Prove the identity: $\sin^2 \left(\alpha + \beta\right) - \sin^2 \alpha - \sin^2 \beta = 2 \sin \alpha \sin \beta \cos \left(\alpha + \beta\right)$
$\begin{aligned}
LHS & = \sin^2 \left(\alpha + \beta\right) - \sin^2 \alpha - \sin^2 \beta \\\\
& = \left(\sin \alpha \cos \beta + \cos \alpha \sin \beta\right)^2 - \sin^2 \alpha - \sin^2 \beta \\\\
& = \sin^2 \alpha \cos^2 \beta + \cos^2 \alpha \sin^2 \beta + 2 \sin \alpha \cos \beta \cos \alpha \sin \beta - \sin^2 \alpha - \sin^2 \beta \\\\
& = \sin^2 \alpha \left(\cos^2 \beta - 1\right) + \sin^2 \beta \left(\cos^2 \alpha - 1\right) + 2 \sin \alpha \cos \beta \cos \alpha \sin \beta \\\\
& = - \sin^2 \alpha \sin^2 \beta - \sin^2 \beta \sin^2 \alpha + 2 \sin \alpha \cos \beta \cos \alpha \sin \beta \\\\
& = 2 \sin \alpha \cos \beta \cos \alpha \sin \beta - 2 \sin^2 \alpha \sin^2 \beta \\\\
& = 2 \sin \alpha \sin \beta \left(\cos \alpha \cos \beta - \sin \alpha \sin \beta\right) \\\\
& = 2 \sin \alpha \sin \beta \cos \left(\alpha + \beta\right) = RHS
\end{aligned}$
Hence proved.