Prove the identity: $\sin^2 \left(45^\circ + \alpha\right) - \sin^2 \left(30^\circ - \alpha\right) - \sin 15^\circ \cos \left(15^\circ + 2 \alpha\right) = \sin 2 \alpha$
$\begin{aligned}
\sin^2 A - \sin^2 B & = \left(\sin A + \sin B\right) \left(\sin A - \sin B\right) \\\\
& = 2 \sin \left(\dfrac{A + B}{2}\right) \cos \left(\dfrac{A - B}{2}\right) \\
& \hspace{2cm} \times 2 \sin \left(\dfrac{A - B}{2}\right) \cos \left(\dfrac{A + B}{2}\right) \\\\
& = 2 \sin \left(\dfrac{A + B}{2}\right) \cos \left(\dfrac{A + B}{2}\right) \\
& \hspace{2cm} \times 2 \sin \left(\dfrac{A - B}{2}\right) \cos \left(\dfrac{A - B}{2}\right) \\\\
& = \sin \left(A + B\right) \times \sin \left(A - B\right)
\end{aligned}$
$\begin{aligned}
\therefore \; \sin^2 \left(45^\circ + \alpha\right) - \sin^2 \left(30^\circ - \alpha\right) & = \sin \left(45^\circ + \alpha + 30^\circ - \alpha\right) \\
& \hspace{2cm} \sin \left(45^\circ + \alpha - 30^\circ + \alpha\right) \\\\
& = \sin 75^\circ \sin \left(15^\circ + 2 \alpha\right) \\\\
& = \sin \left(90^\circ - 15^\circ\right) \times \sin \left(15^\circ + 2 \alpha\right) \\\\
& = \cos 15^\circ \sin \left(15^\circ + 2 \alpha\right)
\end{aligned}$
$\begin{aligned}
\therefore \; LHS & = \sin^2 \left(45^\circ + \alpha\right) - \sin^2 \left(30^\circ - \alpha\right) - \sin 15^\circ \cos \left(15^\circ + 2 \alpha\right) \\\\
& = \cos 15^\circ \sin \left(15^\circ + 2 \alpha\right) - \sin 15^\circ \cos \left(15^\circ + 2 \alpha\right) \\\\
& = \sin \left(15^\circ + 2 \alpha - 15^\circ\right) \\\\
& = \sin 2 \alpha = RHS
\end{aligned}$
Hence proved.