How do I show $\sin^2(x+y)−\sin^2(x−y)≡\sin(2x)\sin(2y)$? I really don't know where I'm going wrong, I use the sum to product formula but always end up far from $\sin(2x)\sin(2y)$. Any help is appreciated, thanks.

Just expand, note that $(a+b)^2-(a-b)^2 = 4 ab$.
Expand $\sin (x+y), \sin (x-y)$ in the usual way. Let $a = \sin x \cos y, b= \cos x \sin y$.
Then $\sin^2(x+y)−\sin^2(x−y)= 4 \sin x \cos y \cos x \sin y$.
Then note that $\sin 2 t = 2 \sin t \cos t$ to finish.
Note that the only trigonometric identity used here is $\sin (s+t) = \sin s \cos t + \sin t \cos s$.
