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

Jan 8, 2016

The given identity is not true.

#### Explanation:

Take $a = \frac{\pi}{2}$ and $b = 0$

Then

$\sin \left(a + b\right) \cos \left(a + b\right) = \sin \left(\frac{\pi}{2}\right) \cos \left(\frac{\pi}{2}\right) = 1 \cdot 0 = 0$

but

${\sin}^{2} \left(a\right) - {\sin}^{2} \left(b\right) = {\sin}^{2} \left(\frac{\pi}{2}\right) - {\sin}^{2} \left(0\right) = {1}^{2} - {0}^{2} = 1$

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