# Verify the identity sin^2b+sin(a+b)*sin(a-b)=sin^2a.

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### 1 Answer

We'll re-write the product sin(a+b)*sin(a-b), using the formulas for the sum and difference of the angles a and b.

sin(a+b) = sina*cosb + sinb*cosa

sin(a-b) = sina*cosb - sinb*cosa

sin(a+b)*sin(a-b)=(sina*cosb + sinb*cosa)(sina*cosb - sinb*cosa)

We notice that the product is the difference of squares:

(sina*cosb + sinb*cosa)(sina*cosb - sinb*cosa) = (sina*cosb)^2 - (sinb*cosa)^2

We'll re-write the identity, substituting the product:

(sinb)^2 + (sina*cosb)^2 - (sinb*cosa)^2 = (sina)^2

We'll subtract (sina*cosb)^2 both sides:

(sinb)^2- (sinb*cosa)^2 = (sina)^2 - (sina*cosb)^2

We'll factorize by sin b to the left side and we'll factorize by sin a, to the right side:

sin b(1 - (cosa)^2) = sin a(1 - (cosb)^2)

sin b(sin a)^2 = sin a(sin b)^2

We'll divide by sin b and sin a both sides:

sina=sinb