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