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.

Apply $\sin^2A-\sin^2B=\sin(A+B)\sin(A-B)$
Or

$\sin^2(x+y)-\sin^2(x-y)=\{\sin(x+y)+\sin(x-y)\}\{\sin(x+y)-\sin(x-y)\}$
Now (i)apply $\sin C+\sin D=2\sin\left(\frac {C+D}2\right)\cos\left(\frac {C-D}2\right)$
and $\sin C-\sin D=2\sin\left(\frac {C-D}2\right)\cos\left(\frac {C+D}2\right)$
Or (ii)use $\sin(A\pm B)=\sin A\cos B\pm \cos A\sin B$
