Answer
$cos^4x=\frac{3}{8}+\frac{1}{2}cos~2x+\frac{1}{8}cos~4x$
Work Step by Step
$sin^2x+cos^2x=1$
$cos^2x-1=-sin^2x$
$cos~2x=cos^2x-sin^2x=2~cos^2x-1$
$2~cos^2x=1+cos~2x$
$cos^2x=\frac{1}{2}+\frac{1}{2}cos~2x$
Also: $cos^22x=\frac{1}{2}+\frac{1}{2}cos~4x$
$cos^4x=(cos^2x)^2=(\frac{1}{2}+\frac{1}{2}cos~2x)^2=\frac{1}{4}+2·\frac{1}{2}·\frac{1}{2}cos~2x+\frac{1}{4}cos^22x=\frac{1}{4}+\frac{1}{2}cos~2x+\frac{1}{4}(\frac{1}{2}+\frac{1}{2}cos~4x)=\frac{1}{4}+\frac{1}{2}cos~2x+\frac{1}{8}+\frac{1}{8}cos~4x=\frac{3}{8}+\frac{1}{2}cos~2x+\frac{1}{8}cos~4x$
