## Precalculus (6th Edition) Blitzer

The value of $8{{\sin }^{2}}x{{\cos }^{2}}x$ is $1-\cos 4x$.
We have to find the value of $8{{\sin }^{2}}x{{\cos }^{2}}x$; various cos and sin formulas are used which represent the answer in the form of cos. \begin{align} & 8{{\sin }^{2}}x{{\cos }^{2}}x=8\left( \frac{1-\cos 2x}{2} \right)\left( \frac{1+\cos 2x}{2} \right) \\ & =\frac{8\left( 1-{{\cos }^{2}}2x \right)}{4} \\ & =\frac{8}{4}-\frac{8\left( {{\cos }^{2}}2x \right)}{4} \\ & =2-2\left( \frac{1+\cos 2\times 2x}{2} \right) \end{align} In step 2, the identity of $\left( a+b \right)\left( a-b \right)={{a}^{2}}+{{b}^{2}}$ is used. \begin{align} & 8{{\sin }^{2}}x{{\cos }^{2}}x=2-1\left( \cos 4x \right) \\ & =1-\cos 4x \end{align}