Answer
See below.
Work Step by Step
$sin^4\theta=(sin^2\theta)^2=(\frac{1-cos(2\theta)}{2})^2=\frac{1}{4}-\frac{1}{2}cos(2\theta)+\frac{1}{4}cos^2(2\theta)=\frac{1}{4}-\frac{1}{2}cos(2\theta)+\frac{1}{4}(\frac{1+cos(4\theta)}{2})=\frac{3}{8}-\frac{1}{2}cos(2\theta)+\frac{1}{8}cos(4\theta)$
