THE LIBRARY OF THE UNIVERSITY OF CALIFORNIA LOS ANGELES "i |C r 'ast date stamped below SOUTHERN BRANCH UNIVERSITY OF CALIFORNIA LIBRARY LOS ANGELES. CALIF, TREATISE ON PLANE AND SPHERICAL TRIGONOMETRY. BY WILLIAM CHAUVENET, PROFESSOR OF MATHEMATICS AND ASTRONOMY IN WASHINGTON UNIVERSITY, SAINT LOUIS. ELEVENTH EDITION. PHILADELPHIA : J. B. LIPPINCOTT COMPANY. 3 Entered according to Act of Congress, in the year 1850, by WILLIAM CHAUVENET, in the Clerk's Office ol the District Court of the Eastern District of Pennsjivamc ELECTROTYPEO AND PRINTED BY J. B. LIPPINCOTT COMPANY, PHILADELPHIA, U. 8. A. C3 taui> - 6 a 17. The SECANT of the angle is the quotient of the hypotenuse divided by the adjacent side. Thus secJ.= 7> 860.6=:- 6 a 18. The COSINE, COTANGENT, and COSECANT of an angle, are re- spectively the SINE, TANGENT, and SECANT of the complement of the angle. Since the sum of the two acute angles of a right triangle is one right angle, or 90, they are, by Art. 12, complements of each other; therefore, according to the preceding definitions, we shall have sin A = cos B = - cos A = sin B = - c c tan A = cotB=- cot A = tan B - 6 a sec A = cosec -6=7 cosec A = sec B=- b a c n 19. Since - is the reciprocal of -> it follows from the first and last a c of these equations, that the sine and cosecant of the same angle are reciprocals ; and from the other equations, also, that the cosine and secant, the tangent and cotangent are reciprocals. That is, (2) or more briefly, sin A cosec A = cos A sec A tan A cot A = 1 (3) Sill A. - cosec A cosec A sin A po^ A 1 sec A 1 sec .4 cos A tan A - 1 cot A I cot A tan A 16 PLANE TRIGONOMETKY. SINES, ETC. OF ARCS. 20. The sine, tangent, and secant of an arc are respectively the sine, tangent, and secant of the angle at the center measured by that arc. Thus, Fig. 6, 7? C 1 - A'" The sine of an arc, therefore, does not depend upon the absolute length of the arc, but upon the ratio of the arc to the whole circum- ference, (Art. 7.) It follows that the relations (2) and (3) are also applicable when A expresses an arc. 21. If the radius = 1, all the trigonometric functions above de- fined may be represented in or about the circle by straight lines. Representing the arc AB, or angle A OB, by x, we have, when OA = OB=1, BC EC = T AT AT = ~=- OT OT sec x = = ~ = OA 1 and from the arc A'B= 90 x we find in the same way cos x = BD=OC cosecz OT' Therefore, in the circle whose radius is unity, the sine of an arc, or of the angle at the center measured by that arc, is the perpendicular let Jail from one extremity of the arc upon the diameter passing through the other extremity. The trigonometric tangent is that part of the tangent drawn at one extremity of the. arc, which is intercepted between that extremity and the diameter (produced) passing through the other extremity. The secant is that part of the produced diameter which is intercepted between the center and the tangent. The cosine is the distance from the center to the foot of (he sine. In a circle of any other radius than unity, the trigonometric FUNDAMENTAL FORMULAE. 17 functions of an arc will be equal to the lines drawn as above, divided by that radius. The properties here stated have heretofore been used by most writers upon trigonometry as definitions, but without limiting the radius to unity ; and it is evidently from this mode of viewing these functions that they have derived their names. 22. Besides the functions already denned, others have been occasionally employed to facilitate particular calculations, as the versed sine, which in the circle is the portion of the diameter intercepted between the extremity of the arc and the foot of the sine ; thus, Fig. 6, the versed sine of A B is A C, or the radius being = 1, versinz = l cos a; (4) by means of which formula we may always substitute versed sines for cosines, and reciprocally. The coversed sine (covers.) is the versed sine of the complement, and suversed sine (suvers.) is the versed sine of the supplement. The chords of arcs have also been used, and may be substituted for sines by the formula ch x = 2 sin J x (5) which is evident from Fig. 6, where if the arc B B' = x, we have chord B B' 2 B C = 2 sin A B. 23. From what has now been stated, the student will perceive that angles are to be subjected to computation by means of the quanti- ties sine, cosine, etc., commonly designated by the comprehensive term trigonometric functions* It becomes necessary, therefore, for the computer to know the values of these functions for any given value of the angle. The trigonometric tables contain these values for every minute, and sometimes for every second, from to 90 ; and with these tables all the numerical computations of trigonometry are carried on. In practice, then, we are not required to compute the functions themselves, and we shall therefore defer the methods for that purpose to a subsequent part of this work, and proceed at once with the investigation of the formulae and methods by which these tables are rendered available. FUNDAMENTAL FORMULAE. 24. Given the sine of an angle, to find the cosine. From the right triangle ABC, Fig: 7, we FlG - 7 - have by geometry a 2 4- b 2 c 2 Dividing by c 2 , this equation becomes * Also trigonometric lines, from the properties explained in Art. 21. 18 PLANE TRIGONOMETRY. or, by the definitions of sine and cosine (1), sin 2 A + cos 2 A = 1 (6) in which the notation sin 2 A signifies " the square of the sine of A." From this formula, if the sine is given, we find cos 2 A = 1 sin 2 A = (1 + sin A) (1 sin A) cosA=y(l sin 2 A) = y [(1 + sin A) (I sin A)] (7) and if the cosine is given, we find sin A=y(l cos 2 A)=y [(1 + cos A) (1 - cos A)] (8) 25. Given the sine and cosine of an angle, to find the tangent. By (1) we have also therefore tan A ==- b sin A cos A tan A = c c sin A cos A And since the cotangent is the reciprocal of the tangent, cos A cot A = sin A (9) (10) 26. Given the tangent of an angle, to find the secant. The right triangle A B C, Fig. 7, gives c=6 2 + a a Dividing by b 2 , this becomes r 2 n z -=14-- 6 2 r 6 2 or, by the definitions of secant and tangent (1), sec 2 A = 1 + tan 2 A This formula applied to the complement of A gives cosec 2 A = 1 + cot 2 A 27. The preceding formulae are also directly obtained from Fig. 8. If the angle A O R, or the arc A B, be denoted by x, the right triangle OJ3C, gives (11) (12) or remembering that the radius is unity, by Art. 21, sin 2 x + cos 2 x = 1 (13) FUNDAMENTAL FORMULAE. 19 The triangle OB Ogives by the definition, Art. 16, sn a; - . or tana;- (14) cos a; Since the angle BOD is the complement of JBOC, tan BOD = cot a*, and the triangle BOD gives BD 00 cos a; /t N or cot x =- (15) sin x In a similar manner the triangles A0 r l\ A' OT' give sec 2 a; 1 -f tan 2 x (16) cosec 2 x 1 + cot 2 x (17) 28. The following equations are easily demonstrated by combining (13), (14), (15), ^16), (17), and employing the property of the reciprocals (2). They are of frequent use. sin z = cosec z sec x cot z 1 cot z sin z /1nx -=cotzsinx = (19) sec z cosec z tan z sin z = i/ (1 cos 1 z), cos z = -j/ (1 sin 1 z) (20) sec z i/ ( 1 -f tan 2 z), cosec z = -j/ (1 + cot 1 z) (21 ) (sec'z 1), cot x = l / (cosec 2 z 1) (22) tan z 1 I/ (1 + tan 1 z) i/ ( 1 + cot 2 z) (23) *=7T7T^~7rr= ,V , . (24) (25) (26) \/ (1 cos 2 z) sinz 29. To find the sine, etc. of 30 and 60. In Fig. 8, let the arc A B = 30, and BB' = 2AB = 60. By Art. 21, sin AB = BO, and by geometry the chord of 60, or of one- sixth of the circumference, is equal to the radius 1 ; therefore 2 sin 30=- 2 BC=^ BB' = 1 whence sin 30 = I = cos 60 (27) 20 PLANE TRIGONOMETRY. and by (7) cos 30 = y [(1 + J) (1 - 1)] = ( whence cos 30 = y 3 = sin 60 Then, by (9) and (2), X J) - cot 60 sin 30 \ 1 tan 30 = - cos 30 \ cot 30 = - - = r/ 3 = tan 60 tan 30 sec 30 = - = = cosec 60 cos 30 ]/3 1 = 2 = sec 60 C (28) (29) (30) (31) (32) cosec 30 = sin 30 30. To find the sine, etc. of 45. Since 45 is the complement of 45, we have sin 45 = cos 45 whence by (13), putting x = 45, sin 2 45 + cos 2 45 = 2 sin 2 45 = 2 cos 2 45 = 1 sin 2 45 = cos 2 45 sin 45 = cos 45 = V\ = \ 1/2 (33) tan 45 - cot 45 == sec 45 - cosec 45 = cos 45 1 = 1 sin 45 (34) (35) These values are readily verified in the circle, Fig. 9, where OAT A' is a square described upon the radius. The diagonal T bisects the right angle, whence AOT= 45, and tan 45 = A T = OA = 1 ; cot 45 = A'T=l; sin 45 = BC = OC=cos45, etc. 31. The sines and cosines of two angles being given, to find the sine and cosine of the sum, and the sine and cosine of the difference of those angles. Fl - Let the two angles be A B and BOC, Figs. 10 and 11. At any point B in the line OB draw BC perp. to OB. Draw BA and CD perp. to OA, and 7? A 1 perp. to CD. FUNDAMENTAL FORMULA. 21 Theii the triangles BCE and BOA are mutually equiangular, the three sides of the one being perp. to the three sides of the other respectively ; therefore the angle B C E = A B. Let x = AOB = BCE y = BOC Then, Fig. 10, x + y = COD Fig. 11, x y = COD and in , CD BA + CE BA.CE O, 8 m(* + 3,)==- = -- - . _ CD BA - CE BA CE 1,31 -y):- - ~ CQ = - CO ~w and in both figures BA BA ^ BO cd = Bd x co^^<^ CE CE CB which being substituted in the above expressions of sin (x + y) and sin (x y) give sin (a; -f- y) = sin x cos y -(- cos x sin y (36) sin (a; y) = sin x cos y cos x sin y (37) Again in . OD OA-EB OA EB F.g.10, oos(* + y Fig. 11, 008^-^ and in both figures, OA OA OB 07) OA + EB OA . EB EB EB , BC OC = BC X OC = * }nX * my therefore cos (x -|- y) cos x cos ?/ sin x sin y (38) cos (re y) = cos x cos y + sin x sin y (39) and (36), (37), (38), and (39) are the required formulae. These may be considered as the fundamental formulae of the trigo- nometric analysis, and will form the basis of our subsequent inves- tigations. They are equally applicable to arcs represented by x and y (Art. 20). CHAPTER III. TRIGONOMETRIC FUNCTIONS OF ANGULAR MAGNITUDE IN GENERAL. 32. THE definitions of sine, etc. given in the preceding chapter apply only to acute angles, since the angle is there assumed to be one of the oblique angles of a right triangle. But we shall now take a more general view of angular magnitude and of the functions by means of which it is subjected to computation. If, Fig. 12, we suppose the line OA to revolve from the position OA to OA' in the direction of the arc AA' (or from right to left), it will describe an angular magnitude of 90 ; when it arrives at OA" it will have described an angular magnitude of 180 ; at OA'", 270; and at OA again, 360. If it now continue its revolution, when it arrives at OA' again, it will have described an angular magnitude of 360 + 90, or 450 ; and thus we may readily conceive of an angular magnitude of any number of degrees. In like manner we may have arcs equal to or greater than one, two, or more circumferences. To obtain trigonometric functions for angles and arcs thus gene- rally considered, we shall avail ourselves of the fundamental formu- lae established in the preceding chapter ; first deducing their values analytically, and then explaining their geometrical signification. 33. To find the sine, etc. of and 90. In (37) and (39) let x = y ; the first members become sin (x #) = sin0 , and cos (x x) = cos ; and by (13) they are reduced to sin = sin x cos x cos x sin x = cos = cos 2 x -+- sin 2 x =1 and since and 90 are complements of each other, Art. 12, sin = cos 90 = cos = sin 90 = 1 from which by (9) and (2) tan = cot 90 snO _^ cos ~ 1 ~ (40) (41) (42) 22 FUNCTIONS OF ANGULAK MAGNITUDE. 23 cotO= *'90=-i-=l= (43) sec = cosec 90 = - \ = l (44) cos 1 cosec 0= sec 90 = - -=^ = 00 ( 45 ) sin 34. To find the sine, etc. of 180. In (36) and (38) let x = y 90 ; these equations become by means of the preceding values sin 180 = 1 X + 0X1 (46) cos 180 = 0X0-1X1 = ! (47) whence by (9) and (2) tan 180= -0 cot 180= - (48) 1 U sec 180 = -^- = 1 cosec 180 = - = oo (49) 35. To find the sine, etc. of 270. In (36) and (38) let x = 180, y = 90, then sin 270 OXO + ( 1) X 1= -1 (50) cos270 = (-l)XO OX 1 = (51) tan 270 = - = oo cot 270 = = (52) oo sec 270 =- = oo cosec 270 = -j- = -1 (53) 36. To find the sine, etc. of 360. In (36) and (38) let x = y = 180; then sin360 = OX(-l) + ( 1)XO = (54) cos 360 = ( 1) x ( 1) -0X0=1 (55) the same values as for 0, whence it follows that all the trig, func- tions of 360 are the same as those of 0. The same process continued will give for 450 ( = 360 + 90), the same trig, functions as those of 90 ; fer 540 the same functions as for 180, etc. 24 PLANE TRIGONOMETRY. 37. The preceding values now furnish us at once with the values of the functions for all possible values of the angle. In (36) and (38) let x , they are reduced to sin y = sin cos y + cos sin y = sin y cos y = cos cos y sin sin y = cos y which are simply identical equations, and reveal no new property. But if in (37) and (39) we put x = 0, we have, after substituting the functions of 0, sin ( y) = siny cos ( y) = cosy (56) whence by (9) and (2) (57) (58) (59) (60) or, the sin., tan., cot., and cosec. of the negative of an angle are the negative of the sin., tan., cot., and cosec. of the angle itself ; and the cos. and see. are the same as those of the angle itself. 38. In (37) and (39) let .r = 90; we find after reduction sin (90 - y) cos y cos (90 -y) = sin y which agree with the definition of cosine, but give no new relations. But in (36) and (38) let x = 90, we find sin (90 + y) = cos y, cos (90 -f y) = - sin y (61) whence by (9) and (2), tan (90 + y) = cot y cot(90 + .y) = - tan y (62) sec (90 + y) = cosec y cosec (90 + y) = sec y (63) or, the sin. and cosec. of an angle are equal to the cos. and sec. of the excess of the angle above 90 ; and the cos., tan., cot., and sec. are equal to the negatives of the sin., cot., tan., and cosec. of the excess of the angle above 90. sin( y)_ \. sin y tan ^ y) cos( y) cos( y) cosy cos ?/ sin( y) 1 1 cot i/ ecy - cosec sec ( y) jsec ( y) = cos( y) 1 cosy 1 sin( y) siny FUNCTIONS OF ANGULAR MAGNITUDE. 25 30. In (37) and (39) let* = 180 ; we find sin (180 y) = sin y cos (180 y) = - cos y (64) tan (180 y)= - tan y cot (1 80 y) = cot y (65) sec (180 y) = secy cosec (180 y) cosec y (66) or, the sin. and cosec. of the supplement of an angle are the same as those of the angle itself; and the cos., tan., cot., and sec. are the nega- tive of those of the angle itself. 40. If y is acute (that is, less than 90), all its trig, functions are positive; and since its supplement 180 y is obtuse (that is, greater than 90), it follows from the preceding article, that the sin. and cosec. of an obtuse angle are positive) while its cos., tan., cot., and sec. are negative. 41. In (36) and (38) let x = 180 ; we find sin (180 + y) = - sin y cos (180 + y) = - cos y (67) tan (180 + y) = tan y cot (180 + y) = cot y (68) sec (180 + y) = - sec y cosec (180 + y) = - cosec y (69) by means of which, if y is acute, we obtain the values of the sines, etc. of angles between 180 and 270. 42. In (37) and (39) let x = 270 ; we find sin (270 y) = cos y cos (270 y) = sin y (70) tan (270 y) == cot y cot .(270 y) = tan y (71) sec (270 y) = cosec y cosec (270 y) = sec y (72) 43. In (36) and (38) let x = 270 ; we find sin (270 + y) = cos y cos (270 + y) = sin y (73) tan(270 + y)= - cot y cot (270 + y) = tan y (74) sec (270 + y) = cosec y cosec (270 -f y) = sec y (76) 44. In (37) and (39) let x = 360 ; we find sin (360 y) = - sin y cos (360 y) = cos y (76) tan (360 y) = tan y cot (360 y) = cot y (77) sec (360 y) = sec y cosec (360 y) = cosec y (78) or the functions of 360 y are the same as those of y (Art. 37). 45. In (36) and (38) let x = 360 ; we find sin (360 + y) = sin y cos (360 + y) = cos y (79) or, the functions of an angle which exceeds 360 are the same as those of the excess above 360. 4 C 26 PLANE TRIGONOMETRY. It follows that the functions of 720 -J- y are the same as tho.se of 360 -j- y, and therefore the same as those of y ; and in like manner for an angle which exceeds any multiple of 360. 46. Since .y 90 is the negative of 90 --y, we obtain from Art. 37, sin (y- 90)= - sin (90 - y) = -cosy ) cos (y 90) = cos (90 y) = sin y C ' whence also tan., etc. ; and in the same manner we may find the func- tions of y 180, y 270, y 360, etc. 47. We shall now give the geometrical interpretation of the pre- ceding results. In Fig. 13, let the radius revolve from the position OA to OA', OA", etc., as in Art. 32, thus describing a continuously increasing an- gular magnitude ; or, which is equivalent, let the arc commencing at A increase contin- uously to AB, A A', AB', etc. Then the changes in the values of the several trigono- metric lines may be traced as follows. 1st. The sine being, by Art. 21, the perpendicular from one extre- mity of the arc upon the diameter drawn through the other extremity, we shall have sin AB = BC, sin AB' = B' C f , sin A A" B" = B"C', sin AA" B'" = B'" C, and if we make AB = A" B' = A" B" = AB'" = y we have sin y = B C sin (180 -y) = B' C' B'" sin (360 y) = B'" C The lines BC, B'C', B"C', B f " C, however, represent only the numerical values of the sines, and are here equal. But the results above obtained from our formulae enable us to distinguish between them by means of their algebraic signs. Thus, by (64), (67), (76), sin (180 y) = siny sin (180 + y)= -siny sin (360 y) = - sin y so that the sines from to 180 are positive, while those from 180 to 360 are negative ; or the sines which are above the diameter A A" are positive, while those which are below this diameter are FUNCTIONS OF ANGULAR MAGNITUDE. 27 negative; or still more generally, the sines that have opposite di- rections, with reference to the fixed diameter from which they are measured, have opposite signs. 2d. The cosine being, by Art. 21, the distance from the center to the foot of the sine, we have cos y = O C cos(180 2/) = cos (360 y) = OO but by (64), (67), (76), cos (180 y) = cos y cos (180 + y)= cos y cos (360 y) = cos y so that the cosines on the right of the diameter A' A'" are positive, while those on the left of this diameter are negative ; or rather the cosines that have opposite directions, with reference to the diameter from which they are measured, have opposite signs. We have here only exhibited a well-known principle in the appli- cation of analysis to geometry, viz. : that all lines measured in opposite directions from a fixed line have opposite signs. To interpret the results (56), it is only necessary to observe that a negative arc will be one reckoned from A towards B" 1 ', or in the opposite direction to that of the positive arc, so that sin A B'" = sm(y) = B" r C = - B C= sin y cos A B'" cos ( y) = C= cos y as in (56). The same principle applies to the tangents, but it will be simpler in practice to obtain their signs (as also those of the secants), ana- lytically, from those of the sine and cosine, as has been already shown. It will be sufficient to bear in mind the following table, which is also expressed by Fig. 13. SINE COSINE 1st QUAD. 2d QUAD. 3d QUAD. 4th QUAD. + + + 28 PLANE TRIGONOMETRY. 48. The particular values of the sine and cosine at A, A r , A", etc., or sin. and cos. of 0, 90, 180, etc., may also be found by Fig. 13, upon the same principles ; but this we leave to the student. 49. GENERAL REMARK. In the demon- stration of the fundamental formulae for sin (xrti/), and cos (xy\ Art. 31, the angles x, y and x y were all taken less than 90 and positive. In this chapter these formulae have been applied to angles of any magnitude, and the resulting functions have been shown to take opposite signs when the lines representing them take opposite directions. It follows that, in deducing trigonometric formulae from geometrical figures, we need not embarrass our demonstrations with the consideration of the various cases of the problem, or of the various values of the angles of the figure. The formula deduced from any supposed position of the lines of the figure will be of general application, provided in the practical application of this formula to the particular cases, we observe those values and signs of the trigonometric functions which have now been determined. 50. The results of this chapter may be expressed by a few general formulae. From (79) it appears that all the trigonometric functions return to the same values after one or more complete revolutions of 3fiO. If we* represent the semi-circumference, or two right angles, hy TT (Art, 11), and let n = any whole number or zero, we shall have sin 4 n = sin (4n + 1) =1 sfn (4 n -f 2) = 4U sin (4 n -f- 3) 1 whence tan 4 n m tan(4n+'2) A ~ 1 cos 4 n = 1 cos (4 n + 2) = 1 35 cos (4 n + 3) = tan (4 n -f- 1 ) = 00 tan (4 n -f- 3) =00 2 (81) (82) (83) (84) or the tan. of the even multiples of =0, and of the odd multiples = oo, so that we SS may write more simply tan 2 n = 2 tan (2 n -\- 1) = 00 z (85) FUNCTIONS OF ANGULAR MAGNITUDE. 29 In these formulae we have only to give n one of the values 0, 1, 2, 3, 4, etc., to obtain the functions of any given multiple of the right angle. Thus, we find sin 450 sin 5 - = sin (4 4 1) ^ = 1 by making n 1, in (82). Since the subtraction of 8 n - from the arc will not change the functions, the above 2t formulae are also true when n is a negative whole number. 51. In a similar manner we obtain sin Hi n ~ 4 yl = sin y cos J4 n ~ 4 yj = cos y (86) sin j (4 n 4 1) 4 y 1 cos y cos J(4 n 4 1) -|- 4 y \ = sin y (87) sin (4n + 2)|-4y]= - sin y cos [ (4 n 4 2) -| 4 yl == cos y (88) | i p- ~^ sin (4n43) 4y = cosy cos (4w 4 3) 4 y Biny (89) L 2 J L 2 J tan I 2 ?i 4 y I = tan 2/ tan I (2.41)+y|= coty (90) in which n may be any whole number, positive or negative, and y any angle, positive or negative. 52. A still more concise form may be given to the formulje of the two preceding articles, as follows: n being, as before, any whole number, positive or negative. sin2n^-:=0 cos 2n- = ( l) n (91) sin (2n + 1)- = ( !) cos (2n + 1) - =0 (92) 2t ** =(-l) n siny cos [2 n |- + yj = (-!) cosy (93) =(-l)co8y cos (2n + 1) -\ !/ = -(-l) B iny(94) and from these (85) and (90) may be directly deduced. 53. We have seen that an angle being given, there is but one corresponding sine. On the other hand, a sine being given, there is an indefinite number of angles corre- sponding; for if a denote the given sine, and y any corresponding angle, then o is also the sine of all the angles if y, 2 TT -f y, 3 IT y, etc. _,r y, _27r4y, _ 3 n- y, etc. or in general a = siny = sin [n?r 4 ( l)"y] (95) c2 30 PLANE TRIGONOMETRY. In like manner if a is a given cosine, and y any corresponding angle, a = cos y = cos (2 n TT y) (96) and if a is a given tangent corresponding to the angle y, a = tan y = tan (n if + y) (97) SINE AND TANGENT OF A SMALL ANGLE OR ARC. 54. When the angle A B = x, Fig. 14, is very small, the sine and tangent are very nearly equal to the arc A B, which measures the angle, the radius being unity ; and the cosine and secant are nearly equal to OA = l (Art. 21). There- fore, to find the sine or tangent of a very small angle approximately, we have only to find the length of the arc by Art. 9 ; thus sin 1" = arc I" = 0000004848137 log. sin 1" = 4-6855749 and x being a small angle, or arc, expressed in seconds, sin x = tan x = x sin 1 " (98) If # is expressed in minutes, sin x = tan x x sin V (99) If x expresses the length of the arc, the radius being unity, sin x = tan x = x (1 00) The employment of these approximate values must be governed by the degree of accuracy required in a particular application. It is found, for example, that they are sufficiently accurate when the nearest second only is required in our results, provided the angle does not much exceed 1. 55. If x and y are any two small angles, it follows from the pre- ceding article that sin x : sin y = x sin V : y sin 1" = x : y that is, the sines (or tangents) of small angles are proportional to the angles themselves. The application of this theorem, however, like that of the preceding, must depend upon the accuracy required in the problem in which it is employed.* *Fora full discussion of the limits under which this theorem maybe employed, see a paper, by the author of this work, in the Astronomical Journal, (Cambridge, Mass.) Vol. i. p. 84. CHAPTER IV. GENERAL FORMULAE. 56. WE have already obtained four fundamental equations, (36), (37), (38), and (39), involving two angles, x and y. From these we shall now deduce a number of formula, either required in the sub- sequent parts of this work, or of general utility in the applications of trigonometry. 57. The sum and difference of the equations (36) and (37) are sin (x -f- y) -\- sin (x y) 2 sin x cos y sin (x + y] sin (x y) 2 cos x sin y and the sum and difference of (38) and (39) are cos (x -j- y) + cos (x y) = 2 cos x cos y (103) cos (x -f- y) cos (x y) 2 sin x sin y (104) 58. If we put x -f- y = x' x y = y' whence 2 x = x' + y', x = | (x' -f- y') 2y = x' y', y= i(x' y') equation (101) will become sin x' + sin y' = 2 sin 1 (#' + y'} cos i (#' y') and (102), (103), and (104) admit of a similar transformation. But since x' and y' admit of all varieties of value, we may omit the accents and apply the formulae to any two angles x and y, we have thus sin x -f sin y 2 sin | (x + y) cos % (x y) (1 05) sin x sin y = 2 cos | (z -f- y) sin f (z y) (106) cos # + cos y = 2 cos | (# + y) cos | (x y) (107) cos x cos t/ = 2 sin | (.r + T/) sin | (a- 2/) (108) Each of these equations may be enunciated as a theorem ; thus 31 32 PLANE TRICxONOMETRY. (105) expresses that "the sum of the sines of any two angles is equal to twice the sine of half the sura of the angles multiplied by the cosine of half their difference." These formulae are of frequent use (especially in computations performed by logarithms), in transforming a sum or difference into a product. 59. Dividing (105) by (106), we have by (14) and (15) or by (2) sin x -f- sin y r-* = tan (x + y) cot $ (a; y) sin x sin y sin x -f sin y _ tan i^_+ y} sin x sin y tan J (x y) and from (107) and (108) we find in the same manner -ten^ + yJtanHs-y) (HO) cos x + cos y We find also sin x -\- sin y * = tan \ (x + y) (111) cos x + cosy sin x sin?/ N -tanifa y) (112) cos x + cos y sin x -f sin ?/ , ~eot|(s y) (113) cos x cos y sin a? sin y . -cot|(a?-f-y) (114) cos x cos y 60. Divide the equations (36), (37), (38) and (39) by cos x cos y ; then by (14) we have sin (.T: -f ?/) J ^ tan x + tan y (11 o) cos a; cos y sin (a; y} = tanar tany (1 16) cos x cos y cos (x + ?/) - z - = 1 tan # tan v cos x cos y cos (# y) = 1 + tan .7; tany (118) cos x cos y GENERAL FORMULAE. 33 61. Divide (36), (37), (38) and (39) by sin.r siny; and by sin x cos y ; then sin (re v) v v> =coty cot:c sin # sin y cos (x y] r v ;-*"- cot x cot y q= 1 (1 20) sin # sin y /ioi\ 1 cot x tan y (121) cos (# y) /., oox = cot cr :+: tan y (122) sin x cos y 62. Divide (115) by (117), and (110) by (118); then by (14) tan x -f- tan y /1 .,, tan (x -{- y) = (1 23) 1 tan x tan y tana; tan y /io,i\ tan (a? y) = - (124) 1 + tan x tan y by which, when the tangents of two angles are given, we may com- pute the tangent of their sum or difference. To find the cotangent of the sum or difference when the cotangents of the angles are given, divide (120) by (119), , N cot y cot.-eq= 1 /io-\ cot(a?:y)=- (12^) cot y cot x 63. Dividing (115) by (116), and (117) by (118), (or from the equations of Art. 61), we have sin (x -f- y) _ tan x + tan y _ cot y -\- cot x . --- -- ! - sin (x y) tan x tan y cot y cot x cos (x -{- y) _ 1 tan x tan y _ cot x cot y 1 /1 _, ( i Z i J cos (x y) 1-4- tan x tan y cot x cot y -f 1 64. Formulae for secants are obtained from those for cosines by means of (2) ; thus we find sec (x y) cos x cos y =F sin x sin y and multiplying numerator and denominator by sec x sec y, / s sec x secy sec(xy) = a 1 =F tan x tan y 34 PLANE TRIGONOMETRY. Also since 1 1 cos v =fc COST sec x sec y - - = - a cos x cos y cos x cos y we find by (107) and (108) 2 OOB *(' + y) " *('-) sec x + sec y = - (1 29 ) cos x cos y sec x - sec y = 2 sin * (z + ^ sin *-l x > (130) cos x cos y In the same manner from (105) and (106) 2 sin J (x -4- 1/) cos i (x V) /ioi\ cosec x -}- cosec y = "- 5 r *"- 1LJ - *' (131) sin x sin y cosec x - cosec y = - *-(*-ti*A(*=3Ll (132) sin x sin y These formulae, although generally omitted in treatises on trigonometry, will be found useful in a subsequent part of this work. 65. The product of (36) and (37), and of (38) and (39), are sin (x -\- y) sin (x y) = sin 2 x cos 2 y cos 2 x sin 2 y cos (x -f- y) cos (x y) cos 2 x cos 2 y sin 2 x sin 2 y By (13) we have cos 2 x 1 sin 2 x and cos 2 y = 1 sin 2 y, which substituted in the preceding equations, give sin (x -f- y] sin (# y] = sin 2 # sin 2 y = cos 2 y cos 2 x (133) cos (x -f- i/) cos (x y) = cos 2 # sin 2 y cos 2 2/ sin 2 # (1 34) 66. In (36), (38) and (123), let y = x, we find sin 2 x 2 sin x cos a; (1 35) cos 2 # = cos 2 z sin 2 x (1 36) 2 tan x tan 2 a? = (130 1 tan 2 x by which the functions of the double angle may be found from those of the simple angle. 67. To find the functions of the half angle from those of the whole angle, we have, from (13) and (136), cos 2 or + sin 2 x = 1 cos 2 x sin 2 x cos 2 x the sum and difference of which are 2 cos 2 x = 1 -f cos 2 x 2 sin 2 x = \ cos 2 x GENERAL FORMULA. 35 As these express the relations of an angle 2 x and its half x, their meaning will not be changed by writing x and J x instead of 2 # and x; whence 2 cos 2 % x = 1+ cos x (138) 2 sin* $a; = l cosx (139) the quotient of which is 1 ~ cosa; 1 + COS X 68. The following may be proposed as exercises. (140) cinx= 2 tan $ x 1 -j- tan 2 \x cot x -f tan $ a; tanz = 2tan *f =- 2 , (142) 1 tan* x cot J x tan j a; tan 2 Jx + 2cotx tan x 1 = tan z J x 2 cosec x tan \ x + 1 = 1 tan* J x cos x = : 1 + tan 9 \ x tan \ x = cosec r cot x = smx cot \ x = cosec x + cot x = 1 + CO8 ? : sn x cos z 69. Several useful formula result from the preceding, by intro- ducing 45 or 30. If z = 45 in (36), (37), (38), and (39), we have, by (33), sin (45 y) = cos (45 + y) = cos V sm V (149) I/ 2 whence tan (45 y) = cot (45 + y) - ( 150 ) cos y =F sin y in which either the tipper signs must be taken throughout, or the lower signs throughout. If we divide the numerator and denominator of (150) by cosy or sin y, 1 tan y cot y 1 tan (45 y) = - = *_: 1 T tan y cot y + 1 36 PLANE TRIGONOMETRY. From this, by (57), tan(y-45)= ta " y ~ (152) tany+1 70. Again, let z = 90 dry in (138), (139), (140) and (146), sin(45 Jy)=co S (4o =FhO = Yv 2 ) (153) tan (45 zb \ y) = -* / /I i sin^\ / ' 1 ^f sin y/ (154) tan (45 % y) = - From the last we find tan (45 + \ y) + tan (45 tan (45 -f $ y) tan (45 cosy l^siny A 1M 9 cop * (155) (156) (157) 5 y i * sec y cosy i\ 2 sin y n . J y) = * = 2 tan y cosy the quotient of which is tan (45 + j y) - tan (45 - $ y) . , 71. In (101), (102), (103) and (104), let x = 30; then by (27) and (28) sin (30 + y) + sin (30 y) = cos y (1 59) sin (30 + y) sin (30 y) = sin y y 3 (160) cos (30 + y) + cos (30 y) = cos T/ y 3 (161) cos(30 + t/) cos(30 y)= -sin?/ (162) and in a similar manner we may introduce 60; but it is unneces- sary to extend these substitutions, as they involve no difficulty, and can be made as occasion demands. FORMULAE FOR MULTIPLE ANGLES. 72. From (101) and (102) we have sin (y -j- x) 2 sin y cos z sin (y z) sin (y -j- x) = 2 cos y sin z -f- sin (y z) in which let y = (m 1) z ; then sin m.e = 2 sin (m 1) z cos z sin (m 2)x (163) sin mz = 2 cos (m 1) z sin z -f- sin (m 2) z (164) which arc the general formula? for computing the sine of any multiple mz, from the lower multiples (m 1) z and (m 2) z, and the simple angle z. FORMULAE FOR MULTIPLE ANGLES. 37 If we make m successively 1, 2, 3, 4, etc., these formulae give sin x = sin x sin x sin 2 x = 2 sin x cos x =2 cos x sin x sin 3 x = 2 sin 2 x cos x sin x =2 cos 2 x sin x -(- sin x sin 4 x = 2 sin 3 x cos x sin 2 x = 2 cos 3 x sin x -f- sin 2 z etc. etc. 73. From (103) and (104) cos (y -\- x) = 2 cos y cos x cos (y x) cos (y -\- *) 2 sin y sin x -(- cos (y x) which, if we put y=(m 1) x, become cos mx = 2 cos (m 1 ) x cos x cos (m 2) x (165) cos mx 2 sin (m 1) x sin x -)- cos (m 2) x (166) If wi is taken successively equal to 1, 2, 3, 4, etc. cos x = cos x cos x cos 2 x = 2 cos x cos a; 1 = 2 sin x sin x + 1 cos 3 x 2 cos 2 x cos x cos x = 2 sin 2 x sin x -f- cos x cos 4 x = 2 cos 3 x cos x cos 2 x = 2 sin 3 x sin x -\- cos 2 x etc. etc. 74, In (123) let y (m 1) x; then tan wix = ta "( m ~ l)x + tanz 1 tan (m 1) x tan x whence tan x = tanx tan 3 x = -* an 2 x + tanx - 1 tan 2 x tan x tan2z = 2tanx - tan4x= 1 tan 1 x 1 tan 3 x tan x etc. 75. If in the expression for sin 3 x, Art. 72, we substitute the value of sin 2 x, we find sin 3 x = 4 sin x cos* x sin x, by which we find the sine of the multiple directly from the functions of the simple angle. If this be substituted in the expression for sin 4x, the latter will also be expressed in terms of the simple angle. By these successive substitutions we easily obtain the following tables : sin x = sin x sin 2 x = 2 sin x cos x sin 3 x = 4 sin x cos* x - - sin x sin 4 x = 8 sin x cos* x 4 sin x cos z etc. 76. cos z = cos x cos 2 x = 2 cos* x 1 cos 3 x = 4 cos 5 x 3 cos x cos 4 x = 8 cos* x 8 cos 1 z -j- 1 etc. D 38 PLANE TRIGONOMETRY. 77. If in these equations we substitute for cos 1 x = 1 sin* x they become sin x = sin z sin 2 x = 2 sin a: j/ ( 1 sin 1 z) sin 3 x = 3 sin z 4 sin 3 z sin 4 x = (4 sin x 8 sin 5 x) j/ (1 sin 1 2) etc. 78. cos x = i/ (1 sin 2 x) cos 2 x = 1 2 sin 2 x cos 3 x = (1 4 sin 2 x) -j/ (1 sin 1 x) cos 4 x = 1 8 sin 2 x -)- 8 sin* x etc. From the preceding tables it appears that the cosine of the multiple angle may always be expressed rationally in terms of the cosine of the simple angle ; but that the sine of only thfe odd multiples and the cosine of only the even multiples can be expressed rationally in terms of the sine of the simple angle. 79. By successive substitutions we find from the formulae of Art. 74. tan x = tan x 2 tan x tan 2 x = tan 3 x = tan 4 x = 1 tan 2 x 3 tan x tan 8 x 1 3 tan 1 x 4 tan x 4 tan 8 x 1 6 tan 2 x -j- tan* x etc. 80. The preceding results are but particular applications of general formulae to be given hereafter (Chapter XV.). They are introduced here for the convenience of reference in elementary applications. The powers of the sine or cosine of the simple angle may also be expressed in the multiples of the angle: but they are most readily obtained from the general formulae of Chapter XV. RELATIONS OF THREE ANGLES. 81. Let x, y, and z be any three angles ; we have, by (36) and (38), sin (x -}- y -\- z) = sin (x -f- y) cos z -f- cos (x + y) sin z = sin x cos y cos z -f- cos x sin y cos z -\- cos x cos y sin z sin x sin y sin z (168) cos (z -|- y + z) = cos (x + y) cos z sin (x + y) sin z = cos x cos y cos z sin z sin y cos z sin x cosy sin z cos x sin y sin z (1G9) and in the same way we may develop the sines and cosines of x + y *, x y + z, etc. ; but we may find these directly from (168) and (169) by changing the sign of z, y, etc., and observing (56). RELATIONS OF THREE ANGLES. 39 The quotient of (168) divided by (169) gives, after dividing the numerator and denominator by cos x cos y cos z, , > tan z + tan v + tan z tanz tan y tanz /tnn\ tan (x + V + z) = ~ ^-^ ( 1 70) 1 tan x tan y tan z tan z tan y tan z 82. Let x, y and z be any three angles, and from the equations sin (x z) = sin z cos z cos z sin z sin (y z) = sin y cos z cos y sin z let cos z be eliminated ; we find sin y sin (z z) sin x sin (y z) = sin z (sin z cos y cos z sin y) = sin z sin (z y) If sin z is eliminated, we find cos y sin (z z) cos z sin (y z) = cos z sin (z y) These equations may be more elegantly expressed, as follows: sin z sin (y z) -f sin V sin ( z x ] + sin z sin ( x y) = (171 ) cosz sin (y z) -f- cosy sin (z x) -f- cosz sin (z y) = (172) A number of similar relations may be deduced from these by substituting 90 : x, etc., for z, etc. 83. Let v = J (x + V + ) we have by (104) 2 sin v sin (v z) = cos z cos (2 v z) = cos z cos (y + z) 2 siu (u y) sin (v z) = cos (y z) cos (2 v y z) = cos (y z) cos z the product of which is 4 sin t> sin (v z) sin (v y) sin (v z) = cos z [cos (y z) + cos (y + z)] cos 2 z cos (y -f- 2 ) cos (y z) Reducing the second member by (103) and (134) ; 4 sin v sin (v z) sin (v y) sin (v z) = 2 cos z cos y cos z cos 1 z cos' y cos 1 z + 1 (173) In the same manner we find 4 cos v cos (v z) cos (v y) cos (v z) 2 cos z cos y cos z -{- cos 3 z + cos 2 y + cos 1 z 1 (174) 84. The following may be proposed as exercises. sin z + sin y + sin z sin (z -f y + z) = 4 sin $(* + !/) sin J (* + 2) sin J (y -f- z) (175) cos z + cos y -|- cos z -f cos (z + y + z) = 4 cos (z + y) cos (z -f z) cos } (y + z) (176) tan z + tan y -f tan z tan z tan y tan z ^ Rin 1?-+J-+-?) (177) cos z cos y cos z cot z + cot y + cot z cot z cot y cot z = eos ( z + y + ?) (178) sin z sin y sin z 40 PLANE TRIGONOMETRY. 4 [sin (x -f y + z ) + 2 sin x sin y sin z]* = 4 [sin (x + y) cos z -f cos (x y) sin z] s ' = [1 cos (2 x + 2 y)] (1 + cos 2 z) + [1 + cos (2 z 2 y)] (1 cos 2 z) f (179) + 2 (sin 2 x + sin 2 y) sin 2 z = 2 (1 + sin 2 a; sin 2 y -f sin 2 z sin 2 z + sin 2 y sin 2 z cos 2 z cos 2 y cos 2 z) . 85. Let the sum of three angles x, y and z be w, or a multiple of r, that is, an even multiple of - a condition which is expressed by the equation Z z + y + z = 2n.