Answer
The value of $8{{\sin }^{2}}x{{\cos }^{2}}x$ is $1-\cos 4x$.
Work Step by Step
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.
$\begin{align}
& 8{{\sin }^{2}}x{{\cos }^{2}}x=2-1\left( \cos 4x \right) \\
& =1-\cos 4x
\end{align}$
