1. anonymous

$\frac{ \sin ^{2}x }{ 2-2\cos x }=\frac{ 1 }{ 2 }*\frac{\sin ^{2}x }{1-\cos x }*\frac{1+\cos x }{1+\cos x }$ $=\frac{ 1 }{ 2 }*\frac{ \sin ^{2}x \left( 1+\cos x \right) }{ 1-\cos ^{2}x }$ you can solve and use $1-\cos ^{2}x=\sin ^{2}x and 1+\cos x=2\cos ^{2}\frac{ x }{ 2 }$

2. anonymous

thanks for the help!!! I never thought about splitting at the beginning into two fractions :P thanks!

3. anonymous

yw