^ (180) 2 then, tan (x + y + z) = 0, and the first member of (170) being thus reduced to zero, the numerator of the second number must be zero, or tan x -f- tan y -f- tan z = tan x tan y tan z (181) an equation, it must be remembered, that is true only under the condition (180). Since x, y and z may be selected in an infinite variety of ways so as to satisfy (180) it follows from (181) that there is an infinite number of solutions of the problem, " to find three numbers whose sum is equal to their product." Let the sum of three angles x, y and z be or an odd multiple of - ; that is, let - x + y + z = (2n + l)^ (182) then, tan (x -j- y -}- z) = oc, and the denominator of (170) must be zero, or tan x tan y + tan x tan 2 -f- tan y tan z = 1 which, divided by tan x tan y tan z, gives cot x + cot y + cot z = cot x cot y cot z (183) a relation that holds only under the condition (182). 86. Let x-fy + z = nTT = 2n- (184) 2 We have by (93) and (91) cos (x -f- y z) = cos (ra TT 2 z) = ( 1)" cos 2 z cos (x y + z) = cos (n T 2 y) = ( 1)" cos 2 y cos (y + z x) = cos (n TT 2 x) = ( l) n cos 2 x cos(y + z + :r) = cosn7r =( 1) B the sums of the first two and of the second two are by (103) 2 cos x cos (y z) = ( 1) B (cos 2 z + cos 2 y) 2 cos x cos (y + 2) = ( l) n (cos 2x -f 1) and the sum and difference of these equations are 4 cos x cos y cos z = ( 1 )" (cos 2 z + cos 2 y -f cos 2 x + 1 ) 4 cos x sin y sin z = ( 1)" (cos 2z-fcos2y cos2x 1) 4 cos x cosy cos 2= cos2x4-cos2y-|-cos2z + l (185) 4 cos x sin y sin z = cos 2 x -\- cos 2 y + cos 2 z 1 (1 86) the upper sign being taken when n in (184) is even, the lower when n is odd. INVERSE TRIGONOMETRIC FUNCTIONS. 41 In the same manner we obtain =f 4 sin x sin y sin z sin 2 x + sin 2y -(- sin 2z (187) =F 4 sin x cos y cos z = sin 2 z -(- sin 2 y -f sin 2 2 (188) the signs being taken as above. Again, let * + y + * = (2n + l)^ (189) we shall 6nd by the same process 4 sin z sin y sin z = cos 2 z -(- cos 2 y -f cos 2 z 1 (190) zb 4 sin x cos y cos z = cos 2 x -+- cos 2 y -f- cos 2 2 -f 1 (191) 4 cosz cosy cosz sin 2x -\- sin 2y -f sin 2z (192) 4 cos x sin y sin z = sin 2 z -f sin 2 y -)- sin 2 z (193) + or according as n in (189) is even or odd. INVERSE TRIGONOMETRIC FUNCTIONS. 87. If ?/ = sin x y is an explicit function of x, and, since x and y are mutually de- pendent, x is an implicit function of y ; but to express x in the form of an explicit function of y, we write * which is read, t( x equal to the angle (or arc) whose sine is y" and x is called the inverse function of y, or of sine x. In like manner tan ~ l y is " the angle or arc whose tangent is y," etc. 88. Many of the formulae already given may be conveniently expressed with the aid of this notation. Thus, by (16), + tan j z) or if we put y = tan x tan - * y = sec ~ l \/ (1 -f y 1 ) * This notation was suggested by the nse of the negative exponents in algebra. If we have y nz, we also have x = n ~ ' y, where y is a function of z, and z is the corresponding inverse function of y. The latter equation might be read "z is a quan- tity which multiplied by n gives y." It may be necessary to caution the beginner against the error of supposing that sin -1 y is equivalent to - sin y For a general view of the nature of inverse functions, see Peirce's Diff. Calc., Arts. 13, ct seq. 42 PLANE TRIGONOMETRY. And in the same way the formulae of Art. 28 give sin - * y = cosec ~ * - cos - 1 i/(l y 7 ) = tan - 1 y i/(i-y') cos ~ l y = sec ~ * = sin - tan~'y= cot-'-^siu- 1 * cos y Formulae (123) and (124) may be written - 1 =F tan x tan y or putting t = tan z, i' tan y, tan- 1 = =acotA (203) 6 tan A E 2 54 PLANE TRIGONOMETRY. EXAMPLES. 1. Given A = 25 18' 48", a = -085623 ; find b. Ans. 6 = -1810278 2. Given B = 39 17' 5", 6 = -01 ; find c. ^4s. c = -0157934 111. CASE V. Given the two sides, or a and 6. To find A and B. We have (204) 6 To find c. We have sin A = - c = - - = a cosec A (205) c sin A We may also find c directly by geometry, from c 2 = a 2 -f b' z whence c = y (a 2 + 6 2 ) but this is not readily computed by logarithms. EXAMPLES. 1. Given a = 30, 6 = 40 ; find c. Ans. c = 50 2. Given a = 8*678912, 6 = 2-463878 ; find A and c. ^ns. ^4 = 74 9' 4" -1 c = 9-021875 ADDITIONAL FORMULA FOR RIGHT TRIANGLES. 112. By inspecting the tables it will be seen that when the angles are very small, the cosines differ very little from each other ; consequently a small angle cannot be found with very great accuracy from its cosine. For a similar reason an angle that is nearly 90 cannot be accurately computed from its sine. It is therefore desirable, when a rf -quired angle is small, to find it by its sine, and when near 90 by its cosine, or in either case by its tangent or cotangent ; and for this purpose special formula are some- times necessary. We shall deduce several such formulae, from which one adapted to a particular case may be selected. 113. From (197) we find, by (139) 1 cos A = 2 sin 2 \ A = 1 c c (206) which may be used instead of (197), when A is small, that is when b is nearly equal to c. It gives also c b - 2 c sin 2 A (207) by which c b may be accurately found when A is small. ADDITIONAL FORMULA FOR RIGHT TRIANGLES. 55 Also from (197), by (140) 1 A //I COS A\ lid - b\ C - b /nno\ tanM = \l7-7 -- -.) = A/(~TT) = - ( 208 ) \ \l-\-ctmAJ \ Vc + 6 / a 114. From the equation sin A = - c we deduce by (153) and (154) sin (45 *A) = ) (209) 2 (210) tan (45 $A)=J M^) (211) \ \c =F af and from tan .4 = y we find by (151), 6 tan (45 A) = ^ (212) 6 =F a 115. By (136) we have cos 2 -4 = cos* J. sin* .4 = c 1 which, since 2 ^ = A + 90 B =. 90 (.6 A), gives By (135) cos (B A)= sin 2^ = 2 sin 4 cos A = (214) c and from (213) and (214) tan (B-A) = MLrj*i_VL=i2L (215) by wliich 5 ^4 is found with great accuracy when 6 and a are nearly equal. EXAMPLE. Given c = 4602'836, 6 = 4602-21059 to find A. By (206). c 6 = 62541 log 9-7961648 2 c = 9205-672 log 3-9640555 2)5-8321093 \A = 28' 20 // -18 log sin \A = 7'9160547 A = 56' 40 /A 36 The ordinary process gives log cos A = 9'9999410, whence A = 56 r 40 // . Tliese results are obtained by Stanley's Tables, in which the log. sines, etc., are given for every 10" for the first 15. A greater discrepancy between the two results would be found by tables in which the functions were given only for each minute. A slight error remains in the value of A = 28' 20 //- 18, on account of the large differences of the log. sines in this part of the table, or rather on account of the 56 PLANE TRIGONOMETRY. rapid change of these differences. We avoid the use of these large differences, and gain somewhat in accuracy, by employing the approximate value of sin. A given by (98), whence sin ^A ^Asml f/ t %A = sm * sin 1" Thus we have found above log sin J A = 7'9160547 Art. 54, log sin 1" = 4*6855749 \ A = 1700"-12 = 28' 20 /A 12 log i ^4 = 3-2304798 But to obtain J A with the utmost precision, recourse must be had to the following process, which is constantly employed in observatories, and wherever small angles are to be computed with extreme accuracy. Special tables are prepared containing for every minute from to 2 the logarithms of *H=h and which do not vary rapidly, and may therefore be taken with accuracy from the tables. Then we have tan x = x . A tan x A table of this kind will be found on page 156 of Stanley's Tables, where the notation used is q = log sin x, n = log x and therefore in the column marked q n we find the log 51IL?. Thus in the above x example we have found log sin J A = q = 7 % 91 60547 and from the table q -n = 4*6855700 } A = 1700"'14 = 28' 20 /A 14 log } A = n = 3'2304847 which is the true value of J A within //- 01. Stanley's Table contains also the values of log - q n X n lug i.) 1B ' ' 7 ~l~ n sin x i X \n(t n 4- n (n Inrr rot, r,. the use of which may easily be inferred from the example just given. CHAPTER VII. FORMULAE FOR THE SOLUTION OF PLANE OBLIQUE TRIANGLES. 116. As every oblique triangle may be resolved into two right triangles by a perpendicular from one of the angles upon the opposite side, we are enabled to deduce all the formulae for their solution from those of the preceding chapter. 117. The sides of a plane triangle are proportional to the sines of their opposite angles. Denote the angles of the triangle ABC, nG ' 16> c Fig. 16, by A, B and C, and the sides oppo- ^^** X T\ site these angles respectively by a, 6 and c. ^^* M \ From C draw C P perp. to A B and put A * p CP = p. Then in the right triangles AGP, BCP, we have, by (195) p = b sin A, p = a sin B whence b sin A ~ a sin B which, converted into a proportion, gives a : b = sin A : sin B (216) and in the same way we may prove that a : c = sin A : sin C b : c sin B : sin C and these three proportions may be written as one, thus : a : b : c = sin A : sin B : sin C or thus, -- = -- = -^ (218) sin A sin B sin C When the perpendicular falls without the c triangle, Fig. 17, the angle CB P is the sup- plement of J5, but by Art. 39, it has the same sine, so that the triangle C B P gives p a sin CB P= a sin B 8 57 58 PLANE TKIGONOMETKY. the same as was found from Fig. 16. The proposition is therefore general in its application.* 118. The sum of any two sides of a plane triangle is to their differ- ence as the tangent of half the sum of the opposite angles is to the tangent of half their difference. For, by the preceding article, a : b = sin A : sin B whence, by composition and division, a + b : a b = sin A + sin B : sin A sin B But from (109) if x = A, y = B we obtain the proportion sin A + sin B : sin A sin .B = tan ^ (A + B) : tan ^ (A B) which, compared with the above, gives a + b:a b = teu$(A + B):tan%(A B) (219) This may also be written o+ ten ^ + 3 a - b tan (A - B) (220) and we may infer the same relation between b, c, B, C and a, c, A, C. 119. The square of any side of a triangle is equal to the sum of the squares of the other two sides diminished by twice the rectangle of these sides multiplied by the cosine of their included angle. In the triangle ABC, Figs. 16 and 17,* we have either BP=c-AP orBP = APc p FlQ 17 but in both cases BP* = AP 2 + c* 2cXAP Adding C P 2 to both members, we find But the triangle A CP gives by (196) A P = b cos A *The consideration of Fig. 17 was not strictly necessary according to the principle stated in Art 49. It may, however, be useful for the student to verify that principle when convenient. FORMULAE FOB OBLIQUE TRIANGLES. 59 which substituted in the preceding equation gives a 2 = b 2 + by which an angle is found when the three sides are given ; but to adapt it for convenient computation by logarithms, the following transformations are necessary : Subtract both members from unity; then 2 be b 2 - ') cos $ C _ sin \ (A + B) sin ^ (.4 jB) _ 2 cos \A sin ^ .5 sin $(A + B) cos i (7 Substituting s = } (a + 6 -f c), these equations become s_ _ cos ^ A cos c sin J (7 sin a c _ sn c sin (7 6 _ sin ^ A cos Tr O "FT O TT 2Ji sin 5 = ?^ sinC=?-f (243) be ac ab The quotient of the first of these divided by the second is sin A ac a sin B be b which brings us back to the theorem of Art. 117. 127. The sum of A, B and C being 180, and the sum of $ A, J B and J C being 90, we have, by Arts. 85 and 86, the following relations among the angles of a plane triangle. tan A + tan B + tan 0= tan A tan B tan C cot i A + cot } B + cot \ C cot J ^4 cot } B cot } C sin J. + sin 1? -|- sin C= 4 cos J .4 cos i 5 cos C sin .4 + sin B sin 0=4 sin \ A sin j- ./? cos J C' cos ^4 + cos B -f- cos C 1 = 4 sin J A sin J 5 sin J (7 cos ^4 -f- cos B cos (7+ 1 = 4 cos \ A cos J J? sin J (7 in the last of which we may interchange A, B and C. These relations may be substi- tuted in the equations of Art. 125. * K is the area of the triangle. See Art. 148. FORMULJ2 FOR OBLIQUE TRIANGLES. 63 128. The following equations are added as exercises. tanM _s-b tan J B a a sin (A B} _ (a + 6) (a 6) sin (A + B) c tan J .4 tan B cot (7= ^ ^ JL oot M + cot } J? + cot } C= sin i ^4 sin J B sin ^ C= cos J A cos J 5 cos (7= abcs tan ooc \B tan J C= - (a 2 6 1 -a 4 6* CHAPTER VIII. SOLUTION OF PLANE OBLIQUE TEIANGLES. FIG - 18 - 129. CASE I. Given two angles and one side, or A, B and a. Fig. 18. To find the third angle. We have C= 180 (A + 3) To find b and c. We apply the theorem of Art 117, and state the proportions thus : the sine of the angle opposite the given side is to the sine of the angle opposite the required side, as the given side is to the required side. Thus we have whence and whence sin A : sin B = a : b , a sin B . ~ 6 = ; - = a sin B cosec A sin A sin A : sin C=a : c a sin C c = sin A = a sin C cosec A (244) (245) EXAMPLES. 1. Given A = 50 38' 52", 5 = 60 7' 25" and a = 412-6708, to find C, 6 and c. ^ + #=110 46' 17" d= 69 13' 43" By (244). By (245). log cosec 0-1 1 16730 log cosec O'l 1 16730 log sin 9-9380702 log sin 9-9708 129 log 2-61 56037 A = 50 38' 52" 72 = 60 7' 25" C = 69 13' 43" a = 412-6708 log 2-61 56037 log 6 2-6653469 6 = 462-7505 log c 2-6980896 c=-498-9875 fi4 SOLUTION OF OBLIQUE TRIANGLES. 65 2. Given 4 = 100 16' 35", = 25 16' 13", and 6 = 29-167, find a and c. Am. a = 67-22857 c=55-59178 130. CASE I. Given A, B and a. Second solution. We find C= 180 (^1 + 5) ; then, by (233) and (234) (246) .. cos J (-B + C) sin } A -.-. !"Li-i*=l = a . Bit(J-0) (247 ) sin * (-B + C) cos M wliich give the sum and difference of the required sides; adding half the difference to half the sum, we find the greater side, and subtracting half the difference from half the sum, we find the less side. 131. CASE I. Given A, B and a. Third Solution. When A and B are nearly equal, and great accuracy is desired, we may compute the difference between o and 6; for we have, from (244), , a sin B sin A sin B a o = a -- = a . -- : - - sin A sin A or a b = 2 a Cos K A + B } sin ^ A ~ B) (248) sin A EXAMPLE. Given A = 35 40' 12"-3, B = 35 37' 48"-6, and a = 26246-948. A = 35 40' 12"-3 log cosec 0-2342442 0'23424 B = 35 37' 48 // -6 log 2 0'3010300 0'30103 l(A+B) = 35 39' 0"-5 log cos 9-9098720 9*90987 I (A B)= V ll"-9 log sin 6-5423038 6.54230 a = 26246-948 log 4-4190788 4-41908 a _ ft 25-499 log 1-4065288 1-40652 6 = 26221-449 One of the advantages of this process is, that a b may be found with sufficient accuracy with five-figure tables, as in the second column of logarithms above. If a had been given to ten figures instead of eight, we should still have been able with the seven-figure logs, to find a 6 to seven figures, and therefore b to ten figures, which could not be done by the ordinary methods without ten-figure tables. 132. CASE II. Given two sides and an angle opposite one of them, or, a, 6 and A. To find B. To find the angle opposite the other given side, we apply Art. 117, and state the proportion thus: the side opposite the given angle is to the side opposite the required angle as the sine of the given angle is to the sine of the required angle. Thus, with the present data, we have a : b = sin A : sin B whence sin B = - (249) a 9 r 2 66 PLANE TRIGONOMETRY. To find C. We have C= 180 (A + B) To find c. Having found C, we now have the data of Case I. ; therefore, by (245) c = a sin C cosec A (250) 133. It is shown in geometry that when two sides and an angle opposite one of them are given, there may be constructed two triangles, as in Fig. 19, whenever the given angle is acute and the given side opposite to it is les* than the other given side. In one of them, the required angle B is acute, and in the other it is obtuse, and the two values are supple- ments of each other ; for B = B B' C= 180 A B' C These two values of B are given in the trigonometric solution by the consideration that sin B found by (249) is at once the sine of an acute angle, and the sine of its supplement, Art. 39. In general, when an angle is determined only by its sine, it admits of two values, supplements of each other, unless the conditions of the problem are such as to exclude one of these values. In the present case, the obtuse value of B is excluded when a is greater than 6, and there is but one triangle whether A is acute or obtuse, as in Fig. 20. 134. If the given parts were such that a = b sin A a would be equal to the perpendicular from C upon the side c, and we should have but one solution, namely, a right triangle, B and its supplement both being 90. 135. If the given parts were such that a 90. 136. When there are two solutions, represent the two values of B by B' and B", then the two values of Cwill be C' = 180 (A + B'} == 180 B' A = B" A (251) C" = 180 (A + B"} = 180 B" - A =' A (252) SOLUTION OF OBLIQUE TRIANGLES. 67 and the two values of c will be c' = a sin C" cosec A c" = a sin C" cosec A (253) EXAMPLES. 1. Given = 31-23879, 6 = 49-001 17 and;! = 32 18'; find 5, C and c. = 31-23879 ar. co. log 8-5053058 b = 49-001 1 7 log 1-6902064 A = 32 18' log sin 9-7278277 B' = 56 56' 56"-3 log sin 9-9233399 .#"=123 3' 3"-7 C' = 90 45' 3"-7 log sin 9-9999627 C" = 24 38' 56"-3 log sin 9-6201962 log cosec A 0-2721723 0-2721723 log a 1-4946942 1-4946942 log c' 1-7668292 log c" 1-3870627 c' = 58-45601 c" = 24-38163 Ana. B = 56 56' 56"-3 *| f = 123 3' 3"-7 C= 90 45' 3"-7 I or \ (7= 24 38' 56"-3 c= 24-38163 2. Given a = -051234, b = -042356, A = 55 ; find B, C and c. Ans. B =6= 42 37' 32"-7 C= 82 22' 27"-3 c= .06199202 ( B = 97 45' 24"-3 C=27 14' 35"-7 3. Given a = -042356, b = -051234, yl = 55 ; find B, Cand c. 4n*. ^ = 82 14' 35"-7 "| C=42 45' 24"-3 lor c= -03510331 (^ c = -02366993 4. Given a = 40, 6 = 50, A = 53 7' 48"-4 ; find A Ans. jB = 90. 5. Given a = 40, 6 = 50, A = 60 ; solve the triangle. Ans. Impossible. 6. Given 6 = 40, c = 50, B = 100 ; solve the triangle. Ans. Impossible. 68 PLANE TRIGONOMETRY. 137. CASE II. Given o, 6 and A. Second Solution. We may solve, separately, the two right triangles AP 0, B PC, Fig. 21, which is a convenient method when there are two solutions. We first find B by (249) ; then we have P AP = bcosA, P=acoaB and c = A P + B P The cosine of the obtuse value of B is negative, (Art. 39), so that B P is then nega- tive, and we have the two values of c from the formula c=APBP There will be but one solution, if B P > A P, for c cannot be negative. 138. CASE III. Given two sides and the included angle, or a, 6 and C. To jind A and B. We have first A + B == 180 C H^ + -B) = 90-iC from which we next find the half difference of A and B by the theorem of Art. 118, which gives a + b : a b = tan (A + B) : tan (A B) tan 4 (4 .B) = - tan(.A+ J3) = -- cot* C (254) a -f- o a -f- 6 The half difference added to the half sum gives the greater angle, (opposite to the greater given side), and the half difference subtracted from the half sum gives the less angle. To find c. We have the data of Case I., and therefore c = a sin (7 cosec A = b sin C cosec B (255) EXAMPLES. 1. Given a = '062387, 6 = -023475, and C= 110 32' ; find A, B and c. A + B = 180 C= 69 28' a+b = -085862 ar. co. log 1-0661990 a b=-. -038912 log 8-5900836 B)=-- 34 44' log tan 9*8409174 (A B) = 17 26' 33" log tan 9-4972000 A= 52 10' 33" B = 17 17' 27" log cosec 0-5269189 C= 110 32' log sin 9-9714931 b = -023475 log 8-3706056 c = -0739635 log 8-8690176 Ans. A = 52 10' 33" 5=17 17' 27" c = -0739635 SOLUTION OF OBLIQUE TRIANGLES. 63 2. Given a = 31-0005, 6 = 15-1101, C= 10 15' ; find A and B. Ans. ^ = 160 17' 13"-7 B = 9 27' 46"-3 3. Given a == 2-463878, 6 = 9-021875 and C= 74 9' 4"-2 ; find A and B. Ans. A = 15 50' 55"-8 5 = 90 0' 0" 4. Given b = 15-1101, c = 31-0005, A = 10 15' ; find B and C. ^ns. J5 = 9 27' 46"-3 C= 160 17' 13"-7 139. Having found A and B as above, the most convenient mode of finding c is by (233) or (234), which give (257 > for we have, from the process of finding A and B, the log. of a -f- b, or of a b, and the values of (.4 + -B) an( i J (^ -^) so tna t we have only two new logs, to find, which are taken out at the same opening of the tables with the tangents of J (A -\- B) and J (A B). 140. CASE III. Given a, 6 and (7. Second Solution. When a and 6 are given by their logarithms, which occurs when they are deduced by a logarithmic process from other data (as, for example, in the computation of a series of triangles in a survey), we proceed as follows. Let x be an auxiliary angle, such that tan* = 7 (258) 6 an assumption always admissible, since a tangent may have any value from to oo . We deduce tan x 1 _ a b tan x -f- 1 a + 6 or by (1 52) tan (a; - 45) = ^ ^ a -f- 6 which substituted in (254) gives tan % (A B) = tan (x - 45) tan % (A + JB) (259) 70 PLANE TRIGONOMETRY. We find x from (258) and employ its value in (259). As this method does not require the preparation of a -f- b and a 6, it is quite as short in practice as (254). EXAMPLE. Given log a = 8-7950941, log b = 8-3706056, and C= 110 32'. (Same as Ex. 1. Art. 138.) log a = 8-7950941 log b = 8-3706056 x = 69 22' 46"-8 log tan 0-4244885 x 45 = 24 22' 46"-8 log tan 9-6562825 $(A + ) = 34 44' log tan 9-8409174 $(A B) = 17 26' 32"-9 log tan 9-4971999 141. CASE III. Given a, b and C. Third Solution. To express A or B directly in terms of the data, we have, from (218) and (224) c sin A a sin C c cos A = b a cos C the quotient of which is and in the same manner tan .4= " sn ^ (260) 6 a cos C ton 3 = (261) a b cos C 142. CASE III. Given a, b and C. Fourth Solution. To find c directly from the data, we have, by (223) & = a* + 6* 2 ab cos C which, however, is not adapted for logarithmic computation. It may be adapted as follows. Substitute by (139) cos (7= 1 2 sin*} C then c* = a '-f& s 2a6 + 4a6sin 2 J (7 = (a 6)* + 4 ab sin 1 J C (262) Let x be an auxiliary angle, such that tan , 4o6 B in'}(7 (a -by siniC a b SOLUTION OF OBLIQUE TRIANGLES. 71 then the radical in the above equation becomes \/ (1 + tan* x) = sec x ; therefore, c=(a. b)secx (264) 143. We may also adapt (223) for logarithmic computation by means of (138), which gives cosC = 1 -f 2 cos 2 J C whence c = o + 6 2 + 2 ab 4 aft cos 1 $ C (265) Let (266) then the radical becomes y (1 sin 1 x) = cos z; therefore, c = (a + 6) cos z (267) 144. It is to be observed, that the supposition (263) is always possible, jjnce a tan- gent may have any value between and oo , and therefore an angle x may always be found having any given number as its tangent. As the greatest value of a sine is unity, it is not so obvious that the supposition (266) is always possible; but whatever the values of a and 6 (a 6) 1 ^ therefore (a + ft) 1 >_ 4 ab 2l/~a5 whence a -f- o = therefore the second member of (266) is never greater than unity. EXAMPLE. Given a = "062387, 6 = -023475, C= 110 32'; (same as Ex. 1, Art. 138) By (266) and (267). a = -062387 log 8-7950941 6 = -023475 log 8-3706056 2)7-1656997 = 8-5828499 a 4- b = -085862 ar. co. log 1-0661990 log 8'9338010 (7= 55 16' 1. cos 9-7556902 log 2 0-3010300 1. sin x 9-7057691 1. cos x 9'9352161 c = -07396344 log S'8690171 72 PLANE TRIGONOMETRY. 145. CASE IV. Given the three sides, or a, b and c. To find A, B and C. We have from (227) and (228X sin i A = ~ ' ' * "- ' . , //(-a)( sin | .S = A/ I ^ \ CLG , ml c= l/( s ~ a ^ s - or by (229) and (230) (268) cos i C' = (269) or by (231) and (232) (270) In these formula s = ^ (a + b -\- c). Either of these three meth- ods may, in general, be employed, but (268) is to be preferred when the half angle is less than 45, and (269) when the half angle is more than 45.* When all the angles are required, (270) will be the simplest, as it requires but four different logs, to be taken from the tables. It is accurate for all values of the angle. * See Art. 112. SOLUTION OF OBLIQUE TRIANGLES. 73 EXAMPLES. 1. Given a = 10, 6 = 12, c = 14 ; find the angles. By (268). a = 10 ar co 1 9-0000000 ar co 1 9-0000000 6=12 ar co 1 8-9208188 ar co 1 8-9208188 c = 14 ar co 1 8*8538720 ar co 1 8-8538720 2 = 36 8 = 18 a a = 8 log 0-9030900 log 0-9030900 s b= 6 log 0-7781513 log 0-7781513 8 c = 4 log 0-6020600 log Q-6020600 _ 2)9-1549021 2)9-3590220 2)9-6020601 log sines -49-5774510 59-6795110 (79-8010300 |^t = 2212'27"-6 B = 28 33' 39"-0 \ (7= 39 13' 53"-5 A = 44 24' 55"-2 5 = 57 7' 18"-0 0= 78 27' 47"-0 Verification. A + 5 + (7= 180 2. Given a = -8706, 6 = -0916, c = -7902 ; find the angles. Am. A = 149 49' 0"-4 B= 3 1' 56"-2 C= 27 9' 3"-4 3. Given a = -5123864, 6 = -3538971, c = -3090507 ; find C. Ans. C=36 18' 10"-2 146. The computation by (270), when all the angles are required, will be much facilitated by the introduction of an auxiliary quantity* (271) from which we find by (270) tan J A = -^, tan J B = ?, tan } 0= r (272) s a s 6 8 c *This quantity r is the radius of the inscribed circle. See (289). 10 Q 74 PLANE TRIGONOMETRY. EXAMPLE. Given o = 6053, b = 4082, c = 7068. We find s = 8601-5 ar. co. log 6 0654258 s a = 2548-5 log 3'4062846 , a b = 4519-5 log 3 6550904 < e 1 533-5 log 3-1856838 2)6-3124846 logr = 3-1562423 J A = 29 20" 54"'47 log tan * A = log T = 9'7499577 s a J J5 = 17 35' 31"-70 log tan } B = log -- = 9'5011519 s b J C= 43 3' 33"'83 log tan J C = log = 9'9705585 Verification. 90 0' 0"'00 147. The case where the three sides are given is some- times solved as follows. From C, Fig. 22, draw CP perp. to c. Then - -* the difference of which is A C* B C" = A P* B P 1 or (4C+.BC) (40 C) = MP+P) (4P- and if .4 P B P= d, this equation gives d(o -f- a) (o a) />7M\ = * -* ' (***) c Then, since A P-f BP = c, and ^4 PBP~d, we have ^P=J(c + d), ^P=J(c d) (274) and in the right triangles A CP, B CP cosA = ^f, c*xB= S - p - (275) o a 30 that (273), (274) and (275) solve the problem. When d > c, B P is negative, cos B is negative, and B is an obtuse angle, (Art. 39). AREA OF A PLANE TRIANGLE. 148. Representing the area by K, and the perpendicular CP, Fig. 22, by p, we have, by geometry, JT=Jcp (276) In the triangle A C P, we have p = b sin A, whence K J be sin A = be sin J A cos } A (277) by which the area is compute*! from two sides and the included angle. Substituting in (277) the values of sin J A and COB } A by (268) and (269), K^^/[s(s a) (s b) (-c)] (278) by which the area is computed from the three sides. CHAPTER IX. MISCELLANEOUS PROBLEMS RELATING TO PLANE TRIANGLES. 149. In a given plane triangle, to find the perpendicular from one of the angles upon the opposite side. Let p be the perpendicular from C upon c. We have p = b sin A (279) or by (239) and (278), the expression for p in terms of the three sides, where s = J (a -f- 6 -f- c) and JT is the area of the triangle. If we substitute in (279) sin A = - sin C, it becomes c = - sin C (281) or, if we substitute the value of b = c sin C sin A sin J? sin A sin J? sin C s When the triangle is right-angled at C, (282) becomes p = c sin A cos A - sin 2 A 2 the expression for the perpendicular upon the hypotenuse. 150. If p' t p // , p /// , denote the perpendiculars upon the sides a, b, e respectively, we have from (280) _J__L 1 i 1 _ + M-_ h " "~ '" ~ 2 A' 76 76 PLANE TRIGONOMETRY. 151. To find the radius of the circle circumscribed about a plane triangle. F 10 - 23- The center of the circle, Fig. 23, lies in the perpen- dicular erected from the middle point of one of the sides, as A B. Let the radius = R. We have, by geometry, and in the triangle A D, 'B _ n AD AO 2R whence R = 1 (284) 2 sin O V ' Substituting the value of sin C from (241), r> abc abc K (285) From (229) and (230), we easily find cos J4 MS 1 .B cos J 0= abc which combined with (285) gives jj & (ftofi} 4 cos A cos J B cos $ C 152. To find the radius of the circle inscribed in a plane triangle. FlG - 24 - The required center 0, Fig. 24, is in the inter- section of the three lines bisecting the angles, and each of the perpendiculars D, E, OF, is equal to the required radius = r. The value of O F in terms of A B = c, A B J A, and sin \ A sin ^ B sin fr A sin | B (9K7\ an}(A+B) ~ cos } G This is reduced by means of (236) to r = (s eJtanJC' (288) Substituting the value of tan J C, r =J(( s a ) (* b "> (* &\=K (289) This is reduced by means of (231) and (232) to r = s tan J A tan J J? tan } C (290) PROBLEMS RELATING TO PLANE TRIANGLES. 77 153. Besides the inscribed circle, strictly so called, there are three other circles that touch the three sides, (or sides produced), and are exterior to the triangle, as in Fig. 25. These have been named escribed circles. Their centers are found geomet- rically, by bisecting the exterior angles CC / , CBB / , etc. Designate the centers of the circles lying within the angles A, B, and C respectively, by / , O", and 0"', and their radii by r', r // , r /// . We find the perpendiculars from O f } etc., upon B (7, etc. by (282) to be / _ cos \ B cos ^ O cos \ B cos \ C ' sin %(B-\-C) cos $ A } (291) tf f cos \ A cos \ B cos sin ^ (.4 -f- B) By means of (235) we reduce those values to r' s tan \ A, -= s tan j B, cos \ O r" f = s tan } C Substituting the values of tan A, etc. r / = / ((-6) \\ c)\ _. r'" = (s o) (s 6)^ _ 8 C (292) (293) 78 PLANE TRIGONOMETRY. Also, by means of (236) applied successively to a, b and e, we may reduce (291) to the following : r' ( o) tan J A cot $ B cot J C r" = ( s b) cot i ^4 tan $ J? cot $ C = (s c) cot J /I cot $ 5 tan $ C (294) 154. Relations between the radii of the circumscribed, inscribed, and thret escribed circles of the preceding article, and the three perpendiculars from the angles upon the opposite sides. The four equations of (289) and (293) give s ( a) (s - b) (s - c) K* Dividing this successively by r 1 , r' 1 , etc. ^~^- =*' ^9^=( 8 -a)' r r' r"' _ r r' r" (295) (296) Again, we have, (Art. 127), tan }A tan B + tan $ A tan \ C + tan 5 tan \ C 1 and substituting in this the value of the tangents from (292) r/ r // _j_ r / r / 1,1,1 1 T-H 77- + -777 = - r r /x r w r (297) From (292) we find tan \A _ T tan \ B ~~ r r" tan } A r' tan } B from which it follows that in Fig. 25, the distances A D and B D', are equal (D, D' being the points of contact of the circles 0', O' f with A B produced), and therefore B D = A D'. Other curious geometrical properties may be traced with the aid of our equations. From (284), n _ __ a _ _ b __ _ _ c _ 4 sin J A cos J A 4 sin J B cos B 4 sin J C cos J (7 which combined with (287) and (291) give, by Art. 127, = 4 sin J >4 sin j B sin j (7 = cos -4 -f cos -B -|- cos C 1 R -. R r" -- = 4 cos } A sin J 5 cos f A -~ JB cos ^4 cos J5 -f- cos (7 -}- 1 cos^l + cos-B cos(7+l (298) PROBLEMS RELATING TO PLANE TRIANGLES. 79 Changing the signs of the first of these equations, the sum of the four is r/ + r// + r/// "^- r - 4, B = t(r' + r" + "' - r) (299) Finally, if p', p", p x// denote the three perpendiculars from the angles upon the sides a, 6, c respectively, we have by (283), (289) and (297) the following relation: JL -4- J_ 4- 1 -L 4. JL 4. _A_ = L gf * p" ^ $'" **'- r"' r (299*} 155. To find the distance between the centers of tlie circumscribed and inscribed circles.* Let P, Fig. 26, be the center of the circumscribed, and O, that of the inscribed circle. Put P O = D. By Arts. 151 and 152, whence FIG. 26. and by (221) r>_ r> therefore A 1 - g -2PA. OAcosPAO _f __ 2-RrcoaH-B-C') sin 2 \ A sin \ A _ 4 Rr sin J? sin J (7 . sin 4 5 vl (J + (300) 156. Let PO / = D / , Fig. 26, O 7 being the center of the escribed circle lying within the angle A. If r' = radius of this circle, we have, as in the preceding article n/z PS _u r/t 2 Rr' cos \ (BG - ~T-'J1J l j j sin 4 Rr' cos JS cos j (7 sin J = J? 4- 2 Kir (301) * Hymers' Trig. Appendix, Art. 58. 80 PLANE TRIGONOMETRY. the expressions for the distances of the centers of the three escribed circles from that of the circumscribed circle. The sum of (300) and (301) gives by (299) Z> 2 + D' J + iy n + D" n = 12 IP (302) 157. Given two sides of a plane triangle and the difference of their opposite angles, (or a, 6, and A ). to solve the triangle. We have \ (A + B) directly from (220), which also solves the case where two angles and the sum or difference of two sides are given. 158. Given the angles and the sum of the sides, (or A, B, C, and a4-6-f-c = 2s). By (235) cos A cos J B and a and b are found by similar formulae. 159. Given one angle, the opposite side, and the sum of the squares of the other two sidei, (or C, c, and a' + b 3 = e 2 ). In the identical equations (a + 6)* = e 2 -f 2 oi, (a b) t = e s 2ab substitute the value of 2 ab given by (223), namely, e 1 c 1 2ab = cos C we find (a + 6) = e 1 + -, (a - 6) = * - = cos (7 cos C which determine a -f- b, and a b, and therefore a and b. To compute these equations by logarithms, let then (a + 6)* = e 2 + f, (a b) t (303) that is a + b is the hypotenuse of a right triangle whose sides are e and g ; and a b is one side of a right triangle whose other side is g, and whose hypotenuse is e. Let the angle opposite g be denoted by x in the first triangle and by x' in the second, then by the formulae of right triangles tan x = -* e sm z' = * o o = e cos z' e (304) so that the problem is solved by logarithms by finding log g from (303) and employing its value in (304). The above may serve as an example of a geometrical method of introducing the auxiliary quantities, which is occasionally useful. The analytical process in the present instance is similar to that of Art. 143; thus PROBLEMS RELATING TO PLANE TRIANGLES. 81 therefore if tan x = -2 we have -/ ( 1 -f- i = sec z e \ \ e 1 / and if sin x r = -^ we have -* I 1 1 I = cos a/ e \ \ e 1 ) whence the same formulae as before. 1 60. Given an angle, its opposite side, and the difference of the squares of the other two sides, (or C, c, and a a 6 2 =/ 2 ). We have by multiplying (233) by (234) ain (A B) _ olmi 2 _ /* sin # # sin (A B) - sin C c 1 whence A B, and since A + B 180 C, the angles are determined. There will be two solutions given by sin ( A B) except where the obtuse value of A Ji is greater than A -\- B. 161. Given the three perpendiculars from the three angles upon the opposite sides. Denote the perps. upon a, b and c respectively by a / , b' and c x , and let " If k = 2 area of the triangle aa' = bb f = cc' = k and therefore a = a" k, b = b" k, c Substituting these values of a, 6 and c in (225), (227), etc. , 6" e" in which 2 s" = a" + 6 r/ + c /r . 162. G-iwcw -2r)] [ (305) cos } (A B) = - 2^zl - 1 /[2 J R(p-2r)] which determine | (A -f JS) and J (.4 5) and therefore A and .B. The sides are then found by the formula c = 2 R sin C Fl0 - 27 - 163. 7ft a given plane triangle ABC, Fig. 27, to ^nd a point P suth that the three lints drawn from this point to the angles A, B and C shall make, given angles with each other. Let the given angh-s be BPC=a a.r\(\APC=(i and the required angles P A (7= x P B C = y _ The sum of the angles of the quadrilateral A CB P is C= 360 whence } (x + y) = 180 J (a + ft -j- C) (306) PROBLEMS RELATING TO PLANE TRIANGLES. In the triangles A PC, B PC, we have 83 PC- b sin . x a sin y sin /? sin a sin x a sin 9 = - . - = =m sin y o sin a from which sin x -f- sin y _ tan | (x -f- y) _ OT -p- 1 sin a; sin y tan J (x y) TO 1 tan H* -y) = tan * (x + y) m + To compute this equation by logarithms, let a sin 8 tan 7 = m = b sin a then by (152), tan J (x y) = tan (7 45) tan % (x + y) so that the angles x and y are found by (306) and (307). 164. The following problems are proposed as exercises. In a plane triangle ABC 1. Given c, the perp. upon c = p and a -f- b = m. (m + c) (m c) a 6 = c cos x 2 pc (m + c) (TO c) 2. Given c, the perp. upon c = p, and a b = n. tan 7 x (c -f n) (c n) a _[. ft c gee a: 2 pc 3. Given C, c, and a& = and the radius of the inscribed circle = r. m e 11. Given c, a b = n, and the radius of the inscribed circle = r. 4r" 12. Given the radii >', r", r fff , of the three escribed circles. (Arts. 153, 154.) r /2 tan* ft A = _ - r / r ff + r / r /// 165. Given the sides of a quadrilateral inscribed in a circle, to find its angles and area. FIG. 28. i n Fig. 28, let AB = a, BC=b, CD = c, DA d. -^ Let 2s a + 6-fc + d and ^T=area of A BCD; then from the triangles ABC, ADC, observing that B = 180 D we find 06 + cd a6 -f cei . c) (s d) = v/ [( - a) (s b) (B -c)(s- d)] (308) CHAPTER X. SOLUTION OF CERTAIN TRIGONOMETRIC EQUATIONS AND OF NU- MERICAL EQUATIONS OF THE SECOND AND THIRD DEGREES. 166. THE solution of a problem in which the unknown quantity is an angle, often depends upon that of one or more equations, in- volving different functions of the angle, which cannot be reduced by merely algebraic transformations. We shall select a few simple ex- amples of such equations from among those that most frequently occur in astronomy. 167. To find z from the equation sin ( -f z} = m sin z (309) in which a and m are given. We have, by (119), sin (a -\- z) = sin a sin z (cot z + cot a) which becomes identical with (309) by taking sin a (cot z -f cot a) = m whence the required solution 7?i cot2 = - -cot a (310) sin a If the proposed equation were sin (a z) msinz (31 1) we should find r*n cot2 = - - + cota (312) sin a Unless 2 is limited by the nature of the problem in which these equations are employed, there will be an indefinite number of solu- tions ; for all the angles z, z + 180, z -f- 360, z + 540, etc., in general all the angles z -\- n TT have the same cotangent. [See (68), (79).] In most cases, however, we consider only the first two of these solutions, taking the values of z always less than 360. H 85 86 PLANE TRIGONOMETRY. Similar remarks apply in all cases where an angle is determined by a single trigonometric function ; but if the problem is such as to give the values of two functions of the required angle, as the sine and cosine, the solution is entirely determinate under 360, since there cannot be two different angles less than 360 that have the same sine and cosine. 168. The solution of the preceding article requires the use of a table of natural cotangents ; to obtain a formula adapted for logar- ithmic computation entirely, we deduce from (309) the following sin (a -f- 2) -|- sin z _ m -f 1 sin (a + z) sin 2 m 1 But by (109), if x = a -f z, y = z, we have sin (a -f- z) -f- sin z _ tan (z -|- ^ a) sin (a -f z) sin z tan | a which substituted above, gives t , i , m -f 1 tan (z -f- IT ) = - - tan i a m1 which determines z -j- ^ a, whence z is found by deducting ^ a. The computation of this equation is facilitated in most cases by introducing an auxiliary angle, such that tan

