos A (1), and b = a • cos C + c • cos A (2), by symmetry ; a = b • cos C + c • cos B (3), by symmetry. Since the cosine of an angle between 90° and 180° is negative, the preceding formulas hold for all triangles, when signs are taken into account. In Figs. 124, 125, and AD + DB = AB, in all cases. EXERCISES. Solve the following examples, and use (1), (2), (3) above to test the correctness of the solutions : 1. Given B, 41° 41'; 6,43.21; c, 49.32. 2. Given ,4,37° 37'; a, 37.37; 6,41.24. 3. Given ,4,135° 24'; c, 45.45; a, 54.54. 4. Given a, 37.38; b, 38.39; c, 40.41. 5. Given B, 137° 37'; 6, 50.51; a, 30.31. 6. Construct some appropriate examples where the sides have one significant figure; two; three; five; six; seven. Solve. Determine in each case how close the test should hold. 126] THE COSINE FAMILY. 227 § 126. Use of the Cosine in Connection with a Projection Proposition of Plane Geometry. In all books on ordinary geometry will be found a proposi- tion reading somewhat like this : The square on the side of a triangle opposite an , [ angle is equal to the sum of ( obtuse J f Hi • • n d ) the squares on the other two sides \ . , by twice I increased J the rectangle contained by one of these sides and the pro- jection of the other on it. The trigonometric statement for this proposition is the same, whether the angle is obtuse or acute, and is this: The square on the side of a triangle is the sum of the squares of the other two sides diminished by twice the product of these two sides and the cosine of their included angle. When the angle is obtuse, its cosine is negative, so that the trigonometric and geometric statements agree. Proof : In Fig. 127, CD is perpen- dicular to AB, and the lengths are as indicated. a 2 = p 2 + y 2 = p 2 + (e — x) 2 = jt? 2 + rc 2 + c 2_2^ = &2 +(? 2_2c.&.COS^. (1) For p 2 + x 2 = b 2 , and x = b • cos A. or, In Fig. 128, taking signs into account, AD + DB = AB, C x + y = c. \ P \b — c — x\ and no change is made in the proof. 228 PLANE TRIGONOMETRY. [§126 We have thus : a2 = & 2 + c 2_2& c . C os^, • (1) 62 _ c 2 + a 2 _ 2 ca . cos B, (2) c 2 = «2 + 52 _ 2 ab ■ cos b (3) Are these formulas adapted to logarithmic calculation ? EXERCISES. 1. Given a 12, b 27, C 30°, find c to two significant figures without using the tables. Do the same when C has the value 45° ; 60° ; 120° ; 135° ; 150°. ' 2. Given a 9, &7, C35°, find c to one significant figure. 3. Given a 12, c 13, £ 48° 30', find b to two significant figures. 4. Given b 23.4, c 54.1, ^4 56°, find a to three significant figures. 5. Determine to one significant figure the cosine of the angles of a triangle whose sides are 7, 8, 9. Can the angles be determined uniquely (§ 71) from these cosines? Why is the case different when the sine of the angle of a triangle is given ? 6. Show that the diagonals of a parallelogram are a 2 + b 2 ± 2 ab cos 6, where 6 is the angle between the sides a, b. 7. Determine the diagonals to the appropriate number of significant figures, given a 3, &5, 035°. 8. Show that twice the length of the median of a triangle (from angle A ) 1S VA 2 +c 2 + 2 6ccos^l. 9. If a, b, c, d are the sides, in order, of a quadrilateral inscribed in a circle, show that if B lies between a, b, (1) a 2 + b 2 -2abcosB = c* + d 2 + 2cdcosB. (2) cosB = a2 + b *- c2 - d2 . v 2(ab + cd) § 127] THE COSINE FAMILY. 229 (3) sin 2 B = ( a + b + c - d )( b + c + d ~ a )( c + d + a - h )(d + a + b-c) 1 ' ±(ab + cdy 2 (4) (ab + cd) sin B = 2 V(s - a) (s - 6) (s - c) (s - rf) , where 2s = a + 6 + c + d. (5) Area of quadrilateral = V(s — a) (s — 6) (s — c) (s — cQ. (6) Find the area when the sides are 3, 5, 7, 9. (7) Show that the sine of the angle between the diagonals is 2 V Q - a) (s - b) Q - c) (s ^~d) ab + cd 10. Show that the sines of the angles which the median of a triangle (from the angle A) makes with sides c, b are asinC asinJ5 Vb 2 + c 2 + 2 be cos A V6 2 + c 2 + 2 6c cos A 11. Two circles of radii a, b cut each other at an angle 0. Prove that the length of their common chord is 2 ab sin Va 2 + b 2 + 2 ab cos 12. The revolving crank-arm R of a stationary engine is connected by a link-bar I to a piston rod, whose stroke S is in a straight line not passing through the centre of revolution of the crank-arm. Draw the positions of the crank at the dead points and show that a triangle will be formed whose sides are S, I — R, I + R. Show that if 6 is the angle between the positions of the crank-arm at the dead points, 5 2 = 2 (I 2 + R 2 )-2 (I 2 - r 2 ) cos 6. Show that S — 2 R when the stroke line passes through the centre of revolution of the crank-arm. § 127. Determination of the Angles of a Triangle from the Cosines of the Half Angles. Since there is only one angle less than 180° and positive which has a given cosine, the formulas of the preceding sec- tion determine the angles of a triangle uniquely from the cosine. However, they are not suitable for use when the sides are expressed by numbers of more than one or two significant figures. The cosine of the half angle is suitable for logarithmic computation. 230 PLANE TRIGONOMETRY. [§ 127 In § 92 it was shown that in Fig. 77 AD = 8 — a, and 2& = Lz2.hc. A _ AD I s(s-a) 2 AO "V 6c Similarly, cos | = ^ ~ & > 2 * «6 What sign must be given to these radicals, A, B, C being the angles of a triangle ? When all three angles are to be calculated, these formulas may be used to best advantage in the forms : A cos- = * abc a(s - - a), B 008 _ = * abc >b(s- -*>, cos- = * aha c(s - -o), corresponding to which we have the following calculation- scheme, the numerals indicating the order of filling in : Calculation-scheme. a= (1) s-c= (8) b= (2) log(«-a) = (9) (3) log (•-*) = (10) 2s= (4) log (•-*)= (11) *= (5) loga= (12) «-a = (6) log 5= (13) «-& = (7) log) = (19), by adding (10), (13) log c (s — (?) = (20), by adding (11), (14) 2 log cos | = (21), by adding (17), (18) 2 log cos - = (22), by adding (17), (19) 2 log cos — = (23), by adding (17), (20) i -A •'• log cos - = (24), by taking half of (21) log cos - = (25), by taking half of (22) log COS - = (26), by taking half of (23) A " 2 (27) B 2 (28) (7 2 (29) .-. A= (30) B= (31) C= (32) Test. A + B+ C-- = 180°, approximately. (33) Note. — Here, as in §93, all logs, (9) to (15) inclusive, are to be looked up before their manipulation begins with (16) and the following numerals. 232 PLANE TRIGONOMETRY. [§ 127 The test, A + B + = 180°, is subject to the same limita- tions here as in the sine-calculation. (See page 175.) EXERCISES. 1. Solve the following by cosines and by sines and compare results : (i) Given a, 54.2 ; b, 59.5 ; c, 67.3. (ii) Given a, 9.847 ; b, 8.352 ; c, 7.283. 2. Solve by cosines the following and test by the formulas of § 125 : a, 543 ; b, 586 ; c, 600. a, 437.2 ; b, 564.3 ; c, 498.3. Test the same by a sin B = b sin A . 3. Is there any convenient way of determining whether the test of § 125 is met with sufficient accuracy ? 4. What is an appropriate logarithmic test for the accuracy of the solution by cosines? If one uses the test a sin B — bsinA, how close ought it to check ? 5. Make up and solve a triangle requiring the use of a five-place table. Check the same. § 128. The Reciprocal Cosine, or Secant, and Graphs. Related to the cosine as the cosecant to the sine is another function of the angle called the secant. Secant = — : — ; cosecant = cosine sine The origin of the words " secant " and " cosecant," as that of the "cosine," is clear on the unit-circle of the old-time trigonometry. Draw a tangent at the initial point A of the unit-circle. Prolong the terminal of the angle to cut the tangent. The secant of the angle, or arc, is the amount cut off from the terminal by the initial tangent. 128] THE COSINE FAMILY. 233 In each of the following diagrams (Figs. 129-132) OS is the secant of the angles of the corresponding terminals. Fig. 129. Fig. 130. Fig. 131. Fig. 132. For ^ = ^ ; but both OA and OP are unity, and OM OA OM J is the cosine. ,-. 0S= — : — , or the secant, cosine By looking up the line picture of the cosecant (§ 96) it will be observed that the cosecant is the secant for the complementary angle, or arc, just as the cosine is the sine of the complementary angle, or arc. Thus the co- of cosecant signifies the secant of the complement. The sign of the secant is the same as that of the cosine. The range of values of the secant is determined by consider- ing that of the cosine and taking the reciprocal range. As the angle runs from 0° to 360°, the cosine runs over the continuum from 1 to to — 1 to to 1. At the same time the secant runs over the reciprocal set of values. It goes from 1 to positive infinity, while the angle passes from to 90°. As the angle passes over 90°, the secant makes a jump from positive infinity to negative infinity. As the angle passes on from 90° to 180°, the secant goes from negative infinity to — 1. As the angle changes from 180° to 270°, the secant changes from — 1 back to negative infinity. As the terminal passes over 270°, there is another break in the continuity of the secant, by a spring from negative infinity to positive infinity. As the terminal moves on from 270° to 360°, the secant passes over the continuum from positive infinity to 1. 234 PLANE TRIGONOMETRY. [§128 r^ w 37T 27T 2 AS - A m Fig. 133. If sin (90° + ^4)= cos J., then cosec (90° -f A) — sec A. Therefore, if we look on the sine and cosecant as running through a set of values, the cosine and secant will be running through the same set of values 90° behind the sine and cosecant, respectively. Thus the graphs of the cosine and secant are the same as those of the sine and co- secant, respec- tively, if we pull these back along the horizontal axis a distance representing 90°. (Compare Figs. 93, 133.) The diagram shown in Fig. 134, the circle being the unit-circle, illustrates the secants of all angles. The distances OS v OS 2 , OS 3 , etc., represent the secants for the ter- minal positions OP v OP v OP s , etc., and their opposites. When the ter- minal is almost the upright vertical, the secant is almost infinite, positive. Just after the terminal has passed the position of the upright vertical, the secant is on the opposite ter- minal, and is thus negative and almost infinite Just what the secant is when the terminal is the upright vertical or downright vertical, it is impos- sible to say. It is customary to take it as -f- x> for 90 and as - oo for 270°. Illustrate on a single diagram the line-picture of the sine, cosine, secant, cosecant of an angle. Give such a diagram for each quadrant. Fig- 134. § 131] THE COSINE FAMILY. 235 § 129. Line Picture of the Secants of All Angles. The secants are (Fig. 134) 0A 1 ••• 0S t - 0S 2 ••• 0S S ~> + oo, then -oo ••• OS r _ 4 ••• 0S_ B ••• 0S_ 2 ••• 0S_ X ••• OA, with repeti- tions. EXERCISES. 1. Change the word " sine " to " cosine," and the word " cosecant " to " secant " in the Exercises of § 97, and solve. 2. By what tilt can the graphs of y = cos -1 x and y = sec -1 x be obtained at once from those for y = cos x and y — sec x ? § 130. Cosine Waves and Cosine Harmonic Motion. Since the graph of the cosine is the same as that of the sine pulled back a distance representing 90°, the cosine may be used in connection with waves, just as the sine. (See §§ 99, 100.) EXERCISES. 1. If y = r • sin (at + a), and y—r- cos (at + ft), are two waves, how far behind one wave is the other at the same time-instant? 2. Draw a diagram illustrating the superposition of the waves of Ex. 1. For harmonic motion on the horizontal axis the cosine plays the same role as does the sine on the vertical axis. (Reread § 100.) § 131. The Versed Sine, Exsecant and Coexsecant. When the line definition was in use, the balance of the radius (unity) beyond the cosine was called the Versed Sine. Andsti11 versing = 1 - c b ; a = b ; a < 6.) (Conchoid of Nicomedes.) Show that the conchoid can be drawn readily by means of a straight line. 5. Make a graph for a nos 1 Show it is a circle. (Use rectangular coordinate paper.) 6. Make a graph for = a cos 6 1 (An ellipse.) {x — a cos 6 1 y = a sin 6 J (x = a cos 1 y = b sin 0J Show that points on this curve are readily located mechanically by means of two concentric circles of radii a, b, prolonging ordinates of the one to meet abscissas of the other. CHAPTER VII. THE SINE AND THE COSINE IN UNION. § 133. Relation of the Sine of Any Angle to its Cosine. In Fig. 135, OM 2 + MP = OP 2 . This is true for all posi- tions of the terminal, OP. Thus, for all positions of OP, and for each quadrant : fOM\ 2 _ L _fMP\ 2 _, {opJ+Vop)- 1 ' or, (cos x) 2 + (sin x) 2 — 1, (1) for all values of x. For any special angle, 0, this is written in the form : cos 2 6 + sin 2 = 1, (2) The relation (2) serves to de- termine the sine of 6 when the cosine is given, or the cosine when the sine is given. (3) (4) 2 2 2 2 2 2 2 2 / n jf Fig. 136. cos 6 = ± Vl _ S in- 9, sin0 = ±v / i _ cos 2 6. Explain on a diagram the significance of the double sign. When the sine (or cosine) is given in the form of a common fraction or a simple decimal (tenths), it is better to calculate the cosine (or sine) from a diagram than from the formulas (3), (4). 237 238 PLANE TRIGONOMETRY. [§133 For example, given the sine = |, what is the cosine ? Construct the terminals (§§ 66 and 111), assuming the modulus as 5 and the ordinate as 4 (Fig. 136); calculate the remaining side. Here it is ± 3. The cosine is thus ± f . Similarly, for sine — f, the adjoining diagram (Fig. 137), where the cosine is ±f. V> \/. t ^ — m ^•"•l*. -L** ^ / \ z s / - ^ r -\ 7 I t \ ^6^- %- 2\ lf\ V ■*■ -/ V A 1 % =& 7 \ 3 f v Zfr ]S - j. \ /X- 3\ ' \ -/ v z ±z V ^ v ^z Fig. 136. Fig. 137. EXERCISES. 1. Determine by constructions the cosines and secants when the sines are the following, taking each fraction both positively and negatively : A> tV, & B, H. 0-2, 0.4, 0.75. 2. Determine by constructions the sines and cosecants when the cosine has the following values, taken with both signs : &, ih lb ti *i 0.3, 0.1, 0.8. 3. Determine by constructions the secants and cosines when the cose- cants have the following values, taken with both signs : II H, W-, -W-, Hi Hi See the table of triangles, page 145, for a large number of integers which may be selected as the sides of right-angled triangles. When the sine (cosine) is given in the form of a decimal of several figures, it is better to get the cosine (sine) from the formulas (3), (4), than from a diagram. 133] THE SINE AND THE COSINE IN UNION. 239 EXERCISES. 1. Find the cosines to two decimal places when the sine has the values (+ or -): 023) a36> a27> 0>21 . 2. Find the sines to two decimal places when the cosine has the values (+ or -): a67> Q ^ Q 45> 0QL It will be recalled that when z is small z Vl — z = 1 — -, approximately (§ 42). ,. Vl-^ = l-|- sin 2 6 ,\ cos = 1 — , approximately, when 6 is small. EXERCISES. 1. Given sin 10" = 0.000,048,481,368 .... Show that sin 2 10" = 0.000,000,002,350,4 ... and that cos 10" = 0.999,999,998,825. 2. Assuming (as will be found true later) that for the number of decimal places given above in sin 10", sin 1" is ^ of the sine of 10", and sin n" is — times the sine 10", when n < 10 and positive, calculate the cosines of 1", 2", 3", ... 9". 3. Eliminate 6 from x — a cos 0,y = a sin 6. What is the correspond- ing graph ? 4. Eliminate 6 from x — a cos 6, y = b sin 6. What is the graph ? 5. Eliminate 6 from \ x = a (° ~ sm *) 1 . What graph is this? [ y = a (1 — cos 0) J ° r 6. Eliminate t from \ y ~ * ^ A !■ • This is the path of a [ x = vt COS0 J r projectile. If you know the laws of falling bodies under gravity, prove that the above equations are true for a projectile started with initial velocity v at an angle 6 with the horizontal line. 240 PLANE TRIGONOMETRY. [§ 134 § 134. Use of the Relation, (cosine jt) 2 + (sine *) 2 = 1, in trans- forming a Trigonometric Expression. Model Examples. (1) To show that J} — ^4 = cosec A (1 - cos 4). * 1 + cos J. Multiplying the quantity under the radical, numerator, and denominator by 1 — cos A, and then writing in place of the new denominator, 1 — cos 2 J., its value, sin 2 A, the radical becomes — \ — - — , or, cosec .4(1 — cos^L). sini (2) To show that cos 4 A - sin 4 A + 1 = 2 cos 2 A. cos 4 A — sin 4 A = (cos 2 A + sin 2 A) (cos 2 A — sin 2 A) = cos 2 A — sin 2 A — cos 2 A — (1 — cos 2 A) = 2cos 2 J.-l. From this the transformation follows at once. In working such examples, we may (i) Transform the first member into the second ; or (ii) Transform the second member into the first ; or (iii) Transform the first and second members into some third expression. EXERCISES. (Following from cos 2 A + sin 2 A = 1.) 1. cos 2 ,4 - sin 2 ,4 = 2 cos 2 .4 - 1. 2. cos 2 A - sin 2 A = 1-2 sin 2 A. 3. sec 2 A + cosec 2 ^4 = sec 2 A cosec 2 A 4. sec 2 A + cosec 2 A = (sin A sec A + cos A cosec A) 2 , 5. sin A sec A + cos A cosec A = sec A cosec A. 6. 2(1 -cos^)-(l -cos^4) 2 = sin 2 ^4. 7 1 -. sin A _ 1 + sin 2 A — 2 sin A 1 + sin A cos 2 A 8. cosec 2 A — 1 = cos 2 A cosec 2 A. 9. (sin J. — cos A) 2 = 1 — 2 sin ^4 cos ^4. §134] THE SINE AND THE COSINE IN UNION. 241 10. sin 8 A + cos 8 A = (sin A + cos A) (1 — sin A cos A). 11. sin 8 A — cos 8 J. = (sin A — cos>4)(l + sin A cos .4). 12. sin 4 A + cos 4 ^4 = 1-2 sin 2 A cos 2 A. 13. sin 4 A — cos 4 A = (sin A + cos A ) (sin .4 — cos A). 14. sin 6 J. + cos 6 A = 1-3 sin 2 A cos 2 A. 15. sin 6 A — cos G A = (sin 2 A — cos 2 A) (1 — sin 2 ^4 cos 2 .4). 16. sinM - cosM = (2 sin 2 A - 1) (1 - sin 2 A + sin 4 A), 17. sinM - cos 8 4 = (sin 2 .4 - cos 2 4) (1 - 2 sin 2 A cos 2 A). 18. = sin A cos A. cos A cosec A + sin A sec ^4 19 sin^ + l + cos^ 2cosec ^ 1 -f cos ^4 sin A and cos J + cosA 2seeA 1 — sin A 1 -f sin ^4 2Q 1 + sin 2 ,4 sec 2 ;! = ^ ^ ^ ^ 1 + cos 2 A cosec 2 .4 21. 1 ~ sin ^ =1 + 2 sin A sec 2 .4 (sin .4 - 1). 1 + sin A 22. sin A sec A + sin B sec B =sin ^ gec ^ gin B gec ^ cos ^4 cosec A + cosB cosec jB 23. sin ^4 (1 + sin yl sec A) + cos ^4 (1 + cos ^4 cosec ^4) = sec A -f cosec vl . 24. (sin 2 J. - cos 2 .4) 2 = 1-4 cos 2 A + 4 cos 4 4. 25. sin A cos yl (sin ^4 sec A + cos J cosec A) = 1. 26 sin,4 sec,4 + sec,4 - 1 = (1 + sin ^ )sec ^. sin A sec vl — sec A + 1 27. / l-sin^ - S ec ^ (1 - sin A). \ 1 + sin A 28. cosec 4 + cosec A =2sec 2 A . cosec ^4 — 1 cosec ^4 + 1 or* * cosec -4 29. cos ^4 = cos A cosec ^4 + sin A sec ^4 30. If sin_ f 4 = m sin ^ = ±m = 2.7. 17. A line was measured by three different men. One man reported the length as 1 mile ; another, as 5280 feet ; the third, as 5280.0 feet. Were the reports indentical, so far as care in measurement is concerned ? Does it mean the same to say a race-track is a mile long as it does to say that the distance between two towns is a mile ? Does it mean the same to say a line is a foot long as to say it is 12 inches, or 12.0 inches, or 12.00 inches? What effect on the sun's distance has a change of one-hundredth of a second in the parallax-angle (angle subtended at the sun by the earth's radius) ? 244 PLANE TRIGONOMETRY. [§136 § 136. The Addition- Subtraction Formulas for Sines, Cosines. sin (A + B) = sin A cos B + cos A sin B ; sin (A - B) = sin A cos B - cos A sin Z? ; cos (^1 + -B) = cos ^1 cos J8 - sin ^1 sin B ; cos (^t - B) = cos ^4 cos B + sin ^4 sin B. (i) (2) (3) (4) It is necessary to prove only (1), (3), since they include (2), (4), when A, B are allowed the double sign. Now (1) is nothing but the trigonometric expression of the geometric fact that if the sides of a certain right-angled triangle, drawn as presently to be indicated, are projected on any vertical line, the projection of the hypothenuse is the algebraic sum of the projections of the other two sides. Similarly (3) represents a projection of the same triangle on a horizontal line. In each case the line on which the triangle is projected is, of course, in the plane of the triangle. When a line OP (Fig. 138), counted positive along the terminal of an angle #, and directed by that angle, is pro- jected on a vertical line, its projec- tionis tfP.sin*. If OP, as directed by the angle x, is for any reason negative, its projec- tion is the opposite of the preceding case, or — OP sin x, or, OP- sin + 180°), where the negative sign is taken up by the sine and OP is now merely a length. /. (a) Thus, when a length I, whose direction is that of the terminal of an angle x, is projected on the vertical, projection = I • sin x. (/3) When a length Z, whose direction is that of the opposite terminal of the angle x, is projected on the vertical, projection = I • sin (a; + 180°), or I times the sine of the directing angle of the opposite terminal. 136] THE SINE AND THE COSINE IN UNION. 