Answer
$1-8\sin^2x\cos^2x$
Work Step by Step
RECALL:
(1) $\cos{(2A)}=\cos^2A−\sin^2A$
(2) $\sin{(2A)}=2\sin{A}\cos{A}$
With $4x=2(2x)$, using the double-angle identity for cosine above gives:
\begin{align*}
\cos{(4x)}&=\cos^2{(2x)}−\sin^2{(2x)}\\
&=\left(\cos{(2x)}\right)^2-\left(\sin{(2x)}\right)^2
\end{align*}
Using the double-angle identities for cosine and sine gives:
\begin{align*}
\cos(4x)&=\left(\cos{(2x)}\right)^2−\left(\sin{(2x)}\right)^2\\
&=(\cos^2x−\sin^2x)^2−(2\sin{x}\cos{x})^2\\
&=\cos^4{x}-2\cos^2{x}\sin^2{x}+\sin^4{x}-4\sin^2x\cos^2x\\
&=\cos^4{x}+\sin^4{x}-6\sin2x\cos^2x\\
&=\cos^4{x}+\sin^4x + (2\sin^2x\cos^2x-8\sin^2x\cos^2x)\\
&=(\cos^4{x}+2\sin^2x\cos^2x\sin^4x)-8\sin^2x\cos^2x\\
&=(\cos^2x+\sin^2x)^2-8\sin^2x\cos^2x\\
&=1^2-8\sin^2x\cos^2x &\text{(use } sin^2x+\cos^2x=1)\\
&=1-8\sin^2x\cos^2x
\end{align*}
