$$\cos (A + B)\cos (A - B) = {\cos ^2}A - {\sin ^2}B$$
I have attempted this question by expanding the left side using the cosine sum and difference formulas and then multiplying, and then simplifying till I replicated the identity on the right. I am not stuck. What's bothering me is that the way I went about this question seemed like a rather "clunky" method. I'm just curious if I've missed some underlying pattern that could have made it easier to reproduce the identity on the right.
The way I did it:
$$\begin{array}{l} \cos (A + B)\cos (A - B)\\ \equiv (\cos A\cos B - \sin A\sin B)(\cos A\cos B + \sin A\sin B)\\ \equiv {\cos ^2}A{\cos ^2}B - {\sin ^2}A{\sin ^2}B\\ \equiv {\cos ^2}A(1 - {\sin ^2}B) - (1 - {\cos ^2}A){\sin ^2}B\\ \equiv {\cos ^2}A - {\cos ^2}A{\sin ^2}B - {\sin ^2}B + {\cos ^2}A{\sin ^2}B\\ \equiv {\cos ^2}A - {\sin ^2}B\end{array}$$