245 Suppose that a, Fig. 139, is the terminal of the angle A, laid out from the right-hand horizontal line, and that b is the terminal of J5, laid out from a as a new initial line. Then b is also the terminal of A-\-B, as laid out from the right-hand horizontal line. When a is the new initial, along it is positive; opposite, negative ; while also A + 90° is a positive direction, and A + 270° is a negative direction. The triangle OQP, whose projec- tions give (1), (3), is in all cases drawn as follows: From any point P, in the terminal 5, of A + B, drop a perpendicular PQ on a. No matter what the size or sign of A, B, the sides of the triangle OPQ can only take four dif- ferent sorts of locations as follows : OQ may be on the terminal of J., or on the opposite terminal. QP may be directed by the angle ^4 + 90°, as in Fig. 139, or by A -f- 270°, as in a diagram like Fig. 140. Let OQ= T, or t, according as it lies on A's terminal or its opposite. Let QP = Z7, or w, according as it is directed by A 4- 90°, or by A + 270°. Let OP = r, which is all the time directed by A + B. Thus, no matter what the size or sign of ^4, B, there can be only four cases for the triangle OQP: (i) Sides r, T, U, (iii) Sides r, fi, U, (ii) Sides r, T, u, (iv) Sides r, t, u. Now, assuming that the projection of the hypothenuse of a right-angled triangle is the algebraic sum of the projec- Fig. 140. 246 PLANE TRIGONOMETRY. [§136 tions of the other two sides (§ 86), and denoting by T v the vertical projection of T, we have for the four possible cases : r 9 = T % + U v , case (i) ; r v = T v + u„ case (ii) ; r v = t v + C/;, case (iii) ; r v = t v + u v , case (iv). By (a), p. 244, 2* = T • sin A, and U 9 = lf. sin (4 + 90°) = U • cos A By (£), p, 244, t v =t -sin 4, and w„ = w • sin (J. -f 90°) = u cos J., for, in the last two cases, the directing angle of the opposite of an opposite is the angle itself. We thus have the four possible cases : r • sin (04 + B) = T • sin A + U- cos A, case (i) ; r • sin ( A + B) = T • sin A + w • cos J., case (ii) ; (i?) r • sin ( A + 5) = £ • sin A + £7 • cos ^4, case (iii) ; r • sin (^4 + B) = t • sin A + u * cos ii, case (iv). Since now a is the initial line for the angle B, T=r cosB t = r • cos B U= r • sin B u = r - sin B > , in the appropri- ate diagram ad- joining. Substituting these values in the equations (.27), we have, after di- viding by r, the single equivalent of them all: sin (J. + B) = sin A cos B+ cos A sin B, (1) Fig. 141. Formula (3) is now readily proven, for when the projection is made on a hori- zontal line, sin A above becomes cos A, and sin (A -f 90°) § 137] THE SINE AND THE COSINE IN UNION. 247 becomes cos (yl+ 90°), or — sin A, and sin 04. + B) becomes cos (A + B). No other change is made. .-. cos ( A + B) = cos A cos B — sin A sin B. (3) No special proofs of (2), (4) are needed, for they are, as already mentioned, covered by (1), (3). When B is nega- tive, it is added to A by laying it out clockwise from A's terminal, and no change whatever is made in the preceding proofs. § 137. Special Cases of the Addition Theorem. The addition -subtraction theorems are so important that the student should follow through the proofs in some special cases. For instance, for case (i), Fig. 142: r v = % + U v ; or, O l P l ^O l Q l + Q.P,; or, r • sin (A + &)=T- sin A + U • sin (A + 90°) ; or, r • sin (A + B) = r • cos B sin A + r • sin B cos A. .*. sin (A + B) = sin A cos 2? + cos A sin i?. (1) Similarly, for cosines : r„ = T h + U k ; *- or, OP 2 =OQ, + Q 2 P 2 . *£. The student must note care- / [\^_ fully that here, as in all cases, / ^£\ the addition of the projections is j^a ' ' algebraic, OP 2 is shorter numeri- o -^ £2 cally than OQ 2 , but the added FlG - 142 - (J 2 P 2 is negative. If A, i?, (7 are any three points on a straight line, then, always, no matter what the order of points, AB = AC+CB. The student, up to this point in his career, has probably had this pre- sented to his mind but seldom. The teacher will find it advisable here to draw a variety of diagrams, showing that the above statement is always true in algebraic addition. When the student has once seen that AB —AC + CB, for all possible arrangements, the significance of the generality of the sin (J. + B) and cos (^4 + B) formulas is clear. 248 PLANE TRIGONOMETRY. [§137 or, In Fig. 142, OP 2 =OQ 2 + Q 2 P 2 , or, r • cos (A + B^ = T- cos^L + U- cos (A + 90°), r • cos (^4 4- B) = r • cos i? • cos J. — r • sin i? • sin A. ,\ cos ( J. + i?) = cos A cos i? — sin A sin Z?. (3) Case (ii), Fig. 143. T v = .Z v -|- W v , or, 0^= 0^+ Q X P X (algebraic addition). Here, since a is the initial line for 5, QP is a negative line. It is directed by the angle A -f 270°. Its opposite is directed by the angle .4 + 90°. Thus, by (/3) of the preceding section, the projection of QP on the vertical is QP sin (A + 90°), or u - cos A. Thus, 1 P 1 = r -sin (A + B~); 0^= T - sin A; Q 1 P 1 = u> cos A. .\ r • sin (A -f i?) = I 7 * sin ^4 + u • cos J., as before. But T= r • cos B and w = r • sin i?. .*. sin (A + B) — sin .4. cos B + cos J. sin B (1), as before. Similarly, in Fig. 143, r h =T h + u h , or, OP 2 = OQ 2 + Q 2 P 2 (algebraic, same as arithmetic), or, r • cos (A + B) = T- cos ^4 + u • cos (^4 + 90°), or, cos ( J. + B) = cos ^4 cos B — sin J. sin B. (3) The student may now make diagrams illustrating cases (iii) and (iv) (bottom of page 245), and carry through the proofs, noting how many different orders the points O v P v Q v and O v Q 2 , P 2 , may take, and noting § 137] THE SINE AND THE COSINE IN UNION. 249 that the addition of segments is always algebraic, the algebraic addition occasionally agreeing with the arithmetic. See Ex. 4, p. 344. EXERCISE IN PROJECTIONS. Show that if a straight line is drawn anywhere in the coordinate plane and a perpendicular from the origin is let fall on it, making an angle 6 with the x-axis, then x cos + y sin 9 = p, when p is the length of the perpendicular, (x, y) being any point on the first line. (If P is (x, y), T the foot of the perpendicular, and MP the ordinate of P, project the broken line OMPTO on OT.) EXERCISES. 1. Given sin A =$, cos B = T \. Find, in the form of a common fraction, the values of &in(A + B), sin (4— B), cos (^4 +B), cos(A-B). 2. Solve Ex. 1 when cos A = ^, cos B = f f. 3. Find the sines and cosines, cosecants and secants, of the following angles in the form of radicals and also to two decimal places, without using the tables : ± 75°, ± 15°, ± 105°, ± 165°. What other angles have the same sine as any one of these angles? The same cosine? The same sine and cosine? Answer in radicals : sin 15° = ; cosec 15° = VQ + V2. 4 cos 15° = V ^t^ ; sec 15° = V6 - V2. 4 sin 75° = cos 15°; cos 75° = sin 15°. 4. Take, from the tables, the sines and cosines of two angles, and from these calculate the sines and cosines of their sum and difference, and compare the results with those of the table. 5. Apply the addition-subtraction formulas to the angles 90° ±A> 180° ± A, 270° ± A, 360° ± A. Compare the results with those of § 108. Prove the following : 6. sin 04 + B) sin (A - B) = sin 2 A - sin 2 B = cos 2 B - cos 2 A. 7. cos (A - B) cos (A + B) = cos 2 A - sin 2 B = cos 2 B - sin 2 A. 8. cos(45 -^4)cos(45 - J B)-sin(45 -^)sin(45°- J B)=sin(^+£). (Do not expand the expressions in the first member here nor in many of the examples following.) 9. sin (45° + A ) cos (45° - B) + cos (45° + A ) sin (45° - B) = cos (A - B) . 10. sin (» + 1) A sin (n - 1) A + cos (n + 1) A cos (n — 1) A = cos 2 A. 250 PLANE TRIGONOMETRY. [§ 137 11. sin (w + 1) A sin (n -f 2) A + cos (n + 1) A cos (n -f 2) v4 = cos ^1 . 12. cos a cos (y — ct) — sin a sin (y — a) = cos y. 13. cos (a + /?) cos y - cos (/? + y) cos a = sin /? sin (y — a). 14. sin (a + /?) cos a — cos (a + /?) sin a = sin /?. 15. cos (a + /?) cos (a - /?) + sin (a + /?) sin (a — /3) = cos 2 /?. 16. sin (n — 1) a cos (n + 1) a + cos (n — 1) a sin (n + 1) a = sin 2 na. 17. sin (135° - 0) + cos (135° + 6) = 0. 18. sin 105° + cos 105° = cos 45°. 19. sin 75° - sin 15° = cos 105° -f cos 15°. 20. cos ,4 + cos (120° - A) + cos (120° + A) = 0. 21. By setting sin (A + B + C) = sin (A + B + C), and applying the addition formulas in succession, show that sin (A + B + C) = sin A cos 5 cos C + smB cos C cos ^4 + sin C cos ^4 cos B — sin A sin 5 sin C. 22. From Ex. 21, show that if A + B + C = 180° or (2 n + 1) 180°, sin A sin 5 sin C = sin ^4 cos B cos C + sin B cos C cos A + sin C cos ^4 cos 5. 23. Find, similarly, cos (A + B + C), and give the results, when here, or in Ex. 22, one or more of the quantities A, B, C, are negative. 24. sin (A + B) + sin {A - B) = 2 sin ^ cos £. 25. sin (^ + JB) - sin (^4 - B) = 2 sin B cos A 26. cos (A + J5) + cos (^4 - 5) = 2 cos A cos 5. 27. cos (A - B) - cos (^4 + B) = 2 sin ^4 sin B. 28. V2 sin (A - 45°) = sin A - cos A. 29. V2 sin (^4 + 45°) = sin A + cos A. 30. sin (45° ± A) = cos (45° T -4). 31. V2 cos (.4 + 45°) = cos A - smA. 32. V2 cos (J. - 45°) = cos A + sin ^4 . 33. sin (0 — <£) cos <£ + cos (0 — ) sin = sin 0. 34. cos (0 + <£) cos + sin (0 + <£) sin = cos <£. 35. 2 sin (a + j\ cos (fi-?\ = cos (a - /?) + sin (a + /?). 36. 2 sin (| - a\ cos (| + ^\ = cos (a - ft) - sin (a + /3). 37. cos(<* + /?) + sin(e* - fS) = 2sin (| + a\ cos (f + £). § 138] THE SINE AND THE COSINE IN UNION. 251 38. cos (a + j3) - sin («-/?)= 2 sin (| - a) cos (^ - /?). 39. cos (n — 1) A • cos A — sin (n — 1) A sin ^4 = cos nA. 40. If ,4 + B + C = 90°, show that sin 2 A + sin 2 £ + sin 2 C = 1 - 2 sin 4 sin B sin C. 41. sin 6 sec + sin sec = sin (0 -f- <£) sec sec <£. 42. sin sec — sin <£ sec = sin (0 — <£) sec $ sec <£. 43. cos $ cosec 6 + sin <£ sec = cos (0 — <£) cosec sec <£. 44. cos 6 cosec — sin sec = cos (0 + ) cosec sec . 45 sinflsecfl + sinfseccfr = gin Q + ^ cogec ^ _ +) sin sec — sin <£ sec <£ 46 sinflsecflsin sec 47 sin sec + cos = cQg ^ _ #) gec ^ + +) cos <£ cosec <£ — sin sec 48 cos cosec + cos cosec cosec <£ 49 1 + cos cosec sin sec § 138. The Addition-Subtraction Formulas in Inverse Notation. If sin ( A + B) = sin A cos B + cos A sin i?, (1) then A + B — sin -1 (sin A cos i? + cos J. sin B). (2) If sin A — x and sin B = y, then, cos .A = ± VI — x 2 and cos B = ± Vl — y 2 . ,\ A = sin -1 x — cos -1 ( ± VI — rr 2 ), j5 = sin -1 y = cos -1 ( ± VI — # 2 ). Thus (2) is equivalent to sin" 1 x + sin -1 y = sin -1 ( ± a? VI -|/ 2 ± y VI - as 2 ) . (3) Similarly, when .A and B are determined by sines, sin (A — i?) = sin A cos i? — cos A sin 5 is equivalent, in inverse notation, to 252 PLANE TRIGONOMETRY. [§138 sin -1 x - sin -1 y = sin -1 (±x Vl - y 2 =F y >/l - as 2 ), while cos ( A + B) = cos ^4 cos B — sin J. sin 5 is equivalent to sin -1 x + sin -1 2/ = cos -1 (± Vl - as 2 VI - 2/ 2 - xy), and cos ( A — B) — cos A cos i? + sin A sin i? is equivalent to sin -1 x - sin -1 y = cos -1 (± VI -x 2 VT-y 2 + a?y). iV.5. What has been said in § 73 concerning the multi- plicity of values of A when A — sin -1 x (similar statements holding when A = cos -1 x) must be borne in mind in connec- tion with the results of this section and the examples. What is meant by the statement in Ex. 3, page 253, sin" 1 1 + sin" 1 T 8 y = sin" 1 £|, is, that among the angles whose sine is |-|- is an angle which can be made up by adding an angle whose sine is f to an angle whose sine is y 8 y. Similarly, with reference to the statement A = sin -1 x = cos -1 Vl — x 2 , the proper sign is to be given the radical if a special value is to be given to A. For example, sin -1 J is not eosW ^— — J, if for sin" 1 ^ the special value 150° is taken. In the exercises following, the radicals are to be taken as having both signs. EXERCISES. 1. Show that if cos A = x and cos B = y, the formulas for sin (/I + B), sin (A — B), cos (A + B), cos (A — B) become, respectively, cos -1 a: + cos -1 y = sin -1 (#Vl — x 2 + xVl — y 2 ), cos -1 a: — cos" 1 ?/ = sin" 1 (yVl — x 2 — xVl — y 2 ), cos -1 a: + cos -1 ?/ = cos -1 (xy — Vl — x 2 • Vl — y 2 ), cos -1 x — cos" 1 y = cos -1 (xy + Vl - x 2 . Vl — y 2 ). 2. Give the corresponding formulas when sin A = x, cos B = y, and when cos A = x and sin B = y. §138] THE SINE AND THE COSINE IN UNION. 253 Prove the following, giving also the general solution, as in the model solution (below) of Ex. 3. Model Solution of Ex. 3. Find the general value of sin -1 § -f- sin -1 T 8 y. Let be the principal angle of the first quadrant whose sine is J. Then cos 6 = | . Similarly, if B is the principal angle of the first quadrant for which sin = ^ T , cos = ^. Then sin" 1 f = 2 mr + 0, or (2w + 1)tt-0, and sin -1 ^ = 2 mir + <£, or (2 m + 1) tt — . Then, sin" 1 f + sin" 1 ^ = 2 sir + + , or 2 *tt - (0 + <£), or (2 8 + l)7r + 0-<£, or (2 s + 1)* -f- (f> - #, where s is any positive or negative integer. Now sin (2 sir + + <£) = sin (0 + <£), sin (2 87T - ((9 + )) = - sin (d + <£), sin ((2 1 + 1) it + (0- £)) = - sin (0 - ) = sin (<£ - 0), sin ((2 « + l)ir + <£ - 0) = - sin (0 - 0) = sin ((9 - <£). And sin(0 + 4>) = f.lf . + |.^ = JJ, sin(^-0) = |. T V-f.lf = H. .•. sin" 1 § -f sin" 1 T ^ = 2 m- + «, or (2 r -f- l)7r — a, where a is some angle whose sine is |~£ ; also some angle whose sine is — J-J ; also some angle whose sine is Jf ; also some angle whose sine is — £§ . 3. sin- 1 + sin- A = sin- H . g _ ^ /T+i +c03 . 1 JIp : 4. sin" 1 T 5 5 + sin- 1 ft = cos" 1 1| |. s 2 ' 2 5. cos- 1 1 + cos-4f = cos- 1 H 10 . C0S -iJg_ C0S -i^"+l ; 6. sin-4 + cos-4 = 00°. 3 2V3 6 7. sin- 1 4= + cos- 1 JL = 45°. 11. sin- 1 1- + sin" 1 i| = 00°. v5 vlO ^ ,0 J, ° 8. cos" 1 f - cos- 1 H = cos" 1 f|. 12. sin" 1 $ - sin- 1 £$ = sin" 1 (?). cos -1 a: 7T 254 PLANE TRIGONOMETRY. [§138 19. 2 sin- 1 x = cos- 1 (1-2 z 2 ). 20. 2 cos- 1 a; = cos" 1 (2x 2 - 1). 13. sin" 1 1 - cos" 1 J? = cos -1 (?). 14. cos -1 ^ — sin -1 T 8 r = sin _1 (?). 15. cos- 1 |$ - sin" 1 if = cos" 1 (?). 21. cos- 1 x = 2 sin- 1 V^T^' 16. sin-iff-sin" 1 ! 1 , = cos" 1 (?). 22. cos -1 x = 2 cos -1 17. 2 sin- 1 x = sin- 1 2 zVl - x' 2 . H± 18. 2 cos -1 x = sin -1 2 zVl - a; 2 . 23. sin-^+sin-iA + sin- 1 !?=?:. 5 17 85 2 24. If sin -1 m + sin -1 n = ^, prove that mVl — n 2 + nVl — m 2 = 1. § 139. To find the Sine and Cosine of twice Any Angle, when the Sine and Cosine of the Angle are known. From sin ( A + B) = sin A cos B + cos A sin i? and cos (A + i?) = cos J. cos i? — sin A sin i? follow, by letting B = A, sin 2 ^4 = 2 sin A cos A (1) cos 2 A = cos 2 ^ - sin 2 A (2) = 2cos 2 ^-l (3) = l-2sin 2 .4 (4) A being any angle. That is, by (1), the sine of any angle is twice the sine of half of it into the cosine of half of it, and by (2), (3), (4), the cosine of any angle is the cosine squared of the half angle minus the sine squared of the half angle, or also twice the co- sine squared of the half angle minus one, or also one minus twice the square of the sine of the half angle. •. sin A = 2 sin 2 VV "2 cos^ = cos 2 ^-sin 2 ^» cos^ = 2cos 2 ^-l> 8 cos^l = l-2sin 2 A § 139] THE SINE AND THE COSINE IN UNION. 255 From which follow the useful formulas : + cos A - -cosA = = 2cos 2 ^, = 2 sin 2 4' 8 staf = ± ^l-C08^, cos| = = ± f + t*A sin^l = = ± f-t»*A, cos4 = ±V 1 + c ^ 2 ^ EXERCISES. 1. Express the sine and cosine of the following angles in terms of their half angles : 3A; 4A] 5A; 6A;7A', 8A; nA; 2nA. 2. If sin 6 = £, find sin 2 and cos 2 6, sin 4 6, cos 4 $, sin 8 6, cos 8 6, each to two decimals. 3. If cos = \, find the values called for in Ex. 2. 4. (sin!±cos^V=l±sin0; (sin 6 ± cos 0) 2 = ? 5. If sin $ = I and cos = \, find sin (0 + <£) and sin (20+2 <£). 6. If a, /? are positive acute angles and cos a = \% and sin /? = f , find the values of sin 2 ^1^ and cos 2 2-ZJS« 2 2 7. If a, /? are positive acute, and cos a = f and cos/J=$, find cos a ~" . Solve by Ex. 8 ; also without using Ex. 8. 8. Prove that (cos a + cos /3) 2 + (sin a - sin £) 2 = 4 cos 2 ^~^j (cos a + cos /?) 2 + (sin a + sin /?) 2 = 4 cos 2 a ~" , (cos a - cos /8) 2 + (sin a - sin /J) 2 = 4 sin 2 a ~P , (cos a — cos /8) 2 + (sin a + sin /?) 2 = 4 sin 2 «+fl . 256 PLANE TRIGONOMETRY. [§ 139 9. Look up the sine and cosine of 36° in the tables and calculate sin 72°, and compare the result with the table value. Calculate cos 72° from the three formulas for cos 2 A, and compare results with each other and with that calculated from sin 72° and that of the table. 10. Cosec2fl= cosece - sec9 ; sec2 6= C0Se f^ " SeC ? fl ; cosec80 = 2 cosec 2 - sec 2 se c 4 - sec 2 ♦ sec • cosec 8 11. p r ove sin2fl an$Bee$ -^* ? 1 + cos 2 1 + cos _ 2 p cos 2 _ cos — sin , cosfl _ « 1 + sin 2 ~ cos + sin ' 1 + sin ~ 13 P e cos 2 _ cos + sin . cosfl _ o 1 — sin 2 cos — sin ' 1 — sin _„ -r> 1 + sec 2 x ^„2 1 + sec a; « 14. Prove — ! = 2 cos 2 a; ; — ■ = t sec 2x sec a; 15. Prove sin A sec A + cos 4 cosec A — 2 cosec 2 ^4 . 16. Prove sin A sec A — cos ^4 cosec A — — 2 cos 2 ^4 cosec 2 ^4. 17. Prove °°^ + ^ - <*»*-**■* = 2sin2Asec2A . cos ^1 — sin A cos -4 -f- sin A 18. Prove cosec 2 ^4 (1 + cos 2 ^4 ) = cos A cosec ^4. 19. p r ove -^° =am(j±*V sec ff±^j -^^=? 1 T sin2 \4 / \4 / l±sin0 sin - + sin 20. Prove - sin * ± sin2 ' = sin sec 0; * =? 1 + cos0 + cos20 l + cos? + cos0 21. Prove cos (4 + 15°) cosec (A + 15°) - sin (4 - 15°) sec {A - 15°) _ 4 cos 2 A ~ 1 + 2 sin 2 ^ ' 22. Prove cosec 2 z (1 — cos 2 a;) = sin a: sec a; ; cosec x (1 — cos a) = ? 23. Prove sec 2 A (1 + sec 2 A ) = 2 sec 2 A. 24. Prove cosec ^4 (1 - cos 2 ^4 ) = 2 sin ^4 . 25. Prove 1 + cos 2 2 = 2 (cos 4 + sin 4 0). 26. Prove cos f^ = sec 2 A cosec 2 A — 2 27. Prove 2 -^A = cos2A; Ll*gM gf secM aec 2 24 28. Prove ^ + s ; p ^- c03 f = singsecg; j ± s j n i| ~ cos ^ = ? 1 + sin + cos 2 2 1 + sin i 6 + cos 4 6 § 139] THE SINE AND THE COSINE IN UNION. 257 (A A A A\ cos — cosec sin — sec — ) : 2 2 2 2/' cos 2 A cosec 2 vl = ? 30. Prove 1 ~ cos2 - 4 = sin 2 ,4 secM ; Lni2iil = ? 1 + cos 2 A 1 + cosJ. 31. Prove 2 Bin A gee il = ain2ii . 1 + sin 2 .4 sec 2 ^1 32. Prove ] ~ s ? n ^ g^4 = cos24. 1 -f- sin 2 ^4 sec 2 ^4 33. cos 2 ^4 (1 + sin A sec 4) 2 = 1 + sin 2 A. 34. sin 2 v4 (cos ^4 cosec A — l) 2 = 1 — sin 2 4. 35 / sin ^4 sec 4 + I V _ 1 + sin 2^4 • Vsinyl sec ^4 — 1/ 1— sin 2 A 36. cosM + 8inM = K2-sin2^)» cos ^4 + sin ^4 37. C08, ^- 8inM = K2 + sin2^). cos ^4 — sin ^4 38. cos 4 ^4 - sin 4 A = cos 2 ^4 ; cos 4 2 A — sin 4 2A — 1 , fl B , . . 6 , 1 + 3cos 2 24 39. cos 6 A + sin 6 A = — 4 40 8in3^4 _ cos 3^4 _ « sin A cos A 41. cos 3 A cosec yl + sin 3 A sec ^4 a 2 cos 2 ^4 cosec 2 i4. 42. sin 4 A cosec 2 ^4 = 2 cos 2 ^4 . 43. sin 5 A cosec A — cos 5 .4 sec A = 4 cos 2 -4. 44. sin 5Z cosec £ - cos — sec £ = 2 V3- 12 12 12 12 45. sin f| + tf) sec (| + *\ - sin (? - $\ sec(| - A = 2sin20sec20. 46. sin (| - fl) sec ( j - *) + cos (| - fl) cosec (f ~ #) = 2 sec 2 0. 47. cos (^ + 45°) sec (A - 45°) = sec 2 ^ (1 - sin 2 4). 49. sin 2 (45° + ^)sec 2 (45 + ^)-l = sin2 ^ t sin 2 (45° + A) sec 2 (45° + A) + 1 50. L^_ + , 3— r-+, L-+ r 1 — ^=? 1 + sin 2 a; 1 + cos 2 x 1 + sec 2 a; 1 + cosec 2 x sin 2 A — sin yi 1 — cos^4 + cos 2^4 K , sin 2^4 — sin^l o . A -^ . 51. : = sin A sec J.. sln| = -V 1 - cos 2 i cos=f = =±v l+ cos 2 A 258 PLANE TRIGONOMETRY. [§140 § 140. To find the Sine and Cosine of the Half Angle when the Sine and Cosine of the Whole Angle are known. Case (a). When the cosine of the whole angle is given. By (3), (4) of the preceding section 0) (2) That is, the sine of the half of an angle is the square root (taken with both sines) of one-half of the expression, one minus the cosine of the angle, and the cosine of the half of an angle is plus or minus the square root of one-half of the sum of one plus the cosine of the angle. EXERCISE. From the value of the cosine of the angle of a triangle in a 2 = b 2 +c 2 - 2 be cos A (§ 126) A A deduce by the preceding formulas the values of the sin — , cos — given in §§ 92 and 127. 2 2 Case (5). When the sine of the whole angle is given. By (1) of § 139. Also A A 2 sin — cos — = sin A. A A sin 2 — -f cos 2 — = 1. 2 2 1 1 (3) (4) . sin ^ + cos ^ = ± VI + sin A. A A 1 1 (5) sin ~ — cos ^- = ± VI — sin A. 2 2 (6) . -«- ^ ± VI + sin A ± VI - sin A (7) "* m 2 - 2 cog A _ ± VI + sin A =f VI - sin A (8) § 141] THE SINE AND THE COSINE IN UNION. 259 § 141. The Significance of the Double Signs in (1), (2), (5), (6), (7), (8) of § 140. When any special angle is taken as A, the signs are not plus and minus, but plus or minus, the proper sign being deter- mined by the quadrant of the given angle. For example, if A = 45°, the signs of (1), (2), (5) are all plus, and (6) is minus. Consequently the signs of (7), (8) are plus and minus, plus and plus. Similarly are the signs settled for (7), (8) in any other special case, by first determining what they are in (5), (6). When no special angle is taken for A, but when A is assumed as given only by its sine or by its cosine, then the reading of signs above is plus and minus and not plus or minus. When the cosine of A is given, there are two positions of the terminal, located symmetric to the horizontal (§ 106), and the angles are 2 n . 