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.

Your question involves the basic algebra identity which says, $(a + b)(a - b) = a^2 - b^2 $. For targeting your question, it is easy to assume $ a = \sin A\cos B $ and $b = \cos A \sin B$. The process becomes easy now.
$$\begin{align}(a + b)(a - b)& =& a^2 - b^2\\
 & = & (\sin A \cos B)^2 - (\cos A \sin B)^2\\ & = & \sin^2A\cos^2 B - \cos^2A\sin^2B
 \\ & = & \sin^2A(1 - \sin^2 B) - \cos^2 A\sin^2B 
\end{align}    $$Proceed.
