Prove the identity: $\sin 2 \alpha + \sin 4 \alpha - \sin 6 \alpha = 4 \sin \alpha \sin 2 \alpha \sin 3 \alpha$
$\begin{aligned}
LHS & = \sin 2 \alpha + \sin 4 \alpha - \sin 6 \alpha \\\\
& = \left(\sin 2 \alpha + \sin 4 \alpha\right) - \sin \left(2 \times 3 \alpha\right) \\\\
& = 2 \sin \left(\dfrac{4 \alpha + 2 \alpha}{2}\right) \cos \left(\dfrac{4 \alpha - 2 \alpha}{2}\right) - 2 \sin 3 \alpha \cos 3 \alpha \\\\
& = 2 \sin 3 \alpha \cos \alpha - 2 \sin 3 \alpha \cos 3 \alpha \\\\
& = 2 \sin 3 \alpha \left(\cos \alpha - \cos 3 \alpha\right) \\\\
& = 2 \sin 3 \alpha \times 2 \sin \left(\dfrac{\alpha + 3 \alpha}{2}\right) \sin \left(\dfrac{3 \alpha - \alpha}{2}\right) \\\\
& = 4 \sin 3 \alpha \sin 2 \alpha \sin \alpha = RHS
\end{aligned}$
Hence proved.