Prove the identity: $\sin \alpha + 2 \sin 3 \alpha + \sin 5 \alpha = 4 \sin 3 \alpha \cos^2 \alpha$
$\begin{aligned}
LHS & = \sin \alpha + 2 \sin 3 \alpha + \sin 5 \alpha \\\\
& = \left(\sin \alpha + \sin 3 \alpha\right) + \left(\sin 3 \alpha + \sin 5 \alpha\right) \\\\
& = 2 \sin \left(\dfrac{3 \alpha + \alpha}{2}\right) \cos \left(\dfrac{3 \alpha - \alpha}{2}\right) + 2 \sin \left(\dfrac{5 \alpha + 3 \alpha}{2}\right) \cos \left(\dfrac{5 \alpha - 3 \alpha}{2}\right) \\\\
& = 2 \sin 2 \alpha \cos \alpha + 2 \sin 4 \alpha \cos \alpha \\\\
& = 2 \cos \alpha \left(\sin 2 \alpha + \sin 4 \alpha\right) \\\\
& = 2 \cos \alpha \left[2 \sin \left(\dfrac{4 \alpha + 2 \alpha}{2}\right) \cos \left(\dfrac{4 \alpha - 2 \alpha}{2}\right)\right] \\\\
& = 4 \sin 3 \alpha \cos^2 \alpha = RHS
\end{aligned}$
Hence proved.