sinAsin(A+2B)-sinBsin(B+2A)=sin(A+B)sin(A-B) please prove this.

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- May 4th 2012, 03:44 AMsrirahulanplease fix it.
sinAsin(A+2B)-sinBsin(B+2A)=sin(A+B)sin(A-B) please prove this.

- May 4th 2012, 04:56 AMemakarovRe: please fix it.
Just use repeatedly on both sides formulas for sin(x + y), sin(2x) and cos(2x).

- May 4th 2012, 05:20 AMsrirahulanRe: please fix it.
I can't understand.

- May 4th 2012, 05:29 AMemakarovRe: please fix it.
The left-hand side contains sin(A + 2B). Use the formula sin(x + y) = sin(x)cos(y) + cos(x)sin(y) to replace sin(A + 2B). The right-hand side contains sin(A - B). Use the formula sin(x - y) = sin(x)cos(y) - cos(x)sin(y) to replace sin(A - B). Keep using these formulas along with sin(2x) = 2sin(x)cos(x) and cos(2x) = cos^2(x) - sin^2(x) until you have no longer sine or cosine of a sum of angles or a double angle.