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.

\begin{align}
\Big(\sin(x+y)\Big)^2 - \Big(\sin(x-y)\Big)^2 & = \Big( \sin x\cos y+\cos x\sin y \Big)^2 - \Big( \sin x\cos y+\cos x\sin y \Big)^2 \\[6pt]
& = \Big( \sin^2\cos^2y + 2\sin x\cos y\cos x\sin y + \cos^2x\sin^2y \Big) \\[6pt]
&\phantom{{}=}  {}- \Big( \sin^2\cos^2y - 2\sin x\cos y\cos x\sin y + \cos^2x\sin^2y \Big) \\[6pt]
& = 4\sin x\cos y\cos x\sin y \\[6pt]
& = (2\sin x\cos x)(2\sin y\cos y) \\[6pt]
& = \sin(2x)\sin(2y).
\end{align}