180° ± A (§ 106). Thus the half angles are n • 180° ± ^. But the terminal for n • 180° + 4 A is that of — when n is even and the opposite of that of A A —when n is odd, and the terminal of n • 180° — — is similarly A that of — — when n is even and the opposite when n is odd. Thus the terminals for the half angles are four lines, namely, A A the terminals of — and — — and their opposites. The ter- A A minals of — and are symmetric to the horizontal, and thus have opposite sines and the same cosine. Opposite terminals give opposite sines and opposite cosines. Thus the four terminals give only two sines (opposites) and two cosines (opposites), as shown in (1), (2). EXERCISE. Illustrate the preceding by a diagram. 260 PLANE TRIGONOMETRY. [§141 When the sine of A is given, there are likewise two termi- nals. They are symmetric to the vertical. The correspond- ing general angles are 2 n • 180° + A and (2 n + 1)180° - A. The half angles are n . 180° + ^andw 180 o + 18Q °~^ , whose A terminals are that of — and its opposite, and that of half the A supplement of A and its opposite. This gives four terminals for the half angle, opposite in pairs, and thus four values of the sine of the half angle, opposite by pairs, this being also the case with the cosines, as shown in (7), (8). EXERCISE. Illustrate the preceding by a diagram. Both the preceding cases are covered by § 53, where it was shown that for a single position of the terminal there are two positions of the terminal of the half angle. § 142. What Signs to Use in the formulas ■ i, A ■ sin - + cos- ± V 1 + sin A, sin — - — cos — = ± Vl — sin A, A A (5) (6) for a special angle, is readily settled by a glance at the sine and cosine on the unit-circle. The quadrantal bisectrices mark the angles for which there is numerical equality in the sines and cosines. For terminals between — 45° and -f 45°, the cosine of the angle is positive and numerically greater than the sine. Then (5) is + and (6) is -. The opposite is the case for ter- minals between the opposites of the FlQ - 144, preceding terminals. Thus, for terminals between those of 135° and — 135°, (5) is — and (6) is + . § 142] THE SINE AND THE COSINE IN UNION. 261 For terminals between those of 45° and 135°, the sine is positive and numerically greater than the cosine. Then (5) is + and so is (6). The opposite is the case between the opposite terminals. The diagram shown in Fig. 144 covers all these cases. In any special case we need only to locate the terminal. After the signs of (5), (6) are settled, those of (7), (8) of § 140 follow. ILLUSTRATIVE EXAMPLES. 1. If A = 580°, what signs have (1), (2), etc., of § 140? .-. ^ = 290°, or 270° + 20°. A The terminal of — lies between OR and OS in Fig. 144, with (1)-, (2) + , (5) — , (6) — , and consequently with the order of sines in (7) as , and that in (8) as — + • 2. Given that the numerical value of the sine and cosine at the V2 quadrantal bisectrices is — , or 0.7071, what signs have (1), (2), etc., when sin A = 0.3572 (with cos A +), when sin A = 0.8624, - 0.4517, or - 0.9653, with cos A - ? In the first case the terminal lies between OS, OP ; in the second, between OP, OQ ; in the third, between OQ, OR ; in the fourth, be- tween OR, OS. The corresponding signs are indicated on the diagram. (Fig. 144.) A 3. Where lies — when the signs of (7) are ? In this case (5) was — and (6) was — . Thus, by Fig. 144, the terminal lies between OR, OS. Therefore the angle lies between 2rttr-^ and 2n7r-^- EXERCISES. 1. State the signs for the formulas considered above when A has the following values in degree measure : 44, 45, 382, 560, - 44, - 840, 2040, - 650, 690, 200. 2. Determine the signs for (5), (6), (7), (8), when A lies between 270° and 315°. 262 PLANE TRIGONOMETRY. [§ 143 3. The same, if possible, when A lies between 270° and 360°. 4. What are the signs in (7) when A lies between 450° and 495° ? 5. Where must — lie when the signs in (7) are -\ — ? 6. Show that if A lies between 585° and 630 c the signs in (8) are — + . 7. Determine where — must lie when the signs in (7) are + +, — H, 8. Determine where — must lie when the signs in (8) are + +, H — , § 143. The Double-angle in Inverse Notation. From sin 2 6 = 2 sin 6 cos it follows, if sin 6 — x and cos = ± Vl — # 2 , that sin2 0=±2zVT--z 2 . •. 20 = sin- 1 (±2zVl-z 2 ). \ 2sm- 1 a? = shr 1 (± 2a?\/l -sc 2 ). That is, among the angles whose sine is ± 2 #Vl — x 2 will fall the angle which is twice any special angle whose sine is x. For example, Let sin 6 = + V2 cos# = C+or — Y If 6 is selected as ^, cos 6 = 4' +V2 Then 2 sin" 1 x = sin" 1 ( ± 2 rr Vl - z 2 ) becomes 2sin- 1 f J = sin' 1 f 4-1), V+V2/ 7T Among the angles, sin" 1 (-|-l), is — , which is twice the _ 2 -j angle — , selected as a special angle whose sine is '•• 4 ^5 +v ^ § 144] THE SINE AND THE COSINE IN UNION. 263 Similarly, from the relations cos 26 = cos 2 6 — sin 2 0, cos2 0=l-2sin 2 0, cos20 = 2cos 2 <9-l, follow 2 sin- 1 x = cos" 1 (1 - 2 a? 2 ), 2 cos" 1 x = cos" 1 (2 x 2 - 1). These relations are subject to the limitations to which attention has already been invited in connection with the multiplicity of values in the inverse notation. EXERCISES. Show that for special values of the angles the following relations are true : 1. 2 sin- 1 * = cos- 1 f 9. 2 sin" 1 ^ = cos" 1 1 if. 2. 2sin- 1 | = cos- 1 (- 2 \). 10. 2 cos" 1 ^ = cos" 1 (-#). 3. 2 sin- 1 1 = sin- 1 |f 11. lbr*§+2 sin" 1 H= sin" 1 ©. 4. cos" 1 1 + 2 sin" 1 | = 120°. 12. cos" 1 | + 2 cos- 1 T 5 2 =cos- 1 (?). 5. cos- 1 x = 2sin- 1 A /^^- 13 ' ^-V^cos-^sin-H?). 2 14. sin- 1 ^-2sin- 1 1 JV = sin- 1 (?). 6. cos- 1 * = 2cos- 1 ^— 1£. is. cos- 1 T \-2cos- 1 T 5 1 = cos- 1 (?). 7. 2sin- 1 ^ = sin-4|f 16. cos- 1 1 - 2 cos" 1 1 = sin" 1 (?). 8. 2sin- 1 T 8 7 =sin- 1 |f^. 17. sin- 1 (.35) + 2 cos" 1 (.37) = ? 144. To find the Sine and Cosine of 3 A in Terms of the Sine and Cosine of A, respectively, and the Use of the Results in solving Cubics. sin (3 A) = sin (2 A + A) = sin 2 A cos A + cos 2 A sin A = 2 sin A cos A cos A + (1 — 2 sin 2 A) sin A = 2 sin A (1 - sin 2 ^1) + (1 - 2 sin 2 A) sin 4. .-. sin 3 A = 3 sin A - 4 sin 3 ^. (1) 264 PLANE TRIGONOMETRY. [§ 144 A similar process, expressing the terms as rapidly as pos- sible in terms of cos A instead of sin A as above when sin 3 A is determined, gives : cos 3 A = 4 cos 3 A - 3 cos A~ (2) Inasmuch as (1), (2) have the 4, 3 terms interchanged, it may be helpful, in remembering them, to note that each term contains a 3, and that they must both remain true for any special angle. Take for A the angle whose sine is 1 in the sin 3 A result, and for A, in the cos 3 A result, the angle whose cosine is 1 ; namely, 90° and 0°, respectively. Then sin 3^4 = sin 270° = - 1, so the 4 must follow the 3 ; and cos 3 A = cos = 1, so the 3 must follow the 4.* The equations (1), (2) show that if sin 3 A, cos 3 A are given, the sine and cosine of A are to be obtained by solving cubics. Thus the ancient problem of "trisecting an angle by means of the ruler and compass," is the same as the alge- braic problem of solving a cubic when the second power of the unknown quantity is absent, as is the case in (1), (2). Any cubic equation, as ax* + bx 2 + ex + d = 0, can be freed of the x 2 term by setting x = y 3 a becomes, on expansion, an expression like aff+py + q = 0. (3) Comparing (3) with 4 cos? 4 - 3 cosJ. - cos 3 J. = 0, (2) * Or this : Next " sin " comes " trin," 3 Next " co " comes *' fo," 4 1 § 144] THE SINE AND THE COSINE IN UNION. 265 cos A might be y, if y were less than 1, numerically. We may set (3) in the form a (ny) z + pn 2 (ny) + qn s = 0, (4) where n may be so selected that ny may be less than 1, and thus the cos A. When an expression like (3) is zero, it remains zero when multiplied by any number. Therefore, when two similar expressions are identical and zero, there is proportionality among the coefficients of corresponding terms. So if (2), (4) are identical, and cos A = ny, a pn 2 4 - 3 - qn s - cos 3 A (5) .2 3 <*. .*. 71 — - — ; 4jo (6) p (?) (6) shows that n is real only when a, p are opposite in sign ; (6) also shows that n may have two values, equal and opposite in sign. However, it is necessary to take only one value of w, since — n in (4) merely changes the sign through- out and gives only the same solutions as + n ; (7) shows that no real angle, 3 A, exists unless 3 qn < jt?, numerically, or 9 5% 2 < jt? 2 , or, by (6), 9 q 2 ( -r-^)

-l)(#+t' + l)*6. ,\ 8-1 = 0, or s 2 + s + 1 = 0. The first supposition, 8 — 1 = 0, gives 8 = 1. Solving the quadratic supposition in the usual way, we get 2 2 The three values of 8 taken with y/% Vv give y three values. Summary for solution of ay z + py + q=0, taking a positive : (i) If p is also positive, use Cardan's process as just given, (ii) If p is negative and 27a4jt? 3 , numerically, use Cardan's process. 268 PLANE TRIGONOMETRY. [§144 (iii) If p is negative and 27 aq 2 < 4 p% numerically, use the trigonometric method as in the example on page 265. (iv) If p is negative and 27 a(f = 4 p s , use either method, or solve by the method of equal roots. In cases (i), (ii) there will be one real root and two imagi- nary; in cases (iii), (iv), three real roots, there being equal roots in case (iv). EXERCISES. Determine by which method each of the following examples, from Ex. 2 to Ex. 10, is best solved, and solve: 1. 2x 8 -3x-l = 0. 6. x 8 -3x-2 = 0. 2. x s + 3 x 2 - 1 = 0. 7. x s - x + 6 = 0. 3. x 8 -24x-32 = 0. 8. x 8 - 9 x- 28 = 0. 4. x 8 - 7 x + 5 = 0. 9. 3 x 8 - 6 x 2 - 2 = 0. 5. x 8 + 4x 2 +2x-l = 0. 10. x 8 - 15 x 2 - 33 x + 847 = 0. 11. Show sin 4 = 4 sin $ cos 8 — 4 cos sin 8 0. 12. Show cos 4 0=1-8 cos 2 + 8 cos 4 0. 13. Show cos5 0=16cos 5 0-2Ocos 8 + 5cos0. 14. Show cos6 = 32cos 6 0-48cos 4 + 18cos 2 0-l. 15. Show sin 3 = (2 cos 2 + 1) sin 0. 16. Show cos 3 = (2 cos 2 - 1) cos 0. 17. Show cos3 + 3cos0 = cos8 fl cosec8 fl. 3 sin - sin 3 18. Show sin 3 - sin = sin (cos 3 + cos 0) sec 0. 19. Show sin 3 - cos 3 = (sin + cos 0)(2 sin 2 - 1). 20. Show (1 + 2 cos 0) 2 (1 - cos 0) = 1 - cos 3 0. 21. Show (sec2 + l) 8 (3sin0-sin3 0) 2 = (sec 2 - l) 8 (3 cos + cos 3 0) 2 . 22. Find expressions for sin 5 and sin 6 in terms of functions of 0. 23. Show that A = 18° is a particular solution of the equation sin 2 A = cos 3 A. Express this equation in terms of sin A and cos A . Then in terms of sin A, and solve, showing sin A = sin 18° = +V ^~ 1 A 1QO +VlO + 2V5 cos A = cos 18° = - 1 . § 145] THE SINE AND THE COSINE IN UNION. 269 24. Give the general solution of the equation sin 2 A = cos 3 A . 25. Show from the values of sin 18°, cos 18°, in Ex. 13, that sin36O = ± Vl0-2Vo. 4 cos 36° =2%!, 4 using sin 2 x = 2 sin x cos x. 26. Show that sin 9° + cos 9° = VI + sin 18°, sin 9° - cos 9° = - Vl - sin 18°. From these equations, show Sin 9 o = V3W5-V5 -V5 4 c08 9° = V ^ g + V?r71 . 4 What are the values of sin 81° and cos 81°? § 145. The Addition-multiplication and Subtraction- multiplication Formulas. It is frequently advisable, either for purposes of simplifica- tion or to set an expression in a form suitable for logarithmic computation, to express the sum or difference of two sines (cosines) in the form of a product. From sin (A 4- B) = sin A cos B 4- cos A sin i?, (1) sin (A — B) = sin A cos B — cos A sin B, (2) cos ( A 4- &) = cos A cos B — sin A sin B, (3) cos QA — i?) = cos J. cos i? 4- sin A sin i?, (4) follow by addition (subtraction) : sin ( A + B) 4- sin (A. - ^) = 2 sin J. cos B, (5) sin (.4 + B) - sin (.A - B) = 2 sin 5 cos A, (6) cos (A. - B) 4- cos (A 4- B) = 2 cos A cos 5, (7) cos (A - B) - cos (i + 5) = 2 sin A. sin £. (8) 270 PLANE TRIGONOMETRY. [§ 145 Here A 4- B and A — B may represent any two angles. Let A + B = x, A-B = y. Then A = X -±M, A 2 Thus, (5), (6), (7), (8) may be written : sin x + sin y = 2 sin £±J£ cos ^=-^> (5') sin a; - sin 2/ = 2 sin ^=^ cos ^^' ( 6 cos 2/ + cos a? = 2 cos ^"t y cos X ~V > (7') cos 2/ - cos oc = 2 sin ^±^ sin ^i^ C 8 ') These formulas are general, being true for all values of x, y. The subtractions in the second members must be carried out, so far as sines are concerned, in the order indicated, since sin (— 0) == — sin 6 ; for the cosines, this is a matter of indif- ference, since cos (— 0) = cos 6. The order in (7'), (8 ; ) is set different from that in (5'), (6'), as relates to x, y, in deference to angles of the first quadrant, since here the larger the angle, the larger the sine, but the larger the angle, the smaller the cosine. EXERCISE. 1. Give verbal statements for (5'), (6'), (7'), (8'), paying attention to the order of subtractions. 2. sin 60° + sin 30° = 2 sin 45° cos 15°. 3. sin 60° - sin 30° = 2 sin 15° cos 45°. 4. sin 25 - sin 45° = - 2 sin 10° cos 35°. 5. cos 60° + cos 30° = 2 cos 45° cos 15°. 6. cos 60° - cos 30° = - 2 sin 45° sin 15°. § 145] THE SINE AND THE COSINE IN UNION. 271 7. cos 25° - cos 55° = 2 sin 40° sin 15°. 8. sin 115° + sin 281° = - 2 sin 18° sin 7°. 9. sin 317° - sin 151° = - 2 sin 36° cos 7°. 10. cos 295° + cos 817° = 2 cos 16° sin 9°. 11. cos 370° - cos 280° = 2 sin 35° sin 45°. 12. Verify some of the preceding results by using the tables. 13. Take a pair of angles in each quadrant and apply (5'), (6'), (7'), (8'), letting (i) x>y in (6'), (8'), and then (ii) x) = ^^^ cos (0 + <£) - 2 cos + cos (0 - ) - A sin + 2 sin 3 + sin 5 . ** nncfnn K a 54. . — — . =smdt/ cosec 5 0. sin 3 + 2 sin o + sin 7 M sin (0 - ) + 2 sin + sin (0 + d>) . ^ 55. - — ±- — 7 ( _ . — „ . ;„ y ,\ = sm cosec 5. sin (/3 - ) + 2 sm£ + sm (/3 + <£) M cc sin — sin 5 + sin 9 — sin 13 A * . a 56. - — pi- — j — — = cos 4 cosec 4 0. cos — cos 5 — cos 9 + cos 13 57. cos 3 x + cos 5 a; + cos 7 a: + cos 15 x = 4 cos 4 x cos 5 a; cos 6 x. 58. sin0 + sin20 + sin40 + sin50 = 4cos-cos — sin30. 59. Show, without using (5'), (6'), (7'), (8'), that sin x -f sin y _ cosy — cos a: cos y + cos x sin y — sin x cos — cos 3 cos 6 — cos 4 60 sin — sin 5 sin 8 + sin 2 61 cos(x-f y+z) + cos(— x+y + z) +cos(x — y-f 2) + cos (ar+y— 2) _ cosy sin (x + y + z) + sin(— x + y + z) — sin (a: — y + z) + sin (a; + y — z) sin y g2 cos + cos 3 _ sec 2 cos + cos 5 "~ 2 - sec 2 § 146. Application of the Addition-multiplication, Subtraction- multiplication Formulas to the Solution of Trigonometric Equations of a Special Type. Model Example. Solve sin 6 + sin 7 6 = sin 4 0. (1) By (5), 2sin40cos30=sin40, (2) or, sin40(2cos30-l) = O. (3) - .-. sin 4 6 = 0, (4) or, 2 cos 3 6 - 1 = 0. (5) By (4), 4 = mr. .*. = w-. By (5), cos 3 = | . (6) A particular solution is 3 = —. 274 PLANE TRIGONOMETRY. [§146 IT ,\ the general solution is . 3 6 = 2 nir ± — , o Thus the general solutions of (1) are The applicability of the process .indicated in this example is limited to the case where a factor is apparent after the simplification made in (2). An example like sin -+ sin 7 6 = sin 13 0, or like 5 sin + 3 cos 6 = 4, or 2 sin 6 + 3 sin 7 = 2, could not be solved by the process given above. Later, a solution of such expressions by the aid of auxiliary angles will be given. (See § 148 ; also § 180.) EXERCISES. Solve the following, giving general values for the angle, : 1. cos + cos 7 6 = 2 cos 4 0. 7. cos $ + cos 2 + cos 3 6 = 0. 2. cos + cos 7 = cos 3 0. 8. sin 6 + sin 2 + sin 3 6 = 0. 3. sin - sin 7 = sin 3 0. 9. cos - cos 2 + cos 3-0 = 0. 4. sin70-sin0 = cos40. 10. sin - sin 2 + sin 3 = 0. 5. cos - cos 7 = 2 sin 4 0. 11. cos + sin 2 - cos 3 = 0. 6. cos 70- cos = sin 3 0. 12. sin - cos 2 - sin 3 = 0. 13. sin 2 - cos 2 - sin + cos = 0. 14. sin + sin 2 = 0. 18. sin mO ± sin n0 = 0. m = n. 15. sin — sin 5 = 0. 19. cos mO ± sin n0 = 0. m^n. 16. cos 20 + cos 30 = 0. 20. sin ± cos 3 = 0. 17. cos 7 — cos 5 = 0. 21. sin mO ± cos n0 = 0. m = n. 22. sin (3 + «) + sin (3 - a) + sin (« - 0) - sin (a -f 0) = cos a. 23. cos n0 = cos (n — 2)0 + sin 0. 24. sintL±i^ = sill !Lzi^ + s in0. §147] THE SINE AND THE COSINE IN UNION. 275 25. sin 2 n0 - sin 2 (n -1)0 = sin 2 0. 26. sin 3 - 4 sin sin (0 + j3) sin (0-/3)= 0. 27. cos30 + 2cos20 = O. 31 cos 2 sin 2 = 1 cos2 28. cos20 + 3cos0 = O. " sin ^ cos ^ sin2 ^ 29 cos_0_sin_0 = 2 32. cos 30 - cos50 = sin 0. ' sin0 cos0 33. cos 5 + cos 3 = V2 cos 4 0. 30. ^ + ^ + ^=0. 34. sin60 + sin40 = 2cos0. cos cos 2 cos 3 35. sin (m + 1)0 + sin (m - 1)0 = cos 0. 36. cos m0 — cos (m — 2)0 = sin 0. 37. sin + sin 2 + sin 3 = cos + cos 2 + cos 3 0. 38. sin 2 - cos 2 = cos - sin 0. 39. cos (« - 0) = sin (a + 0). 40. sin (0 + a) + cos (0 + a) = sin (0 - a) + cos (0 - a) . 41. cos 3 sin 8 + sin 30 cos 8 = 0. 42. sin + sin 2 + sin 3 + sin 4 = 0. 43. cos3 0sin 8 + sin3 0cos 8 = — . 8 44. sin 0(cos 2 + cos 4 + cos 6 + cos 8 0) = sin 4 0. 45. 2 cos 2 cos 3 - cos = 0. 46. 4sin 2 + sin 2 2 0=3. 47. 2 sin 8 + 3 cos 2 + sin = 3. 48. cos 8 - cos sin - sin 8 = 1. 49. sin 8 = 0.2341 sin 2 0. § 147. The Multiplication-addition, Multiplication-subtraction Formulas. The formulas (5), (6), (7), (8) of § 145, when read back- ward, may be called the Multiplication-addition, Multiplica- tion-subtraction formulas. sin A cos B = \ {sin (A + B) + gift (A - B)}, (1) sin B cos .4 = ! {sin (A+B)- sin (-4 - B)}, (2) cos^4cos.B = |{cos(^i- J5) + cos (.4 + .B)}, (3) sin A sin B = \ {cos (-* - B)~ cos (A + JB)}. (4) _ fee For example, < . (si] 276 PLANE TRIGONOMETRY. [§ 147 From (3), it appears that if both words of the product are cosines, the product may be changed to a sum of cosines; from (4), that if both words of the product are sines, the product may be changed to a difference of cosines. cos 40° cos 60° = J (cos 20° + cos 100°), sin 40° sin 60° = \ (cos 20° - cos 100°). In (1), (2), the form of subtraction in the second member assumes A numerically greater than B, so we may say that when the words of the product are one sine and the other cosine, then, if sine is connected with the numerically larger angle, a sum of sines is indicated, and if sine is connected with the smaller angle, a difference of sines is indicated. For example, sin 60° cos 40° = } (sin 100° + sin 20°), sin 40° cos 60° = £ (sin 100° - sin 20°). However, sin x cos y — \ J sin (x + y) + sin (x — y) }, no matter what the relative size of x and y, when the differ- ence x — y is taken in the order indicated. For example, sin 40° cos 60° = J J sin (40° + 60°) + sin (40° - 60°) | = l5sinlOO° + sin(-20°)} = i (sin 100° -sin 20°). EXERCISES. 1. sin 80° cos 30° = ? 9. sin0cos<£ = ? 2. sin 30° cos 80° = ? 10 . cos0cos<£ = ? 3. cos 80° cos 30° = ? S$ 5$ 4. sin 80° sin 30° = ? ^ ^T^T^ 5.^70^69 = 1 12> sin 3i cos 5i = ? 6. sin5 0cos70 = ? 7. cos 5 6 cos 7 = ? 8. sin 5 sin 7 6 = 1 2 2 7. cos50cos70 = ? 13. sin — cos- = ? 2 2 § 147] THE SINE AND THE COSINE IN UNION. 277 14. Simplify 2 cos 2 cos - 2 sin 4 sin 0. -.,= a- nt -50 .90 30 15. Simplify sin — cos - — sin — cos — • 16. Prove 2 sin (45° + A) sin (45° - A) = cos 2 A . 17. Prove 2cos(45° + ^1) cos (45° - A)= cos 2^4. 18. Prove that sin - sin — + sin — sin — — = sin 2 sin 5 0. 19. Prove cos 2 cos - — cos 3 cos -— = sin 5 $ sin — . mm £ 11/9 $ 7/9 3 20. Prove sin — — sin - + sin -— sin — — = sin 2 sin 0. 4 4 4 4 21. Prove 2 sin 2 cos + 2 cos 4 sin = sin 5 + sin 0. 22. Prove cos 2 cos - sin 4 sin = cos 3 cos 2 0. 23. Prove sin a sin («+ 2 /?) -sin /3 sin (/3 + 2a) = sin(a-/?) sin (« + /?)• 24. Prove (sin 3 + sin <£) sin <£ + (cos 3 - cos ) cos = 0. 2 sin (0 - ) cos <£ — sin (0 — 2 <£) _ sin 2 sin (/? — <£) cos — sin (/?- 2)~ sin /3* 26 sin sin 2 + sin 3 sin 6 + sin 4 sin 13 _ sin 9 27. sin cos 20 + sin 3 cos 6 + sin 4 cos 13 cos 9 cos 2 cos 3 — cos 2 cos 7 + cos cos 10 cos 6 cos 5 sin 4 sin 3 - sin 2 sin 5 + sin 4 sin 7 sin 6 sin 5 28. cos sin (j3-) + cos /? sin (<£ - 0) + cos <£ sin (0 - )3) = 0. 29. sin (/? - y) cos (a - 8) + sin (y — a) cos ((3 — 8) + sin (a — /?) cos (y-8)=0. 30. cos (0 - <£) cos (0 + )= cos 2 - sin 2 <£ = cos 2 <£ - sin 2 0. 31. versin {0 + <£) versin (0 - ) = (cos — cos ) 2 . 32. 2cos^cos|f + co S i| + eos|| = 0. 33. cos (36° - A) cos (36° + A) + cos (54° + A) cos (54° - 4) = cos 2 A. grin r sin -*- 34. - 2 2 sin <£ + cos + cos cos— IpE cos — —3C n ^ 35 sin • sin 2 + sin 2 sin 5 + sin 3 sin 10 _ _ s i n 7 # sec 7 cos • sin 2 + sin 2 cos 5 - cos3 sin 10 sin 5 _ sin 3 _ sin 2 _ sin 2 § sin 3 § sin 5 cos 5 cos 3 cos 2 cos 2 cos 3 cos 5 278 PLANE TRIGONOMETRY. [§148 § 148. Use of an Auxiliary, or Subsidiary, Angle in Solving Trigonometric Equations of a Certain Type. Model Example. 