Proving $\cos(A + B)\cos(A - B) = -(\sin A + \cos B)(\sin A - \cos B)$. Could you give me some guidance on proving the following trig identity?
$$\cos(A + B)\cos(A - B) = -(\sin A + \cos B)(\sin A - \cos B).$$

Consider $\cos(x\mp y)=\cos x\cos y\pm\sin x\sin y$, add them together, you'll get $$\cos(x+y)+\cos(x-y)=2\cos x\cos y.$$
Now let $x=A+B,\,y=A-B$, so $2\cos(A+B)\cos(A-B)=\cos(2A)+\cos(2B)=2\cos^2A-1+1-2\sin^2B$,
hence we get the desired result as $\cos^2A-\sin^2B=(\cos A+\sin B)(\cos A-\sin B)$.
