Answer
$\frac{sin^2~x}{2-2~cos~x} = cos^2~\frac{x}{2}$
Work Step by Step
$\frac{sin^2~x}{2-2~cos~x} = \frac{1-cos^2~x}{2(1-cos~x)}$
$\frac{sin^2~x}{2-2~cos~x} = \frac{(1-cos~x)(1+cos~x)}{2(1-cos~x)}$
$\frac{sin^2~x}{2-2~cos~x} = \frac{1+cos~x}{2}$
$\frac{sin^2~x}{2-2~cos~x} = \sqrt{\frac{1+cos~x}{2}}~\sqrt{\frac{1+cos~x}{2}}$
$\frac{sin^2~x}{2-2~cos~x} = (cos~\frac{x}{2})~(cos~\frac{x}{2})$
$\frac{sin^2~x}{2-2~cos~x} = cos^2~\frac{x}{2}$