2 sin# + 3 cos# = 3. (1) Divide by the square root of the sum of the squares of the coefficients of sin x and cos x. 2 • — sin x -f V13 cos# = Since and we may set V13 sin 2 y + cos 2 y ,2 V13 1 (2) 2 2V13 sin y. cos y VI3 13_ 3 3V13 (3) (4) V13 13 Either (3) or (4), when expressed as a decimal, directly or by logarithms, determines y in the tables. Thus (2) becomes 3 3V13 sin;?/ sin x+cosy cos x= If we determine z from the tables by means of 3V13 cos,- - , (5) becomes cos >(x—y) = cos 2; or, x - y = 2 n 180° ± z. ,\ x = 2 n 180° + y ± z By (3), 2V13 Zj (6), .-. y = 3VT3 COS Z as — = 13 (5) (6) (7) (8) (9) (The student may determine «/, 3 from the tables.) Then, by (9), x = 2 n . 180° 4- # ± z (as found), where w is any integer. § 148] THE SINE AND THE COSINE IN UNION. 279 z and y are called auxiliary angles ; sometimes subsidiary angles. Example (1) above belongs to the general type. a sin 6 + b cos = c, (^4) whose solution is, similarly, a sinfl+ h cosfl = g V«2 + J2 Va 2 + #> Va2+~P Let sin = , cos 6 = — , V^+~P Va 2 + 6 2 and cos a Va 2 + b 2 •\ cos (0 — <£) = cos «. .-. = 2 mr ± a -\- (j>. Since no cosines are greater than unity, evidently a cannot be determined if c 2 > a 2 -f- £> 2 . In this case no solution exists. When #, b are large, use § 180. • EXERCISES. A. Show that (^4) above may be solved by the form : sin(0±<£) = sin a. B. Determine the general value of in the following cases : 1. sin ± cos = V2. 12. 5cos0 - 2 sin0 = ± V29. 2. sin30±cos30 = — . 13. 2 cos0 ± 5 sin0 = ± V29. V2 3. sin50±cos50 = -V2. 4. sin 20 ± cos 26 = 0. 5. sin ra0 ± cos nO = 0. 6. sin mO ± cos nO = ± V2. 14. 5.32 cos0-2.32 sin 0=2.41. 15. 2cos0-5sin0 = 2. 16. 5cos0-2sin0 = 6. 17. 2cos3 + 3cos30 = 3. 18. 3sin40±4cos40 = 5. 7. sin m$ ± cos n$ = ± __ . . n . . n . y/2 19. 3 sin 4 ± 4 cos 4 = 4. 8. \/3cos0±sin0 = ±V2. 20. 3sin40 ± 4cos 40= 6. 9. V3 sin ± cos = ± V2. 21. 5 sin 5 ± 7 cos 5 = 3. 10. 1 + sin = V3 cos 0. 22. 5 sin + 2 cos = 5. 11. l-cos0 = sin0. 23. 6 cos0 + 8 sin0 = 9. 280 PLANE TRIGONOMETRY. [§ 149 24. By the aid of an auxiliary angle, show the changes in sign and magnitude of sin + cos 0, as goes from 0° to 360°. Ans. From 1 to + V2 to to - V2 to to 1, as passes from 0° to 45° to 135° to 225° to 315° to 360°. 25. Show, similarly, the changes in sign and magnitude of : (a) sin — cos 6. (c) sin 6 — VS cos 0. (e) 5 cos 6 — 2 sin 0. (b) sin 6 + V3 cos 6. (d) cos 2 $ - sin 2 0. (/) 3 cos 6 + 4 sin 0. 26. Is there any limitation in the possibility of solving by the method of auxiliary angles ? § 149. Use of the Addition-subtraction Formulas in calculating a Table of Sines and Cosines. sin (A + B) = sin A cos B + cos A sin B. (1) sin (J. — 5) = sin ^L cos B — cos J. sin B. (2) .-. sin (A + B) =2 sin -4 cos B - sin (^ - B). (3) .-. sin (60° + A) = sin ^. + sin (60° - 4). (4) Formulas (3), (4) may be used in calculating a table of sines ; (3) being used for all angles, J., below 60°, and (4) to get from these the sines for all angles from 60° to 90°. When the sines are known, so are the cosines, since cos^l = sin (90° -A}. For example, suppose the table is to read to every 10", let B = 10" and A = 10", 20", 30", 40", etc., in turn. Then (3) gives : v sin 20" = 2 sin 10" cos 10", sin 30" = 2 sin 20" cos 10" - sin 10", sin 40" = 2 sin 30" cos 10" - sin 20", sin 50" = 2 sin 40" cos 10" - sin 30", etc., etc., etc. The table may be started with any angle whose sine and cosine are known, when the sine and cosine of the angle- difference of the table are known. For instance, let A = 30°, sin A = 0.5000, cos A = 2~ = 0.8660, for a table reading to minutes. Let B — V. Then sin 30° 1' = sin 30° cos 1' + cos 30° sin 1'. §150] THE SINE AND THE COSINE IN UNION. 281 Thus knowing sin 30° 1', then by (3), sin 30° 2' » 2 sin 30° 1' cos 1' - sin 30°, sin 30° 3' = 2 sin 30° 2' cos 1' - sin 30° 1', sin 30° 4' = 2 sin 30° 3' cos V - sin 30° 2', etc., etc., etc. Evidently from the foregoing it is necessary to calculate the sine and cosine of the angle-difference of the table, whether this be 10', 1', 10", or 1", the usual differences. § 150. Calculating the Sine and Cosine of the Angle-difference of Tables. Let 6 be any small angle in circular measure. {*!*? co.p.107 Then We now show cos < 1. sin0>0-— , 4 cos > 1 — — , 28 from which it will follow that when for the sine of a small angle its circular measure is taken, the error is less than one- fourth the cube of the circular measure, and when for the cosine of a small angle unity is taken, the error is less than one-half the square of the circular measure, DAO (Fig. 145) is a portion of the unit circle. CB, DB are tan- gents to this circle. The broken line DBC> arc BAG tf .-. line BC> arc AC. (1) The number for the arc AC is the same as that for the angle AOC. .-. arc AC =6. Triangles OEC, OCB are similar. CB = UC " OC OE 282 PLANE TRIGONOMETRY. But 00 = 1, EO = sin 0, OE = cos 0. .-. OB = sin6 . cos 6 .'. by (1), sm0 ^ COS0 ,\ sin > • cos 0. [§150 (2) That is, the sine of any small positive angle is greater than the product of its circular measure and cosine. . 0^0 /0 n •*• sin 2 > 2 C ° S 2* ( ^ But sin 0=2 sin - cos - . A -i sin 0>2 {r os i) oos i or, sin > cos 2 - . A .: sin 0> 0^1 -sin 2 ' But 0^ . 2 >Sm 2' 2 . 2 •••4 >Sm V By (5), (6), sin0>0(l-J), or, sin0>0-f. 4 Also cos = 1 — 2 sin 2 - cos > 1 - (4) (5) (6) or, cos > 1 — — . A § 150] THE SINE AND THE COSINE IN UNION. 283 The circular measure of 10' = 18ra radianS = 0.002,908,9, about. .-. sin 10' = 0.002,908,9, with an error less than one-fourth the cube of this number, f0.003) 3 or <^ T -A or < 0.000,000,007, about, or less than 1 in the eighth decimal place. Thus, in constructing any place table, up to and including a seven-place table of sines and cosines, we may take sin 10' as equal to the circular measure of 10' for seven places with- out appreciable error. Similarly, sin 1' =0.000,290,89, or, approximately, sin 1' = 0.0003, or, 3 zeros and 3. Similarly, sin 1" = 0. 000,004,848,136,811, or, approximately, sin 1" = 0.000,005, or, 5 zeros and 5. The error in the above value of sin 1' is less than 0.000,000,000,007, and the error in the value of sin 1" is less than 0.000,000,000,000,000,03. fsin<9<0, From the formulas j • q > q _ & sin 1" < 0.000,004,848,136,811 ; sin 1" > 0.000,004,848,136,807. ■% these agree to 13 figures. .-. sin 1" = 0.000,004,848,136,8, which is correct to 13 figures. cos 6 = Vl - sin 2 d = (1 - sin 2 $)* = 1 - £ sin 2 0, approximately (§ 42). .-. cos 1" = 0.999,999,999,988. 284 PLANE TRIGONOMETRY. [§ 150 02 Since cos 6 < 1 and cos 6 > 1 — — , we may set cos 1" = 1, with an error less than |(0.000,004,848) 2 , or 6 — —• 4 sin 6 ^ 1 -, sin 6 ^ -, 2 ... _l-_. Thus —=— can be made to approach as near as we please u to the limiting value 1, as 6 is made smaller and smaller. This is expressed in the form 5H^ = l, when = 0. 6 Similarly, since cos 6 < 1, and cos 6 > 1 - ^, (§ 150) cos = 1, when = 0. § 155. Differentiating the Sine and Cosine. If F(x) denotes any function of x, and if x is given the value x + Ax, and the difference F(x + Ax} — F(x) is formed and divided by Ax, then the limit toward which the quotient — - — ^ ^— ^ approaches, when it approaches a definite limit as Ax approaches zero, is called the differential coeffi- cient of F(x) with respect to x, also, the first derived func- tion of F(x) with respect to x, and is indicated by F'(x). The process here pointed out to which F(x) is subjected is called differentiating F(x) with respect to x. Treating the first derived function as was the original function is called getting the second differential coefficient, also getting the second derived function; and so on. The second derived function is indicated by F"(x), etc. 155] THE SINE AND THE COSINE IN UNION. 289 For example, (1) If F(x) = x*, F(x + As) « (x + As) 2 = s 2 + 2 x • As + As 2 . ^(s + As)-.F(s) , A .-. — r-^ ^ = 2 s + As. As .-. F'(x)=2x. Similarly, F"(x) = 2 and F"'(x) = 0. (2) If ^(s)=sins, F(x + As) = sin (s + As) . F(x + As) — -F(s) _ sin (s + As) — sin s As As . As ( , As\ 2 sin— -cos(s + — J (•+» This can be set in the form • As ojjj ^(s + As) - F(x) 2 _1 _^ 1_Z sb — COS As As 2 . As sin — By the preceding section, — - — = 1, when As = 0. .*. F'(x) = coss. 2 That is, the differential coefficient of sin s, with respect to s, is cos s, s being in circular measure. (3) If i 7 (s) = coss, JP(s + As) = cos (s + As). F(x + As) — F(x) _ cos (s + As) — cos s As As / As\ • As -2 sin As sin ^ + -2") am— , A ;in f sin s + As As 2 .-. .F'(s) = -sin(s). That is, the differential coefficient of cos x with respect to x is — sin s, s being in circular measure. 290 PLANE TRIGONOMETRY. [§155 (4) We thus have : When F(x) = sin #, F'(x) = cos a?, F"(x) = — sin as, F f "(x)=-cosx, F w (x) — sin x, and then a repeat without end. When F(x) = cos x, F , (x)=-smx, F"(x) = — co8z, F" f (x)=smx, F iv (x)=cosx, and then a repeat without end. (5) If, now, we denote by ^ T (0),1 T, (0), .F"(0), etc., the special values which these functions assume when x = 0, we have When F(x) = sin x, ^(0)=0, J"(0)=1, ^(0)=0, 2^(0)= 0, and a repeat without end. (6) If F(x) = x n , where n is a positive integer, JP(> + Ax) = O + Az) w = x n + w^" 1 Az + - ' w ~ 1 x n ~ 2 Ax~ 2 4- etc., assuming the binomial theorem for positive integral indices. When F (x) = cos x, ^(0)=i; JF'(0) = 0, and a repeat without end. jTg + Agj - JPCa;) _ Ax = nx n ~ x -f powers of Ax. and if .% F'(x)=nx n ~ l . .-. if ^0)=^, _F'0)=3a; 2 , F(x) = aa,- 4 , jF'O) = 4 a^, etc. § 156] THE SINE AND THE COSINE IN UNION. 291 § 156. The Sine-series and Cosine-series. Assuming that sin can be expressed in powers of 0, let _F(0) = sin = a + bd + eft + d0* + efr +/0 5 + etc. .-. F'(ff) = cos = b + 2 c0 + 3 d& + 4 0, or r > JV# 2 + \ — \. Thus the sine-series is convergent. So, likewise, the cosine-series. Both of these series remain convergent if all their terms are taken positive, for each would then be a part of the series for e (§ 8), a convergent series. A part of a convergent series, all of whose terms are positive, must be convergent. Thus the convergency of the sine-series and cosine-series is not dependent on the signs of the terms, and is consequently independent of the order in which the terms are taken. These series are what is termed absolutely convergent. The case is very different with the series for log e (1 + z), § 9, where the convergency is dependent on the signs, and that series converges to the logarithm only when the order of terms is held as given in § 9. For information more in detail on series, consult Osgood's "Infinite Series." § 158. Sine and Cosine Relations for the Angles of a Triangle. A + B + (7=180°. (1) .:A + B = 180°- 0. .'. A + B =90°-Q. 2 2 .\ sin (JL + B) = sin 0; cos (A + B) = — cos 0; . A + B n A + B . n sin — g — = cos (J ; cos — j- — = sin C ; with similar relations for any pair of angles. From these relations, together with the addition-multipli- cation, etc., relations, can be deduced many others. 294 PLANE TRIGONOMETRY. [§ 158 (1) To prove sin 2 A + sin 2 B -f sin 2 (7 = 4 sin J. sin 2? sin 0. sin2J. + sin2^ + sin 2 (7= sin 2^4 + sin aB + sin 2 (7 = 2 sin (A + .£) cos (.4 - B) + 2 sin cos (7 = 2 sin cos (J. - B) + 2 sin (7 \ - cos ( J. + B) I = 2 sin C{cos (A - B) - cos (A + B)\ = 2 sin (7(2 sin ^4 sin B) = 4 sin J. sin B sin C. (2) To prove cos A + cos i? — cos C = — 1 + 4 cos — cos — sin — . AAA cos J. + cos B — cos (7 = cos A + cos B — cos (7 = 2 cos 2 |- 1 + 2 sin C v ) w ^ n similar equations for —, — ; A A sin A = 2 sin — cos — , (vi) with similar equations for B, C. A A 296 PLANE TRIGONOMETRY. [§ 159 From these relations, and the treatment of the angles as in the preceding section, follow many other relations. For example, from (i), a _ sin A b sin B . A + B A-B . -^ 2 sin — ! — ■ cos — - — fl + o_sini + sin5_ 2 2 a — b ~~ sin A — sin B ~ . A — B A + B' 2 sin — - — cos — j- — 2 2 , . n . A-B A + B , ,. . A+B A-B .-. (a + o) sin — - — cos — - — = (a — b) sin — ^ — cos — - — . EXERCISES. 1. (b + c)sm-=acos B 2. 2 2 A B 2 cos — cos - a + b + c 2 2 . C sin — 2 3. (b - c) cos - = a sin ^ ~ C . K J 2 2 4. a(cos B + cos C) = 2(6 + c) sin 2 -. 5. a(cos £ - cos C) = 2(c - &) cos 2 —. 6. a 2 + 6 2 + c 2 = 2(ab cos C + be cos .4 + ca cos £). 7. c 2 = (a - 6) 2 cos 2 - + (a + &) 2 sin 2 -. 8. a sin (5 - C) + b sin (C - A) + c sin (.4 - B) = 0. 9. — ^— .sin(5- C)= - b . S m(C-A) = —£ sin (A-B). b 2 -c 2 K J c 2 -a 2 K J a 2 -b 2 K ' 10. a Bm- sin ^-=-^ + &sin- sin C ~ A +csin-sin A^J* = o. 2 2 2 2 2 2 11. a 2 (cos 2 J5 - cos 2 C) + & 2 (cos 2 C - cosM) + c 2 (cos 2 ^ - cos 2 £) = 0. 12. £Lzig? s in 2 A + gi^-g- 2 sin 2 £ + - ~ ^ sin 2 C = 0. a 2 6 2 c 2 13. a + 6 + c = (a + b) cos C + (b + c) cos ^4 + (c + a) cos B. 14. sin (5 - C) = ^-=-£- 2 sin ^. CHAPTER VIII. 3p ^K N \ ^ ^ V p «•"' v _^^"^ ^- ^^^ M \ 2T S Fig. 149. Fig. 150. ric to the vertical give, for equal moduli, oppositely equal abscissas. In the first case the abscissa is unchanged, and in the second case the ordinate is unchanged. Or, and MP t -MP X _ _MP A OM OM OM' M,P, = M,P, m M^ OM 1 " -OM x OM 2 Therefore, (a) The tangent of a negative angle is the negative tan- gent of the corresponding positive angle. (b) The tangent of the supplement of an angle is the negative of the tangent of the angle. 300 PLANE TRIGONOMETRY. [§ 163 Or, tan^° = -tan(-^°); (1) tan = -tan(-0); (2) tan A° = - tan (180° - A 9 ) ; (3) tan0 =- tan(7r-6>); (4) tan A° = - tan (n . 180° - A°) ; (5) tan = — tan (n • 7r — 0) ; (6) tan O • 180° + 4°) = - tan (m • 180° - A°) ; tan (w • 7r + 0) = — tan (m "tt — 0). EXERCISES. 1. Diagram five angles whose terminals are symmetric to the hori- zontal with the terminal for 30° and having tangent oppositely equal to that of 30°. Solve sin 5 = - tan 10 0. 2. Diagram five angles having their terminals symmetric to the ver- tical with the terminal of 135° and with their tangent oppositely equal to that of 135°. Solve tan 6 = - tan 8 6. § 164. Tangents of all Angles are First Quadrant Tangents. It is left for the student to draw conclusions similar to those made for the sine in § 64, and report them. Give also rules for the terminal positions, similar to those in the same article. EXERCISES. Take those in § 64, replacing the word " sine " by " tangent." § 165. Construction of the Terminals for a Given Value of the Tangent. Example. Draw the terminals when tan A is f . Construct on coordinate paper the point (3, 2) and join it with the origin. Draw also the terminal opposite to this line. EXERCISES. Construct the terminals corresponding to the following values of the tangent : f , - §, } , - §, .3, 3.7, 33.7, 0, oo. Construct terminals for table tangents. Measure the angles and compare with table. § 167] THE TANGENT FAMILY. 301 § 166. Some Angles for which the Tangent is readily calculated without the Use of Tables. These are the same as those for which the sines were calcu- lated in § 75. It is left as an exercise for the student to establish the following results, using diagrams : tan 45° = 1; tan 135° = - 1 ; tan 30° = ^; tan 330° = -^; o 3 tan 60° = V3 ; tan 300° = - V3 ; tan 120° = - V3 ; tan 240° = V3 ; tan 150° = - ^P ; tan 210° = ^ ; o 3 tan 225° = 1 ; tan 315° = - 1. EXERCISES. 1. Give the general solution of the equations, tan 2 x = 1 ; tan 2 x = \ ; tan 2 x = 3; tan 2 x = ; tan 2 a: = sin 2 a;; tan 2 a: = 3 cos 2 a: ; 3 tan 2 a: = sec 2 a:. 2. tan 2 a; - H±^l tana: + ^1=0. 4. tana: L_ = 1 _ ^1. 3 3 tana; 3 3. tan 2 a: - 3 ~ ^ 3 tan x --^ = 0. 5. tan 2 a: + 8 tana: + 7 = 0. 3 3 § 167. Value of the Tangent when the Terminal is on a Border Line of the Quadrants. If the point P of the terminal OP is supposed to describe a circle of radius r about 0, then as OP approaches the posi- tion of the initial line to the right, OM approaches the value r, MP while MP approaches the value zero. The value of ——- OM thus approaches the value zero. This is expressed by say- ing tan 0° = 0. Similarly, when OP approaches the initial line to the left of the origin, OM approaches the value — r and MP approaches the value zero. The tangent thus 302 PLANE TRIGONOMETRY. [§167 approaches the value zero when the angle approaches 180°. Thus tan 180° = 0. When OP approaches the upright vertical from the right, MP approaches the value r and OM approaches zero from MP the positive side. Thus — — - becomes infinitely great, posi- tively. If, however, OP approaches the upright position from the left, MP approaches the value r while OM approaches the value zero from the negative side. Thus the tangent becomes infinitely great, negatively. It is thus impossible to determine the value of the tangent for upright positions of the terminal. It is customary to write tan 90° = + oo. And, similarly, tan 270° = — oo. This is a short way of saying that if the terminal is upright, almost, with a slight tilt to the right, the tangent is a very large positive number, while if the terminal is down- right, almost, with a slight passage beyond the position denoted by 270°, the tangent is a very large negative num- ber. Thus, if we look on the terminal as moving, counter- clockwise, about the origin, there is, when *it passes the positions denoted by 90° and 270°, a sudden spring in the value of the tangent from plus infinity to minus infinity. Thus in the function tan x there is what is called a break in continuity of value when x is 90° or 270°, or, in general, when x is (2 n -+- 1)90°, or (2 n + 1)—, n being any positive or negative integer. § 168. Line Picture of the Tangent. MP If in tan = -pr=-f-> OM is taken as unity, tan 6 is MP. Thus, on any scale on which OM is taken to represent 1, MP will represent the value of the tangent of all angles corresponding to the position OP of the terminal. 168] THE TANGENT FAMILY. 303 If we imagine the point P' to describe the unit-circle, and if through each position of P' on the unit-circle we imagine a line OP' prolonged to meet the tangent at M in the point P, we shall have for each such position of P' a line picture of Pi XT X ± 2L ^ ' ^^M^ s - ^ -/ z V- J -**- V ^ -2 - \ 1 -7 \ - -7 -it / u c 3 f L A V 2 s / S^ A **.- ~Z2Z y i\ M Fig. 151. Fig. 152. the corresponding tangent denoted by MP. Thus, tangent of angle M0P{ is MP X ; tangent of angle MOP 2 ' is MP V and so on. When OP- is coincident with OM, the angle is zero, and so is MP. As the angle increases from zero to 90°, the point P f describing the circle counter-clockwise, the point P ascends the line MP, and the tangent increases. When the terminal is almost upright, the point Jr is high up on the line, cor- responding to' tan 90° = + 00, in the preceding section. When OP 1 passes 90°, P passes from high up on the line to loiv down on the line, corresponding to the spring from plus infinity to minus infinity noted in the preceding section, when the angle passes the value 90°. When OP' moves from 90° to 180°, the motion of the cor- responding point, P, on the line is from negative infinity on the line, up to the point M. As OP', goes from 180° to 270°, P goes on up the line from M to plus infinity. There is again the spring from plus infinity to minus infinity as 304 PLANE TRIGONOMETRY. [§168 OP' passes through the position 270°. That is, as OP' passes through 270° the corresponding point P leaps from high up on the line to low down on the line. As OP' goes from 270° back to the starting point, P moves from minus infinity to zero on the line, that is,, from low down on the line up to the point M. The line-diagram teaches clearly the following propositions : (a) tan 6 = — tan(— 0). (5) tan (180° + A°) = tan A*. O) tan (180° -A°)=- tan 4°. (d) tan 0° = = tan 360°. (e) tan 180° = = tan ( - 180°). (/) tan ft . 180° = 0. (#) tan 90° = -f oo (as already agreed on). (K) tan 270° = — oo (as already agreed on). (i) tan (4^-f- 1)90° = oo (as already agreed on). (/) tan (An + 3)90° = — oo (as already agreed on). (6) As the angle passes from 0° to 90°, or from 180° to 270°, the tangent is positive and an increasing function of the angle. As the angle passes from 90° to 180°, or from 270° to 360°, the tangent is negative, but an increasing func- tion of the angle. The tangent is always an increasing func- tion, except on the instantaneous leap from + oo to — oo. (?) The tangent, like the sine and cosine, is a periodic function of the angle, the period being 360°, or, in circu- lar measure, 2 7r. It has also the period tt, since tan 6 = tan (6 -f mr), for all integral values of n. (m) For all angles whose terminals are in the first or third quadrant, the tangent is positive ; for all angles whose ter- minals are in the second or fourth quadrant, the tangent is negative. LABORATORY EXERCISE. Draw a diagram like the preceding, with the radius as one foot. Measure ten angles and tangents. Compare results with the table- tangents for the same angles. Make a graph for the tangent. §169] THE TANGENT FAMILY. 