Prove $ \sin(A+B)\sin(A-B)=\sin^2A-\sin^2B $ How would I verify the following double angle identity.
$$
\sin(A+B)\sin(A-B)=\sin^2A-\sin^2B
$$
So far I have done this.
$$
(\sin A\cos B+\cos A\sin B)(\sin A\cos B-\cos A\sin B) 
$$But I am not sure how to proceed.

\begin{eqnarray}
\sin(A+B)\sin(A-B) &=& (\sin A\cos B+\cos A\sin B)(\sin A\cos B-\cos A\sin B)\\\\
&=& \sin^2 A \cos^2 B -\sin^2 B \cos^2 A\\\\
&=& \sin^2 A \cos^2 B -\sin^2 B (1-\sin^2 A)\\\\
&=& \sin^2 A (\cos^2 B + \sin^2 B) - \sin^2 B\\\\
&=& \sin^2 A - \sin^2 B
\end{eqnarray}