+ 45) tan a (31 4) SOLUTION OF TRIGONOMETRIC EQUATIONS. 87 169. To find zfrom the equation tan (a -f- 2) = m tan 2 (315) We deduce tan (a -f- g) H~ tan 2 m + 1 tan (a -f- 2) tan 2 m 1 so that by (126) and (152) the solution is tan

) cos a (318) 171. To find z from the equation sin (a 2) sin z = m (319) By (108) we find cos a. cos (a 2 z) = 2 sin (a z) sin z = 2 m whence cos (a rfc 2 z) = cos a =F 2 wi (319 *) which determines a =b 2 z, and hence 2 z. From (319*) we have four values of a 2z between and 720 ; therefore, four values of 2 z between the same limits, and four values of z between and 360. In general, we shall have four solutions under 360 in all cases where the double angle is determined by a single function. The logarithmic solution of (319) varies with the signs of m and z. Thus, if the equation is sin (a -f- z) sin z = m m being essentially positive, we have by (133) cos 1 a cos 2 (z + i ) = sin (a -f- 2) sin z = m cos* (z + i a) = cos* a m and by (133) again this is solved by cos* = m, cos 2 (z -f J o) = sin (^ -f- J a) sin ( } ) and the other cases are solved by similar methods. 88 PLANE TRIGONOMETRY. 172. The preceding examples will suffice to indicate the method to be followed with all the equations of the following table. The solutions of the equations involving cosines may be obtained from those involving sines, by exchanging z for 90 z, or o for 90 a. Logarithmic solutions of the first four will be obtained by imitating the process of Art. 171. EQUATIONS. SOLUTIONS. 1. sin (a db z) sin z = m 2. cos (a z) cos z = m 3. sin (a rb z) cos z = m 4. cos (a rb z) sin z = m 5. sin (o t) = m sin z 6. cos (a rfc z) = m cos z 7. sin (a ;fc z) = TO cos z 8. cos (a z) = m sin z 9. tan (a z) tan a = m 10. tan (a rb z) = m tan z cos (o 2 z) = 2 m cos a sin (a 2 z) = 2 TO sin a sin (a 2 z) = sin a 2 m tan ^ = TO, tan (z o) = cot (^ =F 45) tan } a tan m, tan ( J a z) = tan (45 ) cot $ o tan ^ = m, tan (45 J a =F z) = tan (45 ) tan (45 + a) tan = TO, tan (45 | a =p z) = tan (45 =F 0) tan ( a 45) tan ^ = m, cos (a 2 z) = tan (45 =F ?) cos a tan $ = TO, sin (2 z o) = cot ( =F 45) sin o In the numerical solutions the signs of the angles and their functions must be care- fully observed. The signs of the functions should be prefixed to their logarithms, according to Art. 99. The auxiliary angle (j> may be taken numerically less than 90 in all cases, but posi- tive or negative according to the sign of its tangent. It can easily be shown that we shall thus obtain the same values of z as by taking in the 2d quadrant when its tangent is negative, or in the 3d quadrant when its tangent is positive. EXAMPLE. Find z from (317) when a = 65 and m = 1-5196154. By (318) log tan = log n + 0'1817337 56 39' 9" -11 39' 9* - 9'3143426 -\- 9-6259483 log tan (45 ) cos a a-f 2z 2z 2 45 -0 log tan (45 log cos a log cos (a -f 2 z) 8-9402909 95 or 265 or 455 or 625 65 30 or 200 or 390 or 560 15 or 100 or 195 or 280 * It must be remembered that in this employment of the signs + and , these signs belong to the natural numbers ; and when the logs, are added or subtracted, the sign of the result is to be determined according to the rules of multiplication and division in algebra. SOLUTION OF TRIGONOMETRIC EQUATIONS. 89 173. To find z from the equation sin (a -f- z) = m sin (ft -}- 2). Put z' = ft -f- z, a' = a /?, then this equation becomes sin (a' -\- z') = m sin tf which is of the form (309) and may be solved by (309*) or (311) ; then z = z' ft. In the same manner equations of this form, involving cosines or tangents, may be reduced to those of the preceding table. 174. To find k and z from the equations k sin z = m ' We have, by division, V (320) k cos z = n m tan z = which gives two values of z, one less, the other greater than 180; whence, also, two values of k from either of the equations 7 _ m n k ~ sin 2 cos z The solution becomes entirely determinate (z not exceeding 360) as follows : 1st. When the sign of k is given. For if k is positive, sin z has the sign of m, and cos z the sign of ??, and z must be taken in the quadrant denoted by these signs. If k is negative, the signs of sin z and cos z are the opposite to those of m and n, and z must be taken accordingly. 2d. When z is restricted by either the condition z < 180, or z > 180. For under either of these conditions the tangent gives but one solution. If z < 180, k has the sign of m ; and if z > 180 k has the opposite sign to that of m. 3d. When z is restricted to acute values, positive or negative. For under this condition a positive tangent will give z between and -f 90 ; and a negative tangent, between and 90 ; and k will always have the sign of n. It follows that ??i and n being any given numbers whatever, we may always satisfy the conditions expressed by (320), 1st, by a posi- tive number k and an angle z between and 360 ; 2d, by a num- ber k (unrestricted as to sign) and an angle z < 180 ; 3d, by a number k (unrestricted as to sign) and an angle z > 180 ; and 4th, by a number k, and an angle z in the 1st or 4th quadrant. 12 H 2 90 PLANE TRIGONOMETRY. EXAMPLE. To find k and z from (320), (k being a positive number), when m = - 0-3076258, n = + 0.4278735. k sin z - 0-3076258 k cos z + 0-4278735 (a) log k sin z - 9-4880228 (6) logjfe cos z + 9-6313147 (a) (6) log tan 2 -9-8567081 3 324 17' 6"-6 (c) log sin z - 9-7662280 (a) (c) log/; + 9-7217948 k + 0-5269808 Upon this problem and the deductions we have made from it, rests the method of introducing the auxiliary angles required in solving many of the formulae of spherical trigonometry. It is applicable to any equation that can be reduced to the form of that solved in the following article. 175. To solve the equation m cos z -|- n sin z = q (321) m, ?i and q being given. The first member will be reduced to the form k sin ( -\- z) by as- suming k and

be found from (322) by the preceding article, (k being limited to positive values), we can then find by (323) the value ~H z and therefore of z. There will be two solutions from the two values of

= - 54 34' 40", which gives tp + z= 199 34' 40", or 340 25' 20", whence the same values of z as before. We may repeat the latter part of the work with cos tp for verifi- cation. * The solution is, by (323), impossible when -3 is greater than unity ; and by adding the squares of (322), & m 2 -f- n * J therefore the solution is impossible when 9* > m" -f ?i*. 92 PLANE TRIGONOMETRY. 176. To solve the equation a sin (a -f z) -f b sin (/? -f 2) -f csin (y -f 2) -f etc. = g. (325) Developing by (36) and putting a sin o -\- b sin /3 -f" c sin 7 + e tc. = w a cos o -j- b cos p -\- c cos y -f- etc. = n this becomes m cos z + sin a = j which is solved in the preceding article. The same process applies if any or all of the terms contain cosines. 177. To find k and z from the equations k sin (a -f- 2) = m ) Asin (/ + .)= } (326) The sum and difference of these equations are, by (105) and (106), 2 k sin [J (a + 0) -f z] cos J (a 0) = m -f n 2/fccos[$ (a + /j)+s] sin J (a /J)=m n whence 2 A sin , cos J (a /3) sin i a (327) from which 2 & and J (a -(- /?) + z are determined by Art. 174. The logs, of the second members of these equations should be computed separately, for the purpose of readily discovering the signs of the sine and cosine in the first members. The solution is determinate (according to Art. 174) when the sign of k is given. From (327) we find, by division, tan *(<*-/*) (328) which requires a less number of logs, than the separate computation of (327), but we are obliged to refer to (327) to determine (by an inspection of the second members) the signs of the sine and cosine. If we assume tan0 = ^ m (329) re may compute (328) by the formula tan [J (a -f /?) -f 2] = tan (45 + , tan J (a /?) SOLUTION OF TRIGONOMETRIC EQUATIONS. 93 EXAMPLE. In (326) given a = 200, P 140, TO 0'42345 and n~ 0*20123, to find * and k, k being positive. By (327). m -f n 0-62468 m n 0'22222 J( + 0) 170 i(-/3) 30 log (m + n) - 9-7956576 log cos J (a ft) -f 9-9375306 (a) log 2 k sin [i (a -f ft) -f z] 9'8581270 log (m n) 9-3467831 log sin $ (a 19) -f 9'6989700 (b) log 2 A cos [$ (a + |3) + z] 9'647813l (a) (b) log tan [* (a + ft) + z] + 0'2103140 i (a + /3 + z) 23821'38"-6 z 6821'38"-6 (c) log sin [ (a + ft) + a] - 9'9301171 (a) (c) log 2 i + 9-9280099 2 A 0-8472467 k 0-4236234 178. A more general solution of (326) is the following.* Let y be any angle as- sumed at pleasure, and in (171) let x o z, y (distinguishing the e of (171) by an accent) ; then we shall find sin (a /?) sin (y + z) = sin (a y) sin (/? + ) sin (P y) sin (a -f *) In this let y (whose value is arbitrary) be exchanged for y -f- 90 ; then sin (a P) cos (y -f z) = cos (a y) sin (P -(- z) -f- cos (P y) sin (a -f- 2) Multiplying these equations by k and substituting m and n from (326) A sin (o /?) sin (y + z) = m sin (y /3) n sin (y a) j A sin (a /?) cos (y + z) = rn cos (y /3) n cos (y a) J which (y being assumed at pleasure), determine k and y -f- * If we take y = 0, we find tan z m sin /^ + n sin a m cos P n cos a If y = , A sin (a -(- z) = m - If y = /?, we have a similar result. If y = $ (a -(- j3) we obtain the solution of the preceding article. If k is required, without first finding z, we have, by adding the squares of (330) k sin (a ft) = y [m* -f n' 2 m n cos (o /?)] (332) * GAUSS. Theoria Motus Corporum Oaelestium, Art. 78. 94 PLANE TRIGONOMETRY. 179. To find k and z from the equations k cos (a -(- z) = m k cos (/3 -\- z) = n (333) These are reduced to the form (326) by substituting 90 + a and 90 -f /? for a and /?. We find, however, by a process similar to that of Art. 177, 2 A sin [} (a + /J) + ] = 2 A cos [i (a + 0) + ] = n TO sin (a /?) n 4- TO cos J (a /?) cot [ J (a + 0) -f z] = tan (45 -f ) tan $ (a /?) (334) (335) EXAMPLE. In (333) given o = 280 16', ft = 200 10', m G'62342, and n r= 0'69725, find 2 and k, k being positive. Arts, z = 207 5' 34"'4 A = 1-0273643 180. The more general solution of (333) may be found directly from (172), but it will be simpler to obtain it from (330) by substituting 90 + a for o, and 90 -f P for /J, whence k sin (o /?) sin (y -\- z) = m cos (7 /?) + n cos (y o) k sin (a /?) cos (y + z ) = m s i (y /?) 7l si n (7 ) (336) y being arbitrary as before. If y = 0, we find tan z m cos j$ -\- n cos a m sin /3 -f- sin a If y = a, A sin (a + *) = sin (a COS (tt -f 2) = TO (337) If y = $ (a + /?), we obtain the solution (334). If A is rajuired directly, the sum of the squares of (336) gives k sin (a /?) = y [TO* + n j 2 m n cos (a /?)] as in Art. 178. 181. The solutions of the preceding articles may be applied to a single equation of the form n sin (a -f- z) = wi sin (ft -)- z) which is a more general form of (309). For if we assume k sin (a -\- z) m we have k sin (/? + z) = n whence k and z are found by Arts. 177, 178. EQUATIONS OF THE SECOND AND THIRD DEGREES. 95 182. In like manner, if the proposed equation is n cos (a -j- z) = m cos (/? + 2 ) we assume k cos (o -f z) = m whence k cos (/? -f z) = n and and z are found by Arts. 179, 180. As the sign of k (in this and the preceding article) may be arbitrarily assumed, there will be two solutions. NUMERICAL EQUATIONS OF THE SECOND AND THIRD DEGREES. 183. To solve the equation = (338) when q is essentially positive, and p either positive or negative. We have from (144), exchanging x for 0, tan* i i> 2 cosec tan + 1 = (339) and (338) may be reduced to this form by substituting in which we may take the radical only with the positive sign, since we may assume x and z to have the same sign. We thus reduce (338) to which compared with (339) gives 2 cosec -, z = tan or sin J, x = l/g~tan J (340) P which gives two values of less than 360 and consequently two values of x. If 6 be the least of these two values of less than 360 (= 2 TT), all the values of which have the same sine are 6, it 8, 2 TT -f 0, 3 TT 0, etc. and all the values of tan are tan J #, cot J 0, tan $ 0, cot 0, etc. Hence the two roots of (338) are found by the formulae sin 6 = *lS, Xi i/-qtnn % ft, x 2 = Vo^coi f (341) P in which may be always taken < 90 with the sign of its sine, and \/ q is to be regarded as a positive quantity. As long as 2 j/ q is not greater than p, this solution is possible, but when 2 \/ q > p, sin is not possible, and both roots are imaginary ; which agrees with what is shown in algebra. 96 PLANE TRIGONOMETRY. 184. To solve the equation q = (342) when q is essentially negative, p being eitf^er positive or negative. We have, by (143) tan* J^ + 2cot0tanJ0 1 = and (342) is reduced to this form by substituting whence z 1 -j z 1=0 1/9 The required solution is therefore 2 cot = > z = tan 1/9 or tan = J5, z = V^tan J ^ V 343 ) P If is the least value of 0, all the values of which have the same tangent are 6, -f 0, 2 TT + 0, 3 TT -f 0, etc. and all the values of tan i are tan J 0, cot J 0, tan J 0, cot } 0, etc. Therefore the two roots of (342) are found by the formulae at, = V/^tan J 0, x a = 1/7 cot $ (344) P in which, as before, the radical is to be taken as positive, and < 90 with the sign of its tangent. In this case both roots are real, since tan is always possible. 185. To solve a numerical equation of the third degree. It is shown in algebra that any equation of the third degree may be reduced to one in which the 2d term is wanting ; we need consider therefore only the form (345) To resolve this, put we find Now z may be decomposed into two parts, y and z, in an infinite variety of ways, and we may therefore suppose that y and z are such as to satisfy the condition which reduces the first term of the preceding equation to 0, and gives the two con- ditions Put y* = t lt z* = ty ; then we have EQUATIONS OF THE SECOND AND THIRD DEGREES. 97 so that, by the theory of equations, t^ and t t are the two roots of an equation of the second degree in which the absolute term is ^ mi term is 6 ; that is, they are the roots of the equation second degree in which the absolute term is ^ and the coefficient of the second mi If then we find the two roots t, and t t of (m) by the preceding methods we shall have X = y + 2 =: 1 H + 1 H () It will be necessary to consider the sign of a in the equation (m). 1st. When a is positive, (m) comes under the form (342) and the solution by (344) gives " and by (n) x = ^ * (^ tan J0 - & cot i 0) and if we assume tan ^> i/' tan $ this becomes, by (142) z = 2 -y - cot Collecting these results we have, for the solution of (345), when a is positive, , tan J4> = |^tan J0 _ (346) x = - 2 -^ I cot ^ in which the radicals -J and -J are to be considered positive, and is to be \ 27 \ t> taken < 90 with the sign of the tangent. But two of the three values of j/' tan J 8 being imaginary, the given equation has but one real root.* * If r represent the real value of \/ tan i ^ a d a a s the two imaginary roots of unity, the real value of x is and the imaginary values are or since a 1 a s = 1 IS 98 'PLANE TRIGONOMETRY. 2d. When a is negative and 4 a 3 < 27 6*. Equation (m) becomes and is of the form (338) ; its roots are therefore found by (341) which gives . 27 6 = ~ tan * = - cot "< 7 * = & tan * + ^ cot or if we put, as before, tau J $ -fr tan J 6, the solution of (345), when a is negative, is sin Y A/ tun 4 r^? */ tin b\ 27 (347) which gives one real root, (the other two being imaginary, as above), when sin is possible, i. e. when 4 a* < 27 6 1 .* Substituting the values of a, and a. and also we find r = tan = cot i r z, ^= ^ / _ (cot -(- coeec ^ ;/ 3) * 3 x, = - I ^L (cot -- cosec -j/ 3) v 3 or finally, a:, being the real root, the imaginary roots are TI ~~ 2 2 * The two imaginary roots will be found, by a process similar to that employed in the preceding note, to be x t = - - in which z, is the real root found by (347). EQUATIONS OF THE SECOND AND THIRD DEGREES. 99 3d. When a is negative and 4 a 3 > 27 6*. In this case sin 6, in (347), is impossible and the preceding solution fails. This is the irreducible case of Cardan's rule, the roots appearing under imaginary forms, although it is known that they are all three reaL It is, however, readily solved trigonometrically. In Art. 77, putting for x, we have sin*

