Answer
See the explanation below.
Work Step by Step
We use the identities: $2\sin x\cos x=\sin 2x$ and $1-2{{\sin }^{2}}x={{\cos }^{2}}x-{{\sin }^{2}}x$.
Now, the left side can be written as:
$\begin{align}
& 1-8{{\sin }^{2}}x{{\cos }^{2}}x=1-(2\times 2\sin x\cos x\times 2\sin x\cos x) \\
& =1-2\sin 2x\times \sin 2x \\
& =1-2{{\sin }^{2}}2x \\
& ={{\cos }^{2}}2x-{{\sin }^{2}}2x
\end{align}$
We simplify further and use the identity $\cos \alpha \cos \beta -\sin \alpha \sin \beta =\cos \left( \alpha +\beta \right)$.
$\begin{align}
& {{\cos }^{2}}2x-{{\sin }^{2}}2x=\cos 2x\cos 2x-\sin 2x\sin 2x \\
& =\cos (2x+2x) \\
& =\cos 4x
\end{align}$
Thus, the left side is equal to $\cos 4x$.