305 § 169. Some Simple Calculations with the Tangent. Since the tangent is the ratio of the ordinate to the abscissa, when any two of the three quantities, ordinate, abscissa, tangent, are given, the third can be calculated. (a) Given ordinate and abscissa. Tangent equals ordinate divided by abscissa. (6) Given abscissa and tangent. Ordinate equals abscissa multiplied by tangent. ( = - cot (2> - 270°); cotl> = - tan (2> - 270°). EXERCISE. 1. Show that the tangent of any angle of any size can be expressed as the tangent or else the cotangent of an angle less than 45°. Use the last six formulas for practice in taking tans and cotans and their logs from the tables, taking B, C, D in the second, third, fourth quadrant, respectively. § 171. Calculations with the Cotangent. By the definition, division by the tangent is multiplication by the cotangent, and vice versa. (1) Cotangent equals abscissa divided by ordinate. (2) Abscissa equals ordinate multiplied by cotangent. (3) Ordinate equals abscissa divided by cotangent. § 171] THE TANGENT FAMILY 311 EXERCISES. Use the set on pages 308, 309, replacing the word "tangent" by "cotangent." Cotangents are looked up in the tables by the last six formulas of § 170. THEORETIC EXAMPLES ON THE COTANGENT. 1. A BM is a straight line at right angles to the straight line MP. If AB is of length a, express MP, BM in terms of a and the cotangents of angles MBP, MAP. 2. Make up a problem (numerical) corresponding to the preceding, involving a church steeple on one side of a river, and two points, at a known distance apart, on the other side of the river, and calculate the height of the steeple and width of the river. Make a diagram to scale, and test by measurement. 3. ABC is a horizontal triangle, right angled at C. AP is a vertical line. CB is of length a. Angles A CP, ABP are , 6. Express A P in terms of a, cf>, 0. 4. Make up a kite problem (numerical) corresponding to Ex. 3, and solve it, with a diagram to scale. Test by measurement. 5. Express the radius of the circle inscribed in a regular polygon of n sides in terms of the side of the polygon and half the central angle in circular measure. 6. If the side of the polygon in Ex. 5 is 34.52 ft., calculate to four figures the radius of the inscribed circle when the number of sides is twenty. Test with a diagram to scale. 7. Express the area of a regular polygon of n sides in terms of its side and half the central angle in circular measure. 8. Calculate the area of the regular polygon of eighteen sides, if each side is 4.32 ft. 9. The shadows of two vertical walls, which are at right angles to each other, and are a v a 2 feet in height, are observed, when the sun is due south, to be b v b 2 feet broad. Show that if a be the sun's altitude above the horizon, and /8 the inclination of the first wall to the meridian, cot* a = l -\ + % and cot/3 = ^ tt l a 2 a< P\ 10. Apply Ex. 9 to a numerical example. Can logarithms be used ? 11. Two stations due south of a leaning tower (leaning toward the north) are at distances a, b from its foot. If a, /3 are the elevations at the top of the tower from these stations, show that the inclination of the tower to the vertical is an angle whose cotangent is b cot a — a cot B b-a 312 PLANE TRIGONOMETRY. 12. Apply Ex. 11 to a numerical problem. Can logarithms be used? Make diagram to scale ; test by measurement. EXAMPLES FROM RELATED STUDIES. 1. Show that if a straight line passes through the origin and is not upright, the coordinates of any point on it are connected by the equation y = tan $ • x, where is the angle of the line with the right-hand z-axis. 2. Show that if the preceding line, instead of passing through the origin, cuts off the distance b from the ?/-axis, y =± tan • x + b. 3. Show that if (x, y) is any point on a straight line and (a/, y'), (x", y") are two fixed points, the tangent of the angle which the line makes with the #-axis is any one of the expressions y-ti y - y" x-x"' y' - y" x — x' x' - x" y' -y y" -y y" - y' or X' — X X" — X X" — X' 4. A particle has impressed on it instantaneously a vertical velocity of 50 ft. per second, and at the same instant a horizontal velocity of 75 ft. per second. What is the direction and what the magnitude of the resultant ? 5. A cylindrical stone tower 20 ft. high and 4 ft. in diameter is tilted, by settling of the earth, until just ready to tumble over. What is the tilt? 6. A cube is held on an inclined plane by friction. To what angle can the plane be tilted before the cube rolls over, when the cube is held in the position giving the easiest solution ? 7. Solve Ex. 6, replacing the cube by an upright cylinder and by a parallelopipedon of given dimensions. 8. A body weighing 100 lb. is suspended by a chain. What horizontal force will pull it 10° from the vertical, and what will then be the pull on the chain? 9. PB, PC are two equal rods, held in a vertical plane, hinged at P, with B, C in a horizontal line, B being tied to C by a string, and P being above the line BC. A weight is set at P. What pull will this give the string ? Solve some numerical examples. 10. If (x, y) is a point at a distance b from the lower end of a straight beam which slides in a vertical plane, one end being on a horizontal plane and the other on a vertical plane, show that x = a cos cf>, y = b sin <£, where a is the distance of (x, y) from the upper end of the beam, and the angle the beam makes with a horizontal iine. When is x 2 + y 2 = a 2 ? CHAPTER IX. SINES AND COSINES, TANGENTS AND COTANGENTS. § 172. Relation of the Tangent and Cotangent to Sine and Cosine. ordinate ordinate modulus sine tangent = — = — -. = — ; — : abscissa abcisssa cosine modulus sin^t cos^l ..cotA= c ™ A . sin A And since sin 2 A + cos 2 ^4 = 1, sin 2 A -j _ 1 COS 2 .4. ' COS 2 Jl d 1 cos 2 A _ 1 sin 2 A sin 2 A (4) is 1 + Uui 2 A = sec 2 A. (5) is 1 + cot 2 A = cosec 2 A ...tan ^ = ^4- (1) (2) (») (4) (5) (6) (7) .•.tan 2 ^ = sec 2 ^l-l, (8) and cot 2 A = cosec 2 A-l. (9) From (3) the sine can be calculated when the cosine is given, or the cosine can be calculated when the sine is given. From (6) the secant (or cosine) when the tangent is given. From (7) the cosecant (or sine) when the cotangent is given. From (8) the tangent when the secant (or cosine) is given. From (9) the cotangent when the cosecant (or sine) is given. 313 314 PLANE TRIGONOMETRY. [§172 Note that since squares appear in (3), (4), (6), (7), (8), (9), the calculated function has two opposite values for a given value of the given function. Explain this on a diagram. EXERCISES. 1. From the sine of 17° 23' in the tables, calculate the cosine and com- pare with the table value. 2. From the cosine of 43° 21' in the tables, calculate the sine and compare with the table. 3. From the sine of 15° 32' in the table, calculate the tangent through the cosecant. 4. From the cos 67° 41', calculate the tangent through the secant. 5. From the tan 75° 12', calculate the cosine through the secant. 6. From the cot 345° 54', calculate the sine through the cosecant. 7. From sin 23° in the table, calculate all its other functions and test such as the tables give. From tan 25° calculate the other functions. § 173. Use of a Diagram in Calculating One Function from Another. The formulas of § 172 are used in calculating one function from another when the given function is in the form of a deci- mal of two or more figures. When the given function is in the form of a common fraction, it is best to construct a diagram corresponding to the given function, and then read the other functions from the diagram. This is particularly true if the parts of the given function are integers which with some other integer form a right-angled triangle. (See the set of such figures in the table on page 145.) Example. Given sin A = f , calculate all other functions. Figure 154 shows the two positions of the terminal for sin A = | , and gives ± 4 as the third side. And by the diagram : §174] SINES, COSINES — TANGENTS, COTANGENTS. 315 cosec A = f ; tan^. = ±f ; cot A = ± f ; cos J. = ± | ; sec J. = ± | ; versin A — \ — cos .A = J or -| ; ►versin A. = 1 — sin A. = |. EXERCISES. Plot the dual position for the ter- minal corresponding to each of the following, and calculate the ungiven functions: I \ 5 5 > \ :> . S y r V ? / \ / .1 « 1. cos A = fg. 2. cos A = - & 3. sin ^4 = f $. 4. sin 4 = - £f 13. versin A = ffi. 14. versin A — f|f . 5. tan A = |f. 6. tan A = - ||. 7. cot 4 = |f 8. cot A = - ff Fig. 154. 9. cosec A = ffo. 10. cosec A = - | f . 11. sec ^4 = ff. 12. sec A = - J$l. 15. coversin A = T 9 7 . 16. coversin A = |f . 17. Use the diagram for some similar examples when the numbers are selected at random, without reference to making a right-angled tri- angle whose sides are integers. § 174. Use of a Diagram in Expressing All the Functions in Terms of Some One Function. (a) All in terms of the sine. Let 8 = sine. Lay out the double terminals where s represents the sine to the modulus 1. Then the abscissa is ± Vl — s 2 . From the diagram (Fig. 155) one has directly from the definitions of the functions : tan = vrr esc = - ; sec 6 = s cot 6 = ±V1-* 2 ±V1 coversin = 1 — s. versin = 1 ± Vl — s 2 316 PLANE TRIGONOMETRY. [§174 EXERCISES. 1. The diagrams shown in Figs. 156-160 indicate, in order, given, (1) cosine, (2) secant, (3) tangent, (4) cotangent, (5) cosecant. The Fig. 156. — Cosine given. Fig. 157. — Secant given. student may express, as above with the sine, all functions in terms of each as given, using the diagrams. Fig. 158. — Tangent given. — 1 Fig. 159. — Cotangent given. My -V5*- I 2. Deduce the expressions found in the preceding exercise and those in terms of the Vo*— 4, sine, all directly from the formulas of § 172, without the use of a diagram. Fig. 160. — Cosecant given. 3. Show by a diagram that sin -1 4. Show by a diagram that sin -1 + VI + x 2 1 = tan -1 x. *= cot -1 X. + VI + z 2 5. Show by a diagram that cos -1 = cot" 1 (x) = tan -1 ( - ) . 6. Show by a diagram that tan -1 = sin- 1 x. + Vl - x 2 7. What changes are made in Exs. 3, 4, 5, 6, if the radical is allowed the negative sign ? 8. If 6 = sec -1 x find from a diagram the values of the other functions. ±vx —y §174] SINES, COSINES — TANGENTS, COTANGENTS. 317 9. If 6 = sin -1 , find from a diagram the values of the other functions. ± vy* + x x 10. If 6 = cos- 1 , find from a diagram the values of the other functions. Vy +x x 11. If 6 = tan -1 — . find from a diagram the values of the other functions. ^V ~ x 12. If A = sin -1 -5 — £=, find, in the best way, the other functions. x* + y*' j> 13. If sin A = m2 + 2mn s how, in the best way, that m 2 + 2wn + 2n 2 ' J ' . , ra 2 + 2 wn tan ^4. = ± 2 mn + 2 n 2 14. Prove sec (tan" 1 a;) = ±vl + a; 2 . 15. Prove sin (cos -1 x) = ± Vl — a: 2 . 16. Prove cos (cot -1 a;) 17. Prove cot (sin -1 x) ±Vl + a: 2 ±VT^x* x 18. Select the cosine of some angle in the table of cosines and calcu- late the other functions and compare results with the table. 19. Do the same as indicated in Ex. 18 for each function given in the table, selecting for each a different angle. EXERCISES CONNECTING TANGENT AND COTANGENT WITH THE OTHER FUNCTIONS. 1. tan A + cot A — sec A cosec A. 2. (tan A + cot ^4) 2 = sec 2 A + cosec 2 A. 3. l + tan^ ==sin ^ +cos ^ t sec A 4. (1 + tan ^4)(1 + cot A) = 2 + sec A cosec A. - cosec A nnB A 5. = cos A. cot A + tan A - 1 — tan A cot .4 — 1 o. = • 1 + tan A cot -4+1 7 1 + tan 2 A _ sin 2 A < 1 + cot^-cos 2 ^" 318 PLANE TRIGONOMETRY. [§175 8. =sin^4 cos .4. cot A + tan A 9. cos 8 A tan A + sin 2 A cos A tan A = sin A. sec A 10. (tan A — sin A ) cosec 8 J. 1 + cos ^4 ,, cot J. — tan A _ cos ^4 + sin A cos J. — sin A cos -4 sin A 12. V ! ~ S ^ n A = sec yl - tan .4. * 1 -f am A 13. - -= sec .4 + tan ^4. sec A — tan A 14. sec A - tan A = j __ 2 sec 4 tan ii + 2 tan 2 ,4. sec ^4 + tan A , c tan J. . cot ^4 A ,1,1 15. + ; = sec^l cosec A +1. 1 — cot A 1 — tan A n c cos A , sin A A , . 16. + = sin A + cos A . 1 — tan A 1 — cot A 17. (sin A + cos A) (cot A + tan A) = sec ^4 + cosec A. 18. cot 4 ^4 + cotM = cosec 4 A — cosec 2 A. 19. sec 2 J. cosec 2 A - tan 2 ^4 + cot 2 A + 2. 20. tan 2 A - sin 2 J. = sin 4 J. sec 2 ^4. 21 cot A cos ^4 _ cot A — cos ^4 cot A + cos J. cot A cos -4 22. cot j + tan g = cot A tang. cot i* + tan A 23 tan ^ + sec A — 1 _ 1 -f sin A # tan A — sec A + 1 cos J. 24. j ~ s ? n 4 = 1 + 2 tan ,4 (tan 4 - sec A). 1 + smA 25. (tan4+cosec£) 2 -(cot5-sec^) 2 =2tan^cot£(cosec4 + sec J B). § 175. The Addition-subtraction Formula for Tangents. tan(,l + i?)= sin(A + i?) (1) V } cos(A + £) • W _ sin J. cos B + cos J. sin B w cos A cos ^ — sin .4 sin 1? §175] SINES, COSINES — TANGENTS, COTANGENTS. 319 Dividing numerator and denominator by cos A cos B, this becomes sin^i sin B s a x»n cos^l cos B tan (A + B) = — — — — (3) 1 _ sin ^- sin -" cosJL cosi? or, ten{A + B)= te\ A+ ^ B . (4) 1 - tan A tan B v J Similarly, an f A _ Tt\ — _ 1 + tan ^ tan B' Ni-B)=^^ (5) cot(^ + B)= cot ^ cot ^- 1 > (6) cot-B + cot^fi v cot (4 - B) = gotBcot^ + l (7 n k ; cotJS-cot^l w EXERCISES. 1. From the values of tan 45° and tan 30° show that tan 75° = 2 + V3 = 3.73205-... 2. Similarly, tan 15° = 2 - V3 = 0.26795-.. . 3. Tan R = \, tan B = f. Find tan (R + B), tan (i2- 5), cot (R + B), cot (R-B), tan (2 22 + £), tan (2 # - B). 4. Take D, Z? as some angles in the tables, and calculate tan (D + B), tan (D — B), cot (D +-B), cot (D — B), and compare results with the tables. 5. * » n * ~ ten ? = ten (s-y) tans, cot x + tan y 6. Prove tan 2 x = 2tan * , and find tan (A + B + C). 1 — tan 2 x 7. Prove cot 2 x = cot * x ~ 1 9 an d fi n d value of cot 3 x. 2 cot a: 8. If tan L — \ and tan B = |, find tan (£ + 5), tan (L - JB) ; cot (L + 5) ; cot (L - B). 9. Show by formula (5) that tan 0° = 0, and by (4) that tan 90° = . 320 PLANE TRIGONOMETRY. [§176 § 176. The Addition-subtraction Formula in Inverse Tangents. If tan A = x and tan B = y, A + B = tan" 1 x 4- tan" 1 y. Thus (4), (5), of the preceding article may be written in the forms tan- 1 a; + tan- 1 2/ = tan- 1 ^^. (3') " 1+xy v J Similarly, if cot A = x and cot B = y (6), (7), take the forms : t cot- 1 aj + cot- 1 2/ = cot 1 ^-^. (6') x + y K J cot- 1 *; - cot- 1 ^ = cot 1 ^^. (7') Re-read what has been said about multiplicity of values of sin -1 x, cos -1 x. EXERCISES. 1. tan" 1 \ + tan- 1 T ^ = tan" 1 1. 2. tan- 1 £ + tan- 1 i = mr + j- 3. t«^*_ *»-*•=* «,,, + ». rc m + n 4 4. tan -1 1 + tan -1 - = tan _1 - — -• 1 - 1 2 1 - 3 * 2 5. tan- 1 1 + tan- 1 f s | cos- 1 f . 6. cos" 1 1 + tan- 1 f = tan" 1 § f 7. 2 cos -1 —— + cot -1 1|+ \ cos -1 ^ has 7r for one of its values. Give V13 also the general value. Test formulas above by tables. 8. sin -1 — - + cot -1 3 has j for one value. Give also the general value. ^ 5 4 9. tan- 1 f + tan- 1 1 - tan" 1 T \ has | for one value. 10. If tan -1 1 + tan -1 x = tan -1 f , find the general value of x. 11. tan- 1 1 + tan- 1 } + tan" 1 1 + tan" 1 \ = J (one value). §177] SINES, COSINES — TANGENTS, COTANGENTS. 321 p — ox p 12. tan- 1 1,1, = tan" 1 - - tan" 1 x. q + px q p -\- ox , p 13. tan- 1 - — i- = tan" 1 L + tan" 1 x. q -px q^ 14. tan -1 — ^— = tan -1 a;? — tan -1 a;. l + *i 1 x — 15. cos- 1 J = 2oot- 1 x. (Set a; = cot 0.) a: 16. sin- 1 -^- = 2 tan- 1 x. (Set a: = tan 0.) 1 + x 2 v 17. sec- 1 ^ } • s 2 cos- 1 a:. (Set a; s cos 0.) 2 a: 2 — 1 v § 177. Tangent of the Double Angle. From tan (4 + 2>) . tan ^ + tan ^ 1 — tan .A tan i? follows, if A = B, 2 tan ^ tan 2 ^1 = 1-tuPA Thus the tangent of any angle is twice the tangent of half of it divided by one minus the square of the tangent of half of it. 2tan| .*. tan x = ; l-tan*f tan8*= 2tan f;f ; 1- tan 2 4 a; to^. J^SJg , etc. 1 — tan 2 2 x LABORATORY EXERCISE. On a ten-inch circle, measure tan 5° and tan 10°, and test the pre- ceding formula. 322 PLANE TRIGONOMETRY. [§ 177 EXERCISES. 1. Test tan 42° 18' in the tables as compared with tan 21° 9'. 2. If tan A = §, find tan 2 A, tan 3 A, and tan 4: A. 3 tan A - tan 3 A 3. Show tan 3 A = 1-3 tan 2 ;1 4. Show tan 4 A = H^A-t^A) l-6tanM + tan 4 4 sin 2 ^4 5. Deduce tan 2 ^4 from tan 2 4 = -— - by using the formulas for cos 2 ,4 J h sin 2 A and cos 2 A in terms of functions of A. A « -r. A Jl — cos^l 1 _ tan 2 - 8# Prove tan ^ = ± yj- 6. Prove cos ^ = 4- 2 ' 1+COB * 1 + tan 2 - ^_ l-cos^ 9. Prove tan 2 - sinA 2 tan - „ ™ .-i 2 ,,*t^ , i sin i 7. Prove smi= j- 10. Prove tan -~ = j 1 + tan 2 - J 2 cos 2 -- 2 2 11. Prove that tan — always has the same sign as sin A. Can this be shown on a diagram ? 12. Prove sin ^ =cot-- versin A 2 14 -o • ^4 /versing 13. Prove sin — = -vl 2*2 14. Prove cotA A + ] = sec2A + tan 2 ,4. cot A — 1 ^4 yl 15. Prove tan — + cot — = 2 cosec ^4. — — 16. Prove cot A -tan v4 =2 cot 2 A. 17. Prove 2 cosec 2 A— tan ^4 + cot A. 18. Prove 2 C0 J A = tan 2 ,4. cotM-1 19. Prove cosec 2 A + cot 2 ^4 = cot ^4. 20. Prove cos ^ + sil ^ _ cos ,4- sin ^ = 2 tan 2 At cos A — sin ^4 cos A + sin yl 21. Prove (1 + tan ,4) cot .4 1 — tan A §178] SINES, COSINES — TANGENTS, COTANGENTS. 323 22 . Prove i + i«»'(4y-^) m i = cosec 2 ^ 1 -tan 2 (45°-^l) sin2A 23. 1±8in2 ^ = tan(45 ±^). cos 2 A 24. tan (45° + 2 J ) + tan (45° -24) = 2sec44. 25. If tan 2 x = 1, what is the general value of a;? § 178. Tangent of the Donble Angle in Inverse Notation. From tan 2 A = - 2 tan f . l-tan 2 ^l follows, if tan A = x and .A = tan -1 a?, 2 /■ or, 2 tan -*x = tan _ * 5 • Test by tables. 1-a? 2 EXERCISES. 1. 2 tan -1 1 = tan -1 V (one value). 2. 2 tan" 1 £ + tan" 1 } + 2 tan" 1 J = J (one value). 3. 4 tan -1 1 - tan -1 J* + tan" 1 JU = ^ (one value). 4 4. 3 tan -1 J + tan" 1 ^ = J - tan" 1 T ^ J5 (one value). 5. tan' 1 Vl + x * ~ 1 = £ tan-* x. * + i a? 6. sec -1 =- = 2 cot -1 2r. a; x 7. cos- 1 (2 x 1 - 1) = 2 cos" 1 x. 8. sin- 1 -?-^- = 2tan- 1 a:. 1+z 2 9. sec -1 = 2 cos -1 x. 2x 2 - 1 10. cos- 1 5— — =2tan" 1 a;. 1 + z 2 324 PLANE TRIGONOMETRY. [§ 179 11. tan" 1 p^^ = S tan" 1 x. 12. sin- 1 (3 a; -4 a: 8 ) = 3 tan- 1 y/l-x* 13. cos- 1 (4 x s - 3 x) = 3 tan x 14. sin" 1 2 a; VI - x 2 = 2 tan" 1 vT^iT 2 15. cos- 1 (2 x 2 - 1) = 2 cot" 1 VI -x* 16. cos" 1 (1-2 a; 2 ) = 2 tan- 1 VI -x 2 § 179. Tangent of the Half-angle. '2tan^ Since tan A = 1 _ tan 2 4' 2 if we let tan A = a and tan — = x, x is gotten by solving the quadratic ax 2 -f- 2 x — a = 0. # = a . ^ M ^_ -l=bVl + tan 2 ^ . . i«iii -oj- — t 5 • 2 tan J If tan A is given, there are two positions of the correspond- ing terminal, 180° apart. There are thus two positions of the terminal of the half-angle, 90° apart. This explains the dual sign. When A is itself given and not by its tangent, the proper sign is to be selected. Thus tan22°= - 1+V:L + tan2 -i5 tan 45° and tan 100° = = 1 ~ Vl + ta " 2 20 ±°- tan 200° §179] SINES, COSINES — TANGENTS, COTANGENTS. 325 The tangent of the half-angle can also be expressed in terms of other functions. It has been proven (§ 139) that 1- cos A=2 sin 2 4, (1) l + cos^l = 2cos 2 ^, (2) A A and sin A = 2 sin — cos — • (3) — — -WJ-^Si w ■••Mi To get (4), divide (1) by (2); to get (5), divide (1) by (3); to get (6), divide (3) by (2). The dual sign in (4) means that if an angle is given by its cosine, there are two positions of the terminal (as in § 106). Thus there are two positions for the half-terminal, and two values of the tangent of the half-angle, oppositely equal. When the angle itself is given, there is only one terminal and the proper sign is to be selected in (4). Thus tan 11° = +Vj- 7 C0S oL and tan 100 ° = - Vt"" 1 1 + cos 22 T l + - cos 200° cos 200 c EXERCISES. 1. Express tan 22^° in radicals. Express the radicals in decimals and calculate tan 22^°, and compare with tables. Do the same for tan ( — 22^°). 2. Express tan 15° in radicals and in decimals, and compare with tables. 3. Express tan 7\° in radicals and in decimals, and compare with tables. Test the four formulas for tan — by tables. A ^ 4. If cos A = .23, find tan — , and explain the dual sign. A ^ A 5. If sec A — 1\, find tan — and tan A. Express — in inverse tangents. 326 PLANE TRIGONOMETRY. [§ 180 6. tan ( 45° + ^.)=J ] + sin A =sec A + ta,nA. \ 2 / * 1 - sin A 7. sec/| + $\ sec (| - $\ = 2 sec 2 ft 8. If sin $+smcf>=a, and cos + cos = b, find the value of tan :— x. 9. If sin J. = ■§, find sin 2 A, cos 2 .4, tan 2 ^4, and tan — m 10. If cos ^4 =Hj nn< l sin 2 ^4, cos 2 ^4, tan 2 ^4, and tan — • 11. If tan A =£§, find tan 2 ^4, sin 2 A , and cos 2 A. 12. If sec .4 = f, find all the functions of 2 A. 13. If cosec J. = \S find all the functions of 2 ^4. 1-tan 2 - 2 14. ^- = cos^4. 1 + tan 2 ^ -4 ^4 15. 1 -f cot ^4 cot — = cosec A cot — • - _ 16 .ta„(| + |)ta„(|-*)=l. 17. tan - + 2 sin 2 - cot a: = sin x. 18. tanf J + -) = sec a: + tan a:. 