and the solution is . sin 3 c* li A/ a; = 2 sin tf A/ - ^ ' a* '3 which gives three real roots by the different values of 3 0, which have the same sine. If 6 is the least of these values, all the values of 3 are expressed by 2 n TT -f and (2 n -f 1) n 6 n being any integer or ; and all the values of are expressed by 3 3 Now all integers are included in the forms 3 m, 3 m -f- 1 and 3m 1. If n = 3 m, the above values of are whence sin = sin 0, sin = sin (T 0) If n = 3 m + 1, we find in the same way sin ^ = sin J (ir 5), sin ^ = J * the same as before. If n = 3 wi 1, we find both values to be sin * = sin J (ir -f- 5) 100 PLANE TRIGONOMETRY. so that there are but three different values of sin . Substituting these in (348), the three roots of (345), when a is negative and 4 a 1 > 27 6 2 , are found by 6)= 2 -J J- sin (60 3 . (349) in which < 90 with the sign of its sine, and the radicals are taken with the positive sign. EXAMPLES. 1. Solve (345) when a = 6101315, b = 5'766578. We find log A = 0-002651* tl which being greater than any log. sine, we take its arithmetical complement and pro- ceed by (349). Then log sin = 9-9974349 = 8346 / 44 // 55' 34"-7 9-6705571 60 J0 87 55' 34"-7 9-9997155 (60+i #) 32 4' 2o /x -3 9-7251024 0-4551811 0-4551811 0-4551811 0-1257382 1-335790 log i, 0-4548966 x, = 2-850339 log x, 0-1802835 *, = 1-514549 2. Solve (345) when a = 7, and 6 = 7. Ans. x l = 1-356896, x t = 1*692021, z, = 3'048917. 3. Solve (345) when a = 1'5, and 6 = 45. Ans. The real root = 3'41 63975. It may be observed that the algebraic sum of the three roots is always zero, in con- sequence of the absence of the term in x 1 from the given equation. This is easily shown from (349) where there are three real roots, and from the forms in the notes p. 98, where there are imaginary roots. This principle furnishes a simple verification of the values found by (349). * The sign here belongs to the number of which this is the logarithm. CHAPTER XI. DIFFERENCES AND DIFFERENTIALS OF THE TRIGONOMETRIC FUNCTIONS. 186. IN the applications of trigonometry, it is often required to compute a function of one angle from that of an angle which differs from the first by a small quantity. In such cases it is generally most convenient to compute the difference of the two functions, which may be applied to either to obtain the other. 187. To find the increment of the sine or cosine of an angle corre- sponding to a given increment of the angle. Let the angle x be increased by Ax, (this notation signifying dif- ference, or increment of #), and let the corresponding difference or increment of the sine be expressed by J sin x and of the cosine by J cos x ; we have, by this notation, A sin x = sin (x -f- A x) sin x A cos x = cos (a; -f A x) cos x and by (106) and (108) A sin x = 2 cos (x -)- | A #) sin A x (350) A cos a; = 2 sin (x -}- A or) sin ^ J# (351) which are the required formulae. We here consider the difference always as an increment, i. e. an increase, and give it the positive (algebraic) sign ; its essential sign may, however, be negative, and it will then be in fact a decrement. Thus, in (351) the second member will be negative so long as x < 180, and therefore the increment of the cosine is negative; that is, from to 180 the cosine decreases as the angle increases. In like manner A sin x is negative when x > 90, and < 270. 188. To find the increment of the tangent and cotangent. We have A tan x = tan (x -f- A x) tan x A cot x = cot (x -f- A x) cot x and by (11 6) and (11 9) A tan x = - = sec (x + A x) sec x sin A x (352) cos (x + A x) cos x S! II .-/ OC A cot x = - = cosec (x -f J x} cosec x sin J x (353) sin (x-\- Jx) sin x 1 2 101 102 PLANE TRIGONOMETRY. 189. To find the increment of the secant and cosecant. We have A sec z = sec (z + A z) sec z A cosec z = cosec (z -(- A z) cosec z or by (130) and (132) A sec z = 8in \ X .J^S 5lJ3Ll_~J[ (354) cos (z -f- A z) cos z A cosec z = ~ 2_go8 (z -f i A x)jin J_Ax (355) sin (z -f A z) sin z 190. To find the increment* of the squares of the trigonometric functions corresponding to a given increment of the angle. We Lave A sin* z = sin* (z + A z) sin* z = cos* z cos* (z -f A z) whtn.e by (133) A sin* z = A cos* z = sin (2 z -f A z) sin A x (356) From (115), (116), and (119) we deduce tan* z - tan* y = sin ( x + ?) sin (' ~ V) cos* z cos* y cot* z cot* y = sm ( x ~r y/jij? ^ ~ y) sin* z sin* y whence A tan* z = sin ( x + *]JMI^_Z /gg 7 j cos* (z -f- A z) cos 7 z Acot*z = = (358) sin 1 (x -}- A z) sm*z From (16) we have see* (z -f A z) = tan* (z + A z) -f 1 sec* z = tan* z -f- 1 the difference of which gives A sec* z = A tan* z (359) and in the same manner, from (17), A cosec 1 z = A cot* z (360) and the values of A tan* z, A cot* z, may be substituted in (359) and (360). 191. When the increment of an angle, or arc, is infinitely smalt, it is called the differential of the angle, or arc ; and the correspond- ing increments of the trigonometric functions are the differentials of these functions. The differential of x is denoted by dx of sin x by d sin x, etc. DIFFERENCES AND DIFFERENTIALS. 103 192. To find the differentials of the trigonometinc functions from the differential of the angle. Let the angle x and its increment J x be expressed in the unit of Art. 11 ; or, which is equivalent, let x and Jx be the arcs which measure the angle and its increment in the circle whose radius = 1. It is evident that the less the arc, the more nearly does it coincide with its sine or tangent; therefore, when Ax is infinitely small, or becomes dx, sin dx = dx sin ^ dx = ^ dx This may be demonstrated more rigorously thus. When dx is infinitely small, we have cos dx = 1, whence sin dx -- = cos ax = 1 tan dx sin dx tan dx but the arc cannot be less than the sine, nor greater than the tan- gent, and therefore dx = sin dx = tan dx Again, when Jx is infinitely small, or becomes dx, we must, ac- cording to the principles of the differential calculus, reject it when connected with finite quantities by the signs + or ; thus we must substitute x for x + dx, or for x + ^ dx. Upon these principles we find the differentials directly from the finite differences (350), (351), (352), (353), (354) and (355) as follows : d sin x = cos x dx (361) d cos x = sin x dx (362) d tan x = sec 2 x dx (1 + tan 2 x) dx (363) d cot x = cosec 2 x dx = (1 -+- cot* x) dx (364) d sec x = tan x sec x dx (365) d cosec x = cot x cosec x dx (366) 193. In the same manner the equations (356), (357), (358), (359) and (3*60) give d sin 1 x d cos* x = sin 2 x dx (367} d sec* x = 5!5-? ( fa (368) cos* z (369) COS* X COS* X d cot 1 x = d cosec* x = =^1!* dx (370) * (371) 104 PLANE TRIGONOMETRY. 194. Although the equations (361), (362), (363), (364), (365) and (366), are rigorously true only when dx is infinitely small, they may be used when dx is a finite difference, instead of the equations, (350), (351), (352), (353), (354) and (355), provided dx is sufficiently small to be considered equal to its sine without sensible error, and is also very small in comparison with x. This is very frequently the case in practice, and the differential equations are then preferred on account of their simplicity. It is only necessary to observe that dx must be expressed in arc, i. e. in terms of the unit radius ; if it is given in seconds, it may be reduced to arc by Art. 9. 195. To find the differential of an angle from the differentials of its functions. From (361) we have (372) cos a; but it is convenient in this case to employ the notation of inverse functions, Art. 87. Thus, if y = sin x, x sin ~* y, and the preced- ing equation becomes In the same manner from (362), etc., we find <**>-'? = ~ d (374) dtan- 1 ^-^- (375) 1-rjT dcoi ~ l y = TTy> (376) f-r-f-r 77^- (377) d cosec l y = - (378) CHAPTER XII. DIFFERENCES AND DIFFERENTIALS OF PLANE TRIANGLES. 196. IN trigonometrical investigations it is o , often necessary to determine the effect of a small change in one of the data, upon the com- puted parts. Thus, Fig. 29, if A, AB and A C, of the plane triangle A B C } are the data, A and A C is subject to an error of C C f , the required parts will be subject to errors which are respectively, the differences between A CB and A C' B, A B Cand A B C', B C and B C'. In the same figure, the data may be supposed to be A, A B and AB C, and the angle ABC may be regarded as subject to the error CB C' which produces the corresponding errors in the remaining parts. In the same manner, the data may be A, A B, and ACS, A CB being variable ; or, A, A B, and B C } B C being variable. In all these instances, A and A B are constant, while the remaining four parts are variable, and may be considered as receiving, simultaneously, certain increments which are related to each other. We propose, then, to solve the general problem : In a plane triangle, any two parts being constant, and the rest variable, to determine the relations between the increments of the variable parts. It is evident that the solution of this problem resolves itself into an investigation of the differences of two triangles which have two parts in common. We shall consider the several cases successively ; distin- guishing the triangle formed from the given one by the application of the increments as the derived triangle. 197. CASE I. A and c constant. The six parts of the given triangle, ABC, Fig. 29, being A, B, C, a, b, c, those of the derived triangle formed by varying all but A and c, are A y B -f J B, C + J (7, a -f- Ja, b -f J6, and c. In these two triangles we have A + B+ O= 180 A + B + 4B+ C+ JC= 180 whence JJ5 + JC=0, JB = 4C (379) 14 105 106 PLANE TRIGONOMETRY. Also in the two triangles we have a = c sin A cosec C (m) a -f- Aa = c sin A cosec (C-f- JC) (n) the half difference of which by (355) is i j _ c sin ^. cos ((7+ ^ -^ff) sin ^ JO , , sinCsin(C'+ JC) W , , sin | AB sin | J C sin (C + AC) The half sum of (m) and (n) by (131) is _1_ 1 A " cos sin Csin(<7+ JC) which combined with (p) gives tanJJC tan(C+|JC) From (260) we have ~ c sin tan C= b c cos A whence 6 c cos A = c sin A cot C b -\- Ab c cos A = c sin A cot (C-}- AC) the difference of which by (353) is c sin A sin JC , . sin C sin (C + therefore _^A_=__J^_ = _ (382 ) sin AB si" ^ ~ : - "* ' A V ; DIFFERENCES OF PLANE TRIANGLES. 107 This equation gives by (135) \Ab = a sin | JO cos \AG~ sin (O+ JO) and dividing (380) by this Aa _ cos (C+ It AC) ^^ Ab cos \ AC It is to be observed that the increments (or half increments) of the angles must be deduced from their sines or tangents, since it is only by these functions that a small angle can be accurately determined. Moreover, a small arc being nearly equal to its sine or tangent, the equations (380), (381) and (382) express very nearly the ratios of the increments of the sides to the increments of the angles, or rather to those increments reduced to arc by Art. 9, or Art. 54. 198. CASE II. A and a constant. We have as in the preceding case AB = JO; and in the two triangles b sin A = a sin B (b + Ab) sin A = a sin (B + J B) the difference and sum of which give l J b sin A = a cos (B + $ A B) sin J B (p) (b + | Ab) sin A = a sin (B + \ A B) cos \ AB whence by division 4J6 4J6 64-4J6 tnc*A\ a - 2 = - ^ (384) tanlJi? tan^JO tan( + J) In the same way \A G ^ = i Jc = c + ^Ac , 385 tan ^ JO tan % A B tan (O + J O) From the equations c sin A = a sin O (c -f J c) sin vl = a sin (O + JO) we find \ Ac, sin A a cos (O + \ AC] sin ^ JO ((/) 108 PLANE TRIGONOMETRY. which combined with the equation ( p) gives, since sin JO= sin Jc cos(0 From (p) we also have sin JO sin JS which, when J 6 is to be found from J B } is more convenient than (384). In the same way from (q) c cos JO sin JO sin^J^ sin C 199. CASE III. b and c constant. We have c sin -B 6 sin C c sin (B + J J8) = 6 sin (O-f ^C) the sum and difference of which give c sin (B + i J) cos ^ J5 = 6 sin (C+ i 4C) cos ^ JO (ja) c cos (B + | J B) sin | J 5 = 6 cos (C + JC) sin ^- JO (?) the quotient of these gives By (224) we have a = b cos C-\- c cos B a + J a = & cos (O+ 4 C) + c cos (B + the sum and difference of which give a -f Ja = 6 cos (C-f JO) cos JO+ c cos (5 -f J J) cos -^Ja = 6sin(C+ JO) sin JO+ c si These expressions are reduced by ( p) and (q) to and by division tan AC DIFFERENCES OF PLANE TRIANGLES. 109 In the same way we have |Ja _ a + ^Ja (m) Since the sum of the three angles is constant, AA + AB+ JO=0 therefore by (115) sinA(JJ5+ JO) tan 4- AB + tan A JO ^-* cos 4 J.Bcos4 JO i sin Substituting () in (r) we find sin \ AC sin JJ. a-\- Aa By differencing the equation a 2 = b* + c 2 2 be cos we find instead of (392) and (393) \Aa __ be sin (A 4- sinj A A ~ a + \ K s^ AC which substituted in (s) gives j Ja _ csin(^ + ^ AB) sin^ J^ " cos^JC and in the same manner sin^JJS 6 cos (0 + 4 JO) whence also (395) a-\- % A a 110 PLANE TRIGONOMETRY. 200. CASE IV. A and B constant. We have , sn 6 = - a sin A , , A , sin B , b-\- Ab = - -(a- sm J. whence J 6 - Aa sin J. In this case the third angle is also constant and there are but three variables related by the equation (397) sm ^4. sm B sin (7 This case is not strictly included in the general problem as stated in Art. 196, since the two triangles have not two parts in common. 201. The second members of the equations (380), (381), (382), (383), (384), (385), (386), (387), (388), (389), (390), (391), (392), (393), (394), (395), (396), involve the increments themselves, which are the quantities sought. It is therefore necessary, in many cases, to solve these equations by successive approximations. For a first approximation we consider the increments in the second member to be = 0, employing B for B -f- A JS, etc., and taking cos J B = 1, etc. This will evidently produce but a slight error so long as the increments are small as compared with the entire parts of the triangle. We then obtain a second approximation, by recom- puting the equation in its complete form, employing in the second members the approximate values of the increments. With these second values we may, in the same way, obtain a third approxima- tion, etc. Theoretically, it requires an infinite number of such approximations to arrive at a perfect result ; but in practice, the tenths or hundredths of seconds being the limits of accuracy, it is rare that more than a second approximation is necessary. It is also to be observed that in computing the values of small quantities such as the increments in question, we may employ logar- ithms of only four or five decimal places and take the angles to the nearest minute. This is in fact one of the chief advantages of com- puting by differential formulae, rather than by the direct formulae applied to each of the two triangles successively. DIFFERENTIAL VARIATIONS OF PLANE TRIANGLES. Ill EXAMPLE. In a plane triangle whose parts are a = 6053 -B = 35ll'3"-4 C=867'7"-7 b = 4082 c = 7068 let A and a be constant while b is diminished by 50'5 ; to find the change in the angle B. We have in this case Jb = 50'5 ; and by (387) . , 4T>_ Y A b sin B ~ b cos (B sin IST APPROX. 2o APPROX. \Ab - 25-25 b 4082 35 11' 35 11' i- AB -15' B-\- l A B 35 11' 34 56' log \ A b - 1-4023 ) ar. co. log. b log sin B ar. co. 1. cos (B -\- ^ A B) log. sin ^ A B 6-3891 9-7606 0-0876 - 7-6396 - 15' 0" V 7-5520 0-0863 7-6383 - 14' 56"-8 It is evident that changing the angle B -f i ^ B by only three seconds would not affect the fourth place .of its cosine; a third approximation is therefore unnecessary, and we have finally J J5 = 29' 53"*6. As the log. sines of small angles do not vary proportionally with the angles, it will conduce to accuracy to employ the methods explained in Art. 115. DIFFERENTIAL VARIATIONS OF PLANE TRIANGLES. 202. The equations (380), (381), (382), (383), (384), (385), (386), (387), (388), (389), (390), (391), (392), (393), (394), (395), (396) and (397) become differential by making the increments infinitely small, that is, by omitting the increments when connected with finite qnan- 112 PLANE TRIGONOMETRY. tities by the signs + or , and substituting the increment itself for its sine or tangent, and unity for its cosine, (Art. 192.) The char- acter d must also be substituted for J. These changes being made, we easily deduce the following differential relations. CASE I. A and c constant. da dB da dC sin C = cosC db~ C 8 CASE II. A and a constant. db^ dB cU dC dB = -dC db , D = - = 6 cot B dc dJB d dc c cot C cos B cos C (398) (399) CASE III. b and c constant. dB dC da dC tan B tan C da dB = a tan C da ~d~A c sin B = b sin C dC dA dB dA b n - cos C a (400) DIFFERENTIAL VARIATIONS OF PLANE TRIANGLES. 113 CASE IV. The angles , A t B } C, constant, da d b d c sin A sin B sin C (401) 203. These differential relations are often employed when the in- crements are very small, instead of the equations of finite differences. We have already seen that the equation of differences often requires to be solved by successive approximations, the first approximation being in fact obtained by employing the corresponding differential equation. In all cases therefore where a second approximation in the use of finite differences could not alter the result of the first, it is plain that the differential equation is sufficiently accurate. The increments of the angles must generally be expressed in arc. Thus if dB is given in seconds we must divide it by R"= 206264"-8, or substitute dB sin I" for dB. dA But in such fractions as > this substitution is evidently unne- d B cessary provided the two increments are always expressed in the same unit, as minutes, seconds, etc. EXAMPLE. In a plane triangle whose parts are A = 58 41' 48"-9 B = 35 11' 3"-4 (7=86 V 7"-7 a = 6053 b = 4082 c = 7068 suppose b and c to be constant and the angle A to receive the incre- ment dA = 20"-6 ; find da and dC. From (400) we have da = dA sin 1" c sin B ,~ dA c cos B dC= a log dA 1-3139 log ( dA) 1-3139 log sin 1" 4-6856 log c 3'8493 log c 3-8493 log cos B 9-9124 log sin B 9-7606 ar. co. log a 6-2180 log da 9-6094 log dC 1-2936 da 0-407 dC 19"-7 16 K 2 114 PLANE TRIGONOMETEY. 204. The error of employing the differentials in any case may be determined ap- proximately by developing the equation of finite differences and comparing it with the corresponding differential equation. We shall select a simple example. We have from (387) and its corresponding differential equation in (399) sin B A6 = b cot B A.B sin 1" the first of which when developed gives i A6 = b cot B sin J A 2 b sin ^ + * -^ sin J A.B sin J A sin 2? or substituting sin AJ? = J A.B sin \" ', sin } AJ3 = J A.B sin 1", and also J? for B -\- \ AJB in the second term, which will affect so small a term but slightly, A6 = 6 cot B AJ5 sin 1" - f (A.B sin l x/ )* m Comparing this with the differential equation above, the error of employing the latter is approximately - - (A sin 1")* which for A B = 1 is '000015 b. It appears from this example that the error is expressed by a term involving the square of the increment ; and if we develop all the equations of finite differences we shall find that they differ from the corresponding differential equations by terms in- volving the squares and higher powers of the increment. Hence, employing ilte dif- ferentials instead of the finite differences amounts to neglecting the terms involving the squaret and higher powers of the increments. 205. The differential relations above obtained could have been deduced more di- rectly from the formulae of plane triangles by differentiation, employing the values of the differentials given in Art. 192. Thus in CASE I, A and c being constant, if we differentiate the equation a = c sin A cosec C we have d a = c sin A d cosec C = c sin A cot C cosec C d C = a cot CdC as in (398). The student may exercise himself by deducing the other relations of (398), (399), and (400) in a similar manner. CHAPTER XIII. TRIGONOMETRIC SERIES. DEVELOPMENTS OF THE FUNCTIONS OF AN ARC IN TERMS OF THE ARC, AND RECIPROCALLY.* 206. THE investigation of trigonometric series is most readily carried on with the aid of a few elementary principles of the Differ- ential Calculus. All that will be required here will be no more than is generally given in the first chapter of a treatise on that subject, namely, the differentiation of simple algebraic functions, and Taylor's Theorem. We shall employ the following expression of this theorem : ,, . . d.fy h . d? . fy h? . d? . fy A 3 . h = - - in which fy denotes what / (y -f- A) becomes when h = and - Lf ^') *> etc., are the successive differential coefficients, or de- dy df rivatives of fy. 207. To develop sin x and cos x in terms of x. We shall first develop sin (y -f x) and cos (y + x) by (402). By (361) and (362), if fy = sin y I U>* f U ' ' O1U '/ we have = cos y dy . /V d cos v j j = -T- 2 = sin y dy 2 rfy rf'./v d sin t/ = z = cos y dy d*.fy d cos y - = * = sin y dy< dy *The leading results of this Chapter being of very general utility and constant application are printed in the larger type, but as they are not referred to in the subse- quent large print of this work, and moreover require a limited acquaintance with the Differential Calculus, the student can omit them at the first perusal, and pass directly to Part II. 115 116 PLANE TRIGONOMETRY. so that the values of the coefficients of the series (402) recur in the order -f sin y, + cos y, sin y, cos y, and therefore / (y -f #) = sin (y -f x) sin y -f cos y - - sin y cos y -f etc. (403) 1 \* 2t i'&'o If we commence with fy = COS y the coefficients will recur in the order -f cos y, sin y, cos y, -f sin y, and (402) will give X X 3^ cos (y -f- x) = cos y sin y - cos y - -f- sin y -- f- etc. (404) 1 \' 2> l*.2'o If now we put y = in (403) and (404), sin y = 0, cos y = 1, the alternate terms of the series vanish, and we have -'+-+ f (405) cos x = I -f --- -" -- h etc. (406) 1-2 1-2-3-4 1-2-3-4-5-6 It may be observed that (406) can be deduced from (405) by dif- ferentiation. 208. The series (405) and (406) are directly available for the con- struction of the trigonometric table. For this purpose x in the series must be expressed in arc, since (361) and (362), upon which the pre- ceding demonstration rests, require # to be in arc, Art. 9. EXAMPLE. Find cos. 10. Reducing 10 to arc, by Art. 9, we have x = 10 X -01745329 = -1745329 and computing separately the positive and negative terms of (406), --^ 2 - = -01523086 j^~ - -00003866 _ __ = _ -00000004 1-00003866 - -01523090 -01523090 cos 10 = -98480776 agreeing with the tables, which give -9848078. The student may, for practice, verify any other sine or cosine of his table. TRIGONOMETRIC SERIES. 117 209. To develop tan z in terms of x. Representing the coefficients in the series (405) and (406) by letters, we have tan x = in which 1 a, x 2 -j- a 4 x* a 6 x 6 + etc. as= (407) etc. If we perform the division of the numerator by the denominator, we perceive that the result will be a series containing only odd powers of x, and commencing with the term x. But as the law for the successive formation of the coefficients is not easily shown in this way, we shall resort to the following process. Assume the series to be tan x = ! x -f- c s X s -f- c 5 x 5 -f- etc. and differentiate it ; we find, by (363), after dividing by dx, I + tan* x = Cj -f 3 c 3 x 1 + 5 c 5 x* + etc. or, since from the division of (407) we know that e l = 1, tan 1 x = 3 c, x 1 -f 5 c s x 4 + 7 c 7 x 8 -f 9 c 9 a 8 + etc. The square of (m) is (m) tan 1 x = c t Cj X s x* -f which compared with (71) gives x 8 -|- etc. -fete. + etc. + etc. cc etc. + c 6 a, -f c, etc. where the law of derivation is obvious. We have preserved the factor c v although it is equal to unity, in order to render this law more apparent. Since the first and last terms of these expressions are equal, as also the terms equally distant from them, we may write them as follows : 9 'n=~(2 etc. - 2c 7 , etc. c 6 ) 118 PLANE TRIGONOMETRY. in which form any coefficient , + l when n is even, is expressed by terms all of whose coefficients are = 2 ; and when n i$ odd, by terms all of whose coefficients 2 are 2 except the last, which is 1.* If we now substitute the value of q = 1, and deduce the numerical values of the coefficients successively, we shall find 210. To develop cot 2 in terns of x. If we invert (407) we have (40g) 3-5 s - 2 --' ss --' ^ (409) x a s x 3 + 5 x 5 etc. and the first term of the actual division is , the second term (a. a.) x. and x the succeeding terms evidently involve only the odd powers of x. Therefore let cot x = -- d t x rf s X s djX 5 etc. (o) x Tlie coefficients cannot be determined by the method of the preceding article in con- sequence of the negative exponent in the first term ; but they are directly deducible from those of the series for tan x. We have by (142) tan x = cot x 2 cot 2 x ( p) Now the series (o) being true for any value of x will give cot 2 x by substituting 2 x for x, whence 2 cot 2 x = 2 s d l x 2* d t x 3 2 d b x> etc. x Subtracting this from (o) we have by (p) tanz=(2 l 1) d, x + (2* 1) rf 3 x s + (2 1) Designating the coefficients of (408) by c u Cj, c 6 , etc. we have also tan x = c t x -f c a & + e 6 z* + etc, and the comparison of these two values of tan x gives 2* 1 (2. 1)(2 + 1) 1-3 2* 1 ~ (2* I) (2 J +I) ~~ 3-5 2 1 (2 1)(2+1) 7'9 etc. etc. 2"+ 1 *Euler, and after him Cagnoli and others, make these coefficients depend upon those of the series sin x and cos x, but the number of given quantities by which each coefficient is expressed is double the number required in the method of the text. TRIGONOMETRIC SERIES. 119 Substituting the values from (408) e=l * = | = ^ etc - and reducing the coefficients to their simplest forms, we find the series (o) to be I * a? 2x* x 7 2x ,., A v cot x = ------ -- - ----- etc. ( 410) x 3 3'5 3 5 '5-7 3 s '5'-7 s -'-- 211. By a process similar to that of Art. 209, but which we leave to the student, we find , , x 3 , 5x* , 61 3* , 277 x 8 ,.* . (411) And from (408) and (410) by means of the formula cosec x = (cot J x -f- tan J x) we find 1 i x 7 x 3 31 x 5 127 x' , , 10 v 212. To develop sin" 1 y in terms of y. (See Art. 87). Let x = sin" 1 y (or sin x = y) ; then by (373) dy Developing the second member by the Binomial Theorem, dx , . 1*3 1'3'5 6 t \ ^ = i + iy ! +^'+^r 6 / + *- () As this contains only even powers of y, the series from which it would be obtained by differentiation must contain only odd powers of y ; therefore, let * = a \y + a ^f + Oay 5 + ?/ + etc - ( n ) There will be no term independent of y if we limit x to values be- tween and : 90, for then when y = we must also have x = 0.* Differentiating, we have dx = ! + 3 ff^ 2 + 5 a,# 4 + 7 a^y* -f- ete. y which compared with (ra) gives 1 1-3 1-3-5 0^=1 3a s = - 5a 5 = 7 7 = _ fi ete. * The series (413) obtained under this limitation expresses but one of the values of sin" 1 y, but if we denote the series by s, we shall have by (95) the following expres- sion, including all the values, sin -1 y = n TT -(- ( 1 ) n n being an integer, positive or negative, or zero. 120 PLANE TRIGONOMETRY. therefore (n) becomes y 3 , 1-3 y 5 1-3-5 y 7 , ,c+_.+_r +tlBi (418) It is unnecessary to develop cos" 1 y since we have i - i cos y = -- sin y 213. To develop tan" 1 y. Let x = tan" 1 y, then by (375) from which we infer, as in the preceding problem, that the required series contains only odd powers of y ; therefore let x = a l y-\- atf + agf + o 7 y 7 -f etc. (n) dx then - = i -f 3 c^ -f 5 a^ 4 -|- 7 c^y 6 + etc. dy which, compared with (m), gives a l = l 3 03 = 1 5a 6 = l 7 Oj = 1 etc. so that the series is }y 7 -f etc. (414) 214. To compute the ratio ( n) of the circumference of a circle to its diameter. We have heretofore assumed this ratio to be known from geometry, where it is found by means of circumscribed and inscribed polygons which are made to differ from the circle by as small a quantity as we please ; but (414) enables us to express its value in a series. We have tan - = 1, therefore if we make y = 1 in (414) we have K 1 *. . II 1 I 1 SA1C\ - = l- 3+ --- + --etc. (415) But this series converges too slowly to be of any use. To obtain a rapidly converging series y must be a small fraction. We might em- ploy tan - = -- - (Art. 29), but in consequence of the radical, it is 6 /3 TRIGONOMETRIC SERIES. 121 simpler to resolve '- into two or more arcs whose tangents are known, and to compute the value of each of these arcs by the series. To effect this let then by (123) whence - A = tan- 1 1 + tan" 1 i' (416) 4 7T t + V tan - = 1 = 4 1 tt r 1 t l + t from which, assuming any value of i at pleasure, the corresponding value of t r is found. If we take t = , we find t' =\; therefore by (416) and (414) - tan" 1 | -f tan" 1 A few terms of these series give - = -4636476 + -3217506 = -7853982 4 = 3-14169 more accurately ic = S'14159 26535 89793 1 3 If we take I ":' we find t r = -* but the above supposition is evidently the best adapted for rendering both series sufficiently con- vergent.* 215. To resolve sin x and cos x into factors. The series (405) shows that x is a factor of sin x, and gives sinz = x(l -- - -- 1 -- etc.) (p) 1-2-3 T 1-2-3-4-5 *See NOTE at the end of this chapter, p. 124. 18 L 122 PLANE TRIGONOMETRY. and the factors of the series within the parenthesis must evidently be of the form i-* (.) A being a constant, but having a different value in each factor. The required factors must be such as to reduce the second member of ( p) to zero whenever the first member is zero. Now sin x is zero for the value x = 0, whence x is a factor as already seen, and also for x = n K, n being any integer ; therefore the general value of (q) is ,-^, whence A = r? ir* which, substituted in (7), gives as the general factor i __ *L n J 7T Making n successively = 1, 2, 3, etc., the equation ( p) becomes therefore '"=(' -)('-) ('-) (419) The factors of cos x in (406) must also be of the form (q) ; but cos x is zero for x = (2 n + 1)-' n being any integer or zero, and the general value of (q) is 2t (2n+l)'7r' A.r (2n+ l)*7r* whence A = ' --- - r -^- i - which, substituted in (7), gives the general factor Making n successively = 0, 1, 2, 3, etc., we have . (420) 216. Logarithmic sines and cosines. By means of (419) and (420) the logarithmic sines and cosines of the tables are readily computed. Put z TO , then 2 i - Mid taking the logarithms log .inY^k*! + 1ogm h = log l - TRIGONOMETRIC SERIES. 123 Developing these logs, by the known formula log (1 n) = - M (n -f \ n 2 + \ n -f etc.) (in which M= modulus of common logs.) and arranging according to the powers of m, we have 1 WIT , 7T . . log sin - = log + log m 1- -fte+i+i-H Jf / 1 ,1,1 t \ H --- h etc. I ^ 4 sT 6 6 / log o 3 etc. mr . Jlf / 1 = '- etc. By summing the constant numerical series, and substituting the value of the modulus M = '43429 44819 and also of , these formulae become m log sin = 10-19611 98770 + log tn IB -m 1 X 0-17859 64471 -m 4 X 0-01 4G8 89690 -m X 0-00230 11796 -m 8 X 0-00042 58450 - m 10 X 0-00008 49075 log cos ^ = 10 L -m* X 0-5357893412 - m* X 0-22033 45350 - m X 0-14497 43131 -m 8 X 0-10859 04688 - m' X 0-08686 0376G W X 0-00001 76758 m u X 0-00000 37870 - m 16 X 0-00000 08284 - m 18 X 0-00000 01841 etc. (421) m" X 0-07238 25502 - m 1 * X 0-06204 20818 - m" X 0-05428 68115 - MI" X 04825 49426 etc. (422) In these expressions 10 is added to render the logarithms positive, as is usual in the tables.* *See the preface to Callet's Tables, for the coefficients of these series carried to 20 decimal places, and for other forms given them by which they are rendered still more convenient. 124 PLANE TRIGONOMETRY. EXAMPLE. Compute log sin 9. We have m X 90 = 9 m = -1 log ro = 1 and therefore by (421) log sin 9 = 10-19611 98770 1- - 0-00178 59645 O'OOOOO 14689 O'OOOOO 00023 = 10-19611 98770 1-00178 74357 log sin 9 = 9-19433 24413 217. If in (419) we put x = ~, we have 2 _ ff/2 1\ /4* 1\ /i^iJ\ 2 1 2 1 / \ 4 / \ & )' = JL (2-1) (2+1) (4-1) (4 + 1) (6-1) (6 + 1)... 2 2. 2. 4. 4. 6. 6.... whence JL = A . A . A . . A . A . . . (423) 2 133557 which is Wattisfs expression of IT. NOTE to page 121. Computation ofir. Many other series besides those of Art. 214, may be given for computing IT. One method of obtaining them is to resolve tan" 1 t and tan" 1 i' into two others, and thus make \ n to depend upon three or more arcs. From (194) we easily deduce tan" 1 = tan" 1 - + tan" 1 (a) m m + n m j + m n + 1 tan- 1 -1 = tan- 1 tan- 1 (b) m m n m* TO n + 1 in which m being given, n may be assumed at pleasure. The numerators of the frac- tions in the last terms will reduce to unity when m a + 1 is divisible by n ; if therefore we assume n and p so as to satisfy the condition np TO* + 1 (e) we shall hare tan- 1 = tan- 1 (- tan- 1 ^ (a) TO TO + 71 tan- 1 = tan- 1 tan" 1 (e) m m n p m TRIGONOMETRIC SERIES. 125 For example, let m = 3 ; then m 1 -j- 1 = 10 = 1 X 10 = 2 X 5, so that we may take n = 1, p = 10 ; or n 2, p = 5, whence by (d) and (c) tan- 1 i = tan- 1 - + tan" 1 = tan- 1 - + tan- 1 - Substituting in (418) - = tan- 1 - + tan- 1 - + tan = 2 tan- 1 - tan- 1 - 2 7 = 2 tan- 1 1 + tan- 1 ]- (/) = tan- 1 1 + tan- 1 1 + tan- 1 1 ( s ) The equation (/) was employed by CLAUSEN of Germany, in computing n to 200 decimal places, and (g) was employed by DASE, also of Germany, in computing n to the same number of figures. These computations were carried on independently of each other, and the results when communicated to SCHUMACHER, (who gives them in the Astronomische Nachrichten, No. 589), were found to agree to the last figure. They prove the value previously found by Mr. Rutherford to be erroneous beyond the 150th figure. By means of the formulas (a), (b), (c), (d) and (e) we may again subdivide the arcs as often as we please. Thus, it is easy to deduce - = 2 tan- 1 - + tan-' - + 2 tan- 1 - = 2 tan- 1 - + tan- 1 - + tan" 1 ^ -f- tan" 1 -J- . 4 tan- 1 1 - tan- 1 1 + tan- 1 + tan- = 4 tan tan -1 -j- tan- 1 which last is known as Machin's formula. In deducing it we have reduced the dif- ference of two arcs to a single arc by means of formula (a). Another method is, to find by trial, or otherwise, an arc a multiple of which is nearly equal to > and whose cotangent is a whole number; and then deduce the L 2 126 PLANE TRIGONOMETRY. difference between this multiple and Thus it is known (from the trigonometric 4 tables) that cot 11 15' = 5 nearly ; therefore by the last formula of Art. 79, putting tan x = 5 and by (194) therefore = 4 tan- 1 - tan- 1 JL 4 5 239 as was found above. If we resolve tan -1 - by means of (c), (d) and (e), we have m = 239, m 1 -f- 1 = 57122 = 2 - 13* = n p, which offers several suppositions for n and p; if we take n = 13 = 169 and p = 2-13' = 338, we find by () i = 4 tan- 1 L -tan- 1 1 + tan- 1 which was employed by Rutherford. If we take n = 1, p = 57122, we find by (d) ^=4tan- 1 |-tan- 1 -^-tan- 1 ^. CHAPTER XIV. EXPONENTIAL FORMULAE. TRINOMIAL OR QUADRATIC FACTORS. 218. To demonstrate Euler's formula cosx=% (e " / - 1 + e -"- 1 ) (424) s\nx = - -(e"- 1 e "- 1 ) (425) in which e is the Naperian base of logarithms, or, e = 1 + I + F2 + li3 + etC - It is shown in the theory of logarithms that <-= 1+ (f)+w + ly + w +ete - (426) where for brevity we write (1) = 1 (2) = 1-2 (3) = 1-2-3, etc. We have by (405) and (406), employing the above notation, x 3 x* x 6 "'= 1 -(l) + (4j-(6j+' te x x 3 x* x 7 the terms of which are the same as those of (426), but with alternate signs. If the signs in these two series were all positive, the sum of the two would be equal to (426) ; and it is evident that we shall make them positive by substituting x* = # or x = z y 1 which gives 0031=1 +++ +etc - m 128 PLANE TRIGONOMETRY. --, whence But 1 -1 x y 1, =- - = x y \ 1/--1 y -1 therefore cos a? y 1 sin a; e~ xy '~ 1 (427) If in this equation we substitute x for x, we have, by (56), cos x -f i/ 1 sin z = e' y ~ l (428) The sum and difference of these equations are (429) (430) whence (424) and (425). 219. The quotient of (430) divided by (429) is _ e 220. If we put y = e* y ~ l =cosx-\~i/ 1 sin a; (432) we have y~ l = e~* v ~ l = cos x y 1 sin x (433) and (429) and (430) become 2 cos * = y + y~ l (434) 2 y 1 sin x = y y~ l (435) If mx be substituted for x in these formulae, we have y m =e m * y ~ ] = cosm#+ y -- 1 sin ma; (436) y~ m = e~ m * y ~ l = cos mx y 1 sin mx (437) 2 cos mx = y m + y~ m (438) 2 V 1 sin mx = y m y~ m (439) EXPONENTIAL FORMULAE. 