How do you prove sin(A+B) * sin(A-B) = sin^2 A - sin^2 B?

1 Answer
May 25, 2015

The standard formula for sin(A+B) is:

sin(A+B) = sin(A)cos(B)+cos(A)sin(B)

Now sin(-B) = -sin(B) and cos(-B) = cos(B), so

sin(A-B) = sin(A)cos(B)-cos(A)sin(B)

So:

sin(A+B)*sin(A-B)

= (sin A cos B + cos A sin B)(sin A cos B - cos A sin B)

= (sin A cos B)^2 - (cos A sin B)^2

...using the identity (p+q)(p-q) = p^2-q^2.

= sin^2Acos^2B-sin^2Bcos^2A

= sin^2A(1-sin^2B)-sin^2B(1-sin^2A)

...using sin^2 theta + cos^2 theta = 1 (Pythagoras)

= sin^2A-sin^2B-sin^2Asin^2B+sin^2Bsin^2A

= sin^2A - sin^2B