19. tan 2 |(l + cot 2 ^V=4cosec 2 ^ = 4(1 + cot 2 ,4). § 180. The Tangent in Auxiliary Formulas. When forms like a cos x -f b sin x (§ 148) are to be cal- culated, and logarithms are to be used, we write a cos x + b sin x = Va 2 + & 2 ( — a cos a: -t sin x\ Wa* + b* Va 2 + 6 2 / We may set — a = sin 6 ; — . = cos , where tan = - is more readily calculated than sin or cos <£. .-. aco8x + bsmx= Va 2 + 5 2 sin (x + <£) = Va 2 + 6 2 sin fa + tan" 1 |Y §180] SINES, COSINES — TANGENTS, COTANGENTS. 327 EXERCISES. 1. Find x when 317 sin x + 212 cos x = 321. 2. Find x when 8.314 sin x - 7.215 cos a; = 1.314. 3. Show VB • sin x + cos z = 2 sin ( x + - ] . 4. Show 3 sin x + 4 cos a; = 5 sin f a; + tan -1 - J. 5. Express the preceding results in terms of the cosine and cotangent. 6. Express a cos x — b sin x in terms of the sine and in terms of the cosine. 7. Solve the examples under § 148 when the tangent expresses the auxiliary angle. EXERCISES CONNECTING THE TANGENT AND COTANGENT WITH OTHER FUNCTIONS. x sin A + sin B = ^ A±B ^ ^-^ sin yl — sin 2? 2 2 2 cosyl+cosi? m CQt ^+£ cQt A-B cos i? — cos A 2 2 3 8inil+BinB = tan ii+g cos A + cos 5 2 4 sin yl — sin Z? _ . A + B cos 5 — cos A 2 - sin 7 ^4 — sin 5 .4 ... A 5 - s- ; — - = tan4. cos 7 ^4 + cos 5 ^4 6. sin^+sin3^ =tan2 ^ t cos ^4 + cos 3 A 7. cos2^ + cos2^ ^ 5) ^ _ ^ sin2i*-cos2,4 v y v ' Q sin yl + sin 2 A .A 8. — -=cot— • cos A — cos 2 ^4 2 9. sin5 - 4 - sin3 - 4 = tan^. cos 3 A + cos 5 ^4 10 sin2^ + sin2^ = ta ^ sin 2 ,4 - sin 2 £ v y v y lx cos21?-cos2,4 = (4 _ B) sin 2 5 + sin 2 4 v ' 328 PLANE TRIGONOMETRY. [§ 180 12. si°(^4-2i?) + sin(4£-2,l) A cos(4:A-2B) + 008(45 -2 A) K ' , tan 5A + tan 3 .4 . n 4 A A 13. — !- = 4 cos 2 .4 cos 4 ^4. tan 5 A — tan 3 A 14 cos3^+2cos5^ + cos7^ =cos2 ^_ sin2 ^ tan3 ^ - cos A + 2 cos 3 v4 -f cos 5 ^4 15 s * n -^ 4- sin 3 ^4 + sin 5 J. + sin 7 ^4 _ . ** cos yl + cos 3 A + cos 5 yl + cos 7 ^4 16 s * n fj — sin 5 ^4 + sin 9 A — sin 13 A _ . * * cos A — cos 5 A — cos 9 A + cos 13 ^4 sin 2 ^ 1 + cos 2 ^4 18. sin 2 A , . = cot A. 1 — cos 2 A 19. 1 — cos 2 A A — — = tan 2 A 1 + cos 2 A 20. tan A + cot^4 = 2cosec2yt. 21. tan A — oot A = — 2 cot 2 A. 22. cosec 2 A + cot 2 ^4 = cot A. 23. ° osA -.=**(& ±4) 1 T sin A \ 2 / 24. sec8 ^ -1 -tan8^-cot2, sec 4 ^4 — 1 0( - 1 + tan 2 (45°- ^4) , 25. — ! -v. -f = cosec 2^1. 1- tan 2 (45°-.4) 26. If sin + sin <£ = a, and cos + cos = b, find and . 27. tan(45° + 4) - tan(45° - A) = 2tan2.4. 28> cos.4 ± sin A _ cosA - sin A = 2tan2 ^ t cos J. — sin .4 cos ^4 + sin ^4 29. cot (JL + 15°) - tan (A - 15°) = 4 cos 2 ^ 1 + 2 sin 2 A sin yl — cos A 1 + cos J. + cos 2 ^4 ~" 1 + sin A + cos A ~ '" 2 30. sin^+sin2^ =tKa A. 31. * ± sin ^ ~ cos ^ = tan §180] SINES, COSINES — TANGENTS, COTANGENTS. 329 32 sin (w + 1)4 - sin (n- 1)4 _ A ' cos(n + l)4 + 2cosn4 + cos(n - 1)4 D ~2* „ sin (n + 1) A +2 sin nA + sin (n — 1) A A 33. 2 — — \ 1 = COt-pr- cos (n — 1) A — cos (n + 1) A 2 34. cot- -tan- = 2 cot A. 2 2 35. cot 4 + cot (60° + A ) - cot (60° - 4) = 3 cot 3 4. 36. tan 3 A tan 2 4 tan .4 = tan 3 4 - tan 2 4 - tan4. 2 tan: 1 - tan 2 - 37. sin A = 38. cos A = 1 + tan 2 : 1 + tan 2 |. 39. tan 6° tan 42° tan 66° tan 78° = 1. 40. Given sin (A — x) = cos {A + x), find tan x. 41. Given sin (x + a) + cos (x + a) — sin (x — a) + cos (x — a) , find tan x. 42. Given tan A tan a: = tan 2 (A + x) — tan 2 (4 — x), find cos a\ 43. Given Prove 44. Given Show that m tan {A — x) _ n tan x cos 2 x cos 2 (-4 — x) tan (4 - 2 x) = — - tan A. tan (4 + x) tan (4 — a:) = n + m 1 -2 cos 2 4 l + 2cos24* a: = 30° is a solution. Show 45. Given n sec 2 x tan (4 — x) = m sec 2 (4 — x) tan a\ n sin 2 4 46. Prove tan 2 a: 1 + sin A 1 — sin A m + n cos 2 4 tan 2 f45° + ^V 47. tan (45° + —\ + cot ^45° + ^\ = 2 sec 4 . 48. tan (30° + 4) tan (30°- 1 x_2cos2^4 -1 ; 2cos24+l 49. If sin a; = sin4 sin (B + x), show tanx = sin 4 sin B 1 — sin 4 cos B 50. If m sin B cos (4 — x) — n sin 4 cos (B + ar), find tana;. 330 PLANE TRIGONOMETRY. [§ 181 § 181. Trigonometric Equations involving All the Functions. Process for Solution : Reduce the equation to a single func- tion and solve in the most general manner, using tables if necessary. Sample Example : sin x — tan 2 x, (1) sin 2 x ^ x .-. sma? = — — • (2) cos^a; ••• (1 — sin 2 x) sin x = sin 2 x. (3) .\ sin x (1 — sin 2 x — sin x) = 0. (4) .% sin x = ; (5) or, 1 — sin 2 x — sin x = 0. (6) By (5), x — iwr, or n • 180°. By (6), sin x = ~ 1 ^ V5 = 0.6180, or - (1.6180). St The latter is rejected, since sin x lies between +1, —1. If x = sin" 1 0.6180; x=2n 180° + 38° 10' (tables); or, x = (2 n + 1) 180° - 38° 10'. EXERCISES. Solve in the most general manner : 1. sec 2 x = 4 tan x. 3. 2 cot 2 = cosec 2 0. 2. 4 tan x — cot a: = 3. 4. 4 cos — 3 sec = 2 tan 0. 5. sin 2 - 2 cos + \ = 0. 6. cot 2 + (\/3 + ^) cot + 1 = 0. 7. tan 2 + cot 2 0=2. 8. sec0- l = (\/2-l)tan0. 9. tan 2 0-(l+V3)tan0+V3 = O. 10. 2cos0 = V3cot0. 11. 1.4 tan + 1.7 cot = 2.3. 14. tan 2 + cot 2 = *£- 12. 2134 sin = 1372 tan 0. 15. tan 2 a: + cosec 2 x = 3. 13. 43 tan = 27 sin 2 0. 16. 2 sin 2 + 3cos0 = 0. CHAPTER X. THE TANGENT AND COTANGENT IN THE SOLUTION OF OBLIQUE-ANGLED TRIANGLES. § 182. Given Two Sides and the Included Angle. Since for a solution by sines a pair of opposites must be given, if two sides and the included angle are given, the triangle can- not be solved by sines without splitting the triangle into two right-angled triangles. A solution by the aid of the tan- gent is possible. Giyen b, c, A. sin C c vs J Assume b> c. sin B — sin C _ b — c sin B + sin b + c sin B-0 B + C cos by § 145, COS B or, tan B-0 C .B+C _ . sm __ cot*±^ b-c , b + c 9 b-c b + c Since 2 b + c 2 A + B+ (7=180°, -L> ~\- C QQO ^l 2 2* B+C (*) tan = tan (90°-^= cot (The tangent of an angle is the cotangent of the comple- ment. § 170.) tan B-C b + c 2 331 (D 332 PLANE TRIGONOMETRY. [§182 From formula (X), or from (Y), by logarithms, or by B — O B 4- natural functions, — - — can be found. And since — ^— is known, B= B+C B-0 and 0= 2 2 7 B + B-0 2 2 B and O being thus known, there remains to find a. b or, a = sin A a = sin A sini?' " sinC In a calculation, the values of B, O should be checked. For this purpose there may be used b • sin = c ■ sin B. LOGARITHMS. b= =e sin (7 = = sin B sum sum Check to within 2 in " final " figure. Since a can be found by two sine formulas, that one not used for calculating a may be used as a check on a. Model Example. Given r 6 = 48.3 J b. 4. 6, c, A, with c < b. 2. a, b, C, with a < b. 5. a, c, B, with a > c. 3. b, c, A, with c > b. 6. a, c, Z?, with a < c. 7. What if the two given sides are equal? 8. The student may construct some appropriate examples, solve and test them (§ 77) : (a) With lines to one significant figure and angles reading to the nearest 5°. (b) With lines to two significant figures and angles reading to the nearest half degree. 334 PLANE TRIGONOMETRY. [§ 183 (c) With lines to three significant figures and angles reading to the nearest five minutes. (d) With lines showing four significant figures and angles reading to minutes. (e) With lines showing five significant figures and angles reading to seconds. (/) Six-figured data ; angles to tenths of seconds. (g) Seven -figured data ; angles to hundredths of a second. 9. Show that a can also be calculated from C + B . C + B cos — — — sin ~£ — « = (* + &) _!—=(<._&) C-B v ' . C-B cos sin § 183. To find the Angles of a Triangle by Tangents when the Three Sides are Given. . sin a; bmce tan x = , cos a; we have, from the values of the sines of the half angles of a triangle in § 92 and the values of the cosines in § 127, the fol- lowing values of the tangents of the half angles of a triangle : 2 * s^s - a) (1) 2 ' s(s-b) (2) (3) with their reciprocals for the cotangents of the half angles. For calculation purposes it is best to multiply (1) by under the radical ; (2) by y ; (3) by — — — Then, tan * = -±- .aBSO = JL ; As — a ^ s s — a :?-J_ J O - «X* -&)(«- <0 _ n 2 s-b'v s "s-b 9 tan 2 ~*_* M )(s -£)(*- -(2) tan — = 2 (9) (7) -(3) tan-= 2 (10) The scheme is made before beginning the work, then numbers are entered on the scheme. The logarithms (1), (2), (3), (4) are all looked up before their manipulation begins in (5), etc. The numbers are close enough on the scheme to be added, subtracted, without rewriting. If not, the log (7) may he written on the end of a card and held in turn over (1), (2), (3), and the subtractions be made and written in their appropriate place. Since the tangent solution requires but four logarithms, those of s, s— a, s — b, s — c, it is more convenient for use than 336 PLANE TRIGONOMETRY. [§ 183 the corresponding formulas for sines and cosines, which required six, seven, logarithms. (See §§ 92, 127.) As the tangent, over much of the tables, varies more rapidly than the sine and cosine for equal changes in the angle (notice this in the tables), the tangent formulas are, as a rule, more suitable for calculation than the cosine and sine formulas. This is again taken up in Chapter XI. Be- sides, the log tan will give the angle when it is near 90° as well as when small, even though the tangent is infinite. Checks. A + B + C=180° (1) or a • sin 2? = b • sin A and a • sin C= c • sin J. (2) How close the second check should check we have often had occasion to point out. How close the first check should check depends likewise on the number of significant figures in the data, if they are assumed as measurements and not as exact. When the sides are given to one, or two, or even three, significant figures, and represent measurements, the summa- tion of angles is not an appropriate test. Better, then, test by logarithms of the sine formulas (second test above). It is often said that if a five-place table is used the test should hold to within "a few seconds." And so it should, for exact data. That is absurd, however, on measurements, for no one of the angles may be knowable to minutes. In one-figured data (measurement) the angles cannot be known even to the nearest degree. It cannot be impressed upon the student too often that there is no such measurable triangle as one having its sides as 9, 8, 7. The more accurately one tries in a physical labo- ratory to measure the length of a given line the more will the results differ. One can measure a line roughly and get the same result each time. But if the same line is measured with extreme care each time, the results rarely agree. It is quite absurd, therefore, to say, "if five-placed tables are used, the angles should sum to 180° to within a few sec- §183] TRIANGLES SOLVED BY TANGENTS. 337 onds." It is childish to use a five-place table on one-figured data to the full extent of the table. When a table of more places than the data is used, calculated results should be cut back to the pattern of the data. The severity of a check must in all cases correspond to the character of the data. If one is using a five-place table, with data appropriate to such a table, it is easy to test the closeness with which the summation of the angles should reach 180°. In §§ 22, 24, it is shown that a logarithmic formula like that of the tan- gent of the half-angle of a triangle may be in error, on account of the approximate values of the tangents in the tables, by about 1 in the last place of logarithms. How much this means for the angle will depend upon the size of the angle. The possible error for each angle may be deter- mined by examining the table in its neighborhood. Take the sum for the three angles and double it and we have the allowable error in the check. Consider, for example, an equilateral triangle. The half- angle is 30°. In the neighborhood of 30° in a four-place table a difference of 0.0001 in logarithms of tangents means a difference of about 20" in angles. So in such a triangle, calculated from its sides by tangents of half-angles, the sum of the three angles may differ from 180° by 2 f . The use of a five-place table on data appropriate for the use of such a table to five places, will cut this by the divisor 10 on each angle, making the allowable variation in the check about 12 seconds. How close the check should check has to be determined for each triangle. As a general rule holds for the logarithmic check, it is best to use that. To say, however, that the place of the table used deter- mines how close the check should check, without reference to the character of the data, is to assume the triangle being dealt with a theoretic triangle. Such triangles are of no value to the engineer. 338 PLANE TRIGONOMETRY. [§ 183 Model Example. NUMBERS. a = 15.47 b = 17.39 c= 22.88 2s = 55.74 s = 27.87 s - a = 12.40 s-b = 10.48 r- c= 4.99 LOGARITHMS. s-a= 1.0934 s-5 = 1.0204 s- c = 0.6981 &m = 2.8119 s = 1.4451 r« a = 1.3668 r { = 0.6834 tan 4 =1.5900 2 .-. 4=21°15'(+) tan |=1.6630 .-. - = 24° 43' 2 tan - = 1.9853 2 .-. !Z = 44° 2' Z ^ = 42° 31' . B « 49° 26' (7= 88° 4' 180° 1' At 21° 15', the log difference 0.0001 means about 15". At 24° 43', the log difference 0.0001 means about 20". 90" At 44° 2', the log difference 0.0001 means about ^—. 55 " So the solution may be considered correct. It is clear that the angles may sum to 180° and still the individual angles be incorrect. This is rather unlikely. The use of a five-place table with the data of the foregoing example (representing measurements) could in no wise affect the closeness of the check, since with data to only four sig- nificant figures, a fifth figure in the tangents of angles would § 183] TRIANGLES SOLVED BY TANGENTS. 339 be without significance. However, if we assume the data as exact, representing a theoretic triangle, a five-place table will demand a closer check than a four-place table. See § 77. Logarithmic Test, logarithms. a = 1.1895 1.2403 = 6 nBtm 1.8806 1.8298 = sin A 1.0701 1.0701 They agree too well ! C should also be checked. EXERCISES. 1. The student may select some numerical exercises, solve and test : (a) With sides to one significant figure, calculating angles to the nearest five degrees. (b) With two-figured data ; angles to the nearest half degree. » (c) With three-figured data ; angles to the nearest five minutes. (c?) With four-figured data ; angles to the nearest minute. (e) With five-figured data ; angles to the nearest second. (/) With six-figured data ; angles to tenths of a second. (<7) With seven-figured data ; angles to hundredths of a second. 2. In Fig. 145, line DEC < DAC < BBC. Show from this that b a sin < < tan 0, and - — ^ = 1, when = ; also ^ = 1, when = 0. sin tan 3. In Fig. 145, triangle OEC < sector OAC < triangle OBC. Show from this that sin < < tan 0. 4. Show that (1 + sin $y°™8 = e, when $ = 0. 5. Show that (1 + 3 tan 2 0)2cosecMe ^ g^ w h en = 0. 6. From the series for sin 6 and cos 0, find a series for tan 0. 7. Given r sin = 21.71, r cos = 31.41, find r, <£. 8. Given r cos <£ cos 6 = 3.172, r sin cos = 2.113, r sin = 3.121, find r, 0, cji. 9. Given 3121 tan <£ + 2171 cot <£ = 3141, find <£. 10. Given sin (<£ + 16° 17') = 0.3142 sin , find <£. 11. Given tan (<£ + 31°) = 23 tan <£, find <£. 12. Change va 2 — x 2 and va 2 + x 2 to trigonometric forms. CHAPTER XI. GENERAL REVIEW ON THE SOLUTION OF TRIANGLES. LIST OF FORMULAS TO MEMORIZE. § 184. Solution of Triangles. It is possible, as we have seen, to solve any sort of triangle with the sine alone. Similarly, we might use the cosine alone, or the tangent alone. In fact, since all the trigo- nometric functions are expressible in terms of any one of them, any one of them might be used alone for the solution of triangles. EXERCISES. 1. Show that it is possible to solve all possible cases in right-angled triangles by using : (1) The sine alone. (2) The cosine alone. (3) The tangent alone. 2. Show that for triangles not right-angled it is always possible, either directly, or by breaking the triangle into two right-angled triangles, to solve every case : (1) By the sine alone. (2) By the cosine alone. (3) By the tangent alone. § 185. The Best Method of Solving Triangles. What is the best method in any particular case involves two considerations : (a) reaching an answer with the least labor, (/3) reaching an answer the most accurate possible from the data. Evidently, if the observer is the computer, 340 § 185] GENERAL REVIEW. 341 he should know beforehand what method of computation is contemplated and avoid observations leading to (a) Angles near 90° to be computed from the sine. (5) Small angles to be computed from the cosine. EXERCISES. 1. Consider all possible data in right-angled triangles, and determine what formulas will give answers with the least labor. 2. Do the same with triangles not right-angled. 3. In case the three sides of a triangle are given and all three angles are desired, how many logarithms are required if (i) sines are used, (ii) if cosines are- used, (iii) if tangents are used? 4. When is it better to calculate by the natural functions rather than by logarithms? Investigate as to whether practical engineers of your acquaintance use logarithms or natural functions in ordinary calculations. It is frequently stated that an angle can be calculated better from the log tan than from the log sin or log cos. The table on page 342 will show that, with a five-place logarithmic table, (i) Angles less than about 25° can be calculated as ac- curately from log sin as from log tan. (ii) Angles between about 65° and 90° can be calculated as accurately from log cos as from log tan. (iii) Angles between 25° and 65° can be calculated more accurately from the log tan than from log sin or log cos. (iv) An angle of any size can be calculated as accurately from log tan as from log sin or log cos. (v) Small angles cannot be calculated at all accurately from the log cos. (vi) Angles near 90° cannot be calculated at all accurately from log sin. (vii) Log tan (log cot) varies the same way at both ends of the table, so, therefore, when an angle is near 90° its value is given as readily by log tan as if it were near 0°, even though tan 90° = qo. 342 PLANE TRIGONOMETRY. [§185 A° Sin Tan AND Cot Cos A° d 60" d d 60" d d 60 d" 1° 724 0".08 724 0".08 0.22 273" 89° 2° 362 0.17 362 0.17 0.44 136 88° 3° 241 0.25 242 0.25 0.66 91 87° 4° 181 0.33 182 0.33 0.89 67 86° 5° 144 0.42 145 0.41 1.1 55 85° 10° 72 0.8 74 0.8 2 30 80° 15° 47 1.3 50 1.2 3 20 75° 20° 35 1.7 40 1.5 5 12 70° 25° 27 2.2 33 1.8 6 10 65° 30° 22 2.7 29 2.1 7 9 60° 35° 18 3.3 27 2.2 9 7 55° 40° 15 4.0 26 2.3 11 5 50° 45° 13 4.6 25 2.4 13 5 45° 60" corresponds to — - The columns headed d are the differences in the logarithms for 1' in the neighborhood of the angles given at the side, and since sin A = cos (90 — A), and tan A = cot (90 — A), a difference like 724 for sine, opposite 1°, is the difference for cos 89°, etc. Since d is for 1', a difference of 1 in logarithms The table gives, at various parts of a five-place logarithm table, the difference in seconds corre- sponding to a difference of 1 in logarithms. In the neighborhood of 1° a difference of 1 in logarithms corresponds to a difference of about 0.08" for angles obtained from log sin and for log tan, and to about 273" for angles obtained from log cos. Such an angle can, therefore, be determined with equal accuracy from log sin and log tan, but cannot be deter- mined accurately from log cos, where any interpolation is necessary. 60" Note that up to about 25° the values of are nearly the same for sine as for tangent and cotangent. Therefore angles less than about 25° can be determined almost as accurately from the log sin as from log tan, and, §186] GENERAL REVIEW. 343 consequently, those between 65° and 90° can be determined almost as accurately from the log cos as from log tan. It is thus clear that the angle can be determined at any place as accurately from the log tan as from the log sin or log cos, while between 25° and 65° it can be determined more accurately from the log tan than from either log sin or log cos, since there a difference of 1 in logarithms cor- responds to a smaller difference for tangents than for sines or cosines. In a table of six, seven, ten places the advantage in calcu- lating from the tangent would appear stronger than it does here with a five-place table. § 186. Angles determined when they are Small from Log Sin and Log Tan. Large tables are arranged to give such angles from the formulas log sin x — log x" + S log tan x = log x" + T. See Gauss's or Hussey's Tables. The student may be assigned some exercises. EXERCISES. 1. Explain from the formula tan x = ^LE w hy at the beginning of COS X the table the difference for log sin is about the same as for log tan, and near 90° the difference for log tan is about the same as for log cos. 2. Explain why the difference for log tan is everywhere the same as for log cot. 3. Explain why the differences for log sin decrease from 0° to 90°. 