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sin B", cos B', their values, there will result sin (A B') = sin A sin B — cos A cos B. Hence, because B" = 3 O – B, we have sin (A – B') = sin (A + B – 3 O) = sin [(A + B) – 4 O] = — sin [3 O - (A + B) = — cos (A + B)]. This value of sin (A - B') being substituted for it in the equation above, it becomes cos (A + B) = cos A cos B – sin A sin 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. 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 it sin B cos A cos (A + B) = cos A cos B >|- sin A sin :} (c.) 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 formulae, applicable to cases where the radius has another value R, we

- - - in A ... " have only to substitute in the expression, o for sin A,

cos A tan A --- for cos A,

for tan A, and so on. Or, gene

rally, 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

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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 (p + A) and sin (B - A); that is, add . . . . . . . . sin (B + A) = sin B cos A + cos B sin A to . . . . . . . . . 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) + cos (B - A) = 2 cos A cos B. Whence, cos (B +A) = 2 cos Á 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 2A = 2 cos A sin A sin 3A = 2 cos A sin 2A — sin A sin 4A = 2 cos A sin 8A – sin 2A (1.) sin na = 2 cos A sin (n − 1) A — sin (n − 2) A cos 2A = 2cos A cos A – cos 0 (= 1) cos 3A = 2 cos A cos 2A — cos A cos 4A = 2 cos A cos 3A — cos 2A (K.) cos na = 2 cos Acos (n − 1) A — cos (n − 2) A 21. If the cosine of A be represented by a particular binomial, the formula o will be transformed into a class of very elegant and curious theorems: thus

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A like substitution in the forms for 2 cos 3A, 2 cos 4A, &c. will give

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