Show that: $\sin ^2 (a+b)+\sin ^2 (a-b)=4(1-\cos ^2 a \cos ^2 b)$ Show that: $$\sin ^2 (a+b)+\sin ^2 (a-b)=4(1-\cos ^2 a \cos ^2 b)$$
I tried two approaches.
Approach 1: $$\sin ^2 (a \pm b)=\sin a\cos b \pm \sin b\cos a$$
This reduced to: $$2(\cos ^2 a +\cos ^2 b -2\cos ^2 a\cos ^2b)$$
I can't see where to go from here.
Approach 2: $$\sin ^2 (a \pm b)=1-\cos ^2 (a \pm b)$$
This reduced to: $$2(\cos ^2 a +\cos ^2 b -2\cos ^2 a\cos ^2b)$$
Both approaches reduced to the same expression which is a good sign, but I still can't figure out how to make the final manipulation.

hint:$$\sin ^2 (a+b)+\sin ^2 (a-b)=4(1-\cos ^2 a \cos ^2 b)=\\1-\cos(2a+2b)+1-\cos(2a-2b)=\\2-(\cos(2a+2b)+\cos(2a-2b))= $$use sum to product formula 
$$=2-(\cos(2a+2b)+\cos(2a-2b))=\\2 -2\cos (\frac{2a+2b+2a-2b}{2}).\cos (\frac{2a+2b-(2a-2b)}{2})=\\2-2\cos (2a).\cos (2b)=\\$$