4. Explain why the differences for log tan and log cot decrease from 0° to 45°, where they are least, and then increase to 90°, being the same at equal distances from 0° and 90°, and reconcile this with tan 90°=oo and log tan 90° = go, and explain why it is, in a table running to 45° with function and co-function, like Gauss's or Hussey's, an angle near 89° can be determined as accurately from log tan as an angle near 1°. 344 PLANE TRIGONOMETRY. [§186 WRITTEN REVIEW EXERCISE. 1. Make out a statement for the possible data in right-angled tri- angles, with the formulas best adapted for calculation in each special case, with suggestions as to what should be done in exceptional cases, when certain angles or lines are small. 2. Do the same for triangles not right-angled. The following general conclusions should appear : (a) In right-angled triangles : (i) The cosine (secant) connects the hypothenuse and the side bordering the angle, (ii) The sine (cosecant) connects the hypothenuse and the side opposite the angle, (iii) The tangent (cotangent) connects the two sides. (5) In triangles not right-angled : (i) When a pair of opposites are given, use the sine. (ii) When three sides are given, use tangents of half- angles, (iii) When two sides and the included angle are given, get the two remaining angles by tangents (§ 172), and the third side by sines. 3. Make a tabulated statement of the formulas and processes of trigo- nometry which should be committed to memory. (The extent of this will depend upon the career in view by the class, and the outline is left to the teacher.) 4. Show that c = a cos 2? + b cos A, (1), § 125, becomes, on making a = sin A — - — , b = sin B • - — — , sin C sin C since sin C = sin (A +5), sin (A + B) = sin A cos B + cos A sin B (§ 136). Show that this is a general proof. Show also, from Fig. 127, that -cosc^+n;-/- 62 2 ah .\ cos (A + B) = cos A cos B Show that this proof is general. x _y_-t h a b a smA sin B. CHAPTER XII. THE QUANTITY V^~l IN TRIGONOMETRY. § 187. The Argand Diagram. X ^^~ * >^. 41^ N /_ >s t^^ $ - ^ ^ 1 Q_ . " -A L _i _ A 7 ^ 7 S s J? ^^^^ r^ Fig. 163. If OA, 00 (Fig. 163) are equal in length but opposite in direction, 0O=-0A = (-l)-0A. Thus multiplying a line by (—1) is equivalent to a turn through 180°. Let i (the initial letter of imaginary) stand for V — 1. Then w = -l. ... 00=(-l)'OA = i-i-OA. Thus, multiplying twice by i is equivalent to a turn through 180°. Consequently, a reasonable interpretation of multiply- ing once by i is a turn through 90°. .-. i-OA=OB, i.i-0A=(-l).0A = 0O, i.i'i'OA = (i'i)-(i-OA) = (-l)-OB=OD, i-i-i'i-OA=(i.i)>(i'i).OA = (-l)-(-l)-OA = OA Thus, if numbers expressed in terms of the unit (+1) are laid out to the right on a horizontal line, those in terms of the unit ( — 1) are to the left on the same line, while those in terms of the unit (+ 1) are laid out on the upright vertical and those in terms of the unit (— i) are on the downright vertical. 345 346 PLANE TRIGONOMETRY. [§187 Numbers of the form a + bi, where a, b are in terms of the units (1), (— 1), one or both, as 5 + 5t\ («) -5 + 5*, (/3) -5-5i, (7) 5-5*, (S) are called complex numbers. They are plotted by laying out a, according to sign, from the vertical axis along the horizontal axis, and then laying out 5, according to sign, from the ter- minal of a, parallel to the vertical axis. The points P v P 2 , P 3 , P 4 (in Fig. 164) represent the num- bers (a), (/3), (7), (8) above, a side of one of the small squares being taken as a unit. When a, b are given all real values, positive, negative, ra- tional, irrational, from zero to infinity, the expression a -f bi is a varying quantity, and is repre- sented, in a one to one correspondence, by the points of the plane, each point representing a number and each number represented by a point. Such a diagram-representation of a + bi is called the Argand Diagram, from the French mathematician who first made the use of it prominent, though priority of suggestion is accredited to Gauss. Fig. 164. EXERCISES. On coordinate paper plot the numbers : ±1±»; ±2±4t; ±3±2i; ±-L±-Lt; ±i±^»; a + 5»; V2 V2 2 x + yi ; ± cos 25° ± t sin 25° ; ± cos 30° ± t sin 30° ; ± cos 100° ± • sin 100° ; ± cos 190° ± i sin 190° ; ± cos 280° ± i sin 280° ; 2 (cos 30° + i sin 30°) ; 3 (cos 120° + i sin 120°) ; 5 (cos 103° + i sin 103°) V2 (cos 315° + i sin 315°) ; cos 797° + i sin 797°. 188] THE FOUR UNITS. 347 § 188. The Addition (Subtraction) of Complex Numbers. (a + hi) + (c + di) = (a + c) + (b + d>'. Pj is a + 6i, if Oil^ = a, and Tlf^ = 6, (Fig. 165) and P 2 is c -f- cfa', if OM 2 = c, and ifcf 2 P 2 = d. If the parallelogram on OP v OP 2 is completed as in Fig. 165, evidently, P 3 is (a + c) + (b + <*>', with Oif 3 = a + *, and M Z P 3 = b+d. Thus, to add P 2 to P 2 , start with P v and draw a line, P X P Z , parallel and equal to OP 2 (and in the same direction). The point reached, P 3 , represents the sum, P 1 -f P 2 . ii i / > r< ' :i .- -' [' / •-' -' , / r> / r / - . n , J* / - I - ^ ■ £. j ( ) v. u ' / " Fig. 165. Fig. 166. Similarly, since subtraction is the reverse operation to addition, to subtract P 2 from Pj (Fig. 166), start with P v and lay out P^ s parallel and equal to P 2 (not 0P 2 ). Thus, points are added (subtracted) as forces in mechanics. EXERCISES. The teacher may select some examples for plotting additions, sub- tractions, making the selection so that pairs of points fall in the same quadrant, and in different quadrants (all possible combinations). Use coordinate paper (ten by ten to the inch will be found convenient). 348 PLANE TRIGONOMETRY. [§189 IT ^* ^^ T^ 7 /vf - 2 .1/ \ . _r - —\r -*\ 7 -A V X- 7 n M , I * ■ I L i \ / s z ^ ^ Fig. 167. § 189. Points on the Unit-circle, and on Any Circle. The point P x in the unit-circle evidently represents cos 6 + i sin 6. Thus, if 6 is given all values from to 2 7r, the expression cos 6 + i sin will represent all points on the unit- circle. Similarly, if a, b vary subject to the condition Va 2 + 6 2 remaining constant, the expression a + bi will represent all points on the circle of radius Va 2 -f b 2 . If a, 5 are constants, a + bi represents a point at a distance Va 2 + b 2 from the origin and in the direction tan -1 -. Thus a complex number depends upon two things, the distance and the direction of its representing point from the origin. The former is called the modulus of the number, also its absolute value. The latter is called the ampli- tude, or angle. For the amplitude, the smallest positive angle less than 360° locating the modulus is taken. a + bi = Va 2 + _ __L Z*~ * 71 / X°jt^ Z ^ v J~ ■*' $ ~X ^L c t ^ $£-* - I a . ' - A 1 _r y i L ^ ^ 5*. ,* 7 \ Fig. 168. Va 2 + f f Va 2 + & 2 > = Va 2 + b 2 (cos (9 + * sin 0). Thus, a 4- bi has two factors, a length factor and a direct- ing factor. The radical, + Va 2 + b 2 = r, is the modulus, or length factor. The factor, cos + i sin 0, is the directing factor. The angle is the amplitude. The radical, + Va 2 4- £> 2 , is an absolute number, or a num- ber in terms of the unit (+1). §190] THE FOUR UNITS. 349 Since a 4- hi = (cos 6 4- i sin 0) • Va 2 + 6 2 , and since the first factor rep- resents a point on the unit-circle, we may look upon the point P, represent- ing a + bi, as derived from the corre- sponding point P 1 on the unit-circle by multiplication by the absolute number + Va 2 + b 2 . Thus with • 7 S \&f r >t rf - e """" 5k " ^ \ ± / z y M^~p :M:::^:::: -t s::::z: j_: _^ _:=-s: zL_ mS^:::^::: Fig. 169. a + bi = (cos + i sin 0) • Va 2 4- 6 2 and c + di = (cos <£ 4- i sin <£) • Vc 2 4- d 2 , (a 4- bi) (c 4- ^0 = {(cos 6 4- » sin 0)(cos <£ + t sin <£)| • { Va 2 4 6 2 • V^ + d 2 j. Hence the multiplication of any two complex numbers can be reduced to the multiplication of two numbers (points) on the unit circle, together with the multiplication of two abso- lute numbers as in arithmetic. § 190. Multiplication (Division) of Points on the Unit-circle. (cos 6 l 4- i sin 0j)(cos 6 2 4- i sin # 2 ) = (cos 1 cos 2 — sin 1 sin # 2 ) - 4- 1 (sin 0j cos 2 4- cos 1 sin 2 ) = cos (^ + 2 ) 4 i sin (0 X 4- 2 )- or, P x P t = P z . Thus any two points on the unit- circle are multiplied by adding (alge- braically') their angles. Since division is the reverse operation of multiplication, P 1 on the unit-circle is divided by P 2 on the same circle by subtracting (algebraically) the angle of P 2 from that of P v Similarly, the point representing the product of three such points has its angle equal to the algebraic sum of the angles of the factors. Cor. : In general, (cos 4- i sin 0) n = cos n6 + i sin n 0, when n is a positive integer. IE -■i*7' - - == - ,N "S- /\ a t-fl /L \ -L ^ ^ :::s-:::::;-z::: ___S,----,,Z Fig. 170. 350 PLANE TRIGONOMETRY. [§ 191 EXERCISES. 1. Prove by radicals and direct multiplication that (cos 30° + i sin 30°) 2 = cos 60° + i sin 60°, and that (cos 60° + i sin 60°) 2 = cos 120° + i sin 120°. 2. Prove by dirept multiplication, using the tables, that (cos 10° -f i sin 10°) (cos 20° + i sin 20°) = cos 30° + i sin 30°. 3. Prove by using radicals, with direct multiplication, that (cos 30° + i sin 30°) 3 = i ; (cos 120° + i sin 120°) 3 = 1 ; (cos 60° + i sin 60°) 3 = - 1 ; (cos 300° + i sin 300°) 3 = - 1. 4. Prove by using the tables that (cos 36° + /sin 36°) 2 = cos 72° + i sin 72°. 5. Prove by using radicals, with direct multiplication, that (cos 45°+ i sin 45°)*= - 1 ; (cos 135° + i sin 135°) 4 = - 1 ; (cos 225° + i sin 225°) 4 = - 1 ; (cos 315° + i sin 315°) 4 = - 1. 6. Use a diagram (Groat's coordinate paper is convenient) to solve the preceding examples. 7. Prove directly, and by diagram, = cos 6 + i sin 6, cos 6 — i sin 6 and show that if P x - P% = 1, P v P 2 are symmetric to the horizontal axis. § 191. Multiplication of Points not on the Unit-circle. This may be carried out in three ways, (a) Direct multiplication. Example : 2 + 3 i 4-5t 8 + 2i+ 15 = 23 + 2i. This method is suited to the case when a, b in a + hi are small. (6) Put each number in the form of its factors (modulus and directing factor) . Then the product is evidently the product of the moduli into the directing factor of the product, whose angle, as for points on the unit-circle, is the algebraic sum of the angles of the factors. It is necessary to use the tables, except in those few special cases when the angles are such that their sines and cosines are known without use of the tables, as 30°, 45°, etc. §191] THE FOUR UNITS. 351 (i) Example with tables not necessary. 1 + i = V2(-^ + — zj = V2(cos 45° + i sin 45°). V3 + i = 2^— +| *)= 2(cos 30° + i sin 30°). ... (l + 0(V3 + = 2 V2(cos 75° + t sin 75°). The sine, cosine of 75° may now be taken from the tables, or expressed in radicals. (ii) Example in which tables are advisable. (375 + 274 0(432 + 548 0. 375 + 274 i = V(375) 2 + (274) 2 (cos A + i sin A), where A = tan" 1 §£|. A is best found by the use of logarithms, and the value of the modulus from a table of squares. The second factor is treated in the same way. The product is then r x r 2 J cos ( A + B) + i sin (A + B) \ . The student may complete calculations. (c) Pictorial multiplication. Since the modulus of the product of two factors is the product of the moduli of the factors, and the angle is the sum of the angles, constructive mul- tiplication is readily carried out by the use of similar triangles. Join one of the points, as P t (Fig. 171), representing one of the factors of the product, to the unit point, A. Construct the triangle P 2 0P B similar to the triangle A0P v by making the angle P 2 0P s equal to the angle A0P x and the angle 0P 2 P s equal to the angle 0AP v P 3 represents the product P x -P & Jv l\ *- V * V -i -hp+- T ' - f / p Xr' *dx 17 t ^7 i j w / t t *' t J-^.*- 4 (^ 7 O -A- Fig. 171. P 2 , for r 2 :1. 352 PLANE TRIGONOMETRY. [§ 191 .«. r 3 = r x r v and the angle of P 3 is the sum of those of P v P 2 . A scale of equal parts to measure 0P 3 , a protractor to meas- ure the angle AOP z , and the tables, will give an approximate numerical value for P 3 , if it is desired. Groat's paper can be used very advantageously when the points are located by polar coordinates or by forms like r(cos 6 + i sin 0). EXERCISES. 1. Find the product of (1 4 i) ( - 1 - i) ; (3 + 4 i) (5 + 2 i) ; (1 + (-V3 + 0; (l+V2 + 0(l-V2-i); (21 + 37 i) (64-18 i) ; (273 + 564 i) (613 + 515 i). 2 Simplify ( cos ^ ~ j sinfl) 10 , (cos a + i sin ct) (cos /? + i sin / ?) (cos + i sin 0) 12 ' (cos y -\- i sin y) (cos 8 + i sin 8) (cos 2 - t sin 2 fl) 7 (cos 3 6 + i sin 3 fl)- 5 (cos 4 + i sin 4 0) 12 (cos 5 - i sin 5 6y* Hint: Make use of Ex. 7, § 190. § 192. Division of Points not on the Unit-circle. (a) Since (a + bi)(a — bi) = a 2 4- 5 2 , division by a + fo' ^s multiplication by — — (as in case (a) of the preceding section). Example: §±J" ( 2 + 3 0( 8 ~ 4 Q = ™±i 3 + 4 i 25 25 This is the best method when a, b are small. (5) Ify using the directing angles of the numbers. 1 1 cos 2 + sin 2 1 r(cos 6 +- i sin 0) r cos + i sin r (cos 6 — i sin 0) = i(cos (- 0) + t sin (- 0)). .-. £-±4? = ^ * ( cos *i + * sin ^i)(cos (- 2 ) 4- i sin(- 2 )) c 4- di r 2 = ^{008(^-^)4^8^(^-^)1, where r v r 2 , are the moduli of a 4 fo\ c + di; 6 V 2 , their directing angles. § 192] THE FOUR UNITS. 353 Thus the modulus of the quotient is the quotient of the moduli, and the directing angle of the quotient is that of the dividend minus that of the divisor. Two cases may arise : (1) The directing angles may be apparent without the use of tables, as in V3 (?+!) V3 + t __ V 2 2 J 2 cos 30° + i sin 30° 1 + i ~ A /g/ 1 1 A" Vl ' cos 45° + t sin 45° if— +— <1 W2 V2 ; = V2 • (cos 15° - 1 sin 15°), and the final numerical value can here be obtained either with or without the use of tables. Such examples are of infrequent occurrence, when not made to order. (2) Tables advisable. 213 + 315 % = V(213) 2 + (315) 2 cos x + i sin x 314 + 426 i V(314) 2 + (426) 2 cos# + isiny' where tan x = f if , and tan y = |||. x, y may be found from tables and the work carried out as above. Unless a table of squares is at hand to obtain the values of the radicals, there is no advantage in this method over the direct process of (a). (c) Pictorial division. This is carried out similarly to pictorial multiplication. Instead of advancing the terminal of the dividend let it retro- grade (algebraically) by the angle of the divisor. The tri- angle P s OP 2 (Fig. 172) is made similar to IOP v by turning OP 2 back (algebraically) an angle equal to IOP v and making at P 2 the angle OP 2 P 3 equal to IP 1 (if rectangular paper is p used). Then P 3 represents -^, for the angle of P s is that of i r 1 P 2 minus that of P v and, from similar triangles, — = — , or 354 PLANE TRIGONOMETRY. [§192 r 3 = -^. In the diagram P 2 is (14, 50°), P x is (7, 15°), I r i (the unit-point) is (5, 0°); so P 3 is (10, 35°). Fig. 172. EXERCISES. 1. Divide 1 4- i by — 1 + V— 3 by the first and second methods, without using the tables. Carry out the division also pictorially. 2. Divide 1+Vdi bj -1-V3i, as in Ex. 1. 3. Divide 213 + 321 i by 324 - 245 i. 4. Find the quotient of any point by its opposite point (the point symmetric to the given point with reference to the origin). What is the result when the points are on the unit-circle ? Find also the product of such pairs of points. 5. Carry out pictorial division on pairs of points selected with variety of quadrantal position, using Groat's paper. Pictorial multiplication. § 193] THE FOUR UNITS. 355 § 193. Integral Roots of Numbers (Points) on the Unit-circle. If P v _P 2 , P 3 , three points on the unit-circle whose angles are 4 4 + 120 °> 4 + 2 40°, be cubed > there will result the single point whose angle is A. Thus, there are three cube roots for any point on the unit-circle, and they are 120° apart at the vertices of an equilateral triangle. It is also easily apparent that there are only three cube roots, for if the angle A locates a terminal, the same terminal is located by A + n • 360°, and since any number divided by three can leave only the remainders 0, 1, 2, the integer n above can have only the three forms 3m, 3 m 4- 1, 3m + 2, where m is an integer. Thus — ^^ will be 4 + ™<' 360, or 4 + m • 360° + 120°, or 4 + m • 360° + 240°. o o 3 These angles locate only the three terminals located by 4,4 + 120 °> 4 + 240 °> as above. Thus there are three > and 3 3 o only three, cube roots for a point on the unit-circle. Similarly, if the four points on the unit-circle with the angles 4 4 + 90 °> 4 + 180 °> 4 + 2 ™° are raised to the 4 4 4 4 fourth power, there will result the single point whose angle is A. We have thus four, and only four, fourth roots for the point whose angle is A. And, in general, the wth power of points whose angles are A A 360° A 360° A , ^360° , . _, _ + , _ + 2 , .- - + (n-l) , («) n n n n n 7i v n will be the single point whose angle is A. And these points are the n nth roots of the point whose angle is A. The argu- ment for only three cube roots holds for only n nth roots, for an nth of the angle locating a given terminal, or, , can have only the n values (a) above plus multiples of 360°, locating only n different terminals, since r divided by n can leave only n different remainders. 356 PLANE TRIGONOMETRY. [§194 § 194. Integral Roots of Positive Unity (or Solution of x n - 1 = 0). This is the special case of the preceding paragraph when ^ = 0°. Thus, the three cube roots of unity are located by 0°, 120°, 240°, and are 1, cos 120° + i sin 120°, cos 240° + i sin 240°, 1 1 . VS. 1 V3. rm < 17 o x The four fourth roots of unity are located by 0°, 90°, 180°, 270°, and are 1, t, - 1, - u (Fig. 174.) The five fifth roots of unity are located by 0°, 72°, 144°, 216°, 288°, 432°, and are 1, cos 72° + * sin 72°, etc., and may be calculated by the tables. (Fig. 175.) in 7f •k"" ""*>« /4£*. X y2* ^* V 1 4 V -a. X 4 -5 - 4 Iw A X tzt zz l \4?-' -/ 122- y 2^ — ' ±-££S* i. I ^■?Ps j 7 ' **~ "V^k. T i x ^ ~~l -*~ '- K -it A ii \z -A \ ^ ZCt \ *z Sj i^lZJL*- ^iz:^ 7 <"T~T' J? f "v -,^ 1 ^ / % jL*_ v i N££ ■- i J?v J ^ A -/ § shown by Fig. 177. V2 V2 358 PLANE TRIGONOMETRY. [§ 195 Similarly, the n nth roots of — 1 are the n values which cos 2r + 1 180° + i sin 2r + 1 180° can take when r is 0, 1, 2, n n 3, "-n — 1, these being the only values such an expression can take when r takes all integral values from — oo to + oo. EXERCISES. 1. Solve by algebraic process the equations x 3 + 1 = 0, x* + 1 = 0, and compare the results with the pictorial solution. 2. Solve pictorially the equations x 2 + 1 =0, x 5 + 1 = 0, x 6 + 1 = 0, x 1 + 1 = 0, x 8 + 1 = 0, using the tables when necessary to get the numeri- cal solution. 3. Show pictorially that each root found in the preceding cases is the appropriate root. 4. State and prove for roots of negative unity propositions correspond- ing to Exs. 4-8 inclusive, of the preceding section. § 196. Integral Roots of Any Number, a + bi. Set the given number in the form of modulus times direct- ing factor, the latter being taken in its general form. Then an nth. root is the ordinary nth root of the modulus times any one of the n nth roots of the directing factor, found as already given for points on the unit-circle. a + bi = VW+V* \ cos (2 r • 180° + A) -f * sin (2 r - 180° + A)\, where A is the smallest positive angle locating a -h bi, a\,2,i2^ 2r-180° + ^ •• 2r -180° + A\n and \ (a 2 +b 2 )2« cos ~ p % sin ! \ [ . n n } = a 4- bi. The n nth roots of a + bi are the n different values which the bracketed expression can take when r is given the values 0, 1, 2, 3, ••• n — 1. Thus the roots for points on any circle with radius R are located on the same terminal lines as the corresponding roots for corresponding points on the unit-circle, and all lie on the circle whose radius is the ordinary required root of R. The cube roots of 8 are located on the circle of radius 2, as are the cube roots of unity on the unit-circle. §197] THE FOUR UNITS. 359 EXERCISES. 1. Locate the three cube roots of — 8 ; the four fourth roots of 16 and — 16 ; the five fifth roots of 32 and — 32. Give the numerical values of these roots. 