Prove the identity: $\sin \alpha - \sin 2 \alpha + \sin 3 \alpha = 4 \cos \left(\dfrac{3 \alpha}{2}\right) \cos \alpha \sin \left(\dfrac{\alpha}{2}\right)$
$\begin{aligned}
LHS & = \sin \alpha - \sin 2 \alpha + \sin 3 \alpha \\\\
& = 2 \sin \left(\dfrac{3 \alpha - 2 \alpha}{2}\right) \cos \left(\dfrac{3 \alpha + 2 \alpha}{2}\right) + \sin \alpha \\\\
& = 2 \sin \left(\dfrac{\alpha}{2}\right) \cos \left(\dfrac{5 \alpha}{2}\right) + 2 \sin \left(\dfrac{\alpha}{2}\right) \cos \left(\dfrac{\alpha}{2}\right) \\\\
& = 2 \sin \left(\dfrac{\alpha}{2}\right) \left[\cos \left(\dfrac{5 \alpha}{2}\right) + \cos \left(\dfrac{\alpha}{2}\right)\right] \\\\
& = 2 \sin \left(\dfrac{\alpha}{2}\right) \times 2 \cos \left(\dfrac{\dfrac{5 \alpha}{2} + \dfrac{\alpha}{2}}{2}\right) \cos \left(\dfrac{\dfrac{5 \alpha}{2} - \dfrac{\alpha}{2}}{2}\right) \\\\
& = 4 \cos \left(\dfrac{6 \alpha}{4}\right) \cos \left(\dfrac{4 \alpha}{4}\right) \sin \left(\dfrac{\alpha}{2}\right) \\\\
& = 4 \cos \left(\dfrac{3 \alpha}{2}\right) \cos \alpha \sin \left(\dfrac{\alpha}{2}\right) = RHS
\end{aligned}$
Hence proved.