## Precalculus (6th Edition) Blitzer

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