To prove the identity \(\sin^2 \theta \cos^2 \theta \equiv \frac{1}{8}(1 - \cos 4\theta)\), we start by using the double angle identities.

First, recall the double angle identity for cosine:

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

We can express \(\sin^2 \theta \cos^2 \theta\) in terms of \(\cos 2\theta\):

\(\sin^2 \theta \cos^2 \theta = \left(\frac{1}{2} \sin 2\theta\right)^2 = \frac{1}{4} \sin^2 2\theta.\)

Using the identity \(\sin^2 x = \frac{1 - \cos 2x}{2}\), we have:

\(\sin^2 2\theta = \frac{1 - \cos 4\theta}{2}.\)

Substituting back, we get:

\(\sin^2 \theta \cos^2 \theta = \frac{1}{4} \cdot \frac{1 - \cos 4\theta}{2} = \frac{1}{8}(1 - \cos 4\theta).\)

Thus, the identity is proven.