To prove the identity \(\cos 4\theta - 4 \cos 2\theta + 3 \equiv 8 \sin^4 \theta\), we start by using trigonometric identities.

First, express \(\cos 4\theta\) and \(\cos 2\theta\) using double angle formulas:

\(\cos 4\theta = 2\cos^2 2\theta - 1\) \(\cos 2\theta = 2\cos^2 \theta - 1\)

Substitute \(\cos 2\theta\) into the expression for \(\cos 4\theta\):

\(\cos 4\theta = 2(2\cos^2 \theta - 1)^2 - 1\)

Expand \((2\cos^2 \theta - 1)^2\):

\((2\cos^2 \theta - 1)^2 = 4\cos^4 \theta - 4\cos^2 \theta + 1\)

Substitute back:

\(\cos 4\theta = 2(4\cos^4 \theta - 4\cos^2 \theta + 1) - 1 = 8\cos^4 \theta - 8\cos^2 \theta + 2 - 1 = 8\cos^4 \theta - 8\cos^2 \theta + 1\)

Now substitute \(\cos 4\theta\) and \(\cos 2\theta\) into the left-hand side of the identity:

\(\cos 4\theta - 4\cos 2\theta + 3 = (8\cos^4 \theta - 8\cos^2 \theta + 1) - 4(2\cos^2 \theta - 1) + 3\)

Simplify:

\(= 8\cos^4 \theta - 8\cos^2 \theta + 1 - 8\cos^2 \theta + 4 + 3\) \(= 8\cos^4 \theta - 16\cos^2 \theta + 8\)

Now express in terms of \(\sin^2 \theta\):

\(\cos^2 \theta = 1 - \sin^2 \theta\)

Substitute:

\(8\cos^4 \theta - 16\cos^2 \theta + 8 = 8(1 - \sin^2 \theta)^2 - 16(1 - \sin^2 \theta) + 8\)

Expand:

\(= 8(1 - 2\sin^2 \theta + \sin^4 \theta) - 16 + 16\sin^2 \theta + 8\) \(= 8 - 16\sin^2 \theta + 8\sin^4 \theta - 16 + 16\sin^2 \theta + 8\) \(= 8\sin^4 \theta\)

Thus, the identity is proven.