Prove the identity: $\dfrac{\sin^2 4 \alpha}{2 \cos \alpha + \cos 3 \alpha + \cos 5 \alpha} = 2 \sin \alpha \sin 2 \alpha$
$\begin{aligned}
2 \cos \alpha + \cos 3 \alpha + \cos 5 \alpha & = \left(\cos \alpha + \cos 3 \alpha\right) + \left(\cos \alpha + \cos 5 \alpha\right) \\\\
& = 2 \cos \left(\dfrac{3 \alpha + \alpha}{2}\right) \cos \left(\dfrac{3 \alpha - \alpha}{2}\right) \\
& \hspace{2cm} + 2 \cos \left(\dfrac{5 \alpha + \alpha}{2}\right) \cos \left(\dfrac{5 \alpha - \alpha}{2}\right) \\\\
& = 2 \cos 2 \alpha \cos \alpha + 2 \cos 3 \alpha \cos 2 \alpha \\\\
& = 2 \cos 2 \alpha \left(\cos 3 \alpha + \cos \alpha\right) \\\\
& = 2 \cos 2 \alpha \times 2 \cos 2 \alpha \cos \alpha \\\\
& = 4 \cos^2 2 \alpha \cos \alpha
\end{aligned}$
$\begin{aligned}
\sin^2 4 \alpha & = \left(2 \sin 2 \alpha \cos 2 \alpha\right)^2 \\\\
& = 4 \times \left(2 \sin \alpha \cos \alpha\right)^2 \times \cos^2 2 \alpha \\\\
& = 16 \sin^2 \alpha \cos^2 \alpha \cos^2 2 \alpha
\end{aligned}$
$\begin{aligned}
\therefore \; LHS & = \dfrac{\sin^2 4 \alpha}{2 \cos \alpha + \cos 3 \alpha + \cos 5 \alpha} \\\\
& = \dfrac{16 \sin^2 \alpha \cos^2 \alpha \cos^2 2 \alpha}{4 \cos^2 2 \alpha \cos \alpha} \\\\
& = 4 \sin^2 \alpha \cos \alpha \\\\
& = 2 \sin \alpha \times 2 \sin \alpha \cos \alpha \\\\
& = 2 \sin \alpha \sin 2 \alpha = RHS
\end{aligned}$
Hence proved.