24. Multiplying together the expressions for sin (A + B) and sin (A-B), equa. v, and reducing, there results sin (A + B). sin (A−B) = sin2 A sin2 B. And, in like manner, multiplying together the values of cos (A+B) and cos (A-B), there is produced COS (A+B). COS (A-B) = cos2 A - cos2 B. a Here, since sin2 A sin2 B, is equal to (sin A + sin B) × (sin A sin B), that is, to the rectangle of the sum and difference of the sines-; it follows, that the first of these equations converted into an analogy, becomes sin (A-B): Sin A-sin B :: sin A + sin B : sin (A+B) (XX.) That is to say, the sine of the difference of any two ares or angles, is to the difference of their sines, as the sum of those sines is to the sine of their sum. If A and B be to each other as n + 1 to n, then the preceding proportion will be converted into sin a : sin (n+1)a− sin na :: sin (n+1)a + sin na : sin (2n+1)a. ... (XXI.) These two proportions are highly useful in computing a table of sines; as will be shown in the practical examples at the end of this chapter, = 25. Let us suppose A + B = A', and A-B B'; then the half sum and the half difference of these equations will give respectively A=(A'+B'), and B=(A-B'). Putting these values of A and B, in the expressions of sin A. cos B, sin B.cos A, cos A. cos B, sin A . sin в, obtained in arts. 21, 22, 23, there would arise the following formulæ : Dividing the second of these formulæ by the first, there will factors of the first member of this equation, are tan(A-B', and R R tan (A+) respectively; so that the equation manifestly becomes B')' This equation is readily converted into a very useful proportion, viz, The sum of the sines of two arcs or angles, is to their difference, as the tangent of half the sum of those arcs or angles, is to the tangent of half their difference. VOL. III. F 26. Operat 26. Operating with the third and fourth formula of the preceding article, as we have already done with the first and second, we shall obtain tan {(A' + B′) . tan †(▲’— B′) COS B'-COS A' R2 = COS ACOS B Making B = 0, in one or other of these expressions, there results, These theorems will find their application in some of the investigations of spherical trigonometry. 27. Once more, dividing the expression for sin (A± B). by that for cos (AB), there results then dividing both numerator and denominator of the second fraction, by cos A. cos B, and recollecting that sin tan = we CUS R2 tan A. tan B tan A tan в. R2 which, after a little , 28. We might now proceed to deduce expressions for the tangents, cotangents, secants, &c, of multiple arcs, as well as some some of the usual formulæ of verification in the construction of tables, such as sin (54°+a) +'sin (54°— x)—sin (13°+ a) — sín (189—A) sin (90°—▲); sin + sin (36°— a) + sin (72° + a)=sin (36o + a) + sin (72°—▲). &c. &c. But, as these enquiries would extend this chapter to too great a length, we shall pass them by; and merely investigate a few properties where more than two arcs or angles are concerned, and which may be of use in some subsequent parts of this volume. 29. Let A, B, C, be any three arcs or angles, and suppose radius to be unity; then sin A. sin C+ sin B. sin (A+B+C) sin (A+B) sin (B+c) = For, by equa. v, sin (A+B+C) = sin A. cos (B+c)+cos A. sin (BC), which, (putting cos B. Cos Csin B. sin c for COS (B+c)), is sin A • cos B. cos csin A. sin B sin c+ COS A. Sin (B+e); and, multiplying by sin B, and adding sin A. sin C, there results sin A. sin C+ sin B. sin (A+B+C) sin A COS B cos c. sin B+ sin A. sin c. cOS2 B+ COS▲. sin B. sin (B+c): sin A. cos B. (sin B. cos C + cos B. sin c) +cos A. sin B. sin (B+ c) = sin A. cos B. sin (B + c) + cos A. sin A. sin (B+C) = (sin A. cos B + cos A. Sin E) X sin (BC) sin (A + B). sin (B+c). Consequently, by dividing by sin (A+B), we obtain the expression above given. = In a similar manner it may be shown, that sin (B-C) = sin A. sin c- 30. If A, B, C, D, represent four arcs or angles, then writing c+D for c in the preceding investigation, there will result, sin ▲ . sin (c+ D) + sin B. Sin (A+B+C+D) sin (B+C+D) = sin A. sin (c+D+ E) + sin B. sin (A+B+C+D+E) sin (A+B) sin (B+C+D+E)=? sin (B+C+....L)= sin A. Sin (c+D+..... L) + sin B. Sin (A + B + C + ...1) sin (A+B) 31. Taking again the three A, B, C, we have sin (CA) sin c. cos A - sin A COS C, sin A. cOS B sin B. cos A. Multiplying the first of these equations by sin a, the second F 2 by by sin B, the third by sin c; then adding together the equations thus transformed, and reducing; there will result, sin A. sin (B-C) + sin B. sin (C-A)+ sin c. sin (A—B)=0, COS A. Sin (B-C)+cos B. sin (C-A)+cos C. sin (A—B)=0. These two equations obtaining for any three angles whatever, apply evidently to the three angles of any triangle. COS 32. Let the series of arcs or angles A, B, C, D contemplated, then we have (art. 24), sin (A+B). sin (A — B) = sin2 A sin (B+c). sin (B-c) = sing B sin2 B, sin2 c, sin (c + D). sin (C–D) = sin2 c sin D, &c. &c. &c. sin (LA) sin (LA) = sin2 L - sin2 A. L, be If all these equations be added together, the second member of the equation will vanish, and of consequence we shall have sin (A+B). Sin (A − B ) + sin (B+c). sin (B-C)+&c... + sin (L+A) + sin (L− a) = 0. Proceeding in a similar manner with sin (A-B), cos (A+B), sin (B-C), COS (B+c), &c, there will at length be obtained COS (A+B). Sin (A−B) + cos(B+C).sin (B-C)+ &c... +cos (L+A). sin (L-A) = 0. ... · 33. If the arcs A, B, C, &c L form an arithmetical progression, of which the first term is 0, the common difference D', and the last term L any number n of circumferences; then will B- A=D', C—B=D', &c, A + B = D', B+C = 3D', &c and dividing the whole by sin D', the preceding equations will become sin D' + sin 3D' + sin 5D' + &c = 0, COS D' + COS 3D' + COS 5D' &C=0;} (XXV.) cos + &c 0. If E' were equal 2D', these equations would become sin D ́+ sin(D'+E') + sin (D' + 2E') + sin (D'+3E′) + &c=0, COS D'+cos (D'+E') + cos (D'+2E') + cos (D'+3E)+&c=0. 34. The last equation, however, only shows the sums of sines and cosines of arcs or angles in arithmetical progression, when the common difference is to the first term in the ratio of 2 to 1. To investigate a general expression for an infinite series of this kind, let ssin A+ sin(A+B) + sin (A+2B) + sin (A + 3B)+&c. Then, since this series is a recurring series, whose scale of relation is 2 cos B-1, it will arise from the developement of a fraction whose denominator is 1-2z. cos B + z2, making 2 = 1. Now sin A +z [sin (A + B) — 2 sin ▲ COS B] 1-2%. COS B +22 Now this fraction will be = S= sin A+ sin (A + B) — 2 sin A. COS B COS B = Sin (A A 2-2 cos B ; and this, because 2 sin A. + B) + Sin (A — B). (art. 21), is equal to sin ▲ — sin (^—B). But, since sin A'- sin B'= 2 cos(A'+B'). 2(1- -COS B) sin (A-B'), by art. 25, it follows, that sin a-sin (A—B)= 2 cos (AB) sin B; besides which, we have 1 COS B 2 sin B. Consequently the preceding expression becomes s = sin 4 + sin (A + B) + sin (A+2B) + sin (▲+3B) + &c, COS (A-1B) ad infinitum = 2 sin B (XXVI.) 35. To find the sum of n + 1 terms of this series, we have simply to consider that the sum of the terms past the (n+1)th, that is, the sum of sin [A+(n+1)B] + sin [A+(n+2)B] + sin [A+ (n + 3)] + &c, ad infinitum, is, by the preceding cos [A+ (n+B] theorem,= Deducting this, therefore, from the former expression, there will remain, sin a + sin (A+B) + sin (A + 2B) + sin (A + 3B) + . . . . sin (A + NB) = COS (AB))—cos [A + (n + 1)B] sin (A+B). sin (n+1)B. (XXVII.) 2 sin B 2 sin B = sin B By like means it will be found, that the sums of the cosines of arcs or angles in arithmetical progression, will be COS A+ COS (A+B) + COS (A + 2B) + COS (A + 3B) + &c, sin (A-B) ad infinitum = COS ACOS (A + B) + COS (A+2B) + COS (A + 3B)+.... ..... (cos A+NB) = (COS COS (A+B). Sin (n+ I)в sin B .... (XXIX.) 36. With regard to the tangents of more than two arcs, the following property (the only one we shall here deduce) is a very curious one, which has not yet been inserted in works of Trigonometry, though it has been long known to mathematicians. Let the three arcs A, B, C, together make up the whole circumference, O: then, since tan (A+B) = R2 (tan A+ tan в) (by equa. xxIII), we have R2x (tan A+tan B+ tan c) = R2x [tan A+ tan в-tan (A+B)]=R2 × (tan A + )=, by actual multiplication and re |