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

May 25, 2015

The standard formula for $\sin \left(A + B\right)$ is:

$\sin \left(A + B\right) = \sin \left(A\right) \cos \left(B\right) + \cos \left(A\right) \sin \left(B\right)$

Now $\sin \left(- B\right) = - \sin \left(B\right)$ and $\cos \left(- B\right) = \cos \left(B\right)$, so

$\sin \left(A - B\right) = \sin \left(A\right) \cos \left(B\right) - \cos \left(A\right) \sin \left(B\right)$

So:

$\sin \left(A + B\right) \cdot \sin \left(A - B\right)$

$= \left(\sin A \cos B + \cos A \sin B\right) \left(\sin A \cos B - \cos A \sin B\right)$

$= {\left(\sin A \cos B\right)}^{2} - {\left(\cos A \sin B\right)}^{2}$

...using the identity $\left(p + q\right) \left(p - q\right) = {p}^{2} - {q}^{2}$.

$= {\sin}^{2} A {\cos}^{2} B - {\sin}^{2} B {\cos}^{2} A$

$= {\sin}^{2} A \left(1 - {\sin}^{2} B\right) - {\sin}^{2} B \left(1 - {\sin}^{2} A\right)$

...using ${\sin}^{2} \theta + {\cos}^{2} \theta = 1$ (Pythagoras)

$= {\sin}^{2} A - {\sin}^{2} B - {\sin}^{2} A {\sin}^{2} B + {\sin}^{2} B {\sin}^{2} A$

$= {\sin}^{2} A - {\sin}^{2} B$