129 221. Moivre's Formula. The value of y m from (432), compared with (436) gives (cos x -f y 1 - - 1 sin x) m cos mx -f i/ 1 sin mx (440) which is Moivre's Formula. It shows that the involution of the expression cos x -{- \/ - \ sin x is effected by the multiplication of the angle. Again, if we multiply (432) by cos x' + ]/ - - 1 sin x' e Y-i we have (cos x -\- |/ - - 1 sin x) (cos x f + i/ 1 sin #') = e (z + y) l/ ~ l cos (x -}- x') -}- \/ - - 1 sin (a; -f- x'} which shows that factors of this form are multiplied by the addition of the angles. We have also (cos x-\-\/ 1 sin #) (cos x y 7 1 sin #)=cos 2 x-\-sin 2 x=e = 1 (441) 222. Oeneral form of Moivre's Formula. As long as m is an integer, both members of (440) can have but one value ; but if in = ^ the first member becomes (1 ( cos x -f- i/ 1 sin x)* = i*' (cos x-\- -\/ 1 sin x] f which has q different values* in consequence of the radical of the degree q, while the second member cos ^ x -f- i/ 1 sin 2- a; (a) 1 9 has but one value. In order that both members may have the same generality, as should be the case with every analytical expression, it is necessary to suppose that we take for the arc x not merely the arc less than the circumference which has the given sine and cosine, but also all the arcs which have the same sine and cosine ; that is, c denoting the circumference, all the arcs x, x -\- e, x -f- 2 e, x -f 3 c, etc. Now there is an infinite number of these arcs, but only q of them can give different values to (a) ; for all the values of the arc in (a) will be 9 f ' f P. , + !_, x + [iJLE2, etc * That is q values real and imaginary ; thus it is shown in algebra that y a* = -}- a - a ; = + a, a ~ 1 + V ' - 3 \ and a ~ 1 and - a ; & = + a, a (~ 1 + V ' - 3 \ and a (~ 1 ~ l^r_ 3 \ V' a 4 = -f- a, a, -f- a y' 1, a ^x 1 ; etc. 17 PLANE TRIGONOMETRY. 2 % -}- pc has the same sign and cosine as ^ z ; 99 9 + **-3 ^~? = ( x -f- ^ I + pc the same sine and cosine as % z 4- ^, etc., 9 9 V 9/ 99 so that after the first g terms of the above series, the same values of the sine and cosine return ad infinitum. Representing, therefore, the circumference by 2 n t the equation is entirely general under the form P^ (cos x -\- i/ 1 sin z) = cos 2. (2 w TT -f- x) -f- j/ 1 sin ^ (2 n TT -(- z) (442) 9 9 in which n is any number of the series 0, 1, 2, 3 . . . .9 1. 223. Trigonometric expressions of the real and imaginary roots 'of unity. If z = in (442) it gives (!) = cos^ 2n7r-j- 1 / 1 sin^ 2mr (443) 9 9 or (l) m cos 2 mmr -4- 1/ Isin2win7r (444) m being fractional or integral. If p = 1, (443) gives ^l^ cos l^!i +1 /_i s i n 2jL^ (445^ 9 9 which expresses the q roots of unity by making n successively 0, 1, 2, 3 . . . q 1. For example, let q = 4, ( 445 ) gives for n = 0, i/' 1 = cos + l/ 1 s i n = 1 n = 2, ^ 1 = cos TT -(- |/ 1 sin TT = 1 as found in algebra. If x - in (442), it gives Z (v/ _ 1)- = cos 7ra ( 4n + ] ) 7r + T/ - 1 sin m ( 4n + 1) ? (446) 2t 2t which shows that an imaginary term of any degree can be reduced to a binomial of the form A -\- -\/ 1. If x = TT in (442) we find (!) = cos m (2ri + 1) TT + -\/ \ sin 711 (2 n -f 1) n (447) 224. To reduce an imaginary quantity of the form (a + b I/ l) m to ^' e /orm 4 + .Bi/--l. Let it and x be determined from the equations k cos x = a, k sin z = b by Art. 174 ; then, by Moivre's Formula, (a -f 6 i/ l) w = k m (cos z -4- V' 1 sin x) 1 * = m (cos mo- -{- |/ 1 sin mx) and putting A = k m cos nuc, B = k m sin mx, (o 4- 6 T/ l) m = vl + J B l /-l. TRINOMIAL OR QUADRATIC FACTORS. 131 TRINOMIAL OR QUADRATIC FACTORS. 225. To find the quadratic (trinomial) factors of the expression z tm 2 z m cos + 1 J m being integral. By (438) and (434) we have y lm 2 y m cos mx -f- 1 = y* 2 y cos x +1=0 Therefore if we put y = z, mx = 2 n TT -f- 0, or a; = t^, we have (448) (449) 771 As these two equations exist at the same time, they have common roots, and the second is therefore a divisor or factor of the first; but this factor has m values in consequence of the m values of cos f (Art. 222), found by making m n = 0, 1, 2, 3 . . . m 1. Therefore the m quadratic factors of (448) are all ex- pressed by (449), and we have 2Z" 1 cos i> -f 1 = (z 2 2 z cos 4- + l] \ ml x ... X (z 2 - 2 cos 2("-l) ff + ^ + A (450) V in I 226. To obtain the simple factors of (448), we have only to find the two simple factors of each of the quadratic factors in (450), or to find the two factors of the general quadratic (449). Now, by the theory of equations, if 2 X and z t are the two roots of (449), the first member is equal to (-,) (-%) but we have by (432) . . z = y = cos x 4- T/ 1 sin x = cos m which gives the two values of z by the double sign belonging to |/ 1. Therefore the simple factors of (448) are all included in the form (451) 132 PLANE TRIGONOMETRY. EXAMPLES. 1. Find the quadratic and simple factors of z* 2 z 2 + 1 Here m 2, 2 cos = 2, cos # = 1, ; and by (450), z* 2 z 2 + 1 = ( z* 2 z cos + 1 ) (z- 2 z cos * + 1 ) = ( z 2 2z-fl) by (461) = (cosO + v/ Isin 0)] X [z (cosO |/ 1 sin 0)] X [z (cos T + i/ 1 sin ?r)] X [ (cos TT i/ 1 sin ?r)] = (*-!) (*-!)( + !)(*+!) 2. Find the factors of z* -f 2 2' -f 1. Here m = 2, 2 cos = 2, * = it, and 3. Find the factors of 2* z 1 + 1. * *' + 1 = (," 2 z cos 30 + 1) (z 2 + 2 z cos 30 + 1 ) 4. Find the factors of z 2 2 s + 1. = (2' 2z+ 1) ( Z J +2 227. To /nrf the quadratic factors of z m 1 when m is odd. In (450) let = 0, it becomes 2jr , m X (# 2 z cos -f \ m x . . . n / i\ _ v (452) Now m being odd, m 1 is even, and the number of trinomial factors in (452) exclusive of (z I) 1 , is even ; but so that the first and last of these factors are equal. In the same manner it is shown that any two of these factors equally distant from the first and last are equal ; TRINOMIAL OR QUADRATIC FACTORS. 133 therefore, uniting these equal factors and extracting the square root of both members, we have, when m is odd, a_ 1 = (z 1) X (z 2 2 z cos + \ m (453) / 228. To find the quadratic factors o/z"* - 1, when m is even. When TO is even, m 1 is odd, the number of factors in (452), exclusive of ( I) 1 , is odd, and the middle factor will not combine with any other. This factor is the and contains [ ) cos - = cos JT = 1 m and is therefore equal to so that uniting the remaining factors, and extracting the square root, we have, when m is even, a" 1 = (z 1) (z + 1) X (z 2 2 z cos ~ V TO X(* J 2zcos \ TO X ---- (464) m 229. To find the factors of z m -f- 1, when m is odd. In (450) let = TT, it gives (a* + I) 4 = fz a 2 z cos + l \ TO -- TO and it is easily shown, as in the preceding articles, that the factors equally distant from the first and last are equal, and that the middle term is z 1 -{- 2 z + 1 = (z -f I) 1 M 134 PLANE TRIGONOMETRY. Hence we find, when m is odd, 2 + 1 = (z + 1) X (z 2 2 z cos + V m X # 2 z cos + \ m x.... X (2' - 2 * cos (^ + 1 (465) 230. To /ncf the factors of z m -\ - 1, w/ien m ts even. The same process gives z + 1 = ( 2 2 z cos + l\ \ m I X (* 2 z cos + l) V 7/1 / x.... (456) 231. The simple factors of (453) and (454) are obtained from (451) by putting = 0, and those of (455) and (456) by putting = IT. There will be found pairs of equal factors as in the preceding articles, but all the different simple factors will be found by taking only the positive sign of the radical i/ 1. 232. Any function of the form 2*"' 2 p z -j- q may also be resolved into quadratic factors. It is only necessary to reduce it to one of the preceding forms. By resolving the equation 2 3m _ 2 p z* -f q = (457) we shall find from its two values of z m and if we put the absolute term in one of these factors = dc a m (according to its sign) it becomes 2 a = a m f db 1\ = a" (z" 1) \a m / in which z = az', and the factors of this last expression may be found by one of the preceding articles. If, however, the values of z" 1 in (457) are imaginary, i. e. if p* < q, this method fails to discover the real quadratic factors, and we must proceed as follows. Put q = o 2m , then the proposed function becomes in which z = oz' ; and since in the present case p < a m , -2- is a proper fraction, and a m we may put -2- = cos 0, which reduces the given function to the form (450). a* CHAPTER XV. TKIGONOMETKIC SERIES CONTINUED. MULTIPLE ANGLES. 233. THE true developments of sin mx and cos mx in series, when m is not restricted to integral values, were first obtained by Poinsot, and form the subject of a memoir read by him before the French Academy of Sciences, in 1823.* The fol- lowing problem is the basis of these investigations. 234. To develop (k -f- V" k? l) m , in a series of ascending powers of k. Let z = (k -f V ' K 1 1) m (a) and assume Differentiating (a) and putting / = ok dk we find z' = m (k the square of which gives m z Differentiating this and putting X _fc \_ ma P_i)j-V(*_ dk we find, after dividing by z', m* z ltz f (A 2 l)z //P = Again, differentiating (6) twice, we find, Substituting in (d) the values of z, z x , z", given by (b) and (e), we have (d) = m t A n .2.4, 2.3 ^ 3 A -f- m 2 J. 2 + m t A n nA n (n 1) n A n (n in which each of the coefficients of the powers of k must be zero. To discover the law which governs these coefficients, it will suffice to examine that of the general term, or the coefficient of &", which is whence * _* A n (n + 1) (n + 2) * See the published memoir, " Recherches sur F Analyse des Sections Angulaires," Paris, 1825. 135 136 PLANE TRIGONOMETRY. o that from the first coefficient, A , we find by making n = 0, 2, 4, 6, etc., A - m * A **" 15* A- m '~*A -"V"'-*) * 3'4~ At ~ 1-2-3-4 A A -' ** ___ m(m' 5-6 1-2-3 -4-5 -6 etc. and from the second coefficient, A v we find by making n = 1, 3, 5, etc., 2-3 , _ '- " 2-3-4-5 m' 5* . (OT 1') (m 8) (m a 6') ^~ ~6y~ A " 2" etc. Therefore, if we put the equation (6) becomes 2 = and it only remains to find A and ^4^ In (a), (b), (c) and (e), put i = ; we find =(,/ 1) = 4, 2 ' = m( 1 /-l) m - 1 = A Therefore we have, finally, / - 1)"- 1 m JT' (458) 235. To develop (VI A* + A -j/ l) m in a series of ascending powers of h. We have h v l} m = (v l) m (h + V^f^l) m therefore by (468), exchanging k for h, ( vT=^ + h v - \T = (v !) m \.(v i) m R+ (v i)"- 1 mff/ ] in which H and IP are what K and .K 7 become when h is put for k. Combining the imaginary factors in the second member, observing that (V ~ 1)" X (V ~ l) m = (V !) m = 0)' MULTIPLE ANGLES. 137 (which must not be put equal to unity, since m may be a fraction, and unity has imaginary roots,) and also that (v - !) m x (v - 1)"- 1 = v - 1 ( v - 1) 1 x (v - 1)"- 1 = v- 1 (i) ^ we have tf+ A y - 1) = (1)* H+vl (1) mH' (459) in which - 236. To develop the sine and cosine of the multiple angle in a series of ascending powers of the cosine of the simple angle. When m is an integer, this problem requires us simply to develop sin mx and cos mx in a series of powers of cos x ; but when m is a fraction = JL, the angle mx 9 has q values which have the same sine and cosine, (Art. 222), if we consider x to repre- sent all the angles which have the same sine and cosine as the simple angle. We shall therefore employ Moivre's Formula in its general form (442), or (cos x -f V 1 i n z ) m = eaim (2 nr 4* *) 4* V* 1 sin m (2 n n -f- x) Putting k = cos x we have by (458) and (446), (cos z--f V 1 sin x) m = (k + V'P l) m = (T/ - 1) JT+ (v' I)"- 1 m X"' . /( I sin I v ^ mJt / Comparing the real and imaginary terms of these two values of (cos x -f i/ 1 sin z) m , we have sin m (2 n TT + x) = sin ^ ^ + 1) ^ ^ + g - n ^(m- 1) (4^ -f 1) If m is a fraction = -^, each member of these equations receives q values by taking ? successively for n, or n 7 , the numbers of the series 0, 1, 2, 3, ... q 1 ; but we are now to show what values of n and n' correspond to each other in the two members. Let x = -, then k = 0, K=l, K' 0, and we have 18 M 2 138 PLANE TRIGONOMETRY. therefore these two angles can only differ by some multiple of 2 T, or we must have m (4n 4- 1) TT TO (4 n' -{- 1) TT .. ~2~ T- whence m (n n') = n" hut TO being a fraction ?, and n, n f numbers of the series 0, 1, 2, . . . q 1, we can- not have m (n n x ) equal to an integer n", unless it is zero ;* therefore n n' = 0, n = n' and the above developments are coswi (2n7r + x ) = cos /j(4 + l) ff \ . _g- + cos (( m ~ 1 U 4n + !) ff \ . mjfi : ( 460 ) sinm(2n7r + z)=:sin(-^" o r *> j.^+sin ( v ""~ V ^ ft "^ ^ " ) . mJT / (461) in which .. 1*2 1*23'4 K> = cos x - ?^1' cos* , + ("'- *;) (' -^ cos* x - etc. It hence appears that, in general, it requires the combination of two series to express the cosine and sine of a multiple angle in powers of the cosine of the simple angle, when m is fractional. 237. When m is an integer, one of the terms of (460) and (461) will always become zero, and we shall have but a single series to express the function of the multiple angle. The first members become in all cases cos (2 mn it + mx) =- cos mx sin (2 mn TT -(- mx) = sin mx and the second members vary according to the form of m. In (460), if m = 4 m', cos mx = K m = 4 m/ -f- 1, cos mx = mK f m = 4 m' + 2, cos mx = K m = 4 m' -f- 3, cos mx = mK / and since when m is even, the series K terminates, and when m is odd, the series K' terminates, these four equations are all finite expressions, and will give the equations of Art. 76, by making m = 1, 2, 3, etc. In (461), if m = 4 m', sin mx = mK m = 4 m / -(- 1, sin mx = K m = 4 m' -{- 2, sin mx = mK' m = 4 m f -\- 3, sin mx = K * Since ^ is supposed to be reduced to its lowest terms, p and q are prime to each q other; therefore, if J_ n 1 is not zero, q must divide n n' \ which is impossible, ? since the greatest value of either n or n' is q 1. (462) MULTIPLE ANGLES. 139 In these formulae, however, the series do not terminate, but by differentiating (462) we find for m t 2* m = 4 m', sin mx = m sin x (cos x cos 1 x -f- etc.) (463) m = 4 m' -J- 1, sin mx = sin x (1 - cos* x -j- etc.) 1*2 m = 4 m' -{- 2, sin ma; = m sin a; (cos z ~~~^ cos * z ~f~ etc> ) m9 m t _ Jl m = 4 m' -f- 3, sin mx = sin a; (1 -- cos* * -j- etc.) i i all of which terminate and give the equations of Art. 75. 238. To develop the sine and cosine of the multiple angle in a series of ascending povxra of the sine of the simple angle. We take as before (cos x + i/ 1 sin x) n = cos m (2 n TT -f- x) -f- 1/ 1 sin m (2 n ?r -f x) Putting A = sin x, we have, by (459) and (444), (cos x -f- i/ 1 sin x) m = (V 1 A 1 -+- AT/ l) m = (1) ff -f |/ 1 (l = cos mn' TT . H ' -(- j/ 1 sin m n' K . H + I/ ^ cos ( TO 1) n' TT . mU' sin (wi 1) n' ir . m J Comparing the real and imaginary terms of these equations, cos tn(2n7r-|-x) = cos m n f IT . .ff sin (m 1 ) n f if . m H f sin m (2 n JT -f- a;) = sin m n 7 TT . T + c 08 ( TO 1) n x TT . m IT' and to find what values of n and n f correspond, let x = 0, then h = sin x = 0, JJ= 1, H' = 0, and we have sin 2 m n TT = sin m n' IT from which we infer that 2 m n TT = m n r ir, or 2 n = n', and hence cos t (2 T -f- z) = cos 2 mn TT . H sin 2 (m 1) n IT . mH' (464) sin m (2 n ?r -j- x) = sin 2 mn n . H -\- cos 2 (m 1) n JT . mJiF (465) in which m being a fraction = E, n is any number of the series 0, 1, 2, 3, ... g 1 ; and 1 l - t sin'x-etc. ' a tt) sin'x-etc. 2-3 2-3-4-5 239. When mis an integer, the first members of (464) and (465) become cos mx and sin mx ; and the coefficients of the second members contain only multiples of 2 IT ; therefore we have cos mx = H sin mx = mH' But the series H terminates only when m is even, and the series H' only when m is 140 PLANE TRIGONOMETRY. odd, and we must also employ the derivatives of these equations to obtain finite ex- pressions in all cases ; thus we have also sin mx = cos mx mdx dx Therefore differentiating the series H and H / , we shall have, when m = 2 m', cos mx. = 1 sin 1 x -f- etc. I'm m = 2 m' + 1> cos mx = cos x (1 m ~ sin* x + etc. ) 1 4 m = 2 m f , sin mx = m cos z (sin x m ' sin* x -f etc.) 4 O S -]J TO = 2 m' + 1, sin mx = m (sin x sin 5 x -\- etc.) 2*3 (466) (467) all of which terminate, and give the equations of Arts. 77 and 78. 240. To develop the sine and cosine of the multiple angle in a series of ascending powers of the tangent of the simple angle. We have cos m (2 n n + x) -f i/ 1 sin m (2 n IT -f x) = (cos a; + v 1 sin *)" = cos m x (1 -f- V 1 tan x)* Expanding by the Binomial Theorem, and putting tan'* + etc. 1*4*0 we have cosm (2nT + x) + i/ 1 shim (2 n ir -|- x) = cos 1 " x (T+|/ 1 T x ) But the imaginary and real quantities are not yet distinctly separated in the second member, for m being fractional cos m x has a number of imaginary values. If we desig- nate its real value by cos m x, all its values are included in the expression cos* x (l) m = cos m x (cos 2 mn' IT -f y 1 sin 2 mn' n) which, substituted above for cos"* x gives cos m (2 n IT -f x) -f y 1 sin m (2 WT -f x) = cos" x (co82mn x n . T sin 2mn x ir . 2") + 1/ 1 cos m x (sin 2 mn / ir . T -f- cos 2 mm' n . T') Comparing the real and imaginary terms, we now have cosm (2n7r + x)=cos" > x (cos 2mn'ir . T sin 2 mn' ir . T') sinm(2n7r + x)=co8"x (sin2mn / 7r . T-f cos 2mn'ir.T') nd it is shown as in the preceding problems that n = n', whence cosm (2n7r4-x)=cos m x (cos 2 win* . T sin 2mn7r . T') (468) sinm (2 n TT -(- x) = coe m x (sin 2mn?r . T-f cos 2mnir . 2") (469) MULTIPLE ANGLES. 141 in which m being a fraction = %, n is any number of the series, 0, 1, 2, . . . q 1 ; __ and cos m z denotes only the real value of ^(cos xp. 241. By the division of (469) by (468) x tan 2mn7r. T+ T f IAnK . tan m (2 n ir -f- x) = (470) T tan 2mmr.T' 242. When m in an integer, both the series T and T f terminate, and in all cases cos 2 mn rr = 1, sin 2 mn TT = ; and (468), (469) and (470) give cos mx = cos m x . T (471) sin mx = cos m x . T f (472) tan mx = ~ (473) which last expression embraces all the equations of Art. 79.* 243. Before the memoir of Poinsot, developments were given for the multiple arcs in series of descending powers of the sine or cosine of the simple arc ; but he has shown that these developments are impossible, except when m is integral, and in this case the series are the same as the preceding, with the terms written in inverse order. 244. To develop any power of the cosine of the simple angle in a series of sines or cosines of the multiple angles, the cosine of the simple angle being positive. If y = cos x + |/ 1 sin x, we have, by (434) and the Binomial Theorem, (2 cos z) m = (y + y- 1 )" = y m + my m - 2 -f m ( m ~ 1 I y m-4 _j_ eta % and by Moivre's Formula, y* = cos m (2 n tr -\- x) -\- i/ 1 sin m (2 n TT -j- x) m ym-t m egg ( m 2) (2 n K -j- x) -f- wi i/ 1 sin (m 2) (2 n IT -(- x) ' Isin (m 4)(2n?r-(-x) a etc. Therefore, if we put P? nn+x = cos m (2 n K -\- x) -\- m cos (m 2) (2 n ir -f- *) + etc. P'anTT+z = sin m (2 n ?r -f- x) -f m sin (m 2) (2 n n -f x) -f etc. we have (2 COS X) m = P to , + + V 1 P'tHir+x (<*) Now m being a fraction (2 cos x) M has imaginary values, but when cos x t's positive, it will have at least one real positive value, and then (2 cos x) m being understood to de- note only this real value, all the values are included in the formula (2 cos x) m X (1) = (2 cos x) m (cos 2 mn'tr -)- y 1 sin 2 mn'ir) * Although the formulae for multiple angles require, in general, the combination of two series when m is not an integer, yet there are certain cases, even when m is a frac- tion, in which one or tlie other of the series will disappear. See the memoir of Poinsot, cited at the beginning of this chapter. 142 PLANE TRIGONOMETRY. Therefore we have ( 2 cos z) m (cos 2 mn' T -f i/ Isin2 mn' n) = P 2nir+I -f i/ 1 -P'jnir+B Comparing the real and imaginary terms, (2 cos x) m cos 2 mn' K = P 2nw+I (2 cos x) m sin 2 mn' K = P'zmr+z and to find the corresponding values of n and n x , let z = 0, then (2 cos x) m =2 m , and the series become P Jmr = cos 2mn7r (1 + m + m(m ~ 1 ) -f etc.) m = cos 2 mn if (1 -f- I) 1 * = 2" 1 cos 2 mn TT and in the same way P / znir = 2 m sin 2 mn TT Therefore our formulae become 2 m cos 2 win' TT = 2 cos 2 mn ir 2 sin 2 mn x TT = 2 m sin 2 mn JT and as in former cases, it is shown that n = n', so that we have finally (2 cos z) = ****+-- (474) = n -^ (475) sin 2 mn tr From this it appears that the real and positive value of (2 cos x) m may be expressed either by a series of cosines or by one of sines of the multiple angles, and by comparing (474) and (475), we have the following constant relation between these series. P'tnir+x _ sin 2 mn TT Pjnir+z COS 2 mn 7T 245. If n = 0, (474) gives (2 cos x) m = P x = cos mx -f m cos (m - 2) x + *(* ^ cos (m 4) x + etc. (476) 2 which may be employed as the general development of the real value of (2 cos x) m , when x < - 246. The same supposition of n = 0, gives sin 2 mn n = 0, and (475) gives there- fore, = P'x = sin mx + m sin (m 2) x + " (* 1) s j n ( m _ 4) x + etc . (477) 2t a remarkable property of this series of sines of multiple arcs, which holds for all values of m, provided z < 2t 247. To develop any power of the cosine of the simple angle in a scries of sines or cosine^ of the multiple angles, the cosine <>/ the simple angle being negative, MULTIPLE ANGLES. 143 If the denominator of n is even, there is no real value of (2 cos x) m when cos x is negative ; but we may put (2 cos x)" = ( 2 cos x) ( !) = ( 2 cos x) m [cos m (2 n' + 1) K + ^/ 1 sin m (2 n' + 1) IT] which, substituted in equation (a) of Art. 244, gives ( 2cosx) m cosm (2 n' + 1) T = Pj + . ( 2 cos x) m sin m (2 n' + 1) TT = P' Swir + , Making x = TT, cos x = 1, ( 2 cos x) m = 2, and the series become, by the process shown in Art. 244, P( + ijir = 2 m cos TO (2 n -|- 1) TT P / (ii+)ir=2 sinm (2n + l)*r and we have 2 m cos TO(2n / -f-l)7r = 2 m cos m (2 n -f- 1) w 2 m sinm (2 n' + 1) IT = 2 sin m (2 n -f 1) ir whence, as before, n = n', and our formulae are ( 2 cos z)" = J !lnir + " - (478) cos TO (2 + l) TT (- 2 cos x)" = - P ^"^' - (479) sin m (2 n + 1) IT by which it appears that the real value of ( 2 cos x) m is also expressed either by a series of cosines or of sines of multiple arcs, which series have the constant re- lation P'g n IT + x sin m (2 n -\- 1) TT Plnir + x COS m (2 -|- 1) JT 24ft. If n = 0, (478) and (479) give ( 2 cos x) = P .* = ~ (cos rox-fmcos(ro 2)z + etc.) (480) cos m TT cos m TT ( 2 cos x) m = ~ x ~ - (sin mx + m sin (m 2) x + etc.) (481) sin m TT sin TO TT In this case sin m TT is not zero, unless m is an integer, so that the series P f x does not become zero when x > ~, and both (480) and (481) may be employed as the true m developments of ( 2 cos x) m . 249. When m is an integer, the series (476) and (480) always terminate at the (m -f- l)th term ; and, since in (480) cos nur = db 1, according as TO is even or odd, and ( 2 cos x) m = (2 cos z) m in the same cases, both (476) and (480) become (2 cos x) = cos mx + m cos (TO 2) x + 5^? ill cos (m 4) x + etc. (482) 2 But the series (481) becomes zero, so that (482) is the only series by which (2 cos x) m can be developed in functions of the multiple arcs, when m is integral. 144 PLANE TEIGONOMETEY. 250. To develop any power of the sine of the simple angle, in a series of sines or cosines of the multiple angles. If y = cos x -j- y 1 sin x, we have, by (435) and the Binomial Theorem, (l/ l) m (2 sin z) m = (y y ) _ ete> in which \f*, y m ~*, etc. have the same values as in Art. 244, but the signs of the coefficients are alternately + * n d so that if we put Qt*w + x = cos m (2 n it -f x) m cos (m 2) (2 n TT -(- x) -f etc. Q'inw + x = sin m (2 n it + x) m sin (m 2) (2 n TC -f x) + etc. we have (V l) m (2 Bin z)" =$,, + ,+ y 1 ' 2nir + * Substituting the value of (^/ l) m by (446), and comparing the real and imaginary terms, we find (2 sin ,)- sin ' = V ,. + . 8 and if we make a: = -, we shall find by the process frequently employed above m that n = n x ; whence (2sin Je ) = - -Slpi^ (483) cos m (4 n -f- 1) *r (2 sin x) m = - Q'** + * (484) sin m (4 -f 1 ) ?r so that the real value of (2 sin x) m may be developed in either the cosines or sines of the multiples. The two series have the constant relation g^ng^fs _ sin ft m (4 n -f 1) it Qtnn + x cos Jm (4n + l)r 251. If n = in (483) and (484), (2 sin x) m = - ^* - = - - - (cos mx m cos (m 2) * + etc.) (485) (2 sin z) m = - = -r r - (sin tnz m sin (m 2) x + etc.) (486) both of which series are applicable when m is fractional. 252. When m is an integer, one or the other of the series (485), (486), will always be zero, according to the form of m, and there will be but one series to express (2 sin x) m . If m 4 m', (2 sin z) m = cos mx m cos (m 2) a; -f etc. (487) m = 4 m' -f 1, (2 sin z) w = sin ma; m sin (m 2) z + etc. (488) m = 4 m r + 2, (2 sin z) m = (cos mx m cos (m 2) x -f etc.) (489) m = 4 m' +3, (2-sin x) m = (sin mz m sin (m 2) z -f etc. ) (490) MULTIPLE ANGLES. 145 253. The series (485) and (486) become zero when m is an integer, as follows : If m 2m', = sinnu; msin(m 2) x -f etc. (491) m =z 2 m' -f- 1, = cos mx m cos (m 2) x + etc. (492) The reason why these series are zero is obvious, since they terminate at the (m + l)th term, the terms equally distant from the first and last are equal with oppo- site signs, and the middle term of (491) is zero. 254. Given the equation (493) to express x y in a series of multiples of y. Substituting the values of tan x and tan y given by (431) _ ' P -1 -f- whence or putting fs =rZll (494) e (z-v)V'-l = Taking the Naperian logarithms of both members, 2 (x y) i/ 1 = log (1 q e~ J'*'- 1 ) log (1 and developing the second member by the formula log (1 n) = n J n* J n s etc. we have 2 ( 2 _ y ) y 1 = q e - J vv- 1 _ 9 e-^^-i J g e-'vv'- 1 etc. Substituting in the second member by (430), x y q sin 2 y -f \ (f sin 4 y -f- J g* sin 6 y -f- etc. (6) The equation (a) might have been put under the form 1 L vV i 1 1- 9 from which, by taking the logarithms and substituting as before, sin 2 y sin 4 v sin 6 y /\ x + y = * z z etc. (c) 9 27 s Sg 3 In this investigation, we have, in effect, used Moivre's formula, in its limited or less general form ; but the requisite generality may be given to our results, by observ- ing, that (493) would hold if we were to substitute tan x tan (n'n -f *), tan y = tan (n /x 7r -(- y), and therefore we may substitute for the first member of (6), 19 N 146 PLANE TRIGONOMETRY. n'Tr -)- z (n"T -(- y) = x y (n" n'} it = x y wr, n being (like n' and n") an arbitrary integer or zero. Hence, the required general development of z y in series is z y = nir -f q sin 2 y + J 9* sin 4 y + 9* sin 6 y -f- etc. (495) In like manner, since tan z = tan (z n /7r ), tan y = tan (y TI^JT), we may sub- stitute in the first member of (c), z n'n -\- y n f/ ^ = x -f- y w, and the general development of z -j- y in series is sin2y sin 4 y sin 6 y , (AQC\ x -f- y = TIT * z- - -|- etc. (496) q 2q* 3 q 3 In these formulae z and y are supposed to be expressed in arc, and to obtain z ^ y in seconds, the terms of the series must be divided by sin l x/ . 255. The preceding problem is particularly useful in finding z when p and y are given, and z is nearly equal to y ; in which case p is nearly equal to unity, either q or is a small fraction, and one of the series (495), (496) converges rapidly. EXAMPLES. 1. Given y = 50 and p = 1-00065, to find z from (493). Taking only the first term of the series (495), and assuming n = 0, 2 y = 100 log sin 2 y 9'99335 00065 ,.. - 7 , a = log q o O1174 2-00065 ar co log sin 1" 5-31443 x y = 65"-995 log (z y) 1-81952 z = 50 l'5"-995 2. Given y = 50 and p = 1-00065, to find z from (493). In this case _ 2-00065 00065 and the computation by (496), if we assume n = 0, is 2 y = 100 log sin 2 y 9'99335 _1 = '00065 loj-^l- 6-51174 q 2-00065 ar co log sin 1" 5-31443 z -f y = 65"-995 log (z + y) 1-81952 z = y 65 /A 995 = 50 I' 5 "'995 or, if n = 1, z = 180 50 1 ' 5 /A 995 = 129 58' 54 /A 005. In general, (493) is to be solved by (495) when p is positive, and by (496) when p is negative. 256. Given the equation sin (a -f- 2) =wisinz (497) to express z in a series of multiples of a. We deduce as in Art. 168, an (z + J a) _ ^ ^ MULTIPLE ANGLES. 147 which is reduced to (493) by putting x = z + $a y = * p= m _ 1 whence 9 = p -f- 1 m and (495) becomes i sin a i sin Z> a , sin o ct \ / A(\Q\ z = HTT H H -) + etc. (4yo) 03 o Vn 3 which is to be employed when m > 1 ; and (496) becomes z -f- a = TIT m sin a J m* sin 2 a m* sin 3 a etc. (499) which is to be employed when m < 1, n being any integer or zero. 257. Given the equation (500) 1 -(- m cos a to express z tn a series of multiples of a. This equation in the form sin z m sin a cos z 1 + m cos a gives sin z -\- m sin z cos a = m cos z sin o sin z = m sin (a z) tan (a J a) = OT ~ 1 tan } a m -f 1 which is reduced to (493) by substituting p _ whence g = i p + 1 m and the series (495) and (496) become (501) m 2 m? 3 wi s z n TT -(- m sin a J m* sin 2 a + J m 8 sin 3 o etc. (502) 258. Given the equation tanz= msina (503) 1 m cos o to express z in a series of multiples of a. The equation (500) becomes (503) by changing the signs of both m and a: the same changes in (501) and (502) give _etc. (504) m 2 nr 3 m' = n?r-|-Tnsin a -\-\rn* sin 2 o -|~ J m s sin 3 o -f etc. (505) 148 PLANE TRIGONOMETRY. 259. In a plane triangle ABC, given a, b and C, to find A or B by a series of multiples of C. By (260) ~- sin C tan A = 1 - cos (7 b which, compared with (503), gives, by (505), (506) n being necessarily = in this case. B is found by the same series, interchanging a and b. 260. In a plane triangle, ABC, given a, b and C, to find c by a series of multiples of C. We have (507) 4= * 2*008(7+1 a* a 1 a by (451) -(cosC + i/ lsinC)~| X F - (cos C i/ - 1 sin C)~\ --= [l-y (cosC+i/ IsinC)] X [l - |- (cos (7- T/ - 1 sin <7)"| Taking the common logarithms, employing in the second member the formula log (1 n) = M (n + $ n + J n* + etc.) and applying Moivre's Formula (440) in expressing the powers of cos \/ 1 sin C, we have 2 log c 2 log 6 = -M Py (cosC+V lsin(7) + -^ (cos 2C-f ^-1 sin 2C) +etc.l JJf P- (cosC V lsinC)+-^;r (cos2C !/ 1 sin 2 C) -f etc.] L b 2o* J i i vr i a n i O 1 COS 2 C i O* COS 3 (7 , . \ /CAO\ Iogc = log6 M { coeC+ -- - -- J- ^ -- (-etc.j (508) This series was first given by Legendre. The series (495) and (496), upon which are based those of the subsequent articles, (Arts. 256, 257, 258 and 259), are due to Lagrange. PART II. CHAPTER I. GENERAL FORMULAE. 1. SPHERICAL TRIGONOMETRY treats of the methods of computing the unknown from the known parts of a spherical triangle. It is shown in geometry,* that a spherical triangle may, in gene- ral, be constructed when any three of its six parts are given, (not excepting the case where the three angles are given). We are now to investigate the methods by which, in the same cases, the unknown parts may be computed. We shall at first confine our attention to such triangles only as are treated of in geometry, namely, those whose sides are each less than a semicircumference, and whose angles are each less than two right angles ; that is, those in which every part is less than 180. 2. It is shown in geometry, that if a solid angle is formed at the center of a sphere by three planes, the three arcs in which these planes intersect the surface of the sphere form a spherical triangle. Now the real objects of investigation in spherical trigonometry are the mutual relations of the angles of inclination of the faces and edges of a solid angle ; but, for convenience, the spherical triangle which forms the base of the solid angle is substituted for it. The sides of the triangle being proportional to the angles of inclination of the edges of the solid angle, are taken to represent those angles ; and the angles which those sides form with each other are regarded * The student is here supposed to be acquainted with Spherical Geometry, at least so much of it as is to be found in Legendre's treatise, or in that of Prof. Peirce, of Harvard University. N 2 149 J50 SPHERICAL TRIGONOMETRY. as identical with the angles of inclination of the faces of the solid angle. But, since varying the radius of the sphere would not, in any respect, change the solid angle, or the values of the angles which enter into it, the mutual relations in question ought to be deduced without any reference to the magnitude of the radius of the sphere. In fact, we shall deduce our fundamental formulae from a direct con- sideration of the solid angle itself. 3. In a spherical triangle, the sines of the sides are proportional to the sines of the opposite angles. FlG . lt Let A B C, Fig. 1, be a spherical triangle, the center of the sphere. The angles of the triangle are the inclinations of the planes A B, A C and B C, to each other, and will be designated by A, B and C; their opposite sides respectively will be designated by a, b and e, as in plane triangles. The trigonometric functions of these sides will be the same as those of the angles B C, A C, A B, which they subtend at the center of the sphere. (PI. Trig. Art. 20.) From any point B' in OB, let fall B' P perpendicular to the plane AOC; and through B' P let the planes B'PA', B f P C' be drawn perpendicular to A and C, intersecting the plane OAC in the lines PA', P C', and the planes A OB, BOC in the lines A' B', B'C'. The plane triangles A'PB f , B' PC' are right angled at P and OA'B', C'B' are right angled at A' and C'. The angle B' A' P, being formed by two lines perpendicular to OA, is the measure of the inclination of the planes AOB, AOC, or of the angle A', and B' C' P is the measure of the angle C. We have therefore, by PI. Trig. Art. 15, B' P sin A = sin B'A'P= ~ whence sin A B'P B f C' _ B[C' } tinC B'A' IV P IV A' GENERAL FORMULAE. 151 Again, sin c = sin B'OA' = B'A' B'O whence sin a sin c B'O B'C' B'O B'A' B'A' Comparing (m) and (n), sin sin A sin c sin C (n) (1) which in the form of a proportion is sin a : sin c sin A : sin C which is the theorem that was to be proved. 4. In Fig. 1, A, a, Cand c, are each less than 90, but the con- struction would not vary if any of these parts were greater than 90, except that the points A' and C' might be found in the lines AO, CO, produced through ; and one or more of the right triangles A'B'P, etc., would contain the supplements of A, a, C, or c instead of these quantities themselves. But the sine of an angle and of its supple- ment being the same, the preceding demonstration would still be valid, so that the theorem is applicable to any spherical triangle. Indeed, according to PI. Trig. Art. 49, this result follows from the nature of the trigonometric functions themselves, and the demon- stration of the preceding theorem might therefore be considered as general, without requiring a special examination of the various posi- tions of the lines of the diagram. 5. In a spherical triangle, the cosine of any side is equal to the product of the cosines of the other two sides, plus the continued product of the sines of those sides and the cosine of the included angle. Let the plane B'A'C', Fig. 2, be drawn perp. to O A, intersecting the planes AOB, BOC and AOC 1 , in the lines A'B', B' C' and A'C'. Then the angle B'A'C' = A, and B'OC' = a, and by PI. Trig. Art. 119, in the triangles A' B'C', OB'C', we have FIG. 2. B'C"* =- A'B' 2 + A'C' 2 2 A'B' . B'C' 2 = OB' 2 + O C' 2 - 2 O B' . A'C' cos A C' cos a 152 SPHERICAL TRIGONOMETRY. Subtracting the first of these equations from the second, and observ- ing that in the right triangles OA'B', OA'C', O B' 2 A' B n = A' 2 , C" 2 A' C' 2 = O A' 2 we have = 2 OA' 2 + 2 A'B' . A'C' cos A 2 B r . C' cos a OA' . OA' A'B' .A'C' whence cos a = Qf Q ^ + Q , ^ cos A Substituting the trigonometric functions derived from the right tri- angles OA'B', OA'C', cos a = cos 6 cos c -J- sin 6 sin c cos A (2) which is the theorem to be proved. It may be regarded as the fun- damental theorem, for the preceding (1) can be deduced from it, but as the process is somewhat circuitous, we have preferred deducing the two theorems from independent constructions. 6. In the construction of Fig. 2, both 6 and c are supposed less than 90, while no restriction is placed upon A and a but the equa- tion (2) is no less applicable to all the other cases if the principle of PI. Trig. Art. 49 be granted. As that principle may not be suffi- ciently evident to the student unacquainted with analyti- cal geometry, we shall verify it in this case, as follows.* 1st. In the triangle ABC, (Fig. 3), let b < 90 and \ c c> 90. Produce BA, BCto meet in B', forming the lune BB' ; then AB' 180 c, and 6 are both < 90, and the preceding demonstration would apply to the triangle AB'C. Therefore, applying (2) to AB'C, we have cos (180 a)=cos 6 cos (1 80 c) + sin 6 sin (180 c) cos (180 4) or by PI. Trig. (64), cos a = cos b cos c sin b sin c cos A and changing all the signs cos a = cos b cos c + sin b sin c cos A the same result that would have been found by applying (2) directly to ABC. * Hymer's Spherical Trigonometry. Cambridge, 1841. GENERAL FORMULA. 153 2d. In the triangle ABC, Fig. 4, let b > 90, c > 90 ; produce AB and AC to meet in A' ; then A 'B and A'C being both less than 90, the formula (2) is applicable to A' EG. Therefore cos a = cos (180 6) cos (180 c) + sin (180 6) sin (180 c) cos A = ( cos 6) ( cos c) -f sin b sin c cos A = cos 6 cos c + sin 6 sin c cos A the same result as before. 