How do you prove $sin(a+b)cos(a+b) = (sina)^2 - (sinb)^2$ ?

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Answer 1 · 2016-01-08T21:17:19

The given identity is not true.

Take $a = pi/2$ and $b = 0$

Then

$sin(a+b)cos(a+b) = sin(pi/2)cos(pi/2) = 1*0 = 0$

but

$sin^2(a)-sin^2(b) = sin^2(pi/2)-sin^2(0) = 1^2 - 0^2 = 1$

thus the left hand side and right hand side do not always match, and so the identity is not correct.

How do you prove sin(a+b)cos(a+b) = (sina)^2 - (sinb)^2 sin(a+b)cos(a+b) = (sina)^2 - (sinb)^2 ?