# How do you prove sin^2(x)cos^2(x) = (1-cos(4x))/8?

May 14, 2015

Let's start form the Left Hand Side($\text{LHS}$):

$\implies \text{ LHS } = {\sin}^{2} x {\cos}^{2} x = {\left(\frac{2}{2} \sin x \cos x\right)}^{2}$

Remember that, $2 \sin x \cos x = \sin 2 x$

$\implies \text{ LHS } = {\left(\frac{1}{2} \sin 2 x\right)}^{2} = \frac{1}{4} {\sin}^{2} \left(2 x\right)$

Use the half angle identity, $\cos 2 \theta = 1 - 2 {\sin}^{2} \theta$

Replace $\theta$ by $2 x$ $\implies \cos 4 x = 1 - 2 {\sin}^{2} \left(2 x\right)$

Rearrange that $\text{ ; } {\sin}^{2} \left(2 x\right) = \frac{1}{2} \left(1 - \cos 4 x\right)$

$\implies \text{ LHS} = \frac{1}{4} \left(\frac{1}{2} \left(1 - \cos 4 x\right)\right)$

$= \frac{1}{8} \left(1 - \cos 4\right) x = \frac{1 - \cos \left(4 x\right)}{8} = \text{ RHS}$

And that's it!