7. The theorems expressed by (1) and (2) being applied succes- sively to the several parts of the triangle, give the two following groups : sin a sin B = sin 6 sin A sin b sin C = sin c sin B > (3) sin c sin A = sin a sin C cos a = cos 6 cos c + sin 6 sin c cos .4 cos 6 = cos c cos a -\- sin c sin a cos J? f (4) cos c = cos a cos 6 + sin a sin 6 cos C 8. Let A'B'C', Fig. 5, be the polar triangle of ABC, and designate its angles and sides by A', B', C', a', b' and c'. Then, by geometry, a' = 180 4 6' = 180 B c ' = 180 C ^' = 180 a, 5' = 180 - 6, C' = 180 c, and applying the first equation of (4) to A'B'C', cos a f = cos b' cos c' -f- sin b' sin c' cos A or by PL Trig. (64), cos A = ( cos B) ( cos (7) + sin B sin C ( cos a) cos .4 = cos B cos C sin B sin G cos a Changing the signs of this, we have the first of the following group : cos A = cos B cos (7+ sin B sin Ccos a cos B = cos C cos A + sin C sin A cos b cos 0= cos A cos j5 + sin A sin 5 cos c (6) 20 154 SPHERICAL TRIGONOMETRY. It is thus that, by means of the polar triangle, any formula of a spherical triangle may be immediately transformed into another, in which angles take the place of sides, and sides of angles. 9. Several other important fundamental groups of formula? are obtained from the preceding with the greatest ease. The first of (4) multiplied by cos c is cos a cos c = cos 6 cos 2 c -f- sin 6 sin c cos c cos A and the second of (4) is the same as cos a cos c + sin a sin c cos B = cos b the difference of which is sin a sin c cos B = (1 cos 2 c) cos b sin 6 sin c cos c cos A Since 1 cos'c = sin 2 c, this may be divided by sin c, and gives sin a cos B = sin c cos b cos c sin 6 cos A whence sin b cos C= sin a cos c cos a sin c cos B (6) sin c cos A = sin 6 cos a cos 6 sin a cos C If we interchange jB and C, and therefore also b and c, the group becomes sin a cos C = sin b cos c cos b sin c cos A sin 6 cos A = sin c cos a cos c sin a cos J5 V (7) sin c cos /> = sin a cos 6 cos a sin 6 cos C 10. If (6) and (7) are applied to the polar triangle, they give, after changing the signs of all the terms, sin A cos 6 = sin C cos B -\- cos C sin B cos a sin 7? cos c = sin J. cos C -f cos ^4 sin C cos 6 I (g) sin C cos a sin B cos .4 + cos B sin ^4 cos c and sin ^4 cos c = s\n B cos C -f- cos B sin 6' cos a sin B cos a = sin Ccos A -f- cos (7 sin ^4 cos 6 j> (9) sin Ccos b = sin A cos 5 -f~ cos A sin J? cos c 11. Dividing the first of (6) by the following derived from (3), sin a sin B , , r - = sin 6 sin A GENERAL FORMULAE. 155 we find the first of the following group sin A cot B sin c cot 6 cos c cos A ^j sin jB cot C = sin a cot c cos a cos B > (10) sin C cot A = sin 6 cot a cos 6 cos C } and in the same way from (7), or by interchanging the letters B and (?, b and c in (10), we find sin A cot C = sin b cot c cos 6 cos ^1 ^ sin B cot .A = sin c cot a cos c cos .B > (11) sin C cot jB = sin a cot 6 cos a cos (7 J If (10) are applied to the polar triangle, we find (11), so that no new relations are elicited. 12. The preceding formulae are sufficient to furnish a theoretical solution for every case of spherical triangles, but some transforma- tions are required to facilitate their application in practice. In the first of (4) substitute, by PL Trig. (1 39), cos A = l 2 sin 2 A we find, by PI. Trig. (39), cos a = cos (6 c) 2 sin b sin c sin 2 A (12) and we have similar expressions for cos b and cos c. If we substitute in (4), by PL Trig. (138), cos A = 1 -f- 2 cos 2 A we find, by PL Trig. (38), cos a = cos (b + c) -f- 2 sin b sin c cos 2 J A (13) and, of course, similar expressions for cos b and cos c. 13. Substituting in (5) cos a = 1 2 sin 2 J a = I + 2 cos 2 ^ a we find by the same process cos A = - cos (B + C) 2 sin B sin C sin 2 1 a (14) cos ^4 = - cos (j5 C) + 2 sin 5 sin C cos 2 } a (15) which might have been obtained by applying (12) and (13) to the polar triangle. 156 SPHERICAL TRIGONOMETRY. 14. If in ( 12) we substitute cos o = 1 2 sin* J a, cos (b c) = 1 2 sin* J (6 c), we obtain the first of the following equations ; and the others are obtained by a similar process from (12j, (13), (14) and (15). sin* j a = sin" J (b c) + sin b sin c sin* J A (16) sin* | o = sin* (6 -f c) sin 6 sin c cos* $ ^4 (17) cos* a = cos 2 J (6 c) sin b sin c sin* J A (18) cos* J o = cos* (6 + c) -f- sin 6 sin c cos* .4 (19) sin* } A = cos* J (B + C ) -f sin B sin C' sin* J a (20) sin* } A = cos* i (5 - 6') sin B sin C cos 2 $ a (21 ) cos* i .-1 = sin* J (J3 + C) sin sin C sin* o (22) cos 1 M = sin* } (JS C) + sin sin (7 cos* * a (23) 15. By PL Trig, we have 1 = cos* J A + sin* } ^4 cos A = cos* i ^4 sin* J .4 whence cos b cos c = cos 6 cos c cos* J A + cos 6 cos c sin* f .4 sin 6 sin c cos ^4 = sin 6 sin c cos* \ A sin 6 sin c sin* \ A the sum of which is, by (4), cos o = cos (6 c) cos 1 \ A -f- cos (6 + c ) sin 1 J A (24) and substituting 1 2 sin* \ a, etc., for cos a, etc. sin* } a = sin* $(& ) cos* } ^ + sin* J (6 + c) sin 1 J A (25) cos* i a = cos* } (6 c) cos* A + cos* J (6 + ) sin* i ^ (26) la the same manner we deduce from (5) cos^ = cos(B C)sin Jo cos ( + n 2 J (-P - C 1 ) si" 11 i + '"' *(& + . . - sin s sin (s 6) , , r, sin (s a) sin (s 6) tan 2 i(?=-. sin s sin (s c) 19. From (14) we find Bin * a = cos A + cos (B -|- 180, $>90, cos S is negative, and cos S is positive. 20. From (15) we find , , cos A -f cos (B C} cos 2 1 a = - 2 sin B sin C from which we deduce, by a process similar to the preceding, cos* a = r> n sin B sin C GENERAL FORMULA. 159 COS ai _cos(SB)cos(S C) i -5- a . sin E sin 6 cos cos 2 j , cos (8 C) cos (ff A) sin C sin A _cos(SA)cos(SB) I If C . J T- sin J. sin B (38). 21. From (36) and (38) 2 1 cos 8 cos (/$ J.) tan -a- i + #) Q s i C'cos i (a + 6) r = sin c sin i (.4 JB) J2 = cos Csin J (a 6) = sin J c cos i ( .4 .Z?) 5 = sin $ C sin (a + 6) 9 X r, ? X , *" X *, are respectively equal to the products PXQ,PXR,PXS,QXR,QXS,RXS. First. From (3) we have sin c (sin ^4. :fc sin J5) = sin C (sin a sin 6) which, by PI. Trig. (105), (106) and (135), are reduced to sin Jccos Jcsin J (A + -B) cos J ( A J5) =siu J C'cos \ Csin \ (a -\-b) cos J (a 6) sin Jccosjccos J(^4 + J5) sin (4 JS) =sinj Ccos J Ccos J (a + 6) sin J (a 6) or ps = PS and qr=QR Second. From (6) and (7) sin c (cos JB cos A) = (1 =F cos C) sin (a b) which, by PI. Trig, are reduced to sin Jccosjccos J (A + J5) cos J (A _B)=sin J (7sin J (7sin J (a -j- 6) cos J (a + 6) sin J c cos J c sin % (A -\- B) sin J (^4 JB) = cos ^ Ccos J (7 sin i (a 6) cos J (a b) or qsQS and pr PR Third. From (8) and (9) (1 cos c) sin (A J5) = sin C (cos 6 cos a) which, by PI. Trig, are reduced to cos J c cos J c sin J ( .4 + J5) cos J (-4 + J5) = g i n i C'cos J C'cos J (a + &) cos J (a 6) sin } c sin J c sin J (^4 J5) cos $ (.4 J5) = sin J (7 cos i (7 sin J (a -f- 6) sin J (a 6) or pq = PQ and rs = RS 26. TAe notation of the preceding article being still employed, the quantities p* t q* t r f , *, are respectively equal to P s , ^ 2 , fi*, S*. We have and qr Q R the quotient of which is p* = P 1 whence p = db P and in the same way q* = Q 2 g = Q 21 o2 162 SPHERICAL TRIGONOMETRY. 27. In these last equations, the positive sign must be used in all the second members, or the negative sign in all of them. For if we take P = + P the equations pq = PQ, pr = PE, ps = PS being divided by this, give q=+Q, r= + R, s = + S and if we take p = -P the same equations, divided by this, give q = Q, r = R, s = S We have therefore the following, which are generally cited as Gauss's Equations. cos \ c sin J (A -)- B) = cos \ C cos (a 6) cos \ c cos ( A + B) = sin \ C cos \ (a -f- b) sin c sin (4 E) = cos (7 sin (a 6) sin \ e cos ^ (A B) = sin ^ C sin (a -f- b) cos | c sin J (.4 + .B) = cos J (7 cos J (a 6) cos c cos (A -f- 5) = sin J C cos J (a -f- 6) sin J c sin J (J. B) = cos $ C sin (a b) sin J c cos \ (-4 -B) = sin (7 sin J (a + 6) (44) (45) If, however, we consider only those triangles whose parts are all less than 180, the first of these groups, (44), is alone applicable, for we must then have p = -\- P; since cos c, sin (A + ), cos C, cos J (a 6) are then all positive quantities. The use of (45) will be seen in the chapter on the solution of the general spherical triangle. Napier's Analogies, (40), (41), (42) and (43) can be deduced directly from (44). ADDITIONAL FORMULAS. 28. We shall here add some formulae which, though not so frequently used as the preceding, are either remarkable for their elegance and symmetry, or of importance in certain inquiries of astronomy and geodesy. 29. The product of (30) and (32) gives . . A 4 sin s sin (s a) sin ( 6) sin (a c) (AR ^. sm A ' . (to) sin 1 o sin' e Put n* = sin s sin (s a) sin (s 6) sin (s c) (47) then sin A = - - sin o sin c and in the same manner (48) 2n sin B = sin a sin c the quotient of which is sin A _sin a sin B sin 6 which is our first theorem, Art. 3. As (48) was obtained from (30) and (32), and these from (4) without the aid of (3), we may consider the whole fabric of spherical trigo- nometry as resting upon the fundamental formulae (4). ADDITIONAL FORMULAE. 30. We have also from (35) and (37) gin2 o ^ 4 cos S cos (S A) cos (S - B) cos (8 C) and if sin 2 B sin 2 C * = cosS cos (S A) cos (S B)coa(S O) 2N From (48) and (51), n N sin B sin O sin a sin 6 sin c sin A sin B sin (7 163 (49) (50) (51) (52) 31. If we develop (47) and (50) by PI. Trig. (173) and (174) 4 ri* =1 cos 2 a cos* 6 cos 2 c -{- 2 cos a cos 6 cos c 4 jY 2 = 1 cos" A cos 2 B cos 2 C 2 cos A cos B cos C (53) (54) 32. The following simple results are easily deduced from the equations (31 to 38) cos \ A cos ^ B sin a . sin J sin c $ ^ sin $ .B _ sin (s a) cos J C sin c sn cos (55) (56) cos (7 sin | .4 sin ^ J? _ sin (s c) sin \ C sin c sin \ a sin | 6 _ cos S cos c sin (7 sin ^ a cos | 6 _ cos (S A) sin \ c sin C cos | a sin \ b _ cos (S B) sin ^ c sin C cos | a cos ^ 6 _ cos (S C) cos ^ c sin C 33. By means of (55) and (56) we can deduce expressions for the functions of s, s a, etc., in terms of the angles, or of S, S A, etc., in terms of the sides. We have, from (51), sinc = _ 2JV = _ N _ sin A sin B 2 sin A cos \ A sin \ B cos J B which, substituted in (55), gives sin s = N sin (s c) = 2 sin A sin ^ B sin (7 N 2 cos i .4 cos sin J C whence, by interchanging the letters, we have also sin (s a) and sin ( b). (57) (58) 164 SPHERICAL TRIGONOMETRY. Again, we have sin (a e) = sin s cos c cos s sin c whence _ sin s cos c sin (s c) , * 1 cos s- sin c which, by (55), is reduced to cos i A cos i B cos c sin i A sin 4 B cos s = * * * sin $ (7 and from the equation cos (s c) = cos s cos c + sin s sin c we find, by substituting (55) and (59), f, \ _ sin fr ^4 sin J B cos c + cos J -4 cos i S sinTo To eliminate c from the second members of (59) and (60), we have, by (58) cos C + cos A cos .B sin A sin 5 whence cos J .4 cos J B cos c = : i /< ID cos (7 + cos A cos sin J .4 sin } /j cos c = -- ! - $ A cos % B 2 sin i a sin $ 6 cos $ c . o _ sin \ a sin ^ 6 cos C + cos ^ a cos } fe which, substituted in (59) and (60), give _ cos A -\- cos B -f- cos C 1 _ 1 sin' \ A sin a \ B sin* ^ C //,. ^ 4 sin J A sin J sin JO 2 sin .4 sin sin J (7 , __ v _ cos^t-j-cos/? oos(74-l_ cs a | A -|- cos' ^ P cos' J (7 /go) = From the preceding we easily deduce tan , = _ 2JV _ = _ siiLC _ (63) cos A + cos B + cos (7 1 cos c tan J A tan B tan (s c) == _ 2 N __ = _ 8m c _ . (64) COt %A COS J 5 COS C 34. The equations (57 to 64) applied to the polar triangle, give, _cosS=- - " 3- (65) 2 cos ^ o cos ^ 6 cos i c (66) ADDITIONAL FORMULAE. 165 . ] + cos a -f-^osj> j = _cog_c _ cos 2 a -f cos 2 6 + cos 2 c ^j_ ^ ^ 4 cos J a cos 6 cos \ c, 2 cos \ a cos 6 cos c / o ,r<\ cos i a cos J b cos C+ sin j sin fr 6 / fi q\ sin ( o C/ 1 == V"*/ cos j c <-v_ (71 1 ~ cos a ~~ c gA+-gjLg gin!-Lg-+ sin * 3 fc ~ sinit ^ c (70) 4 sin i a sin 6 cos c " 2 sin i a sin | 6 cos J c 2 sin C (71) ~ I -j- cos a + cos b -j- cos c cos (7 -f- cot J a cot J 6 .,.> 2n sin O ^ ' 1 cos o cos 6 + cos c cos C + tan $ a tan J 6 35. From (68) we find . o 2 cos i a cos i 6 cos i c cos" \ a cos" b cos" c + 1 , 7 o\ 1 sino = , , ; (i*) 2 cos 5 a cos o cos c - . . 2 cos } a cos ^ 6 cos \ c + cos" | a + cos 2 ^ 6 + cos 8 ^ e 1 ,_^ 2 cos a cos J 6 cos \ c the numerators of which may be reduced by PI. Trig. (173) and (174), by making x = |a, j/ = J6, 2 = Jc, whence v = } (a -f 6 + c) = \ s, v x = \ (s o), etc. : therefore, .. . ^ _ 2 sin s sin % (s a) sin % (s b) sin | (s c) , 7 -^ J. Sill o - -tit \ ) cos a cos o cos } c .. , . o 2 cos | s cos fr (s a) cos (a b) cos ( c) /-g-, cos J a cos \ b cos J c The product of these equations reproduces (65) ; their quotient is, by PI. Trig. (154), tan 2 (45 J S) = tan J s tan J (s a) tan J (s 6) tan J (s c) (77) 36. CagnoWs Equation. Multiplying the first equation of (4) by cos A, we find cos a cos A = cos b cos c cos A -f sin 6 sin c sin b sin c sin* A and from (5) in a similar manner, cos a cos A = cos B cos C cos a -f- sin B sin sin B sin (7 sin 1 o Observing that by (3) we have sin 6 sin c sin 2 A = sin B sin C sin 2 a, these two equations give, sin 6 sin c -f cos b cos c cos A = sin 5 sin (7 cos B cos (7 cos a (78) a relation between the six parts of the triangle, first given by CAGNOLI. It is a property of this equation that either member is a function which Acw the same value in a given spherical tnangle and its polar triangle. Thus, if we distinguish the sides and angles of the polar triangle by accents, we have* sin 6 sin c -{- cos b cos c cos A = sin &' sin c' -(- cos b f cos c' cos A f (79) * See Mathematical Monthly, (Cambridge, Mass.,) vol. i. p. 282. 166 SPHERICAL TRIGONOMETRY. 37. To deduce the formulae of plnne triangles from those of spherical triangles. The analogy of many of the preceding formulae with those of plane triangles is sufficiently obvious. We can, in fact, deduce the plane formula; from those of this chapter, by regarding the plane triangle as described upon a sphere whose radius is in- finite, the triangle being an infinitely small portion of the sphere. The quantities a, b and c must, in this case, express the absolute lengths of the sides ; and the angles which they subtend at the center of the sphere, expressed in arc, will be , , , r be- r r r ing the radius of the sphere. When r is very large, , , , are very small, r r r and we may express the values of sin , cos , etc. approximately, by one or two r r terms of their expansions in series, PI. Trig. (405) and (400), and if their values be substituted in our spherical formulae, we shall obtain approximate relations between the sides and angles of the triangle. If we then make r infinite we shall obtain exact relations between the sides and angles of a plane triangle. Thus we have . a a a 8 a 5 sin h etc. a h etc. sin A r r 2.3 r s 2.3 r 2 sin B in _6. 1 ^ + etc. b - -*- + etc. r r 2.3 r 3 2.3 r 3 and making r infinite, we find the formula of PI. Trig. sin A ji sin B b In the same manner cos A = a b c -, a 2 , / , 6 2 c 2 , 6 2 c 2 , \ cos cos cos \- etc. (1 etc. ] r r r _ 2 r 2 I 2?-* 2r* 4r* _f . b . c sin sin r r - + etc. and making r infinite, we have the formula of PI. Trig. c a ~r~~ 2 be Formulae that involve only the sines or tangents of the sides may be reduced im- mediately to the plane formula? by substituting a, 6, etc., for sin a, tan a, etc. Thus, (31 to 34) give the corresponding formula? of PI. Trig, by omitting the symbol sin. ; and (40), (41), by omitting the symbol tan. when these symbols are prefixed to sides. CHAPTER II. SOLUTION OF SPHERICAL EIGHT TRIANGLES. 38. WHEN one of the angles of a spherical triangle is a right angle, the general formulae of the preceding chapter assume forms that are remarkably analogous to the relations established for the solution of plane right triangles, and equally simple in their appli- cation. 39. Let C=9Q, Fig. 6. From (3) we Fl0 - 6 - have . , sin a . ~ sm A = . sm C sin c but since (7= 90, sin C 1 ; therefore, sin A = and, in the same manner, sin J5 = sin a sin G sin b sm c (80) that is, the sine of either oblique angle of a spherical right triangle is equal to the quotient of the sine of the opposite side divided by the sine of the hypotenuse. Compare PI. Trig. (1). 40. From (11), we find cos A = sin b cot c sin A cot C cos 6 but if 0= 90, cot C= ; therefore, sin b cot c cos A = - - = tan b cot c cos b or cos A = tan b tan c COS f = tan a tan c (81) 167 168 SPHERICAL TRIGONOMETRY. that is, the cosine of either angle is equal to the tangent of the adjacent side, divided by the tangent of the hypotenuse. Compare PI. Trig. (1). 41. From (10), we have, sin b cot a cos b cos O cotA = - . n sin C which, when C 90, becomes . , sin b cot A = sin b cot a = tana or, taking the reciprocals, A tan a ^ tan b /oox tan A = - tan B = - (82) sin 6 sin a that is, the tangent of either angle is equal to the tangent of the opposite side, divided by the sine of the adjacent side. Compare PI. Trig. (1). 42. From (5), we find, cos B -\- cos C cos A sm A = . cos 6 sm C and if C= 90, cos B . cos A /eQ >. sm A = sm B = (83) cos b cos a that is, the cosine of either angle, divided by the cosine of its opposite side, is equal to the sine of the other angle. In PL Trig, we have sin A = cos B. 43. From (4), we have, cos c = cos a cos b -f- sin a sin b cos C or, when C= 90, cos c = cos a cos b (84) that is, the cosine of the hypotenuse is equal to the product of the cosines of the two sides. In PL Trig, c 2 = a 2 + b 2 . 44. From (5), cos C + cos A cos B cos c = - - TT- sm yl sm 75 or, when 0=90, COS J. COS J? ,,, / QK x cos c = - cot -4 cot B (85) sm J. sm B SOLUTION OF SPHERICAL EIGHT TRIANGLES. 169 that is, the cosine of the hypotenuse is equal to the product of the cotan- gents of the two angles. In PI. Trig., 1 = cot A cot B. 45. No difficulty will be found in remembering the preceding for- mulae for spherical right triangles, if they are associated with the corresponding ones for plane triangles : thus, In plane right triangles. In spherical right triangles. . . sin a r> 8 i n ^ sin A sin B = c c sin A. . sin Jj . sm c sin c A tan 6 D tana cos A - cos B c c !,, 4 a 4 011 P " CuS A COS JL> tan c tan c A tana D tan& tan A tan Jo b a tan A . . tan Jo . Bin 6 sin a . A cos B . jj cos A sin A - cos B, sin B cos A c* = a z + b 2 1 = cot A cot B sin A - sin JJ cos 6 cos a cos c = cos a cos 6 cos c = cot A cot B 46. Napier's Rules. By putting these ten equations under a different form, Napier contrived to express them all in two rules, which, though artificial, are very generally employed as aids to the memory. In these rules, the complements of the hypotenuse and of the two oblique angles are employed instead of the hypotenuse and the angles themselves. The right angle not entering into the formula, they express the relations of five parts, but in the rules the five parts considered are a, b, co. c, co. A and co. B. Any one of these parts being called a middle part, the two immediately adjacent may be called adjacent parts, and the remaining two, opposite parts. The right angle not being considered, the two sides including it are regarded as adjacent parts. The rules are : I. The sine of the middle part is equal to the product of the tangents of the adjacent parts. II. The sine of the middle part is equal to the product of the cosines of the opposite parts. The correctness of these rules will be shown by taking each of the five parts as middle part, and comparing the equations thus found with those already demon- strated. 1st. Let co. c be the middle part ; then co. A and co. B are the adjacent parts, o and b the opposite parts, and the rules give sin (co. c) = tan (co. A) tan (co. B) sin (co. c) = cos o cos 6 or cos c = cot A cot B cos c = cos a cos & which are (85) and (84). 2d. Let co. A be the middle part; then co. c and b are the adjacent parts, co. B and o the opposite parts, and the rules give sin (co. A) = tan (co. c) tan 6 sin (co. A) = cos (co. B) cos o cos A = cot c tan b cos A = sin B cos a 22 170 SPHERICAL TRIGONOMETRY. In the same manner, if co. B is taken as the middle part, sin (co. B) = tan (co. c) tan a or cos B = cot c tan a sin (co. B) = cos (co. A) cos 6 cos B = sin A cos b and these four equations are the same as (81) and (83). 3d. Let o be the middle part ; then co. B and 6 are the adjacent parts, co. A and co. c the opposite parts, and the rules give, sin a = tan (co. B) tan b or sin a = cot B tan 6 sin a = cos (co. A) cos (co. c) sin a = sin A sin c In the same manner, if b is taken as the middle part, sin b = tan (co. A) tan a or sin b = cot A tan o sin b = cos (co. B) cos (co. c) sin b = sin B sin c and these four equations are the same as (80) and (82). It appears, therefore, that these rules include all the ten equations previously proved ; and they include no others, since we have taken each part successively as the middle part. In the application of these rules, it is unnecessary to use the notation co. A, co. B, co. c, since we may write down at once sin A for cos (co. A ), etc.* 47. In order to solve a spherical right triangle, two parts must be given, and from the equations of Art. 45, that equation must be selected which expresses the relation between these two parts and the required part. When Napier's Rules are employed, it is only necessary to determine which of the three parts the two given and the one required is to be taken as the middle part. "These three parts are either all adjacent to each other, in which case the middle one is taken as the middle part, and the other two are adjacent parts; or one is separated from the other two, and then the part which stands by itself is the middle part, and the other two are opposite parts." f 48. In order to distinguish the functions of parts less than 90 from those greater than 90, it will be necessary carefully to observe their algebraic signs, according to PI. Trig. Art. 40. But when a required part is determined by its sine, since the sine of an angle and of its supplement are the same, there will be two angles, both of which may be regarded as solutions, except when this ambiguity is removed by either of the following principles. * If we employ as the five parts, the hypotenuse, the two angles, and the comple- ments of the two sides including the right angle, these parts will be the complements of those used in Napier's Rules, and we shall have MAUBUIT'S RULES. I. The cosine of the middle part is equal to the product of the cotangents of the adjacent parts. II. The, cosine of the middle part is equal to the product of the sines of the opposite parts. With a little attention at the commencement, however, and by observing the ana- logy exhibited in Art. 45, the student will find that he will have little use for either of these artificial rules. t Peirce's Spherical Trigonometry. SOLUTION OF SPHERICAL RIGHT TRIANGLES. 171 49. In a right spherical triangle, an angle and its opposite side are always in the same quadrant, that is, either both less or both greater than 90. For, by (83), cos B sin A = cos b in which, since sin A is always positive, (A < 180), cos B and cos b must have the same sign ; that is, and 6 must be either both less or both greater than 90. 50. When the two sides including the right angle are in the same quadrant, the hypotenuse is less than 90, and when the two sides are in different quadrants, the hypotenuse is greater than 90. For, by (84) , cos c = cos a cos 6 in which, if a and b are in the same quadrant, cos a and cos 6 have like signs, and cos c is positive, that is, c < 90 ; but if a and 6 are in different quadrants, cos a and cos 6 have different signs, and cos c is negative, that is, c > 90. We proceed now to the solution of the several cases. 51. CASE I. Given the hypotenuse and one angh, or c and A, Fig. 6. To find a. The relation among the three parts, c, A, and a, (as in PI. Trig, with the same data), is given by the sine of A ; and by Art. 45, FIG. 6. B , sm a sm A = sm c from which we find* sin a = sin c sin A (86) There will be two values of a corresponding to the same sine, but, by Art. 49, the true value is that which is in the same quadrant as A. To find b. The relation among the three parts, c, A, and b, (as in PI. Trig, with the same data), is given by the cosine of A, or, tan b cos A = - tan c from which t tan b tan c cos A (87) *This equation would be found by Napier's Rules, taking a as the middle part, t We find the same result by Napier's Rules, taking co. A as the middle part. 172 SPHERICAL TRIGONOMETRY. To find B. We have, by (85),* cos c = cot A cot B COS C from which cot B = - = cos c tan A (88") cot^l The quadrants in which b and B are to be taken, will be deter- mined by means of the signs of tan 6 and cot B } according to PI. Trig. Art, 40. Check. To guard against numerical errors, it is often expedient to compute the same quantity by two different and independent methods. In many cases, however, we may test the accuracy of several operations by a single formula, which may be called the check. In the present instance, when the three parts, a, 6, and J3, have been found, we should have, by (82), the relation sin a = tan b cot B so that if the work is correct, we shall find log sin a = log tan 6 + log cot B EXAMPLES. 1. Given c = 110 46' 20", A = 80 10' 30", to solve the triangle. By (86). By (87). By (88). c, log sin 9-9708106 log tan 0-4210061 log cos 9*5498045 A y log sin 9-9935833 log cos + 9-2320794 log tan + 07615038 log sin a 9-9643939 log tan b 9-6530855 log cot B 0-3113083 log tan 6 9-6530855 Check, log sin a + 9-9643938 -4ns. a = 67 6' 52"-7, b = 155 46' 42"-7, B= 153 58' 24"-5 2. Given c = 120, A = 120 ; solve the triangle. Ana. a = 13124'34"-7 6 = 40 53' 36"-2 .5 = 49 6' 23"-8 52. If A = 90, we must also have, by (85), c = 90, and then tan 6 = ^ tan S = ~- eo that b and B are both indeterminate; that is, there is an indefinite number of tri- angles which satisfy the given values of c and A ; but since cos.B = cos6sin.4=cos& we always have B = b', and since sin a = sin c sin A = 1 we have a = 90, and all the parts of the triangle are equal to 90, except b and B. If only c is given = 90, all the parts of the triangle are equal to 90, except A and a ; and we have A a. * Or by Napier's Rules, taking co. c as the middle part. SOLUTION OF SPHERICAL EIGHT TRIANGLES. 173 53. CASE II. Given the hypotenuse and a side, or c and a. To find A. We have by (80), sin a . /onx sm A = = cosec c sin a (89) sin c To find B. By (81), cos B = - = cot c tan a (90) tanc To find b. By (84), cos c = cos a cos 6 from which cos b== -'= cos c sec a (91) cos a Check. We have between A, B, and 6, the relation cos B = sin A cos 6 EXAMPLES. 1. Given c = 140, a = 20 ; solve the triangle. By (89). By (90). By (91). c, log cosec 0-1919325 log cot 0-0761865 log cos 9-8842540 a, log sin 9-5340517 log tan + 9*5610659 log sec + 0-0270142 log sin A 9-7259842 log cos B 9-6372524 log cos b 9-9112682 log sin A -f 9-7259842 Check, log cos B 9-6372524 Ans. A = 32 8' 48"-l J? = 115 42' 23"-8 b = 144 36' 28"-4 2. Given c = 101 16' 16"'7, b= 115 42' 38"-5; find A. Ans. A = 65 32' 56"'4 54. When a = c and consequently both = 90, sin A = 1, .4 90, and T> ~"0 ~~0 so that B = b, but both are indeterminate as in Art. 52. p 2 174 SPHERICAL TEIGONOMETRY. 55. CASE III. Given one angle and its opposite side, or A and a. We shall have sin A = tan A = sin 1 == sm a sin c tan a sin 6 cos A cos a whence sin c = cosec A sin a (92) ./ sin 6 = cot J[ tan a (93) sin J5 = cos A sec a (94) Check, sin 6 = sin c sin jB In this case, there are always two solutions, all the required parts being determined by their sines, and the ambiguity not being removed by either Art. 49 or Art. 50. This also appears from Fig. 7. If AB and AC be produced to meet in A', ABA' and AC A' are semicircumferences and A = A' ; the triangles ABC and A'BC both contain the given parts A and a, c but c', b' and B' are respectively the supplements of c, b and B. It must not be inferred that in every case all the required parts are less than 90 in one triangle, and greater than 90 in the other; but the proper values for each triangle must be selected by Arts. 49 and 50. EXAMPLES. 1. Given A = 100, a 112 ; solve the triangle. Am. c= 70 18' 10"-2 6 = 154 7'26"-5 = 152 23' l"-3 or c=10941' b= 25 52' B= 27 36' 2. Given A = 80, a = 68; solve the triangle. Ans. c= 70 18' 10"-2 ^ f c = 109 41' b= 2552'33"-5 V or I = 27 36' 58"-7 6 = 154 7' B =152 23' 3. Given B = 150, 6 = 160; solve the triangle. Ans. c = 136 50' 23"'3 >, f c= 43 9' a= 39 4' 50"-7 V or I a =140 55' ^1= 67 9'42"-7 J i 4 = 112 50' 49"-8 33"-5 58"-7 49"*8 26"-5 1"'3 9"'3 17"-3 SOLUTION OF SPHERICAL RIGHT TRIANGLES. 175 56. CASE IV. Given one angle and its adjacent side, or A and b. We shall find the required parts by the equations cos B sin A cos b (95) tan a = tan A sin b (96) cot c = cos A cot b (97) Check, cos B = tan a cot c EXAMPLES. 1. Given ^. = 80 10' 30", 6 = 155 46' 42"-7; solve the triangle. Ans. 5 = 153 58' 24"-5 a= 67 6'52"-6 c=110 46' 20"-0 2. Given 5 = 152 23' l"-3, a= 112 0' 0"; solve the triangle. Ana. A = 100 b = 154 7' 26"-5 c= 70 18' 10"-2 57. CASE V. Given the two sides, a and b. We find the required parts by the equations cos c = cos a cos b (98) cot A = cot a sin b (99) cot 5 = sin a cot 6 (100) Check, cos c = cot .4 cot 5 EXAMPLE. Given a = 116, 6 = 16; solve the triangle. 4n*. fe = 114 55' 20"-4 A= 97 39' 24"-4 = 17 41'39"-9 58. CASE VI. Given the two angles, A and B. The required parts are found by the formulae cos c = cot A cot J5 (101) cos a = cos A cosec J? (1 02) cos b = cosec A cos 5 (103) Check, cos c = cos a cos b 176 SPHERICAL TRIGONOMETRY. EXAMPLE. Given A = 60 47' 24"'3, B = 57 16' 20"-2; solve the triangle. Ans. c = 68 56' 28"'9 a = 54 32' 32"-l b= 5143'36"-1 ADDITIONAL FORMULA FOB THE SOLUTION OF SPHERICAL RIGHT TRIANGLES. 59. As in plane trigonometry, cases occur in which particular solutions of greater accuracy than the ordinary ones are required. (PI. Trig. Art. 112.) 60. From (89) we find 1 Bin A sin c sin a 1 + sin A sin c -\- sin a which by PL Trig. (154) and (109) is reduced to tan' (45 M) = ~ (104) tan i (c + a) which will give a more accurate result than (89), when A is nearly 90. 61. From (91) we find 1 cos b cos a cos c 1 -\- cos 6 cos a -f- cos c or tan* b = tan (c + ) tan (c a) (105) which may be employed instead of (91) when b is small, or nearly 180. . 62. From (90) we find 1 cos B tan c tan a 1 -j" cos B tan c -f- tan a tan 1 B = sm v c *v (106) sin (c -j- a) which may be employed instead of (90) when B is small, or nearly 180. 63. By similar transformations the formulas (101), (102) and (103) become tMl * e = ~c^i-~l*f ) (1 7) tan 1 J a = tan [J (A + 5) 45] tan [45 + J U B )l ( 108 ) tan 1 J 6 = tan [J ( J + B) 45] tan [45 } (A B)] (109) We have also, by (14), i (A -f- J?)^ cos (/ Bin*ie = _.,. 2 sin A sin /j which, when C = 90, becomes "si COS ( s\ ~T~ -D I / 1 1 /\\ Bin 1 # c = l (11") 2 sin A sin B and from (15), in the same manner, 2 sin ^4 sin P QUADRANTAL AND ISOSCELES TRIANGLES. 177 of which (110) may be used when c is small, and (111) when c is nearly 180, instead of (101). 64. The equations (92), (93), and (94), of CASE III. give tan. (45 - * c) = ^"^7} (H2) tan % (A + a) tan' (45 - * b) = s j n ( A ~ $ (113) sin ( A -f a) tan 1 (45 $ B) = tan i (A a) tan J (A + a) (114) The roots of these equations having the double sign, we may take the angles 450 J c , etc. either with the positive or negative sign, whence the two solutions of the problem, as in Art. 55. 65. Some of the solutions may be adapted for computation by the table of natural sines. Thus from (86), (95), and (98), sin a = i [cos (c A) cos (c + ^4)] (115) cos B = * [sin (b + A) sin (b A)] (116) cos c = J [cos (o + 6) -f- cos (o &)] (117) 66. The following relations are occasionally useful : From (83) we have cos a sin A cos A sin 2 A f 118) cos b sin B cos B sin 2 -B From (80) and (83), sin J? sin A cos A __ sin 2 A sin c sin a cos a sin 2 a From (80) and (84), sin.<4 sin a cos a sin 2 a cos 6 sin c cos c sin 2 c (119) (120) 67. Various relations may be deduced from the general formula of the preceding chapter by making C = 90. The following are easily obtained : sin (c a) = cos c tan 6 tan \ B cos a sin b tan $ B sin (c -f o) = cos c tan b cot $ B = cos a sin 6 cot J B cos (c a) = cos 6 -\- sin a sin b tan $ B cos (c -{- a) = cos 6 sin a sin b cot $ B sin (a 6) = 2 sin c sin $ (.4 -f J5) sin $ (A B) sin (a + b) = 2 sin c cos J ( J. -f- B) cos (^4 B) g}n 5 _ cosj_acosj_6 C03 ^> = sin } a sin \b cos c cos c Ian S = cot J a cot J b (121) QUADRANTAL AND ISOSCELES TRIANGLES. 68. The polar triangle of the right triangle is a quadrantol triangle, one side (the side opposite the angle C) being equal to 90. The solution of such triangles is as simple as that of right triangles, the formulae for the purpose being obtained from the preceding, by the process of Art. 8. It is unnecessary to produce them here, as quad- rantal triangles are generally avoided in practice, and when unavoidable are readily solved by means of the polar triangle. An isosceles triangle is easily solved by dividing it into two right triangles by a per- pendicular from the angle included by the equal sides. 23 CHAPTER III. SOLUTION OF SPHERICAL OBLIQUE TRIANGLES. 69. IN the solution of spherical oblique triangles, a required part may sometimes be found by its sine, in which case there will be two values of that part, answering to the conditions, unless the proper value can be determined by other considerations. In certain cases, the true value can be selected by applying one or more of the follow- ing principles, some of which are demonstrated in geometry. We still consider only those triangles each of whose parts is less than 180. I. The greater side is opposite the greater angle, and conversely. II. Each side is less than the sum of the other two. III. Tlie sum of the sides is less than 360. IV. The sum of the angles is greater than 1 80. V. Each angle is greater than the difference between 180 and the sum of the other two angles. For, by IV., A + B + C > 180 whence, A > 180 - (B + C) But if B + C > 180, we have, in the polar triangle, A'B'C', Fig. 8, by II., a' < V + c' or 180 A< 180 5+180 C c , -;1<180 -(J5+C) A>(B + C)-180 VI. A side which differs more from 90 than another side, is in the same quadrant as its opposite angle. For, by (4), we have cos a cos b cos c cos A = sin 6 sin c in which the denominator is always positive. If, then, a differs 178 SOLUTION OF SPHERICAL OBLIQUE TRIANGLES. 179 more from 90 than b or than c } we have, (neglecting the signs for a moment), cos a > cos 6 or > cos c and still more cos a > cos 6 cos c Hence cos a being numerically greater than cos b cos c, the sign of the whole numerator, and therefore the sign of cos A, is the same as that of cos a; that is, A and a are in the same quadrant. VII. An angle which differs more from 90 than another angle, is in the same quadrant as its opposite side. For, by (5), cos A + cos B cos C cos a = sin sin C in which, if A differs more from 90 than B, or than C, cos A deter- mines the sign of the whole fraction, and therefore the sign of cos a. VIII. In every spherical triangle there are at least two sides which are in the same quadrants as their opposite angles respectively. This follows from VI. and VII. IX. The sum of two sides is greater than, equal to, or less than, 180, according as the sum of the two opposite angles is greater than, equal to, or less than, 180. In other words, the half sum of two sides is in the same quadrant as the half sum of the opposite angles. For, by (41), tan \ (a -f 6) cos J (A + B) tan | c cos %(A B} the second member of which is always positive, so that tan (a + 6) and cos (-4 -f- J5) must have the same sign. 70. CASE I. Given two sides and the in- cluded angle, or b, c and A. (Fig. 9.) First Solution; when the third side and one of the remaining angles are required. To find a. The relation between the given parts b, c, A and the required part a is ex- pressed by the first equation of (4), cos a = cos c cos b + sin c sin b cos A (M) by which a. may be found by computing separately the two terms of the second member and adding their values to form the natural cosine of a ; but we should thus be required to use, besides the table of log. sines, also the table of logarithms of numbers, and the table of natural sines and cosines. To adapt it for logarithmic computa- tion by the table of log. sines exclusively, we employ the process of ) > 180 SPHERICAL TRIGONOMETRY. PI. Trig., Arts. 174, 175. Thus, let & be a number and y> an aux- iliary angle such that k sin