2. Use the tables to calculate the six sixth roots of 64 (cos 234° + i sin 234°) ; the three cube roots of 213 + 432 i. § 197. De Moivre's Theorem in General. The multiplications, divisions, powers, and roots, given in the preceding paragraphs are all special cases of what is known as De Moivre's theorem, which has two forms : (a) (cos a -f- i sin a) (cos /3 + i sin /3) (cos 7 -f i sin 7) ( ) ()••• = cos(« +/3 + 7 + •••)+* sin O + £+ 7+--). (5) (cos -f- i sin 6) n = cos nQ -f- i sin n6. Form (a) is only the multiplication of complex numbers on the unit-circle (§ 190). Form (6) can be proven readily for all real values of w, — integers, fractions, irrational numbers, — positive, negative. We shall not consider the case when n is itself complex. (1) n a positive integer. The proof is the generalization of the multiplication in § 190, for n factors all equal. (2) n a negative integer. Let n = — m, so that m is a positive integer. Then 1 (cos + i sin 0) n = (cos 6 + i sin 0)-™ = (cos 6 + i sin 0) m cos 2 m6 -f- sin 2 mQ cos mO + i sin m cos m0 + i sin m6 = cos m6 — i sin m6 = cos ( — rn)6 + i sin ( — ni) = cos n6 -J- i sin nd. (3) n a positive common fraction with numerator 1, as 1 n= — 9. In this case, instead of writing \ 6 (cos -f i sin 0)? = cos - + i sin -, 360 PLANE TRIGONOMETRY. [§ 197 the general angle 2 rir + 6 locating the same terminal that 6 does should be taken, since in this case (cos 6 + i sin 6y is A A ^-valued, while cos - -f- i sin - is single-valued. By (1), * » (cos 2r7r + 6 + i sin 2r7r + <9 Y = cos (2 rvr + 0) + e sin (2 rir + 6) = cos 6 + » sin 0. 2nr + . . 2r7r + i9 . . i .-. cos h z sm = (cos v + i sin 0)' 7 . The left-hand member can take q, and only 0-M'sin|?# = (cos 6 + i sin fl)? .*. cos — + * sin ±— = (cos 6 + i sm 0)?. The ^-values which the left-hand member can take when r = 0, 1, 2, 3 •••# — 1, are the #- values of the expression p (cos -H sin 0)*. (6) W a negative common fraction. The proof follows from (5) as (2) from (1). (• § 197] THE FOUR UNITS. 361 (7) n an irrational number. An irrational number like 3.14159 •••, can be looked upon as the limit toward which the numbers 3, 3.1, 3.14, 3.141, 3.1415, etc., are approaching. Thus, by taking more and more figures of the irrational number a common fraction can be found from which the irrational number differs by an amount as small as we please. Between two such fractions, one above the irrational number and one below it, lies any irrational number. The theorem holds for the common fractions, and thus in the limit may be proved to hold for the irrational number. EXERCISES. (A "cyclic group.") Prove by direct multiplication, and also on Groat's polar coordinate paper, the following : 1. Given the six roots of x 6 = 1 : 1 ; cos - + i sin - ; cos — + i sin ?; cos^ + isin^; 3 3 3 3 cos 7r + i sin it ; cos -£ + i sin -^j cos -^ + i sin -^ • O o o o (a) Show that the product of any two is a third one of the set. (b) Show that ( cos - + i sin ^ ) where x = 0, 1, 2, 3, 4, 5 are the six given numbers. Show that the same is true of cos h i sin Is it true of any other one of the set ? (c) To what power must each be raised to produce 1 ? (d) Show that (cos f + s sin -)~ l = cos — + • sin — , and that \ o o / 3 3 ( cos — + i sin — ) = cos - + i sin - • Is this true of any other two ? V o o I o 3 What ones produce themselves when raised to the power -1 ? (e) Show that all the six numbers can be obtained from cos tt + i sin 7r and cos ^ + i sin ^ by forming their powers and successive products. o o Is this true of any other two of the set ? (/) The above numbers are an example of a cyclic group. 2. Write down the roots of a; 10 = 1, and of x 12 = 1 and discuss them as the above numbers are discussed. 3. Assuming De Moivre's Theorem, prove that sin (a + /? + y) = sin a cos ft cos y + sin ft cos y cos a + sin y cos a cos /? — sin a sin ft sin y. 362 PLANE TRIGONOMETRY. [§198 § 198. Use of De Moivre's Theorem to Express Cosines and Sines of Multiples of an Angle in Terms of Powers of Cosines and Sines of the Angle. If a + bi= c + di, then a = c, and b = d, for each complex number locates but a single point on the Argand diagram. By (6), § 197, cos 2 + i sin 2 = (cos + i sin 0) 2 = cos 2 — sin 2 + i • 2 sin cos 0. Equating reals on opposite sides, also imaginaries on oppo- site sides : C os 2 = cos 2 (9 - sin 2 0. sin 2 0=2 sin cos 0, as in § 139. The general process is illustrated in this special example. Similarly : cos 3 + i sin 3 0= (cos + 1 sin 0) 3 = cos 3 - 3 cos (9 sin 2 + 1 (3 cos 2 sin0— sin 3 0). •\ cos 3 = cos 3 — 3 cos sin 2 ; sin 3 = 3 cos 2 sin - sin 3 0. (§ 144) EXERCISES. 1. Express similarly : cos 4 6, sin 4 0, cos 5 0, sin 5 0, cos 6 6, sin 6 0, cosn0, sinnd. 2. Express in terms of tan 0, the values of tan 2 0, tan 3 0, tan 4 0, tan 5 0, tan 6 0, tan 7 0. § 199. Use of De Moivre's Theorem to Express Powers of Cosines, Sines in Terms of Cosines, Sines of Multiple Angles. Let cos + i sin = z. Then cos — i sin = -, 25 J and cos n0 -f i sin n0 = z n . 1 cos — i sin w0 = 2 cos = 25 + -, 25 2 i sin = 2 , z J (i) (2) and 2 COS W0 = 25" + -, z n 2 i sin w0 z 7 ' — 4» (3) (4) § 199 c] THE FOUR UNITS. 363 (1), with the aid of (3), will express any power of a co- sine in terms of cosines of multiple angles. (2), with the aid of (3) or (4), according as an even or odd power is given, will express any power of a sine in terms of cosines or sines of multiple angles. Thus, for cos 3 0, by (1), 23 cos 3 = (z + -) S = z 3 + 3 z 2 • - + 3 z (*Y+ 1 \ zj z \zj z 3 = 2 cos 3 e + 3 • 2 cos 0. .-. cos 3 6 = { {cos 3 + 3 cos 0j. Similarly, for sin 3 0, by (2), 2 3 i 3 sin 3 6 = (z - Vfm z 3 - i - 3 fz - *\ = 2 i sin 3 6 - 3 • 2 i sin 0. .-. - sin 3 d = \ (sin 3 d - 3 sin 0). For sin 4 0, by (2), 2n'*.sin^ = ^-iy=^ + i)-4^2 + y + 6 = 2 cos 4 6 - 4 • 2 cos 2 + 6 .\ sin 4 0=| (cos 4 6 - 4 cos 2 6 + 3). A little study of the preceding examples will make it pos- sible to write the value for any power of sine or cosine with- out making the actual expansion by the binomial theorem : (a) The division factor for cubes is 4, or 2 2 ; for fourth powers it is 2 3 ; for nth. powers it is 2 n_1 . (5) For powers of cosines there are only cosines in the second member, starting with the same multiple angle as the power given, dropping two each term, the coefficients being those of the first half of the binomial expansion. (e) Even powers of sines turn into cosines, odd powers into sines, with alternation of sign, the coefficients being those of the first half of the binomial expansion. The sign of the first member is determined by that of the corresponding power of i. Since i 4 = 1, the value of any power of i is i, i 2 , 364 PLANE TRIGONOMETRY. [§ 199 rf z 3 , or 1. Divide the given power by 4; sign is -f- for remainders 0, 1 ; — , for remainders 2, 3. (d) The binomial coefficients (the first half) are formed by Pascal's rule, as indicated in the following set : 1, 2 1, 3, 3 1, 4, 6 1, 5, 10, 10 1, 6, 15, 20 1, T, 21, 35, 35 1, 8, 28, 56, 70, etc. A number under a number in any line is formed from that immediately above it by the addition to the latter of the number immediately preceding it. From such a table we can write immediately the value of any power covered by the table. For example : cos 8 d = -1 (cos 8 6 + 8 cos 6 6 + 28 cos 4 6 + 56 cos 2 6 + 35), sin 8 = i(cos80-8cos60+28cos40-56cos2<9 + 35), - sin 7 6 = 4(sm 7 6 - 7 sin 5 6 -h 21 sin 3 6 - 35 sin 0). 2 6 EXERCISES. 1. Fill out the preceding table of binomial coefficients to the twelfth power, and use it to write down the values of the first twelve powers of sine and cosine. 2. Carry out the expansions direct for cos 6 0, sin 6 0. 3. Show that an expression for cos 5 sin 7 can be obtained by multiply- ing together the expansions for (x 2 ] and Ix ) , giving - 2 11 cos 6 sin 7 = sin 12 $ - 2 sin 10 $ - 4 sin 8 + 10 sin 6 + 5 sin 4 - 20 sin 2 0. 4. Show that sin w cos n 0, for any positive integral values of m, n, can be changed to multiple angles, by changing each factor, multiplying results together, and then changing products of functions of multiple angles to sums (differences) by § 147. ' Change sin 8 cos lo 0. §200] THE FOUK UNITS. 365 5. Expand x cos v + y sin v + iz to the second, third, fourth, fifth powers, and show that in each case the coefficients of the sines (cosines) of the multiple angles are homogeneous functions of x, y, z, and that in no case are the coefficients linearly connected, that is, connected in a first degree additive (subtractive) relation. § 200. Relation of Sines, Cosines to the Exponential Imaginary. We have shown in §§ 8, 156 that /y3t zyO nA T^ V& f^ Assuming that (3) holds for x = i6, and [2 + [4 + Y ia + !5 ..). .-. e** = cos + i sin 0. (4) e-«0 = cos 6 - i sin 9. (5) .'. 5 = COS0. (6) e i9- e -i9 2i = sm9 ' (?) * " *V*>t0 i *>-»0A - lan W * (8) The preceding equations, (4), (5), (6), (7), (8), give the connection between the trigonometric and exponential func- tions. EXERCISES. 1. Use (6), (7) to prove the addition-subtraction formulas. 2. Use (6), (7) to prove sin x + sin y = 2 sin x ^ cos x ~ y , and the other corresponding formulas. 366 PLANE TRIGONOMETRY. [§201 § 201. Periodicity of the Exponential Function. By equation (4) of § 200, e ie = cos 6 + * sin 6 = cos (2 nir -f 0) +* sin (2 mr + 0) = e 1 ' 6 ,2«7ri 1. Thus, exponential #,(V'), is periodic, the period being 2 iri. § 202. Every Number has an Infinite Number of Logarithms, (a) Base e. y = e x = ^+ 2 ™", by § 201. .-. a? + 2 W7rt, n taking all integer values from minus in- finity to plus infinity, is the infinite set of logarithms of y, x being the ordinary table-logarithm. (6) Any base. If a x = y and e log e a = a, then y = e x log « a = e x log e a+2ra7ri , and # log e a + 2 mri makes up, as before, the infinite set of logarithms. § 203. Logarithmic Picture of the Points in a Plane. The numbers a + hi, when a, b take all real values, cover a plane completely (§ 187). We may now take the logarithms of all numbers in one plane and plot these logarithms on another plane. Since t 1 7 r ~l -i t B- 4 7 2ViE rj Fig. 178. — a, b plane. a + bi = Va 2 + 6 2 (cos + i sin 0), where tan # = -, anv number can a be set in the form R - e id , where R is its modulus. .-. N= Re ie = R- e i0+2nn \ log, iV= log e R + i(0 + 2 nir) = x + iy. 204] THE FOUR UNITS. 367 Here y is infinite-valued, as n goes from — oo to -f 00. If N is a point on the unit-circle, log e B = 0, and thus points on the unit- circle of one plane plot into points on the ?/-axis of the other plane. Any point of one plane plots into an infinite number of points, 2 7r distant from each other, on a line parallel to the ?/-axis, at a distance x, or log e B from it. The point P thus plots into p p p ... pr pt ... ■*■ V ■*■ 2' -*• 3' > ■*■ V 3' ' where 0M=\og e B, MP 1 = 0, P X P 2 = 2 7T, etc. — pH-^J^4 1, Af ;g' : 0=*2 ir '°L 2k" 5 7r 1~ 1 Fia. 179. — x, y plane. § 204. Mercator's Projection. If A, B are the extremities of a diameter of a sphere, and if a plane touches the sphere at A, and if a straight line is then drawn from B through each point of the sphere back to the plane, we have what is called a polar projection of the sphere on the plane. If A, B represent the north and south poles of the earth, the earth's meridian curves become rays from A on the plane and the latitude curves become concentric cir- cles about A. If now, we take the logarithm of this polar projection, we have, when these logarithms are plotted, Mercator's Projection. Which circle on the polar projection is taken for the unit-circle is, of course, quite arbitrary. This unit-circle plots (§ 203) into the y-axis. By § 203, circles concentric with and outside the unit-circle X x K~ X --*s +-«= ^xJ5t - v v -7 2* V Z. ^^ 5 4 ^ ^ t ^V 5 - f- ^ A 4. ^ ^R il T3 X J \ / X 7- \ / — --_ ^-^^ Fig. 180. 368 PLANE TRIGONOMETRY. [§204 Fig. 181. — Polar projection. plot into lines parallel to the y-axis, and to the right of it, while those within the unit-circle plot into lines to the left of the ?/-axis and par- allel to it. Radial lines on the polar projection plot into lines parallel to the rr-axis on the Mercator map. The radius whose directing angle is a radian goes a unit's distance up, etc. Thus, if we take the logarithm of the numbers repre- sented by a sheet of Groat's polar coordinate paper in radian measure (a picture of a polar projection of the earth) we have the numbers represented by a sheet of rectangular coordinate paper (a picture of Mercator's projection). On a polar projection the curve representing a direct sail from Liverpool to New York would be represented by a curve cutting all meridian circles (the radii) at the same angle. This curve is called an equiangu- lar spiral. The logarithms of points on this spiral pass into a straight line on the Mercator map, namely, the line joining the point which is the logarithm of the point representing Liverpool to that which is the logarithm of the point repre- senting New York. - - c 3 J Fig. 182. — Mercator's projection. §205] THE FOUR UNITS. 369 If log (a + bi) = log, R + i(d + 2 mr) any point on the unit-circle will become, in the loga- rithmic-picture, an infinite number of points on the ?/-axis, separated from each other by a distance 2 it. If we im- agine a point to start at the initial line and move around the unit-circle, how would the points representing the logarithms move ? Circles outside the unit-circle and con- centric with it will turn into what? If R is less than 1, log, R is negative ; thus a point within the unit-circle will lie to the left of the y-axis in the logarithmic-picture. A circle within the unit-circle, and concentric with it, will become a line parallel to the #-axis and to the left of it. Since for points on a ray through the origin in the a, b picture, is a constant, the corresponding points in the logarithmic-picture will lie on lines parallel to the #-axis, at distances + 2 nir from it. Thus any such ray will turn into an infinite number of parallel lines. The part of a ray beyond the unit-circle goes into what ? What represents the part of the ray within the unit-circle ? § 205. The /th Power and /th Roots of Numbers. Since a + bi can be put in the form R • e id , every number can be set in the form of two factors, the real factor giving the distance of its representing point from the origin and the exponential imaginary factor indicating its angle. .-. (a + biy = (R • e ie Y= e~ e • R l = i . e il0 *e*. Thus, if the ith. power of a number on one plane is plotted on another plane, the new modulus is the reciprocal of the exponential of the old angle, while the new angle is the loga- rithm of the old modulus. Thus the ith power of all points on the unit-circle will lie on the initial line. i ii Similarly, (a + biy = (R • e 19 )' = e 9 • R 1 = e 9 -R-* = e 9 -e- ilogR . 370 PLANE TRIGONOMETRY. [§205 Example : Find the ith. power of •'. The angle of i is — and its modulus is 1. *=* — s £ 7 /^ J J& A t ^tivVA /°± X ~X -t l \ t -^ z ^ 2? Fig. 183. EXERCISE. Make maps of the ith power and ith roots of all points on a selected curve on a plane. § 206. The Hyperbolic Functions. If x, y are the coordinates of any point on a circle of radius a, as in Fig. 183, x 2 -\-y 2 = a 2 . Thus, if we call - the cosine of 6 and U. a a the sine of 0, then characteristic of the trigonometry of the circle is . cos 2 0+ sin 2 (9=1. Similarly, if x, y are connected by the relation x*-y*=a% or (ff- (£f = 1, _ I the point x, y (Fig. 184) will lie on the curve which has the name equilateral hyperbola, or rectangular hyperbola, a curve obtained by making a sketch of points (in a plane) such that the distance of each from a fixed point in that plane bears to its distance from a fixed line (vertical) of that plane the ratio of the hypothenuse of a 45° right- angled triangle to a side of the same triangle. In correspondence to the trigonometry of the circle, we may call -, as related to the a \ z \ 7 \ 7 \ / \ J 3 t \ 2 - v -T-v X 1 4-Ih A t - *- X :> «■ -g -^ 3C t t X ■J K- L ± 5 J S y \ _/ \ _/ ^ 7 \ 7 \ Fig. 184. §207] THE FOUR UNITS. 371 hyperbola, the hyperbolic cosine, and V- the hyperbolic sine, so that (hyperbolic cosine) 2 — (hyperbolic sine) 2 = 1. There is thus a trigonometry related to such an hyperbola as is the ordinary trigonometry to the circle. Abbreviated names of the functions in this trigonometry are the same as in those of the circle, with an h interpolated : cosh for cos ; sinh for sin ; sech for sec, etc. These names may be read " cosine hyperbolic," " sine hyperbolic," etc. It is becoming customary, for brevity's sake, to throw the h in the pronunciation where it will make the abbreviated form pronounceable. Thus cosh x is pronounced as here written, " cosh " x. Sinh x is " shin " x. Sech x is " shec " x. Tanh x is " than " x (th having its sound in thing), etc. gx _i_ g-w: e ix e -ix We found cos x = and sin x = — — (§ 200) A At As defining relations for cosh x and sinh x, the i is omitted, pX _i_ p- 3D &X p — OC and coshx = — ^ and sinh x = Whence cosh 2 x — sinh 2 x = l, corresponding to the hyperbola — — £- = 1 . § 207. Relation of Hyperbolic Trigonometry to Circular Trigonometry. From the exponential relations of §§ 200, 206 follow at once : cosh x = cos (ix) (*■) i • sinh x = sin (ix). (2) Thus, assuming that circular trigonometry holds for quan- tity of the form (ix), the corresponding hyperbolic trigo- nometry comes at once by changing the word cosine to cosh and the word sine to t • sinh. For example : sin (x + y) = sin x cos y -f- cos x sin y . .*. i • sinh (x + y} — (i . sinh a;) (cosh y) -f- (cosh x)(i • sinh y), or, sinh (x -f y) = sinh x cosh y + cosh x sinh y. 372 PLANE TRIGONOMETRY. [§207 A second example is cos 2 # + sin 2 a; = 1. .*. (cosh a;) 2 + (i • sinhz) 2 = 1. .-. cosh 2 # — sinh 2 # = 1. From (1), (2) above it follows that tan changes to i • tanh, cot changes to — i- coth, sec changes to sech, cosec changes to — i- cosech. EXERCISES. Prove from the defining relations for the hyperbolic functions, or from the changes in circular functions noted above, as may seem best, the following : 1. sinh 2 x = 2 sinh x cosh x. 2. cosh 2 x = 2 cosh 2 x - 1 = 2 sinh 2 x + 1. 3. cosh (x + y) = cosh x cosh y + sinh x sinh y. 4. cosh (x + y) — cosh (x — y) = 2 sinh x sinh y. 5. cosh 3 x = 4 cosh 3 # — 3 cosh a\ 6. sinh 3 x = 3 sinh a; + 4 sinh 3 a;. 7. tanh<> + y) = tanh x + tanh SS . 1 + tanh a; tanh # 8. sinh (a: + y) cosh (x — ?/) = £ (sinh 2 x + sinh 2 y). 9. tanh2*= 2 tanh * • 1 + tanh 2 a: 10. sinh = 0; cosh = 1. 11. sinh S — i ; cosh 52 = ; sinh 7ri = ; cosh vi = — 1- 12. sech 2 a; + tanh 2 x = 1 ; coth 2 a; — cosech 2 x = 1. 13. sinh ar - sinh v = 2 sinh ^^ cosh'^tll. 2 2 14. sinh x + sinh y = 2 sinh - ~*~ - y cosh 2 2 15. cosh x — cosh y = 2 sinh x ~ y sinh x + y . * 2 2 16. cosh x + cosh y = 2 cosh x ~ & cosh ar + ^ « ■ . * 2 2 17. Plot the curves y = e x and y — e~ x and thus obtain graphs for y = sinh a: and y — cosh a;. Show that the latter has the form taken by a string hung on two pegs in the same horizontal line (catenary). 208] THE FOUR UNITS. 373 § 208. Relation of the Anti-Hyperbolic Functions to Logarithms. sinh x = ; cosh x = ^ e ■ (§ 206) Let sinh x = y, so that z = sinh -1 ?/. .-. e* -e~ x = 2 y. .'. e z - \ = 2 y. €■ .'. e 2x -2ye x -l = 0. .-. e z = y ±-\fy 2 4- 1. .-. a? = sinh- 1 y = loge (y ± Vy« + l). (1) Similarly, cosh- 1 y = log e (y ± V y 2 _ 1} . (2) Since the quantity in the bracket after log in (1) is nega- tive when the minus sign is given the radical, and since negative numbers have no real logarithms, it is customary, when the generality of the logarithm has not been considered, to take the sign of the radicals in (1), (2) as plus. Here, as the student is familiar with log (-.l±JL e 2 * + 1 * 1-y •\ x = tanh ~ V as - log _ M > U 2 *l-y Thus, in general, making y a a z ± V«a - a? cosh- 1 — = log a a tanh-i^log^?. a 2 z-a EXERCISES. 1. Show that the hyperbolic functions are periodic. 2. Show that the anti-hyperbolic functions are many-valued. 374 PLANE TRIGONOMETRY. [§209 § 209. Relation of the Hyperbolic Functions to the Circular Func- tions of the Eccentric Angle of the Hyperbola. For the hyperbola, x 2 — y 2 = a 2 . ... £!_2*-l. a 2 a 2 In the diagram, MT being a tangent to the circle, - = sec being called the eccentric angle. V E wm A M And since Also and (1) (2) Fig. 185. y sec 2 $ — tan 2 <£ = 1, .*. - = tan . cosh 2 0-sinh 2 = l. •\ cosh 6 = sec , sinh 6 = tan . ,\ (f> = sec -1 (cosh 0) = tan _1 (sinh 0). The angle as defined by these relations is called the G-udermannian of 6. (1), (2) may be put in the forms cos n sech 0, sin <£ = cos • sinh = tanh 0. ,\ gd = cos" 1 sech == sin -1 tanh 6 = tan -1 sinh 0. EXERCISE. Show that if u = log tan f J + 1\ w = gd -1 a\ 14 DAY USE RETURN TO DESK FROM WHICH BORROWED LOAN DEPT. This book is due on the last date stamped below, or on the date to which renewed. Renewed books are subject to immediate recall. 22May'53AJ ■■ ■ ■ . ■ ■ D m^ m 9 Nov'59FK ■' ' ' ■ ocm m ^«/Vov'eofj REffn i „e W H 196 TOEC?» sep at LD 21A-50m-9,'58 (6889sl0)476B S General Library University of California Berkeley J £*••'•* ■ UNIVERSITY OF CALIFORNIA LIBRARY Cffef*