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

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Answer 1 · 2015-05-25T08:47:03

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#