sin B', cos B', their values, there will result sin (A - B) = sin a sin B - COS A COS B. Hence, because в' = -B, we have sin (A - B') = sin (A + B -10) = sin [(A + B) - = - sin(A + B) = - COS (A + B)]. This value of sin (A - B') being substituted for it in the equation above, it becomes sin a sin B. COS (A + B) = COS A COS B 16. If B in this latter equation be made subtractive, sin B will become sin B, while cos B will not change (art. 3, 4). The equation will consequently be transformed into this: viz. COS (A - B) = COS A COS B + sin a sin B. (c.) If A and B be regarded as arcs instead of the angles which they measure, the results will be equally conclusive and correct. They may be expressed generally for the sines and cosines of the sums or differences of any two arcs or angles, by these two equations, viz. sin (A B) = sin A COS B± sin B CoS A COS (AB) = COS A COS B Sin A Sin B 17. The actual value of the sine, cosine, &c. depends, obviously, not only upon the magnitude of the arc, but also upon that of the assumed radius. In the preceding investigation we have supposed it to be unity. If we wish to make the above or any other formulæ, applicable to cases where the radius has another value R, we have only to substitute in the expression, COS A R tan A sin A R for sin A, for cos A, for tan a, and so on. Or, generally, we must so distribute the several powers of R, as to make all the terms homogeneous, as to the number of lines multiplied: this is effected by multiplying each term by such a power of R as shall make it of the same dimension, as that term in the equation which has the highest dimension. Thus the expression, sin 3A 3 sin A - 4 sin 3A (rad = 1) when radius is assumed = R, becomes 18. Supposing that in formulæ (c) A = B, and taking the superior sign, we shall have sin 2A = 2 sin A cos A = 2 sin A (1 sin 2A) COS 2A = COS 2A sin 2A 2 cos 2A 1 by substituting 1 - cos 2A for sin 2A, and 1 cos 2A. (D) sin 2A for So, again, by supposing B to be successively equal to 2A, 3A, 4A, &c. we shall find sin 3A = sin A cos 2A + cos A sin 2A sin 5A = sin A cos 4A + cos A sin 4A Or, for the successive cosines, COS A COS 2A - sin a sin 2A } (ε.) }(2.) 19. If in the above expressions for the several multiples of the sines, we introduce for the cosines their values in terms of the sines, and in those for the multiples of the cosines the values of the sines in terms of the cosines, the following expressions will be obtained, in which each quantity is expressed in terms of its own kind. sin 6A = (6s-32s3 + 32s5) (1 - s2) (G.) &c. &c. &c. 20. Other useful expressions for the sines and cosines of multiple arcs, may be found thus: Take the sum of the expanded expressions for sin (B + A) and sin (B - A); that is, add... to... .... .. sin (B + A) = sin B COS A + cos B sin A sin (B A) = sin B cos A - cos B sin A there results, sin (B + A) + sin (B-A) = 2 cos A sin B. So again, the sum of the expressions for cos (B + A) and cos (B-A), is COS (B+A) A) + COS (B-A) = 2 COS A COS B. Whence, cos (B + A) = 2 COS A COS B - COS (B-A). Substituting in these expressions for the sine and cosine of B + A, the successive values of A, 2A, 3A, &c. instead of B, we shall have, sin 2A2 cos A sin A sin 3A2 cos A sin 2A sin 4A sin A 2 cos A sifi 3A - sin 2A sin na = 2 cos A sin (n - 1) COS 2A = 2 COS A COSA - COS 0 (=1) }(1) - sin (n-2) A (κ.) 1) A-cos (n- 2) A COS 3A2 COS A COS 2A - COS A COS 4A 2 COS A COS 3A - COS 2A COS NA = 2 COS A cos (n 21. If the cosine of a be represented by a particular binomial, the formula (K) will be transformed into a class of very elegant and curious theorems: thus A like substitution in the forms for 2 cos 3A, 2 cos 4A, &c. will give 22. By an inverted process, the sine and cosine of a single arc may be inferred from those of a double are; or, which comes to the same, those of a half are from those of a whole arc. Letz - z = √(1-cos 2A), or x2 - 2 + 2 = 1 - cos 2A, and assume x2 + z = 1; then 2xx = cos 2A; COS2A and, exterminating z, x2 + = 1. 4. Hence x - x2+=(1-cos2 2A) = 1 sin3 2A, x2= and z2 = Hence, sin A = COSA = Or, for the half sin A = sin 2A, and x = √(1± sin 2A) (1 + sin 2A) + √(1- sin 2A). arc, (1 + sin A) COSA = (1 + sin A) + * This remarkable class of formulæ may be of use in the summation of series. Thus, cos A + cos 2A + cos 3A + cos 4A + &c. ..cos na, is equal to the sum of the two series (3 + 2 + 3 + ployed to determine the sum of all the natural cosines of every minute in the quadrant, gives sin 45 x cos 45° 0′ 30′′ sin 30" 3437-2470374. 23. Otherwise, in order to obtain expressions for the sine and cosine of a half arc, take 1 = cos 2A + sin 2A COS 2A cos 2A sin 2A (see formulæ D), the sum and difference of these respectively, give 2 cos 2A = 1 + cos 2A, and 2 sin 2A = 1 - cos 2A. Changing A and 2A into A and A, we shall have 2 cos2 A = 1 + cos A, and 2 sin2 A = 1 - cos A, Suppose, for example, a = 60°, then cos A = sin 30 =, cos sin and cos 30°=√(2+1)=√3=8660254, sin 30°=(2-1) = 1, as it ought to be : 15°=√(2+3)=(√6+2) = 9659258 15°=(2-3)=(√6-√2) =2588190: and so on, by continual bisections, as low as we please. 24. If the formulæ (c) be divided one by the other, there will result, Here dividing the two terms of the second member of the equation by cos a cos B, and recollecting that If a in this formula be = 45°, we shall have tan A = 1, and consequently, If A = B, the formula (0) will give for the double are, 2 tan A tan 2A 1-tanA .... (Q.) |