being found from (m) we may find a by (w'). But we may eliminate k by dividing the first equation of (m) by the second, and substituting in m' the value of k = - , whence we have, for cosy finding a, tan

) (124) by which the values of a and B, found by (122) and (123), may be verified. 71. If a and C were required, the solution would evidently be similar, only interchanging 6 and c, B and C. By the fundamental formulae we should have cos = cos 6 cos c -\- sin 6 sin c cos A n sin b cos c cos b sin c cos A f (O) cot o - - sin c sin A and denoting the auxiliary angle in this case by , the logarithmic solution would be tan / tan c cos A cos (b y] cos c cos a = -- > - L1 cosy 025) sin/ Check, tan a cos C = tan (6 /) Q 182 SPHERICAL TRIGONOMETRY. EXAMPLES. 1. Given 6 = 120 30' 30", c = 70 20' 20", A =- 50 10' 10"; find a and B. By (122). 6 = 120 30' 30" log tan b 0-2297071 A = 50 10' 10" log cos A -f 9-8065322 y = 132 36' 44"-2* log tan y 0-0362393 c = 70 20' 20"-0 c y = -- 62 16' 24"-2 By (122). By (123). By (124). log cos (e 0) + 9-6676893 log sin (c ) 9'9470304 log tan (c ) 0'2793410 ar co log cos ^ 0-1693898 ar co log sin -f G'1331505 log tan a + 0-4291648 log cos 6 9-7055761 log cot A + 9'9212038 log cos B 9-8501762 log cos a + 9-5426552 log cot B 0'0013847 Check. 0'2793410 a = 69 34' 55"-9 B = 135 5' 28"'8 2. Given b = 120 30' 30", c = 70 20' 20", A = 50 10' 10"; find a and C. Ans. a = 69 34' 55"-9 C=5030' 8"-4 3. Given b = 99 40' 48", c = 100 49' 30", A = 65 33' 10"; find a and B. Ans. a = 64 23' 15"'0 B = 95 38' 4"-0 4. Given b = 99 40' 48", c = 100 49' 30", A = 65 33' 10"; find a and O. Ans. a = 64 23' 15"-0 C= 97 26' 29"-l 5. Given b = 98 2' 20", c = 80 35' 40", A = 10 16' 30" ; find a and C. Ans. a = 20 13' 30"'l C= 30 35' 56"-7 72. If B, C and a were all required, we might find a and C by (125), and then B by Art. 3, which gives sin a : sin b = sin A : sin 1? sin b sin .4 or sin .B = sin a * We may also take ^ = 47 23 X 15"'8, whence c = 117 43' 35"'8. which will give the same values of a and B as found in the text. SOLUTION OF SPHERICAL OBLIQUE TRIANGLES. 183 Of the two values of B less than 180 given by this formula, the proper one may generally be selected by the principles of Art. 69. There are cases, however, in which all the conditions there given are satisfied by both values of B,* and on this account it is preferable, in general, to combine (123) and (125), or to employ the following solution, when the three unknown parts are all to be found. 73. CASE I. Given 6, c and A. Second Solution; when the two remaining angles are required, or when the three unknown parts are all required. We have, by Napier's Analogies, (42) and (43), sin (b + c) : sin \ (b c) = cot \A : tan % (B C) cos | (6 + c) : cos (6 c) = cot A : tan (B + C) whence tan K* - C7) = S !"~1 cot * A (126) sin tan I (B + C) = ^- Cot * A COS which determine J (B (?) and J (B + C) ; then the half difference added to the half sum gives the greater angle, and the half difference subtracted from the half sum gives the less angle. If c > 6, we may write c 6, C B, in the place of 6 c, B C. We may now find a by either of Napier's Analogies, (40), (41), which give f 4 < 128 > '> (129 > * By Art. 69, VI., if 6 differs more from 90 than c, B is in the same quadrant as b, and all ambiguity is removed. If c differs more from 90 than 6. we may find a and B by (122) and (123), and then C by the formula sin c sin A sin C sin a C being taken in the same quadrant as c. f We may also find a from any one of Gauss's Equations (44), which become, in th* present case, cos J a sin J (B -f- C) = cos J A cos \ (6 c) cos \ a cos ( B 4- (7) = sin J A cos (6 -f- c) sin a sin J (5 C) = cos J ,4 sin J (6 c) sin } a cos } (J5 (7) = sin J .4 sin (6 -{- c) 184 SPHERICAL TRIGONOMETRY. EXAMPLES. 1. Given 6 = 120 30' 30", c = 70 20' 20", find J5, C and a. We have (& + c) = 95 25' 25" i tfo c \ _ _ 25 5' 5" 4 = 25 5' 5" By (126). ar co log sin (6 + c) + 0-0019487 log sin % (b c) + 9-6273228 log cot 4 + 0-3296529 log tan |(5 C) + 9-9589244 <7)= 42 17'40"-2 = 50 10' 10"; By (127). ar co log cos (6 + c) 1-0244829 log cos % (b c) + 9-9569757 log cot A + 0-3296529 5 = 135 5' 28"-8 By (128). ar co log sin ^(5 C)+0l 720227 log sin (JS+ C)+9-9994824 log tan (6 c)+9-6703471 log tan a+9-841 8522 a = 3447'28"-0 log tan ^ (5 -f (7) 1-3111115 ( + 46. * The computation of (130) is facilitated by the use of a special table (given in many treatises on navigation), from which, with the argument A, is taken the logarithm of 2 sin* J A = versin A. [PI. Trig. (4) and (139)]. 24 we have, tor finding , cos v cot ^ = tan A cos 6 cos (C #) cot 6 } ( 135 ) cot a = - As in the preceding case, we may either take $ always between and 180, less or greater than 90 according as its tangent is posi- SOLUTION OF SPHERICAL OBLIQUE TRIANGLES. 187 tive or negative; or we may take & numerically less than 90 in all cases, positive or negative, according to the sign of its tangent. (PI. Trig. Art. 174.) Check. The quotient of (n) by (w') is cot a _ cot (C #) cos B sin A sin 6 which, multiplied by sin sin a = sin A sin b gives tan B cos a = cot (C #) (136) by which the values of B and a, found by (134) and (135), may be verified. 78. If B and c were required, the solution would be similar, only interchanging a and c, A and C. By the fundamental formulae, we should have, cos B = cos A cos (7+ sin A sin Ccos b sin A cos (7+ cos A sin C cos 6 cot c = - - r- . -- sin C sin 6 and denoting the auxiliary angle by , the logarithmic solution would be cot = tan C cos 6 sin (A f) cos C B = sin r cos (-4 C) cot 6 - -^ --- (137) n.ns i A r \ r>nr. n cot c COS Check, tan B cos c = cot ( A ) EXAMPLES. 1. Given A = 135 5' 28"-8, C= 50 30' 8"-4, 6 = 69 34' 55"-9; find B and a. By (134). A = 135 5' 28"-8 log tan ^1 9-9986154 6 = 69 34' 55"-9 log cos b -f 9-5426553 tf = 109 10' 31"-0* log cot d 9-5412707 C= 50 30' 8"-4 C& = - 5840'22"-6 * We may also take # = 70 49' 29 /A 0, whence (7 * = 121 19 r 37 /A 4, which will evidently give the same results as those obtained in the text. 188 SPHERICAL TRIGONOMETRY. By (134). By (135). By (136). log sin (C tf) 9-9315664 log cos ((?) + 9'7159386 log cot (C *) 9'7843722 ar co log sin $ + 0*0247897 ar co log cos # 0-4835187 log tan B + 0-0787962 log cos A 9-8501762 log cot 6 -f 9'5708352 log cos a 9*7055757 log COB + 9-8065323 log cot a 9'7702925 Check. 9'7843719 B = 50 10' 10"'0 a = 120 30'29"-9 Ans. B = 50 10' 10"-0 2. Givenjt = 1355'28"-8, (7=50 30' 8"-4, b = 69 34' 55"-9; find P> and c. Ana. B = 50 10' 10"-0 c= 7020'20"-0 3. Given A 65 33' 10", C= 95 38' 4", b = 100 49' 30" ; find B and a. Am. B = 97 26' 29" o= 64 23' 15" 4. Given 4 = 97 26' 29" (7= 95 38' 4", 6 = 64 23' 15" ; find B and a. 4n*. B = 65 33' 10" a = 100 49' 30" 79. If a, c and J2 were all required, we might find B and c by (137), and then a by Art. 3, which gives, sin B : sin A = sin b : sin a sin ^4 sin 6 7 . sin a -- (138) sm .B Of the two values of a given by this equation, the proper one is to be selected, if possible, by the principles of Art. 69.* But as cases occur in which all the conditions there given are satisfied by both values of a, it is preferable, in general, to combine (135) and (137), or to employ the following solution when the three unknown parts are all to be found. *By Art. 69, VII., when A differs more from 90 than C, a must be taken in the same quadrant with A, and all ambiguity is removed. If, then, by A we always denote that angle which differs more from 90 than the other given angle, we may always solve this case by means of (137) and (138), without meeting with any difficulty in determining the quadrant in which a is to be taken. SOLUTION OF SPHERICAL OBLIQUE TKIANGLES. 189 80. CASE II. Given A, C and 6. Second Solution; when the two remaining sides, or when the three unknown parts are all required. We have, by Napier's Analogies, (40) and (41), sin (A + (7) : sin \ (A C) = tan -| b : tan |- (a c) cos (A + C) : cos (A C) = tan 1 6 : tan $ (a + c) whence C) j , (139) ** which determine -^ (a c) and ^ (a + c) ; then the half difference added to the half sum gives the greater side, and the half difference subtracted from the half sum gives the less side. If C>A, we may write C A, c a in the place of A (7, a c. We may now find B by either of Napier's Analogies, (42) and (43), which give* sn (a -f c) . , . ^ x ,., A . J - ( tan |( A 0) (140) sin (a c) tan %(A+C) (141) cos (a c) EXAMPLES. 1. Given J. 135 5' 28"-6, C= 50 30' 8"-6, 6 = 69 34' 56' r '2 ; find a, c and J?. We have A+C = 92 47' 48"'6 ^ 6 = 34 47' 28"-l Then, by (139), ar co log sin (A+ C)4-0'0005176 ar cologcos|(^4+C)-l'3116286 log sin %(A C)+ 9-8279768 log cos $ (A C)+9'8690535 log tan | b + 9-841 8527 log tan ^6 + 9-8418527 log tan $ (a c) + 9-6703471 log tan |(a + c) 1-0225348 %(a c )= 25 5' 5"-0 ^ (a + c) = 95 25' 25"'0 a = 1 20 30' 30"-0 c = 70 20' 20"-0 * We may also find B by any one of Gauss's Equations, (44), interchanging B and C, b and c. 190 SPHERICAL TRIGONOMETRY. By (140). By (141). ar co log sin J (o c) + 0-3726772 ar co log cos J (a c) + 0-0430243 log sin J (a + c) + 9'9980523 log cos J (a + c) 8-9755171 log tan I (A C)+ 9-9589234 log tan $(A+C) 1-3111110 log cot B + 0-3296529 * log cot | + 0-3296524 5 = 25 5' 5"-0 4n*. a = 120 30' 30" c = 70 20' 20" B = 50 10' 10" 2. Given 4 == 95 38' 4", C= 97 26' 29", 6 64 23' 15"; find a, c and B. Ans. a= 99 40' 48" c = 100 49' 30" B = 65 33' 10" 81. CASE II. Given A, C and b. Third Solution. When the third angle B is alone required, the computation by (134) is in most cases as convenient as any other, but there are other methods (cor- responding to those given in Art. 75 for finding a) which may occa- sionally be serviceable. By (14) and (15) we have cos B cos (A -+- C) 2 sin A sin C sin 2 6 (142) cos B = - cos (A C) + 2 sin A sin Ccos 2 ^6 (1 43) the computation of which is similar to that of (130) and (131). EXAMPLE. Given A = 95 38' 4", C 97 26' 29", 6 = 64 23' 15" ; find . By (142). 6= 32ll'37"-5 log sin 2 ^6 = 2 log sin 16 9-4531022 A -f C= 193 4' 33" log sin A 9-9978967 log sin C 9-9963268 log 2 0-3010300 - 2 sin A sin C sin 2 6 = - 0-56021 62 log 9-7483557 - nat cos (A + C) == + Q'9740715 nat cos B = + 0.4138553 B 65 33' 9"'9 *For the reasons given in Art. 74, (141) is, in this example, not so accurate as (140). SOLUTION OF SPHERICAL OBLIQUE TKIAISGLES. 191 82. In Art. 14, several formulae are given, by which B way be computed. By (21) and (22) we have sin 2 \ B = cos 2 J (A C) sin A sin C cos 2 6 cos 1 B =- sin 2 (A + C) sin A sin C sin 2 J 6 which may be adapted for logarithms, thus : sin 2 <]> = sin A sin C cos 2 6 "j sin* = cos 2 J ( J. (7) sin 2 I (144) sin 1 # = sin A sin (7 sin 2 J 6 cos 2 IB = sin 2 \ (A + C) sin 1 = sin [J (4 + C) + *] sin (145) of which (144) is to be preferred when + (7) - 0] < 45, and (145) when J J? > 45. 83. Case II. might have been reduced to Case I. by means of the polar triangle, Art. 8 ; for there will be known in the polar triangle, two sides and an angle oppo- site one of them, being the supplements of the given angles and side of the pro- posed triangle. The polar triangle being solved, therefore, by Case I., and its two remaining angles and third side found, the supplements of these parts would be the two sides and third angle required in the proposed triangle. It is easily seen, also, that all the formulae above given for this case might have been obtained by these considerations. 84. CASE III. Given two sides and an FlG - 9 - o angle opposite one of them; or a, 6, and A. Fig. 9. First Solution, in which each required part is deduced directly from fundamental for- mulae independently of the other two parts. To find c. We have, by (4),* cos c cos 6 -j- sin c sin 6 cos A = cos a to solve which, let k sin

(p 1 , k cos (p 1 = cos a ) c = -f and i? may have the same values throughout. (M) (m) 192 SPHERICAL TRIGONOMETRY. The auxiliary

f in (m'} either with the positive or the negative sign, so that c = y we have, then, for finding C } cos v SOLUTION OF SPHERICAL OBLIQUE TRIANGLES. 193 cot & = tan A cos 6 ^ cos ft' cos & tan b cot a I (147) c=&#> To find B. We have several methods: 1st, directly by (3), . D sin A sin 6 , . sin B - (148 ) sin a which gives two values of B, supplements of each other, correspond- ing respectively to the two values of c and C. We shall presently see how to determine which are the corresponding values of c, C and B. 2d. In (123),

180, and

sin a, since, by (148), we shall then have sin B > 1. EXAMPLES. 1. Given a = 40 16', 6 = 47 44', A = 52 30'; find B. By (148). a= 40 16' ar co log sin a 0*1895350 b = 47 44' log sin 6 9-8692449 A = 52 30' log sin A 9-8994667 B= 65 16' 35" log sin B 9-9582466 or 7? ==11 4 43' 25" 2. With the same data, find c and B. By (146). a= 40 16' logcos a+9-8825499 b= 47 44' log tan 6+0-0414996 ar co log cos 6+0-1722547 A = 52 30' log cos .4+9-7844471 tp= 33 48' 51 "-4 logtan^p+ 9-8259467 log cos ^+9-9195204 \ B} (154) sin i (A -f- 5) , . / , N tan 4 c = - f tan i (a 6) sin ^ (A B) .^ cot \ C , 1 / A tan i ^1 (155) sm ^ ( a 6 ) in which we employ successively the two values of B, and obtain two solutions, except when for one of these values the second mem- bers become negative, for ^ c and % C being less than 90, their tangents must be positive. SOLUTION OP SPHERICAL OBLIQUE TRIANGLES. 197 We leave it to the student to apply these formulae to the preceding examples. 88. To determine by inspection of the data a, 6 and A, whether there are two solutions, or but one. 1st. It has already been seen, Art. 85, that when b differs more from 90 than a, B must be in the same quadrant as 6, and there can be but one solution. It remains to show, 2d. That when a differs more from 90 than b, there will necessarily be two solu- tions. We have, by the first of (4), cos a cos 6 cos c sin c = : sin 6 cos A Two solutions exist so long as both values of c are positive, and less than 180, that is, so long as sin c is positive. Now when a differs more from 90 than 6, we have, (neg- lecting the signs for a moment), cos a > cos b > cos b cos c herefore the numerator of the above value of sin c has the sign of cos o. But by Art. 19, VI., o and A are in the same quadrant, and cos a and cos A have the same sign ; consequently also, the numerator and denominator have the same sign, and the value of the fraction, or of sin c, is positive, as was to be proved.* Hence, there is but one solution when the side opposite the given angle differs less from 90 than the other given side, and two solutions when the side opposite the given angle differs more from 90 than the other given side. 89. CASE IV. Given two angles and a side opposite one of them, or A, B and 6. (Fig. 9). First Solution, in which each required part is deduced directly from the funda- mental formulae. To find c. We have, by (10), sin c cot 6 cos c cos A = sin A cot B or multiplying by sin 6, sin c cos b cos c sin 6 cos A = sin A cot B sin b (M) to solve which, we take k sin

= cos b I * The same proposition may be otherwise proved thus. By the equations (m) and (m f ) Art. 84, we have k sin from the third of which we see that k has the sign of cos A ; if then a differs more from 90 than 6, that is, if cos a and cos A have the same sign, cos ' is positive, and $' < 90. Also since, (neglecting signs), cos a > cos 6, we have cos $ f > cos , or 1>' differs more from 90 than . Hence 0' < and 0' < 180 , or ^ f > and -f- 0' < 180, or both values of c are between and 180. R2 (m') 198 SPHERICAL TRIGONOMETRY. then, putting c (p =

' = sin tp tan ^4 cot B C = (p -\- (D 1 Here

, give two values of c. When the second member of the formula sin

' COS 6 cos a ~ 3d. By (150), /-i ~r>\ (Io9) cos

' is negative, cos a and cos A must have different signs. A like result follows from the first of (157) and (161) with reference to #'. Hence, that value of a which is in the same quadrant with A belongs to the triangle in which

90, &' > 90. This precept enables us to employ (158) with- out ambiguity. In the use of (159), (160), (161), and (162), it is only necessary to observe the algebraic signs of the several terms. ChecJcs. Of the various formulae above given for finding a, one or more may be employed for the purpose of verification. When c and C have been found, however, the most simple check is 200 SPHERICAL TRIGONOMETRY. sin C sin B sin c sin 6 (163) which might have been employed for finding C after c was found, or reciprocally, but for the ambiguity attaching to the sines. 90. According to Art. 69, VII., if A differs more from 90 than B, a must be in the same quadrant with A. But since the two values of a are supplements of each other, only one of them can satisfy this condition, and there will then be but one solution. In such case c and C will each be found to have but one admissible value. 91. The problem will be impossible when B differs more from 90 than A, and yet is not in the same quadrant with b. In such case we should find both values of c (and both values of (?) to be greater than 180, or both negative. The problem will also be impossible when sin 6 sin A > sin B, since by (158) we shall then have sin a > 1. EXAMPLES. 1. Given A = 132 16', B = 139 44', b = 127 30'; find a. By (158). B = 139 44' 0" ar co log sin B 0-1895350 A = 132 16' 0" log sin A 9-8692449 b = 127 30' 0" log sin b 9*8994667 a= 65 16' 35"'l log sin a 9-9582466 or a =1U 43' 24"-9 2. With the same data, find C and a. By (157). B= 139 44' 0" log cos 9-8825499 A= 132 16' O'Mogtanvl 0-0414996 ar co log cos ^0-1722547 b = 127 30' 0" log cos 69-7844471 56 11' 8"-6 log cot #+ 9-8259467 log sin #+9'91 95204 70 29' 31--OJ sin*'+9^74327o 10930'29"-OJ 126 40' 39"-6 165 41' 37"-6 SOLUTION OF SPHERICAL OBLIQUE TRIANGLES. 201 By (161). Check. (162). & = 56 11' 8"-6 ar co log cos $+0-2545328 log cot J3 0-0720848 &'= 7029'31"-0) *V= 109 30' 29"-0 j logcos^'9-5236676 log cot ^'9j5 493427 b =127 30' 0" log cot b 9-8849805 + 9-6214275 log cot a +9-6631809 log cos a+ 9-6214275 = 6516'34"-9 l Ans. C= 126 40' 39"-6 J j C= 165 41' 37"-6 a = 114 43' 25"'l I 1 a = 65 16' 34"'9 3. Given A = 110, B = 60, 6 = 50 ; find c and a. By (156). = 60 log cot +9-9614394 A= 110 log cos^l 9-5340517 logtan^l 0-4389341 b = 50 log tan b + 0-0761865 y= 15749'26"-4 log tan

