Prove the identity: $3 - 4 \cos 2 \alpha + \cos 4 \alpha = 8 \sin^4 \alpha$
$\begin{aligned}
LHS & = 3 - 4 \cos 2 \alpha + \cos 4 \alpha \\\\
& = 3 - 4 \left(1 - 2 \sin^2 \alpha\right) + \cos \left[2 \left(2 \alpha\right)\right] \\\\
& = 3 - 4 + 8 \sin^2 \alpha + 1 - 2 \sin^2 2 \alpha \\\\
& = 8 \sin^2 \alpha - 2 \sin^2 2 \alpha \\\\
& = 8 \sin^2 \alpha - 2 \left(2 \sin \alpha \cos \alpha\right)^2 \\\\
& = 8 \sin^2 \alpha - 2 \times 4 \sin^2 \alpha \cos^2 \alpha \\\\
& = 8 \sin^2 \alpha \left(1 - \cos^2 \alpha\right) \\\\
& = 8 \sin^2 \alpha \times \sin^2 \alpha \\\\
& = 8 \sin^4 \alpha = RHS
\end{aligned}$
Hence proved.