<s>[INST] 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.
 [/INST] I will prove the result by starting with the right hand side of the identity:
$$\begin{align}\sin^2A-\sin^2B&=(\sin A+\sin B )(\sin A-\sin B)\\
&=(2\sin\frac{A+B}{2}\cos\frac{A-B}{2})(2\sin\frac{A-B}{2}\cos\frac{A+B}{2})\\
&=(2\sin\frac{A+B}{2}\cos\frac{A+B}{2})(2\sin\frac{A-B}{2}\cos\frac{A-B}{2})\\
&=\sin(A+B)\sin(A-B)
\end{align}$$
as required,  on using the sum to product formulae in the second line of working and the double angle formulae in  the final line. I hope you found that interesting :)
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