0-0333755 logcos+9-6989700 cos A > cos A cos C therefore the -numerator of the value of sin C has the sign of cos 2?. But by Art. 69, VII., B and 6 are in the same quadrant, consequently the numerator and denominator have the same sign, and the value of the fraction, or of sin C is always positive, as was to be proved.* Hence, there rs but one solution when the angle opposite the given side differs less from 90 than the other given angle ; and two solutions when the angle opposite the given side differs more from 90 than the other given angle. 94. CASE IV. might have been reduced to Case III. by means of the polar triangle of Art. 8. For there will be known in the polar triangle two sides and an angle opposite one of them, being the supplements of the given angles and side of the pro- posed triangle. The polar triangfe being solved, therefore, by Case III., and its two remaining angles and third side found, the supplements of these parts will be the required sides and third angle of the proposed triangle. * It may be shown that both values of C will be admissible, by a process of reason- ing similar to that employed in the note on page 197, applied to the equations of Art. 89. SOLUTION OF SPHERICAL OBLIQUE TRIANGLES. 95. CASE V. Given the three sides, or a, b FlG - 9 - and c. (Fig. 9.) We have three methods for computing the half angles : 1st. By the sines, from (31), remembering that s = ^ (a + 6 -f- c) . , / / sin (s 6) sin (s c) \ sm % A = \ I [ * \ sin 6 sin c / sin B = J ( S1 " ' T C S } * \ sin c sin a / 203 sin A \ sin i n I ( sm ( s a ) sm ( s b) C _ / / ** ' sin a sin 6 / 2d. By the cosines, from (33), cos - cos cos // sin s sin (s a) \ sin 6 sin c / / / * " / J\\ \ ' \ sin c sin a / iC = J/2El_!SJ_! ^ \ sin a sin 6 3d. By the tangents, from (34), | / sin (g 6) sin (s c) Al \ sin s sin (s a) / / sin (s c) sin (s a)\ tan 4 jS = A / 1 . ' \ sm s sin (s 6) / \ i ,' / \ ' ( j\\ \ tan ^ C = \ i - / , * r1 I V \ sm s sin (s c) / (164) (165) (166) When only one of the angles is required, the simplest method will be by (16,5), but if the required uugle is less than 90, it will be found more accurately by (164), for then ^ A < 45, and the sine varies more rapidly than the cosine. And, for a similar reason, if the angle is greater than 90, we should prefer (165). By (166) we always have an accurate result, although the formula is not quite so simple. When the three angles are required, (166) will require the least labor, since sin a, sin 6, and sin c, are not then required. 204 SPHERICAL TRIGONOMETRY. No ambiguity can arise in these solutions, since the half angles must be less than 90 ; they require therefore no attention to the algebraic signs. EXAMPLES. 1. Given a = 100, 6 = 50, c = 60 ; find A. a =100 b= 50 logcosec 0-1157460 c= 60 logcosec 0-0624694 2s = 210 s=105 log sin 9.9849438 8 a = 5 log sin 8*9402960 2)9-1034552 %A= 69 7' 52"-7 log cos 9-5517276 A = 138 15' 45"-4 2. With the same data, find all the angles. By (166). a = 105 1. cosec 0-0150562 1. cosec 0-0150562 1. cosec 0-0150562 s a = 5 1. cosec 1-0597040 1. sin 8-9402960 1. sin 8-9402960 * b = 55 1. sin 9-9133645 1. cosec 0-0866355 1. sin 9-9133645 8 _ c = 45 1. sin 9-8494850 1. sin 9-8494850 1. cosec Q-l 505150 2)0-^376097 2)8-8914727 2)9-01 923T7 1. tan 0-4188049 1. tan 9-4457364 1. tan 9-5096159 Ans. ^ = 138 15' 45"-4 B = 31 11' 14"-0 C = 35 49' 58"-2 3. Given a = 10, 6 = 7, c = 4 ; find the angles. Ans. ^4=128 44' 45"- 1 B= 33 11' 12"-0 C= 18 15'31"-1 96. The method by (166) may be put under the following convenient form. Let p I /sin (s a) sin (a b) sin (g c)\ \ V sin * / then (167) UnM = - ^ rt tan* = - , tanJC = sin (s a) sin (s 6) sin [ c) which are similar to the formulae of PI. Trig. Art. 146, and are computed in the same manner. SOLUTION OF SPHERICAL OBLIQUE TRIANGLES. 205 97. CASE V. Given a, b and c. Second Solution. If the whole angle is required directly,* we have A _ cos a cos b cos c sin 6 sin c which may be adapted for logarithms by an auxiliary thus : cos = cos b cos c Or thus, sin 6 sin c . . cos b cos c COt 9 = ; sin a (169) FIG. 9. sin b sin c sin 98. CASE VI. Given the three angles, or A, B and C. (Fig. 9). We have three methods of finding the half sides : 1st. By the sines, (36). 2d. By the cosines, (38). 3d. By the tangents, (39). The computations are conducted precisely in the same form as those of the preceding case. EXAMPLE. Given A = 120, B = 130, C= 80 ; find c. Ans. c = 41 44' 14"-6 99. The formulae (39) may be arranged for convenient use in the same manner as the corresponding formulae of the preceding case, Art. 96. 100. CASE VI. Given A, B and C. Second Solution. We have, by (5), cos A -4- cos B cos C cos a = 7- 1 sin B sin C which may be adapted for logarithms by an auxiliary, thus: cos cos B cos (170) or, tan sin B sin C cos B cos C sin A cos (A sin B sin O cos (171) *See NOTE at the end of this chapter, p. 211, for the method of computing many of the general formulie of spherical trigonometry directly, without the aid of auxiliary angles. S 206 SPHERICAL TRIGONOMETRY. 101. All the cases of oblique spherical triangles may be solved by dividing the triangle into two right triangles by a perpendicular from one of the vertices to the opposite side, and solving these partial triangles by the methods of the preceding chapter. Bowditch has given two rules, based upon Napier's Rules, (Art. 46), by which the application of this method is facilitated. FlG - 10 - 102. Bowditch' s Rules for Oblique Triangles. " If in a spherical triangle, (Fig. 10), two right triangles are formed by a perpendicular let fall from one of its ver- tices upon the opposite side ; and if, in the two right triangles, the middle parts are so taken that the perpen- dicular is an adjacent part in both of them ; then The sines of the middle parts in the two triangles are proportional to the tangents of the adjacent parts. But if the perpendicular is an opposite part in both the triangles, then The sines of the middle parts are proportional to the cosines of the opposite parts. To prove which rules, let M denote the middle part in one of the right triangles, A an adjacent part, and an opposite part. Also, let m denote the middle part in the other triangle, a an adjacent part, and o an opposite part; and let p denote the perpendicular. First. If the perpendicular is an adjacent part in both triangles, we have, by Napier's Rules, (Art 46,) sin M = tan A tan p sin m = tan o tan p whence sin M tan A tan p _ tan A sin m tan o tan p tan a or sin M: sin m = tan A : tan a Secondly. If the perpendicular is an opposite part in both triangles, we have, by Napier's Rules sin M cos O cos p sin m = cos o cos p whence sin M cos cos p cos O sin m cos o cos p cos o or sin M : sin m = cos : cos o" * We proceed to solve the six cases of spherical triangles with the aid of a perpen- dicular. It will be seen, however, that Bowditch's Rules are applicable but in the first four cases. *Peirce's Spherical Trigonometry, Art. 44. SOLUTION OF SPHERICAL OBLIQUE TRIANGLES. 207 103. CASE I. Given b, c and A. Let the perpendicular C P, Fig. 10, be drawn frcm C, (that is, in such a manner as to put two given parts in one of the right tri- angles). Then the right triangle A C P gives, by Napier's Rules, if we put A P = 0, tan (j> = tan 6 cos A (172) then taking co. 6 and co. a as middle parts in the two triangles, A P= and B P = c * are the opposite parts, whence, by Bowditch's Rules, cos : cos (c ) = cos 6 : cos a whence cos? Again, taking A P and P B as middle parts, co. A and co. B are adjacent parts, whence, by Bowditch's Rules, sin : sin (c ) ( 175) which agrees with (124). By drawing the perpendicular from B, we may in the same manner obtain the formulae (125). The angle C may be found by the proportion sin a : sin c = sin A : sin C or if C has been found by means of a perpendicular from B, B may be found by a similar proportion, as in Art. 72 ; and the quadrant in which the angle is to be taken must be determined by the principles of Art. 69. 104. CASE II. Given A, G and b. Let the perpendicular be drawn as before, Fig. 10, and let then, by Napier's Rules, cot & = tan A cos 6 (176) and by Bowditch's Rules, taking co. A and co. B as middle parts, and therefore co. A C P and co. B C P as opposite parts, sin 1? : sin (C #) = cos A : cos B whence sin (C #) cos A /i77\ = - *If A P should exceed A B, (that is, if the perpendicular should fall without the triangle), BP would be equal to AP A B = c, and the solution could be modified accordingly. But the true results will always be obtained by regarding B P as negative ; that is, by still taking B P=c and attending to the signs of all the terms as already exemplified, p. 182. flf A C P> A C B, B CP= CA C P will become negative, but the true results are still found by attending to the signs, as already shown, p. 187. 208 SPHEKICAL TRIGONOMETRY. Again, taking co. A CP and co. -B CP as middle parts, and therefore co. b and co. o as adjacent parts, Bowditch's Rules give cos & : cos (C #) = cot b : cot a whence / X^ O\ 1 (178) and (176), (177), (178), agree entirely with (134) and (135). The triangle B C P gives tan ^ cos a = cot (C #) (179) which agrees with (136). By drawing the perpendicular from A, we may in the same manner obtain the formula (137). The side c may be found from the proportion sin A : sin C= sin a : sin c and Art. 69; or c being found by means of a perpendicular from A, we may find o by a similar proportion. 105. CASE III. Given a, b and A. Let the per- pendicular be drawn from C, Fig. 10, as in the preced- ing cases, and let A P = , P = <$>' ; then, by Na- pier's Kules, tan = tan b cos A (180) and, by Bowditch's Rules, cos b : cos a = cos ? : cos ' FIG. 10. whence and then cos r = cos 6 (181) In Art. 84, we have found, from analytical considerations, that this case admits of two solutions, and that the general expression for c is c = = h^ / (182) In fact, let us attempt to construct the triangle with the data a, b and A. Having constructed A equal to the given angle, and b equal to the adjacent side, Fig. 11, let !! a small circle be described about C as a pole, with a (circular) radius = a; this circle in- tersects the great circle A B in two points, B and B', and both triangles, A CB and A CB' contain the same data a, 6 and A. If the perpendicular CP is drawn, we have BP B'P, so that in one of the triangles, the side c = AB = A P + P B = + 0', and in the other, c = A B' = A P B' P = f . If both points of intersection, B and B', fall on the same side of A, and within 180 of A, both solutions will be admissible. To find C, let A CP= #, and BCP= #', then by Napier's Rules, cot & = tan A cos b (183) SOLUTION OF SPHERICAL OBLIQUE TRIANGLES. 209 and by Bowditch's Kules, cot 6 : cot a = cos i? : cos $' whence cos $' cos & tan b cot a (184) and since in Fig. 11, C=ACB = ACP+BCP=d + #', or C=ACB' = ACP B'CP=$ tf', we have C=$$' (185) and the formula: (180), (181), (182), (183), (184), (185), agree entirely with (146) and (147). After c was found, we might have found C from the proportion sin a : sin c = sin A : sin C (186) and B is found from the proportion sin a : sin 6 = sin A : sin B (187) The two values of B determined by (187), are both admissible when c has two values as above. It is also evident, from Fig. 11, that the two values of B are supplemental. To determine the corresponding values of c and B, we observe that, by Art. 49, the perpendicular OP is in the same quadrant with A and -with OBP and CB'P, and therefore CB'A is in a different quadrant from A. Hence, that value of B which is in the same quadrant as A corresponds to the value of c = + ^, and that value of B which is in a different quadrant from A corresponds to the value of c = ' ; which agrees with what is shown in Art. 84. In computing (186), the two values of c must be employed successively, and the formula computed twice. At each computation we shall have two values of C found from the sine, one of which must be selected by Art. 69. But as the application of the principles of Art. 69 is tedious and embarrassing, it is better to find C by (184) and (185). The formulae (149), (150), (151), (152), for finding B, may easily be deduced by Napier's and Bowditch's Rules. 106. CASE IV. Given A, B and b. Let the perpendicular be drawn as before, Fig. 10, and let AP = , BP=' whence sin tan A cot B which agree with (156). But <]/ having two supplemental values determined by the sine, c has two values, as already explained in Art. To show the same geometrically, let BP, Fig. 12, be the acute value of $' ', and about C as a pole, let a small circle be described passing through B, and intersecting the great circle AB again in B". Let B ff C be drawn, and produced to meet AB again in B f , forming the lime B"B'. Then we have B' = B" = CBA 27 s2 210 SPHERICAL TRIGONOMETRY. so that in both triangles, ACB and ACS', the value of the angle opposite the side b is the same, that is, both triangles contain the same data, A, JB and 6. Now 180 = B"B'= B' ; P + B'P= BP + B'P, so that .BPand B'P are supplements of each other. In the triangle A CB we have and in the triangle A CB' we have = AP+B'P and hence the two values of c are found by giving $' its acute and obtuse values suc- cessively, as already shown analytically. By Art. 49, CP must be in the same quadrant with A; hence, if B is in the same quadrant with A, P falls between A and B, as in the figure, and for the same reason, between A and B'. But if A and B were in different quadrants, both points, B and B', might fall between A and P. The two values of c would then be found by the formula c = 0-/ 0' taking, successively, its acute and obtuse values. In that case, tan A and cot B would have opposite signs in (189), sin ' would be negative, which would make 0' negative, so that the true results will be obtained, without reference to a diagram, by attending to the signs of the several terms, as already fully exemplified, p. 201. To find C; let ACP= &, BCP=$', then we have, as' before, cot $ = tan A cos 6 (190) and by Bowditch's Rules cos A : cos B = sin & : sin &' whence . . sin # cos B sm $' = cos A (191) which agree with (157). It is evident from Fig. 12, that BCP and B'CP, are sup- plemental, and that the remarks above made with reference to ' apply also to &'. After c was found, we might have found C by the proportion sin 6 : sin c = sin B : sin C (192) and a is found by the proportion sin B : sin A = sin 6 : sin a (193) The two values of a found by (193) are both admissible when c has two values. From Art. 50, it follows that when BP is acute, a must be in the same quadrant with CP, that is, (Art. 49), in the same quadrant with A ; and when BP is obtuse, o must be in a different quadrant from A. That is, that value of a vihich is in thi' .-unm: quadrant with A, belongs to the triangle in which 0' < 90, and that value of a which is in a different quadrant from A, belongs to the triangle in which 0' > 90 ; which agrees with Art. 89. The formulae (159), (160), (161), and (162), for finding a may easily be deduced by Napier's and Bowditch's Rules. SOLUTION OF SPHERICAL OBLIQUE TRIANGLES. 211 107. CASE V. Given a, b and c. The perpendicular cannot be drawn, in this case, so that two of the given parts shall be in one triangle ; nevertheless the case can be solved by means of a perpendicular. Let the perp. be drawn from any angle, as C, Fig. 13, and as before, put AP=, EP = '; then by Bowditch's Rules, cos : cos $' = cos 6 : cos o whence cos $' cos ft _ cos a cos 6 cos 0' -f- cos ft cos a + cos 6 or, by PI. Trig. (110), tan i ( + f) tan $ ( ') = tan $ (b -f a) tan J (6 a) whence, since -f- $' =~- c, tan J (ft ft') = tan $ (b + a) tan (6 a) cot J c 1 / 1 Q ^\ = / ' which determine (ft ft') and (ft + ft') whence ft and ft'. The angles A and are then determined by Napier's Rules. 108. CASE VI. Given A, B and C. In Fig. 13, let ACP=&, ECP=&'; then, by Bowditch's Rules, sin $ : sin #' = cos J. : cos B whence sin 1? sin #' _ cos A cos B sin # -)- sin i9 / cos A -f- cos 5 or, by PI. Trig. (109) and (110), whence, since # + i? r = (7, tan i (# *0 = tan (B + A) tan J (B A) tan | C7 }(* + ^) = iC7 which determine $ (# ^ r ) and i (i^ + ^ x ) an< l therefore i^ and <9 / . The sides a and 6 are then found by Napier's Rules. NOTE REFERRED TO ON PAGE 205. Computation of Spherical Formulae by the Gaussian Table. The Gaussian Table is a table, first suggested by Gauss, for readily computing the logarithm of the sum or difference of two quantities, when the logarithms of these quantities are given. If p and q are the two numbers whose logarithms are given, p being the greater number, (or log p the greater logarithm), we have, in the first place If, then, we put x = %-, we have 9 log x = log p log q log ( p + q) = log q + log (1 -f *) or log ( p + q) = log p + log (l + \ 212 SPHERICAL TRIGONOMETRY. Downes's Table XXII., with the argument log x, the difference of the given logar- ithms, gives log (1 + x), which being added to log q, the less logarithm, gives the required log. sum, or log(p-\-q). Table XXIII., with the argument log x, gives logll + | which, being added to log p, the greater logarithm, gives the required log. sum. Either table may, in general, be employed, but one or the other may be found more convenient in a particular application, and therefore both are given. Again, we have so that, putting, as before, x = 1L, we have 9 log x log p log q log ( P q) = log p + log (l j Downes's Table XXIV., with the argument, log x, gives log II _| which, being added to the greater logarithm, gives the required log. difference, or log ( p q), With these tables, then, we may readily compute any of the preceding formulae which contain two terms in the second member, without the aid of auxiliary angles. EXAMPLES. 1. Given 6 = 120 3180, so long as B, c are each < 180. In the triangle A'B'C' we have ^4/<180, while , C">180. FIG. 15. FIG. 16. B' Again, Fig. 15. C,b and If we next suppose two of the sides to exceed 180, as a and b, Fig. 10, these sides intersecting in two points whose distance is 180, the figure ceases to present the 214 SOLUTION OF THE GENERAL SPHERICAL TRIANGLE. 215 triangle as an enclosed surface, but it will presently appear that such triangles are solved by the same general methods that apply in other cases. To form a just con- ception of the triangle in this case, we may conceive Fig. 16 to be obtained from Fig. 15 by carrying the point A along the arc OA produced until it crosses the side a; the points A and B may then be joined either by an arc less than 180, as in Fig. 16, or by its supplement to 360, as in Fig. 17, in which last case every side exceeds 180. In these figures, to avoid confusion, the point A is not placed in its true position according to perspective. In each figure we have two triangles, whose sides are common, and whose angles are sup- plements to 360. It will be easy to trace the two triangles signified by A BC and A'B'C', by remarking that the letters in each case are all on the same side of the perimeter of the triangle. We may go farther, and suppose the arc joining A and B to be a circumference -f- the arc AB, or any number of circumferences + AB; and similarly the angles may be supposed to be altogether unlimited ; but since the relative positions of any three points of the sphere must be fully determined by arcs and angles less than 360, nothing is gained by passing beyond this limit. 111. All the formuke of Chapter I. are applicable to the general spherical triangle. This proposition might be considered as established by the principle of PI. Trig. Art. 49, but it is also very easily established by a continuation of the process of Spher. Trig. Art. 6, where the fundamental equation was shown to apply to all triangles whose parts are less than 180. It was proved in Art. 29, that all the equations of Chap. I. may be deduced from the fundamental one, cos o = cos b cos c -f- sin 6 sin c cos A (M) We have then only to prove the generality of this single equation. 1st. Let all the sides be < 180, but A / > 180, Fig. 14. The formula being true for the triangle ABC, we have cos o = cos 6 cos c + si Q b sin c cos (360 A') or in the triangle A'B'C', by PL Trig. (76), cos a' = cos V cos c' -f- sin b f sin c' cos A' 2d. Let o>180, Fig. 15, and produce a to complete the great circle. The triangles ABO and A'B'C' are respectively the difference and sum of a hemisphere and the triangle A'ik, all of whose parts are < 180. In the triangle A'ik we have, in terms of the parts of ABC, cos (360 o) = cos 6 cos c + sin 6 sin c cos (360 A] and in terms of the parts of A'B'C', cos (360 a') = cos V cos c' + sin b' sin c,' cos A' both of which reduce to the form (M). But it is here necessary to show that the formula may also be applied to each of the other angles: thus the triangle A'ik gives cos b = cos (360 a ) cos c + sin (360 a ) sin c cos ( 180 B) cos b' = cos (360 a') cos c / -f sin (360 a') sin c / cos ( B' 180) both of which reduce to the form (M). 216 SPHERICAL TRIGONOMETRY. 3d. Let a > 180, b > 180, Fig. 18 ; these arcs intersect at i, and the triangle A'B'i gives FIG. 18. cos (a 180) = cos (6 180) cos c + sin (6 180) sin c cos (360 A) cos (a' 180) = cos (b f 180) cos c' + sin (V 180) sin c 7 cos A' which reduce to the form (M) ; and in the same way the formula applies to the angle B. We have also cos e = cos (a 180) cos (b 180) + sin (a 180) sin (6 180) cos i and since cos t = cos C= cos (360 C'} = cos C", this also reduces to the form (M) for both A B C and A' B' C'. 4th. Let a > 180, b > 180, e > 180, Fig. 19 ; the side c being produced to com- plete the circle, the triangle i k I gives FIG. 19. C> cos (a 180) = cos (6 180) cos (360 c) + sin (b 180) sin (360 c) cos I and since cos I = cos (180 A) = cos A = cos (A' 180) = cos A', this reduces to the form (M) for both ABC and A' B f G f ; and in the same way the for- mula applies to the angle B. We have also cos (360 c) = cos (o 180) cos (b 180) + sin (a 180) sin (b 180) cos i and since cos i = cos (7= cos C f , this reduces to the form (M) for both ABC and A' B' C'. The cases in which the angles or sides ezceed 360 are included in the preceding, in consequence of PL Trig. Art. 45. 112. The preceding demonstration, though tedious, has the advantage of giving a definite conception of the figures which our formulae represent. But perhaps the most satisfactory (as it is the most elegant) method, is to rest the demonstration of our fundamental equations themselves upon the principles of analytical geometry, and, for the sake of those who are acquainted with that subject, we add the following investigation : Any point of the sphere may be referred by rectangular co-ordinates to three planes passing through the centre of the sphere at right angles to each other. Let be the centre of the sphere, Fig. 20, and A B C & spherical triangle upon its surface. Let one of the co-ordinate planes, as X Y, coincide with the great circle A B, and let the axis of X pass through B. HOP be drawn perpendicular SOLUTION OF THE GENERAL SPHERICAL TRIANGLE. 217 to the plaue -X" Y, and CP' and PP' to the axis OX, the co-ordinates of the point G are z P', y = PP', z=CP FIG. 20. FIG. 21. the values of which (0 C being taken = 1) are x = cos a y sin a cos B z = sin a sin B s Ofi' If now the axis of -X" be made to pass through A, Fig. 21, without changing the position of the plane X Y we shall have for x f , y', z', the co-ordinates of C referred to the new axes, x f = cos b y f = sin 6 cos A z' = sin 6 sin A The axis of z being unchanged, the relations between x', y f , and z, y, are expressed simply by the formulae for the transformation of co-ordinates in a plane ; the in- clination of the new axes to the first is here expressed by c, and the formulae of trans- formation are therefore x = x' cos c y sin c y x f sin c + y f cos c z=z' substituting the values of the co-ordinates, we have at once the three following funda- mental equations : cos a = cos c cos 6 + sin c sin 6 cos A sin a cos B = sin c cos b cos c sin 5 cos A sin a sin B = sin 6 sin A (N) which are identical with (4), (6), and (3). 113. Having established the complete generality of our fundamental equations, we may now employ for the solution of the general triangle any of those deduced from them in Chap. I. As a single trigonometric function is not sufficient to determine an unlimited angle or arc, (PI. Trig. Art. 53), it becomes necessary in most cases to deduce expressions for both the sine and cosine of the required part. It will be found that all the six cases of the general triangle admit of two solutions, but that they all become determinate, when, in addition to the other data, the sign of the sine or cosine of one of the required parts is given. In the practical applications in astronomy, it mostly happens that the conditions of the problem supply this sign. 28 T 218 SPHERICAL TRIGONOMETRY. 114. CASE I. Given 6, c and A. First Solution ; when one of the remaining angles, as B, and the third side a are required. The relations between the given and required parts are cos a = cos c cos 6 + sin c sin 6 cos A sin a cos B = sin c cos 6 cos c sin b cos A sin a sin B = sin 6 sin A (196) The signs of the second members will be known from their computed numerical values ; the sign of cos a is therefore known. If the sign of sin a is also given, the quadrant in which a must be taken will be known ; the second and third equations will determine the sign of the sine and cosine of B, and therefore the quadrant in which B is to be taken. In like manner, if the sign of either cos B or sin B is given, that of sin a becomes known, and the problem is determinate. If no conditions are attached to the required parts, there must be two solutions. The numerical solution will be conducted as follows : The values of the second members (or simply their logarithms) are to be separately computed, and their signs carefully noted ; then the quotient of the 3d by the 2d (or the difference of their logs.) will give tnn B, and hence B, which will be taken in the quadrant indicated by the signs of the sine and cosine. Then the 3d divided by sin B, or the 2d by cos B, will give sin a, which, agreeing with the value from the 1st equation, will serve to verify the correctness of the whole process. This solution may be adapted for logarithms by the methods employed in the pre- ceding chapter. 1st. Let k and be determined by the equations It sin $ = sin b cos A k cos = cos b k being a positive number (PI. Trig. Art. 174) ; then cos a = k cos (c 0) sin a cos B = k sin (c ) sin a sin B = sin b sin A (197) 2d. Eliminating k, and taking 0< 180, (PI. Trig. Art. 174), cos A ( tan b cos A cos 6 sin a COB B = ^- - sin (c 0) cos sin a sin B sin b sin A (198) 3d. If the quadrant in which & is to be taken is given, we may give the preceding equa- tions the following form : tan = tan b cos A tan a cos B = tan (c 0) D sin tan A tan a sin B = cos (c 0) OK 180) (199) SOLUTION OF THE GENERAL SPHERICAL TRIANGLE. 219 4th. If both a and b are less than 180, as not unfrequently happens iu the appli- cations of this problem, let sin a n = (200) sin b k then m and n are both positive numbers (k being positive) and (197) gives TO sin ft = cos A m cos ft = cot 6 n sin B = sin ft tan A n cos B = sin (c ft) cos a = cot (c ft) cos B Check. We find sin (c ft) sin o cos B tan A sin ft sin 6 cos A tan B cos (c ft) cos a cos ft cos 6 besides which we may employ, in connection with ( 200), the equation sin o sin B = sin 6 sin A; or in connection with (197) or (198) the equation tan a cos B = tan (c ft). Or when (197) and (198) are employed, we may find a both by its sine and its cosine. 115. The angle C may be found in the same manner as B, interchanging B and C, b and c, in the preceding formulae. But when B and O are both required, the Second Solution to be given presently is preferable. EXAMPLE. Given A = 261 16', b = 45 54', c = 138 32', and a < 180 ; to find a and B. We shall first employ (197). The first column of the following computation, con- taining the symbols expressing the operations to be performed, should be prepared before opening the tables : A 261 16' 6 45 54' c 138 32' log sin A 9-9949352 log cos A 9-1813744 log sin b + 9-8562008 log cos 6 = log k cos ft -f 9-8425548 log sin 6 cos A = log k sin 9*0375752 log tan ft 9-1950204 log cos + 9-9947336 log k + 9-8478212 ft 351 5'42"-6 * c 147 26'17"-4 * As > c, we take c = 138 32 r + 360, so that c may be a positive angle ; but it would be equally convenient to take c = 212 33 X 42 // '6. 320 SPHERICAL TRIGONOMETRY. log sin (c ) + 9-7309514 log cos (c 0) 9-9257303 log k cos (c 0) = log cos a 9'7735515 a 12625'6"-6 (1) log sin 6 sin A = log sin a sin B 9'8511360 log k sin (c 0) = log sin a cos B -j- 9'5787726 log tan B 0-2723634 B 2986'26"-8 log sin a + 9'9056351 log sin B 9-9455009 (1 ) Check, log sin a sin B 9'8511360 If a were not limited, we should have two solutions, the second being a = 233 34' 53" -4, B = 118 6' 26" -8. We shall neit give the computation by (200), which is applicable to this example, since both a and 6 are less than 180. A 261 16' b 45 54' c 138 32' log cos A = log m sin - - 9'1813744 log cot 6 = log m cos + 9'9863540 log tan ^ - - 9-1950204 0351 5'42"-6 c 147 26' 17"'4 log tan A + 0-8135608 log sin ^ -- 9-1897534 log tan A sin log n sin B 0033142 log sin (c 0) = log n cos B -f 97309514 log tan B 0-2723628 B 2986'26 // -9 log cos -B + 9-6731379 log cot (c 0) -- 0-1947789 log cos B cot (c 0) = log cot a -- 9-8679168 a 12625'6"-7 116. CASE I. Given b, c and A. Second Solution ; when the two angles B and C, or when all the remaining parts are required. We have, by Gauss's Equations (44), cos a sin (B -)- C) = cos $ (b c) cos $ A cos i a cos $ (1? -f (7) = cos J (6 + c) sin J ^4 sin J a sin ^ (5 C) = sin (6 c) cos J ^4 sin J a cos J (J? C) = sin J (6 -f c) sin J ^ From the first two we deduce \ (B -f- (7) and cos \ a, and from the second two i (B C) and sin J a, whence 5, C and a. The problem becomes determinate, as before ; that is, when a is limited by one of the conditions a<180, or a > 180 for then the signs of both cos J a and sin J a will be known.* * By Art. 27, Gauss's Equations may also be taken with the negative sign when the triangle is unlimited, as in the group (45), but the same final results are obtained from either (44) or (45). See note at the end of this chapter, p. 227. (202) SOLUTION OF THE GENERAL SPHERICAL TRIANGLE. 221 EXAMPLE. Same as in Art. 115. a< 180. A 261 16' b 45 54' c 138 32' $(b c) 46 19' J (6 + e) 92 13' \ A 130 38' log d = log cos $ (b c) + 9-8392719 log e = log sin | (6 c) 9'8592393 log / = log cos J (6 -f e) 8-5874694 log g = log sin J (6 + c) + 9-9996749 log cos }A 9-8137250 log sin I A + 9-8801803 log d cos J A log cos a sin J (!' + C) 9'6529969 log / sin \ A log cos \ a cos \ (B -f- C) 8'4676497 log tan } (B + C) + 1-1858472 log sin \ (S + (7) 9-9990771 log cos a -f 9-6539198 Ja 6312'33"-3 log e cos $ A log sin a sin J (5 C) + 9'6729643 log gsinl A log sin } a COB } (1? (7) -f 9'8798552 log tan j (B 0) + 9-7931091 l(B C) 3150'28"-7 log sin $ (B C) + 97222788 log sin a + 9'9506855 la 6312'33"'3 = 298 6'26"7 Ans. 1 C = 234 25' 29"'3 a = 126 25' 6"-6 117. If a only is required, we may find it by one of the methods of Arts. 75 and 76 ; and if the sign of sin a is given, the solution is determinate. If the sign of sin B or of sin C is given, we find that of sin a by inspecting the equation _ sin A sin 6 _ sin A sin c sin B sin C 118. CASE II. Given A, C and b. First Solution; when the third angle B, and one of the remaining sides (as ) are required. The general relations between the given and required parts are cos B = cos C cos A -\- sin C sin A cos 6 ~) sin B cos a = sin C cos A + cos Csin A cos 6 ( (203) sin B sin o = sin A sin 6 j which are solved in the same manner as (196). The problem is determinate when the sign of either sin , cos a, or sin a is given. T2 222 SPHERICAL TRIGONOMETRY. Adapting (203) for logarithms, we find 1st. * k sin & = cos A k cos tf = sin .4 cos b cos B = k sin (C&) sin B cos a = k cos ( C &) sin J5 sin a sin A sin 6 (k positive) 2d. cot i? = tan A cos 6 cos B sn cos a = CO s (G sin B sin a = sin A sin 6 3d. When the quadrant in which B is to be taken is given : cot # = tan A cos 6 tan B cos a = cot (C $) tanJ3sina= tan 6 cos * sin (<7 #) 4th. JP&en A and B are both less than 180, let k 9 = then p and q are positive numbers, and we have from (204), p sin $ = cot J. p cos i? = cos 6 q sin a = tan 6 cos # g cos a = cos (C i?) cot 5 = tan ((7 tf) cos o Checks. We have cos ((7 tf) _ sin jB cos a _ tan 6 cos * sin A cos 6 sin (C #) cos B tan a (204) (205) (206) (207) (208) * The same factor k is used here and in (197), although the auxiliaries and $ are different. To show that k lias the same value in (197) and (204), let the squares of the equations k sin = sin 6 cos A k cos = cos b be added ; we find P (sin* -f cos 5 $) = If cos 2 6 + sin 2 6 cos 2 ^4 = 1 sin 2 6 sin 2 A and in the same manner, from the equations k sin i? cos A k cos & = sin -4 cos 6 we find k* = 1 sin" 6 sin 2 A and therefore, in both cases, k = \/ (1 sin 2 6 sin 2 A) SOLUTION OF THE GENERAL SPHERICAL TRIANGLE. 223 besides which we may employ, with (207), the equation sin B sin a = sin A sin 6; or with (204) and (205), the equation tan B cos a = cot (G #). Also, when (204) or (205) is employed, we may find B both by its sine and its cosine. These formulae are computed in the same manner as those of preceding case. 119. CASE II. Given A, O and 6. Second Solution; when the two sides, a and c, or all the remaining parts, are required. We employ Gauss's Equations in the fol- lowing form : sin J B sin (a + c) = cos ( A C) sin b sin \ B cos J (a + c) = cos } (A + C) cos J b cos * B sin J (a ) = J (^ c ') sin t 6 cos i 5 cos $ (a c) = sin \ (A -\- C) cos J b which are solved in the same manner as (202). EXAMPLE. Given A = 121 36' 19"'8, C = 42 15' 13"-7, 6 = 40 V 10", and 5 > 180. By (209). b 40 0'10"-0 A 121 36' 19"-8 c 42 15' 13" -7 \(A-0) $(A + C) 39 40' 33"-0 81 55' 46"-7 20 0' 5"-0 log d = log cos (.4 C) log e log sin ^ (A C) log / = log cos $ (A -{- C) + 9-8863038 + 9-8051224 + 9-1473326 + 9-9956775 log sin $ 6 log cos J 6 + 9-5340806 + 9-9729820 log 180, and cos J B is negative. f It was necessary to increase J (a -f- c) by 360, to obtain e. The corresponding value of 6 would be 616 35' 36"'2. See note at the end of this chapter, p. 227. 224 SPHERICAL TRIGONOMETRY. 120. When B only is required, we may employ the methods of Arts. 81 and 82, which are determinate when the sign of sin B is given ; or when that of either sin a or sin c is given, since we may then find that of sin B by inspecting the equations n sin A sin b sin C sin b sin .o = = . sin a sin c 121. CASE III. Given a, 6 and A. First Solution; when the three remaining parts B, C, and c are all required. We find B by the equation (210) which is determinate when the sign of cos B is given. Then, to find C, we have cos C cos A -j- sin C sin A cos b = cos B sin Ccos A -j- cos (7 sin .4 cos 6 = sin B cos a which have already been employed and adapted for logarithms in Art. 118. If we denote the auxiliary by $, and put C & = #', we find, from (204), k sin # = cos A (k positive) k cos & = sin .4 cos 6 A sin $' = cos B k cos #' = sin 5 cos a To find c, we have (211) cos c cos 6 sin c cos 6 8m c sm cos - = cos a cos c sin b cos .4 = sin a cos which have already been employed and adapted for logarithms in Art. 113. If we denote the auxiliary by ^, and put c = 0', we find, from (197), jt sin ^ = sin 6 cos A (k positive) k cos = cos 6 Checks. We have k sin 0' = sin a cos A cos $' = cos a cos A cos (212) cos # x sn sin tan tan a tan 6 cos b tan A (213) One of which may be used as a check when either C or c has been alone computed. 1 When both C and c have been found, the obvious check is sin C sin A (214) *The following relations deserve a passing notice: 8in * * r = sin 5 sin B cos sin 0' = sin a sn tan tan ^ _ g ; tan i? tan # ' sin 2 sin 2 sin 1 6 sin 2 1? sin 2 ' sin 2 a SOLUTION OF THE GENERAL SPHERICAL TRIANGLE. 225 EXAMPLE. Given a = 126 25' 6"'6, b= 138 32' 0", A = 261 16' 0", and cos B negative. 126 25' 6"-6 138 32' 0"'0 261 16' 0"'0 By (210), By (211), log sin a + 9'9056351 log sin b +9-8209788 log sin A 9-9949352 log sin B 9-9102789 B 23425'29"-3 log cos 6 9-8746795 log sin A cos 6 = log k cos & + 9'8696147 -- log k sin log tan 9-1813744 & 9-3117597 tf 34824'53"-0 By <212), log cos a 9-7735515 log sin B cos a = log k cos #' + 9-6838304 log cos B log k sin #' 97647520 log tan &' 0-0809216 $' 30941'33"-7 # + #' = C 298 6'26"-7 log sin 6 cos A log * sin 9'0023532 log cos b = log k cos 9'8746795 log tan + 9-1276737 18738'31"-3 log sin a cos B == log k sin $ log cos a = log It cos ' log tan $ - 9-6703871 - 9-7735515 + 9-8968356 21815 / 28 // -6 4553 / 59 /A 9 log sin C 9-9455010 log sin c -f 9-8562006 log sin c i Check. - 0-0893004 0-0893001 In this example, both i? -f i? x and <& -f- ' exceed 360, and consequently we have to deduct 360 from each of them. We might have avoided this, however, by taking $' = 50 18' 26 // -3, ' has the sign of sin a cos B, so that we have the fol- lowing- formulae : k sin = sin b cos A (k positive) k cos = cos 6 (216) cos 6 (' < 180 with the sign of sin a cos B) Check. The equation (214). 123. CASE IV. Given A, B and 6. First Solution; when the three remaining parts a } e and Care all required. We find a by the equation sin A sin 6 sin B (217) which is determinate when the sign of cos a is given. The remainder of the solution is by (211) and (212). 124. CASE IV. Given A, B and 6. Second Solution; when c and C are required, without finding a. We easily find, from (211), k sin & = cos A k cos & = sin A cos 6 a, sin # cos JS sin w - cos .4 positive) (cos $' and sin -B cos a to have the same sign) (218) And from (212), k sin = sin 6 cos A (k positive) k cos $ = cos b sin ?/ = sin tan ^4 cot 5 (cos d' and cos a to have the same sign ) c = <(> + +' Check. The equation (214). 125. CASE V. Given a, 6 and c. The formula cos a cos b cos c cos A sin o sin c (219) (220) determines A when the sign of sin A is known. If the sign of sin B or of sin C is given, that of sin A becomes known by the equation sin A sin B sin C sin a sin 6 sin c SOLUTION OF THE GENERAL SPHERICAL TRIANGLE. 227 The formulae (31), (33), (34), may be used, each of which will become determinate when the sign of either sin A, sin B, or sin C is known. 126. CASE VI. Given A, B and C. The formula (221) sin B sin C determines a when the sign of sin o is given. If the sign of sin 6 or of sin c is given, that of sin a becomes known by the equation sin a _ sin b _ sin c sin A sin B sin O The formulae (36), (38), (39), may be used, each of which will be determinate when the sign of either sin a, sin b, or sin c is known. NOTE UPON GAUSS'S EQUATIONS. In the unlimited spherical triangle, we may consider any part, as a, to have an infinite number of values, viz. a, a -f 360, a + 720, etc., expressed generally by the formula a -f- 2 n TT, n being any whole number or zero ; and since sin a = sin (o -f- 2 n TT) cos a = cos (o -f- 2 n K) all those equations of Chap. I. that involve only sin a and cos a will not be changed by the substitution of a + 2 n TT for a. A similar substitution may be made for each of the parts, or for all of them, at the same time, so that there is an infinite series of tri- angles to which these equations are applicable. But the substitution of a -f 360 for a, in Gauss's Equations, (202), will change the sign of all of them, since sin J (o + 360) = sin J a cos J (o + 360) = cos J a while the substitution of a + 720 for a will not change their sign, since sin J ( + 720) = sin } a cos (a -f 720) = cos J o In general, their sign is changed by the substitution of o -f- (4 n -f- 2) ir for a, and it is not changed by the substitution of a 4- 4n ir. The same results follow like substitu- tions for each of the parts. It follows that these equations taken only with the posi- tive sign, do not include all the triangles of the infinite series above spoken of, and that they are complete only when taken with the double sign, and expressed in two distinct groups, as (44) and (45) of Art. 27. In practice, however, we may take them with the positive sign only ; for they will then give at least one of the triangles of the series, from which .ill the others, (and particu- larly that whose parts are less than 360), may be directly deduced by the application of 360.* This will be illustrated by the example of Art. 119, p. 223 ; we there find } (a + c) = 63 23 / 3"'3 } (a c) = 193 12' 32"-9 or rather, since (a -(- c) should be greater than (a c), $ (a + c) = 423 23' 3" 3 (a c) = 19312'32"-9 * Gauss (Theoria Molus Corp. Ccel. Art. 54) recommends the use of the positive sign only, observing that any side or angle may be diminished or increased by 360, as the case may require, but confines himself to the statement of this practical precept, with- out explaining the grounds upon which it rests. 228 SPHERICAL TRIGONOMETRY. whence a = 616 35' 36"-2 c = 230 10' 30 /A 4 which is the proper solution of the equations taken with the positive sign. If now we deduct 360 from a, and take, as on p. 223, o = 25635 / 36"-6 c = 230 10' 30"-4 we have the solution that would have been obtained by taking the negative sign in all the equations ; for we now have $ (a -f c) = 243 23' 3"'3 J (a )= 13 12 / 32 /A 9 which, differing from the former values by 180, must change the sign of all the equations. I have given some further particulars respecting unlimited spherical triangles, and a fuller discussion of Gauss's Equations, in an essay which the reader will find in the Astronomical Journal, Vol. I., published at Cambridge, Mass. CHAPTER V. AREA OF A SPHERICAL TRIANGLE. 127. Given the three angles of a spherical triangle, to compute the area. This problem is solved in geometry, where it is proved that the surface of a spherical triangle is measured by the excess of the sum of its three angles over two right angles, by which is meant, that the area is as many times the area of the tri-rectangular triangle as there are right angles in the excess of the sum of the angles over two right angles. To express this analytically, let r = radius of the sphere T surface of the tri-rectangular triangle = ^ surface of a sphere = ^nr 2 2S=A + B + C K= area of the triangle ABC. Also, let the angles A, B and C be expressed in the unit of Art. 11, that is, let A, B, C denote the arcs which measure the angles in a circle whose radius is unity. The right angle expressed in the same unit is -> therefore the number of right angles in 2 8 is 2 2S^- = 2 it and we have, according to the above theorem of geometry, ic) x or K=i*(2S it) (222) and if the radius of the sphere is taken = 1 K=2S it (223) 128. In a plane triangle the sum of the angles is equal to TT, and in a spherical triangle the sum exceeds it by K ; hence this quantity, K t is commonly called the spherical excess. U 229 230 SPHERICAL TRIGONOMETRY. 129. Given the three sides, to find the area. By (223), we have sin J K= sin Is ^J = cos S = cos IS - J = sin : ; TT 1 4 cos i a cos J 6 cos $ c cos* ^ a 4- cos 8 j b 4~ cos* ^ c 1 2 cos J a cos 6 cos c The numerator of (225) being denoted by n, we find, , , JT- 14- cos a 4- cos 6 4" c 08 c COt * A 2n which is known as De Gua's formula. (225) (226) (227) Again, from the formulae of Art. 35, since 1 sin 5 = 2 sin *\ K, 1 -f- sin S 2 cos * J K, we find sn = cos a cos cos c = fcos } a cos ( a) cos H 6) cos } ( c) K = v/ fc L cos \ a cos J 6 cos r, J tan J -ST= i/ L tan J s tan i ( s a ) tan $ ( s ^) tan i ( s c )3 the last of which is known as LhuiUier's formula. 130. Given two sides and the included angle, (or a, 6 and C) to find the area. We have, from (224), by (71), (228) . i ,- _ cos C sin C i !> tan i tan i 6 sin C tan 5 A = 1 -f- tan J (i tan j 6 cos C (229) AREA OF A SPHERICAL TRIANGLE. 231 131. If we admit more than three parts of the triangle into the expression of K, we have, by (56) and (67), i -fT- cos $ a cos \ b -f- sin ^ a sin $ b cos (7 cos 5 A - * (230) the quotient of which gives (229). 132. Since there are always two triangles upon the surface of the sphere which have the same three sides, (Art. 110), the angles not being limited to values less than 180, the formula (225), (226), (227) should give the areas of both of them, and their sum should be equal to the surface of the sphere 4 ir. In fact, by (225), sin K may be either positive or negative, while by (226) the cosine is fully determined, so that these formulae give two values of i K whose sum is 2 TT, and therefore two values of A", whose sum is 4 TT. It follows that (225) alone is not sufficiently determinate when the triangle is un- limited, since it gives four solutions. The most convenient formula is therefore (228), for we must always have J K < TT, and the double sign of the radical gives the two values of J K, one less and the other greater than CHAPTER VI. DIFFERENCES AND DIFFERENTIALS OF SPHERICAL TRIANGLES. 133. Two parts of a spherical triangle being constant, and a third receiving an increment, it is required to deduce the corresponding increments of the remaining three parts. As in plane triangles, (PI. Trig. Chap. XII.), this will be effected by a comparison of two triangles having two parts in common. The triangle formed from the given one by applying the increments to the variable parts will be distinguished as the derived triangle. We shall first consider the increments as finite differences, and give them the positive sign, (PL Trig. Art. 187). 134. CASE I. A and c constant. The parts of FlG - *%, ABC, Fig. 22, being A, c, B, C, a, 6, those of the derived triangle ABO' are A, c, B + J.B, C+ JC, a-i-Ja, 6 + J6; and the parts of the differential triangle BCC' are a, a + Ja, J6, 180 (7, \ tan ^ J jB sin b sin c sin (A + and from (244) and (246), in the same manner, sinJ L JJ[^ sin 2 (a -f A a] tan (Jg tan^JC sin 6 sin c sin ( J. + 137. CASE IV. B and (7 constant. The equations of the preced- ing case (243 to 248), applied to the polar triangle, give sin(^i-f | AA) ~ tan ^ Ab cot (c -f- \ Ac) sin^ A A _ sin(^ tan ^ Ac cot(b tan 1 Ab tan (6 - - tan ^Jc tan (c + % Ac) sin B sin Csin (a -f- ^ A a) sin % A a ~ sin "(A + 236 SPHERICAL TRIGONOMETRY. sin|- A a sin 2 (A + j A A) tan (c -f \ A tan \Ab sin B sin C sin (a -f -| Ad) sin ^ Ja sin 2 ( A -f 1 ^^) tan (6 -f ^ tan^Jc sin ^ sin (7 sin (a -f- FINITE DIFFERENCES OF SPHERICAL RIGHT TRIANGLES. 138. All the preceding equations are, of course, applicable to right triangles, or to quadrantal triangles, and in some cases they assume simpler forms. Thus in Case I., if the variable O= 90 (231) and (232) become sin Ab = sin (a -f- A a) sin A B tan % Aa = tan \ Alt tan \ AC and similar modifications take place in other cases. 139. When one of the constants is 90, the preceding equations do not generally assume any simpler forms, but they may be trans- formed so as to involve the same variables in both members, which is generally desirable in their practical applications.* The method that we shall follow is so simple that it will be un- necessary to repeat it in every case. A single example will suffice to explain it. Let C (= 90) and 6 be the constants; to find the relation of Ac and AB, we have between the two variables and the constant 6, the equations sin B = sin b cosec c sin (B -\- AB) = sin b cosec (c + ^c) the difference and sum of which, by PI. Trig. (105), (106), (131), and (132), are IT> . 1 AT>\ i A n 2 sin 6 cos(c+ \ Ae) sin \ Ac 2 cos (B + A A B) sm i AB = - smcsm(c+ Jo) o / T> . 1 A r>\ 1 A T> 2 sin 6 sin (c + i- Jc) cos 4- Jc 2sm(J5 + i JJ5)cosA AB= - sm csm(c+ Jc) * Cagnoli gives these equations reduced so as to involve the same variables in both members ; but in almost every instance his formulae involve two factors more than are necessary, and are far less simple and convenient than those here given. DIFFERENCES OF SPHERICAL TRIANGLES. 237 and the quotient of these is tan 4 AB tan i- Ac ^ l tan (c -f ^ Ac) which gives the first equation of the following article. This process always eliminates the constant, and is applicable in every case. When the equation to be differenced involves cosines, we employ PI. Trig. (107) and (108); if tangents, (115) and (116); if cotan- gents, (122) ; if secants, (129) and (130). The results are as follows : 140. CASE I. C= 90 and 6 constant. tan^Jc _ tan(c + ^- Ac) tan|^ Ac _ . cot (c-)-^ Ac) /occ\ tan%AB~ ~tan(B + %AB) tanAa~ cot(a + ^Ja) ^ ' sin Aa sin (2 a -|- A a) tan ^ A a, _ tan (a -f j A a) / 9t - R \ lfa~AA ' sin(2.A + J^) sin AB ~ ~sm(2B-\-AB) sin Ac sin (2 c -j- Ac) 141. CASE II. C= 90 and c constant. sin A A sin(2A-j- A A) tan^Aa _ cot (a-f j Aa) sinAB s\n(2B-\-AB) tan|J6 cot(6-{- tan ^ Aa _ tan(q+ ^ A a) tenjM^. _ tan (6 + \ Ab} (>iO9j sinJq __ sm(2a+Aa) JJb_. sin (26+ Ab) ._ fi , 142. CASE III. C = 90 and A constant. tan^ Ac __ tan(c+ | Ac) sin Aa sin ( 2 a + J a) , . tan^Ja tan(a+|Ja) tan|J6 tan(6 + JJi) *?B_I^ cot(c + ^- Ac) ian^Ab _ cot (6 + ^- J6) ,_ . Sin J.B gJTi^9 7?_L ,f Z?\ *~ 1 /\ V " / ^HL^_ = sin (2 c + Ac} tanJ-Ja = cot^a J-^) / 2fi oN sinJ6 sin (26 -f J6) tan^J^B tan(B-\-^AB) 143. If a constant side is 90, the equations of finite differences for the triangle may be obtained by applying the preceding equations to the polar triangle. 238 SPHERICAL TRIGONOMETRY. DIFFERENTIAL VARIATIONS OF SPHERICAL OBLIQUE TRIANGLES. 144. To obtain the differential variations, we have only to make the increments infinitely small in the equations of finite differences, observing the principles of PI. Trig. Art. 192. Or we may differ- entiate. the equations of spherical triangles directly, employing the differentials of the trigonometric functions given in PI. Trig. Art. 192. For example, A and c being constant, to find the relation of d a and d E, we have sin A sin c = sin a sin (7 the differential of which is = sin a d sin C-f sin Cd sin a = sin a cos Cd C -f- cos a sin Cd a da _ tan a JC~ ~ tan C and to find the relation of da and d 6, we have cos a = cos b cos c -f- sin 6 sin c cos A sin ada= sin 6 cos c d b + cos b sin c cos Adb da __ sin 6 cos c cos b sin c cos A db sin a or by (7), d a ^ - = cos C a 6 results which agree with those found from (236) and (232), by making Ja, Ab and JC infinitely small. By either method, then, the fol- lowing equations may be readily verified. 145. CASE I. A and c constant. da _ _ tan a db _ sin a dC~ tan C dB sin C (264) da db tana , 9fif .x - = cos C = ~r~^, ( Zbb ) db d C sin C d^ = sina^ dC _ coga (266) dJ5 tan (7 DIFFERENTIAL VARIATIONS OF SPHERICAL TRIANGLES. 239 146. CASE II. A and a constant. f*f\a r (267) (268) (269) rvf fi (279) (280) (281) dA sin 2 A d a cot a dB sin 2 B d b cot b d a tan a d b tan 6 dA tan A dB~ tanJ5 d a sin 2 a d b sin 2 b dB 2 cot dA 2 cot ^4 151. CASE III. C=90 and A constant. d c tan c d a sin 2 a d a tan a d b 2 tan 6 d c 2 cot c d b cot b dB sin 2 J5 dB cotB d c sin 2 c d a cot a db sin 2 6 dB tan ^ (282) (283) (284) 152. The differential variations are often employed for approxi- mate results, instead of the equations of finite differences, when the increments are very small. The remarks of PI. Trig. Art. 203, apply here also, but it is not necessary to introduce the radius in seconds, since all the parts of a spherical triangle are expressed in the same unit. DIFFERENTIAL VARIATIONS OF SPHERICAL TRIANGLES WHEN ALL THE PARTS ARE VARIABLE. 153. Let the equation cos a = cos b cos c + sin b sin c cos A be differentiated, all the parts being variable ; we find sin a d a = (sin 6 cos c cos 6 sin c cos A) db -f- (sin c cos b cos c sin 6 cos A ) d e -\- sin b sin c sin Ad A Dividing by sin a, this becomes, by (7) and (3), d a = cos C d b + cos B d c -{- sin 6 sin C d A (285 and in the same manner from the 2d and 3d equations of (4) we find db = cos A d c + cos Cd a + sin c sin A dB (286) dc = cos B d a -f- cos A d b -(- sin a sin B d C (287) From these three equations, any three of the six differentials da, db, dc, dA, dB, dC, being given, the other three may be determined by the usual processes of elimination. If any one of the parts be supposed constant, its differential will become zero, and these equations will assume simpler forms. If two of the parts be supposed constant, we can easily deduce all the equations of Arts. 145, 146, 147 and 148. CHAPTER VII. APPEOXIMATE SOLUTION OF SPHERICAL TRIANGLES IN CERTAIN CASES. 154. WHEN some of the parts of the triangle are small, or nearly 90, or nearly 180, approximate solutions may be employed with advantage. These are generally found by means of series. 155. In a spherical right triangle (the right angle being C), given A and c, to find b. We have tan b = cos A tan e (288) which is of the form in PI. Trig. (493), and may therefore be developed by (495) and (496) by putting x = b, y = c, p = cos A, whence ! i j = tan A. p + 1 1 + cos A and (495) and (496) become, [taking n = in (495), and n 1 in (496)], b e tan 1 A sin 2 c + J tan* J A sin 4 e etc. (289) b = IT c + cot 1 \ A sin 2 c J cot* J A sin 4 c -f etc. (290) If A is small, cos A is nearly equal to unity, and 6 exceeds c by a small quantity, which is approximately found by one or more terms of the series (289). If A is nearly 180, or cos A nearly = 1, 6 exceeds K c by a small quantity, which is found by (290). For examples of the mode of computation, see PI. Trig. Art. 255. 156. Although these solutions are termed approximate, it must not be inferred that they are less accurate in practice than the direct solution of (288) by the tables ; for the logarithmic tables are themselves only approximate, and the neglect of the higher powers in such series as (289) and (290) may involve a less theoretical error than the similar neglect of the higher powers in the series by which the tables are computed. In the examples of PI. Trig. Art. 255, the thousandths of a second were found with accuracy, which could not have been effected by a direct solution with less than eight decimal places in the logarithms. These considerations lead to the frequent employment of approximate solutions in astronomy. 157. If A and b are given, to find c, we have tan c = sec A tan 6 which is reduced to PL Trig. (493), by putting x = c, y = b, p = sec A, sec A -\- 1 1 + cos A and the series will be c= 6 + tan 2 J^ sin 26 + J tan* J .4 sin 4 6 + etc. (291) c = IT b cot* \ A sin 2 6 cot* A sin 4 6 etc. (292) 31 V on 242 SPHERICAL TRIGONOMETRY. 158. Similar solutions apply to the equations of right triangles, tan a = sin b tan A cot B = cos c tan A the last being solved under the form tan (90 B) = cos e tan A We may also compute, in the same manner, the auxiliaries and # in (122) and (134), so frequently employed in the solutions of oblique triangles. 159. In a right spherical triangle, given c and A, to faid a, when A is nearly 90. We have from which we deduce sin a = sin A sin c (293) tan i (c a) = tan* (45 } A) tan J (e + a) (294) From this we may find c a, which is supposed very small, by successive approxima- tions. For a first approximation, let o = c in the second member, and find thence the value of c a and of a; for a second approximation substitute in the second member the value of a just found ; and so on until two successive values agree as nearly as may be desired. EXAMPLE. Given A = 89, c = 87 ; find a. Here 45 J A = 30', and for the first approximation (c + o) = 87 log tan J (c -f- a) 1-28060 log tan 2 (45 M) 5-88172 ar co log sin \" 5'31443 \(c a) = 299"74 \ogl(c a) 2'47675 a = 87 9' 59"'48 = 86 50 / 0"'52 2o APPROX. 3D APPROX. 4xH APPROX. !(+) log tan \ (c -(- a) , tan' (45- M) 86 55' 0" 1-26868 8655 / 8" 1-26899 1. 1 op i c 86 55' 8 /A 17 1-26900 1. 1 Q i e sin 1" lyoio log i (c o) |(e-) c a a 2-46483 291 / '-63 9' 43"-26 86 50' 16"-74 2-46514 291 // -83 9''43"-66 86 50' 16' A 34 2-46515 291"-84 9' 43 /A 68 86 50 X 16' A 32 The direct solution of (293) gives a 86 50' W, but cannot give the fractions of a second without tables of more than seven figure logs. We have given this pro- blem, however, not so much on account of its particular utility, as for the purpose of introducing the method of approximation to which it leads, and which is often employed. The process here explained may obviously be applied to any equation of the form sin x = m sin y when m is nearly equal to unity. APPROXIMATE SOLUTION OF SPHERICAL TRIANGLES. 243 160. In a spherical oblique triangle, given two sides and the included angle, to find the other angles and side by series. If a, 6 and C are the data, to find c, we have cos c cos a cos 6 -f- sin a sin 6 cos C Substituting half arcs, sin* i c = sin 2 $ a cos* J b + cos* i a sin 3 J 6 2 sin a cos J 6 cos a sin b cos C which is of the form PI. Trig. (507), and may be developed by (508) by substituting sin $ c for c, sin $ a cos 6 for a, and cos a sin 6 for 6 ; so that (508) becomes ,, ["tan i a ri /tan \ a\ 2 cos 2 (7 , "! /oa*\ logsin $c = logcosjasm 6 M\ cos (7 -f I - + etc. (295) |_tan o Uan o/ Z J To find 4 and 5, we have, tan lA + B) i / t n\ sin * (u o) , i >-y tan J (-A -B) = -^ -^ cot i (7 sin i (a + 6) Comparing these equations with PL Trig. (493), and developing by (495), n = 0, etc. (296) tan ^ a Uan J a If we develop by (496), we find cot J a Vcot J a/ 2 ^. +etc. (297) S_\ sin (7- (52Li^) n_2 + etc (298) tan ^ &/ vtan n (7- 5L H_ _ etc . (299) tan ^ 67 \tan ^ o/ 2 from which a selection will be made in any particular case, according to the con- vergency of the series. The terms of the series are in arc, and must be reduced to seconds, by dividing by sin V. This solution may be applied to the case where two angles and the included side are the data, by means of the polar triangle. 161. To express the area of a spherical triangle in series. Comparing (229) with PI. Trig. (500), and developing by (502), we find J K= tan J a tan 6 sin C J tan* J a tan* \ b sin 2 C+ etc. (300) 244 SPHERICAL TRIGONOMETRY. 162. LEGENDRE'S THEOREM. If the sides of a spherical triangle are very small com- pared with the, radius of the sphere, and a plane triangle be formed whose sides are equal to those of the spherical triangle, then each angle of the plane triangle is equal to the corre- sponding angle of the spherical triangle minus one-third of the spherical excess. Let a, 6 and c be the sides of the spherical triangle expressed in arc, the radius of the sphere being unity; and let A', B f and G' be the angles of the plane triangle whose sides are a, b and c. Then we have, in the spherical triangle, _ i _ cos a cos 6 cos c COS -o. --- sin 6 sin c Substitute in the second member of this, the values of cos a, etc., in series, by PL Trig. (405) and (406), neglecting only powers above the fourth, viz. cos a = 1 i a 3 + zV a * cos6 = l J6 2 + ^6* 8in& = 6 $& cos c = 1 J c 1 + A c* sin c = c c* we find cos A = * (** + *-<*') + A (a 4 -* 4 -- ct -6ye') & c[l-i(6' + <)] Multiplying the numerator and denominator by 1 -f i (& 1 + c*), and neglecting terms of a higher order than the fourth, as before, we have , A b' + c* a' , q* + 6* + c* 2a'6 2o*c* 26V COS ^.l - - ~T ' ------ .. . - 2 b c 24 6 c which, by PI. Trig. (225) and (239), becomes cos A = cos A' \ b c sin 2 A' Let A = A' -f x, then since x is small, we may put cos z = 1, so that, by PI. Trig. (38), cos A = cos A / x sin A' whence x = ^ b c sin A / But J 6 c sin A' = area of the plane triangle = very nearly area of the spherical triangle = K, whence x=\K A' = A \K The same reasoning applies to each of the other angles, so that which proves the theorem/ 163. This theorem is applied in geodetical surveying, and is found to be sufficiently accurate for triangles whose sides are considerably greater than 1. It is to be remem- bered that the sides are to be expressed in arc; and if they are given in fet (for example), they must be reduced to arc by dividing by the radius in feet, or, which is equivalent, the area must be divided by the square of this radius. If then r = radius of the earth in units of any kind, a, b and c the sides of the triangle in units of the same kind, and k the area of the plane triangle, we shall have K in seconds, by the equation jr k r 1 sin 1" EXAMPLE. In a triangle upon the earth's surface, given b = 183496'2 feet, c = 156122'l feetj and A ~ 48 V 32"'35; to find the remaining parts. APPROXIMATE SOLUTION OF SPHERICAL TRIANGLES. 245 We have k = J 6 c sin A, and the mean value of r = 20888780 feet. Hence log 6 5-26363 log c 5-19346 log sin A 9-87159 ar co log 2 r 1 sin 1" Q'37356 K=5" -04 logJT 0-70224 It is evident that great accuracy in the value of r and of the other data is not required in computing K. We now have j K 1"'68, A' = 48 4' 30"'67, and by solving the plane triangle with the data A', b and c, we find a = 140580-0 feet B' = 76 12' 22"'19 C' = 55 43' 7"13 Adding } K to each of these angles, the angles of the spherical triangle are B = 76 12' 23"-87 C = 55 43' 8"-81 For further details respecting geodetical triangles, and for the methods of solving spheroidal triangles, special works upon geodesy must be consulted, such as Legendre's Analyse des Triangles traces sur la surface d'une spheroide ; Puissant' s Traite de Geodesie ; Puissant's Nouvel essai de trigonometric spheroidique ; Fischer's Lehrbuch der hoheren Geoddsie; various papers by Gauss, Bessel, etc. 164. To solve a spherical triangle when two of ite sides are nearly 90. If a and b are nearly 90, c and are nearly equal, and it will be expedient to com- pute the small quantity C c by an approximate method. We have, by (25), sin 3 c = sin' }( + *) sin 1 } C + sin 2 J (a b) cos 2 } C and by PL Trig. sin 2 } 0= [sin 2 } (a + 6) + cos 1 J ( + 6)] sin 2 } the difference of which equations is sin } (C+c)sin } (0 c) = cos 1 } (o + b) sin 2 } (7 sin 2 J (0 6) cos 1 J C Let a' = 90 a 6' = 90 b a' and 6' beiug very small : also, since O and c are nearly equal, put then the above equation becomes sin C sin J (C c) = sin 1 } (a' + 6') sin 2 } (7 sin* } (a' &') cos 2 } (7 Dividing by sin (7=2 sin J C cos (7, and substituting the arcs J (Oe), J (a' + 6'), j (a' 6'), for their sines, we find C- c = sin 1" [ (^^) t tan J C - ( a/ ~ 6/ )' cot J (?] (301) which is the required approximate formula for the case when a', 6' and C are given to find c. If a', 6' and c are given, to find (7, we may exchange C for c in the second member, whence tan Je-cotic (302) v 2 CHAPTER VIII. MISCELLANEOUS PROBLEMS OF SPHERICAL TRIGONOMETRY. 165. In a given spherical triangle, to find the perpendicular from one of the angles upon the opposite side. Fl - ^ Denoting the perpendicular upon the side c (Fig. 25) C by p, we have sin p = sin b sin A (303) If the three sides or the three angles are given, we find by (48), or (51), and (303), A<^T ^^f sin p = -7-^- = - (304) sin c sin C in which n and N are given by (47) and (50). If we admit more than three parts of the triangle into the expression of p, we have, by (55), (56), and (303), 8n - _ sn s = cos C sin 5 c (305) 166. To _/Jnd an d *" = radius of the inscribed circle, we have A P'" + B P f 4- GP f = A P"' 4 a = s, ^4 P /x/ = a a and the right triangle A P"' gives tan r = sin (s a) tan J A (318) corresponding with the formula of PI. Trig. (288). Substituting, in (318), the value of tan A, tan r _ //sin (s a) sin (s 6) sin (s c) \ \ V sin s / or tan r = -- (319) sin s Substituting, in (318), the value of sin (s a) given by (58), tan r = - -^- (320) 2 cos A cos f Jj cos C Also, by (51), we have N ~ sin B sin (7 sin a, which reduces (320) to tan r = Bin j JB Bin } C sin a (321 } cos i ^4 MISCELLANEOUS PROBLEMS. 249 173. Let the radii of the circles inscribed in the three triangles A'BC, B'AC, C'AB of Fig. 27, be r', r" and r' ff . Then if s / denote the half sura of the sides of A'BC, we have 2 s' = 2 r 6 c-fa j/ a = TT J(a + 6 + c so that (318) applied to the three triangles, gives tan r' = sin s tan A tan 7 -// = sin s tan J B tan r' ff = sin s tan J C Substituting in these the values of tan A, etc., or of sin s, n N tan r f = tanr" = sin (s a) 2 cos % A sin _B sin \ G n = N sin (s b) 2 sin A cos i J5 sin $ C Y , ,./// _ Itlll T - Also, by (321), ~~ ; . - rt i j i i /"* sm (s c) 2 sm j .4 sin cos J C , cos ^ B cos i C . tan r r = ^ r 1 sin a tan ,.// = tan r /// = cos i ^ cos M cos cos ^ C 174. The product of (319) and (323) gives tan r tan r' tan r /x tan r /x/ = - = n 1 /x/ = - = 1 n 1 \vlience, as in Art. 170, cot 7- tan r' tan r" tan r //x = sin 1 s tan r cot r' tan r" tan r'" = sin* (s a) tan r tan r' cot r" tan r //x = sin* (s b) tan r tan r x tan ?- // cot r" f = sin* (s c) 175. We find from (319) and (323), as in Art. 171, ... , // i , /// 4 sin J a sin \ 6 sin ^ c cot r + cot r + cot r" + cot r" , i , ,, L /// cot r cot r' + cot r" -f- cot r x// = , , // , L /// cot r -f cot r' cot r" + cot r" f = 4 sin i a cos i 6 cos i c 4 cos J o sin J 6 cos ^ c . .. , ,,, 4 cos \ a cos i 6 sin J c cot r + cot r' + cot T" cot r" = 2 2 cos a cos 6 cos c (322) 1- (323) (324) (325) (326) (327) 250 SPHERICAL TRIGONOMETRY. 176. From (309) and (321), we find tan = 4 sin A sin B sin J (7 cos a cos J 6 cos e From (307) and the first of (327), cot r + cot r' + cot r" + cot r'" = 2 tan R From (315) and (320), tan R + tan .R' + tan H" + tan .R"' = 2 cot r (328) (329) (330) and other similar relations are found by comparing (312) with (327), and (316) with (323). 177. The following relations are also worth remarking. If p is the perpendicular from C upon c, cot r sin == sin } (7 178. The pole of the circle inscribed in a spherical triangle is also the pole of the circle circumscribed about the polar triangle; and the radii of these circles are complements of each other. The arcs bisecting the angles of a given triangle will evidently bisect the sides of the polar triangle, and will be perpendicular to those sides respectively ; the common intersection of these arcs is therefore at once the pole of the circle inscribed in the first and circumscribed about the second. Again, if we join the angular points of the polar triangle with this common pole, the arcs thus drawn, being produced to meet the sides of the first triangle, are perpen- dicular to those sides, and therefore pass through the points of contact of the inscribed circle. Each of these arcs = 90, and is at the same time the sum of the two radii of the circles in question. This latter property is also obvious from the analytical expressions of the two radii. By means of it, we might have deduced all the formula for the inscribed from those for the circumscribed circle, or vice versa. 179. To find the arc joining the poles of the circles inscribed in, and circumscribed about a given spherical triangle.* 3- 29. Let be the pole of the circumscribed circle, Fig. 29, and / that of the inscribed circle. Put 00' = D; then cos D = cos A cos A 0' + sin A sin A 0' cos A O' By Art. 166, we have OAB S C, whence OA 0' = S C J A = %(B C) We have also COB AO' = cos O'P cos AP = cos r cos (s a) sin .40' = sin / P sin OMP sin % A *Hymer's Spherical Trigonometry. MISCELLANEOUS PROBLEMS. 251 Therefore, cos -^ = cot r cos (s a) + tan R cc cos R sin r sin j A Substituting by (319), (307), and (44), cos D _ sin S cos (s a) -f- 2 sin \ b sin ^ c sin ^ (b -f c ) cos .R sin r n sin a -f- sin b -\- sin c ~2 whence, by (53), / co D Y i 1 + sin o sin 6 -f- sin a sin c -f sin b sin c cos a cos 6 cos c Vcos R sin r/ 2 n 2 by PI. Trig. (179), _ /sin s + 2 sin j a_sin_t_6_sinj_c\ \ n / = (cot r -f tan J2) 2 cos" D = cos 2 (.R r) + cos 2 R sin 2 r sin 1 D = sin 2 (JZ r) cos 2 R sin 2 r (332) If the inscribed circle is inscribed in A'BC, Fig. 27, and its radius r', we have, by a similar process, sin 2 D' = sin 2 (R + r'} cos 2 R sin 2 r' (333) 180. To find the equilateral spherical triangle inscribed in a given circle. If R radius of the given circle, and ^1 = one of the angles of the equilateral triangle, we have, by (310), and PI. Trig. Art. 76, ., r> cos | A 3 cos i A 4 cos 8 1 .4. tan 1 R = e - = *- r-r, \ cos* j A cos 3 ^ A whence cos \A = J ( 3 , _ \ (334) \ V 4 + tan 2 .R / 181. To find the equilateral spherical triangle circumscribed about a given circle. If r = radius of the given circle, and a = one of the sides of the triangle, we find BinJa = J( r 3 - ) (335) \ \4 + cot 2 ? / 182. Given the base and area of a spherical triangle, to find the locus of the vertex. FlG - s - Let a = the given base, and K = area of ABC, Fig. 30. Produce AB and AC to meet in ^l 7 . Let O be the pole of the circle described about A'BC. The radius of this circle is given by the first equation of (311), which, by (224) becomes tanJB' = -^i-|; (336) sin j K The second member of this equation, being constant for all the triangles of the same base a, and the same area A", shows that R' is also constant, and consequently, that the 252 SPHERICAL TRIGONOMETRY. point A is always found upon the circumference of the same small circle A'BC. But A and A / being the extremities of the same diameter of the sphere, A is also found upon a small circle, equal and parallel to the circle A'BC. The perpendicular distance (p') of from the base BC, is found by the equation , cos R' cos p' = cos a and the pole of the locus of A is in the same perpendicular, at a distance from BO K p f = p, whence -*^. (337) The equations (336) and (337) determine the radius and position of the pole of the required locus, which may therefore be constructed. This elegant proposition is due to Lexell. 183. To find the angle between the chords of two sides of a spherical triangle. FIG. 31. In Fig. 31, being the center of the circumscribed circle, the angle between the chords of the sides AC and BO is half the spherical angle A OB. If, then, G l angle between the chords of a and 6 we have cos Ci = cos AOP = sin OA P cos AP or, by Art. 166, cos G l = sin (S C) cos J c (338) By (72) this becomes cos Ci = sin a sin J b + cos a cos J b cos C (339) 184. The preceding problem is employed for geodetical triangles, in which Ci differs very little from C, in which case it is expedient to compute the small difference C (7, = x. We easily reduce (339) to the following : cos GI = cos J (a 6) cos 2 \ C cos J (a + 6) sin 2 C = cos 2 } C 2 sin 2 \(ab) cos 2 } C sin 2 J C+ 2 sin 2 J (a -f- 6)sin l i (7 Subtracting cos C = cos 2 $ (7 sin 2 (7, we have, sin H (7+ Ci) sin } ((7 d) = sin 2 J (a + 6) sin 2 } (7 sin 2 i (a 6) cos 2 * (7 or approximately, taking sin J (C 1 4- Ci) = sin C = 2 sin J C cos J C and sin J ((7 - Ci) = J x sin I" t being expressed in seconds, x = ^ sin' \ (a + 6) tan } (7 - ! sin 2 J (<* *) cot | C (340) sin 1" sin l //r MISCELLANEOUS PROBLEMS. 253 185. If a great circle (DE, Fig. 32) bisect the base of a spherical triangle at right angles, any great circle (FG), perpendicular to it, divides tlie sides (AC, BC) into segments whose sines are proportional; that is, sin FA ; sin FC= sin GB : sin GC (341) Let P be the pole of ED, (DP =90), and PGF any great circle drawn through P, and therefore perpendicular to DE. Then, since PB + PA = 2 PD = 180 we have, by (3), sin F sin FA = sin P sin PA = sin Psin PB = sin G sin GB sin F sin FC = sin G sin GC whence, by division, the theorem (341). The arc FG is analogous to the parallel to the base in plane triangles. 186. If two arcs of great circles, (AB, CD, Fig. 33), terminated by any circle, intersect, the products of the tangents of the semi-segments are equal to one another; that is, tan $ AE tan EB tan CE tan J ED (342) FIG. 33. Let P be the pole of the circle DACB. Join PE and draw the perpendiculars PF, PG, bisecting the arcs AB and CD. Then we have co8FE_ cos PE _ cos PE _ cos GE cosFB cosPB cosPD cosGD cosFEcosFB _ cosGE cosGD coaFE-}- cos FB ~ cos GE + cos GD which, by PI. Trig. (110), gives (342). 187. If three arcs be drarcn from the angles of a spherical triangle through the same point, to meet the opposite sides, the products of the sines of the alternate segments of the sides will be equal. Thus, in Fig. 34, we shall have sin AB' sin CA' sin BC' sin CB' sin BA' sin AC' (343) For we easily find sin AP sin APB' sin CB' sin CP sin CPB' sin CA' sin CP sin CPA' sin BA' sin BP sin BPA' sin BC' sin BP sin BPC' sin^lC" sin sin APC' Multiplying these equations together, the product of the second members is unity, whence (343). The same property is easily extended to the segments of the angles. 188. It follows, that when three arcs are drawn from the three angles, so as to satisfy the condition (343), they must intersect in the same point. This occurs in the same cases as in plane triangles, that is, when the angles are bisected ; when the sides are bisected; when the three arcs are drawn from the angles to the points of contact W 254 SPHERICAL TRIGONOMETRY. of the inscribed circle ; and when the three arcs are the three perpendiculars upon the sides. The first three of these cases are obvious. To prove the last, if A', B' and C', Fig. 34, are right angles, we have cos AB' ^ cos CA' cosBC' cos AB cos CA cos BC 1 cos GE' ' cos EA' ' cos^C" ~~ cos CE ' cos BA ' cos AC~ whence cos AB' cos CA' cos BC' = cos GE' cos EA' cos AC' and in the same manner we find tan AB' tan CA' tan BC f = tan CB' tan BA' tan AC' The product of these two equations gives the condition (343), and therefore the per- pendiculars intersect in the same point. 189. To find the arc drawn from any angle of a spherical triangle to a given point in the opposite side. In the triangle PA A", Fig. 35, let PA' be drawn; we have (344) coe P A' sin A A" = cos P A' sin (AA' + A'A") = cos P A' cos A f A" sin A A' -f cos P A' cos AA' sin A'A /f But in the triangles P A A', PA' A" we have, by (4), cos PA' cos A A' = cos PA sin PA' sin A A' cos PA' A cos PA' cos A' A" = cos PA" -f sin PA' sin A' A" cos PA' A which snbstituted above give cos PA' sin A A" = cos PA sin A' A" + cos PA" sin A A' which determines P A', the sides PA and PA" and the segments of the side A A" being given. 190. Let three arcs PA, P A', PA", Fig. 35, passing through the same point P, be intersected by two others A A" and B B" whose intersection is Q; we have several symmetrical relations among the parts of the figure which find their application in astronomy. Let the points A, A', A" be given in position by their distances from Q, and put A Q = a A B =(3 P B = y A' Q = a' A' B' =p' P E' = y' A"Q = a" A"B"=P" P B" = r" By PI. Trig. (171), we have sin o sin (a' a") -f- sin a' sin (a" a) -f- sin a" sin (a a') = and in Fig. 35, sin Q sn y MISCELLANEOUS PROBLEMS. whence and similarly sin ft sm y" sin E" sm a = -*- - sin ) sm Q . . sin P' sin y" sin B" sm a.' -r^-. --- . ~A - sin Y sin Q sin/?" sin j" sin E" _ in a"= - --- : ~ smy which, substituted above, give sin /3_ : _ ,_, _, j. sin^ gin (a -,_ o) 4. ^n ( CTIM -v/ em v'' (345) sin y sin y' sm y' Again, if we express (344) in the notation of this article, it becomes cos (3+ y) sin (a'-a") +cos (p'+ y') sin (a"-a)+cos (/J"+ y") sin (a-a')=0 (346) which, added to (345), gives ->)=0 (347) tan y tan y x 191. If P is the pole of A Q, we have tan y x/ and (345) and (347) both give tan sin (a' a") + tan 8' sin (a" o) + tan p" sin (a a') = (348) 192. To yinii X CD Suppose a sphere to be described about O, whose intersections with the planes BOO, AOC, AOB and DOC are B' C' = a, A' C' = /?, 4'J?' = 7, and C''. The triangle A' C' D' is right-angled at /X, whence CD = c sin C'D' = c sin /? sin or by (46), if