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.

$$S=\sin ^2 (a+b)+\sin ^2 (a-b)=[\sin (a+b)+\sin (a-b)]^2-2\sin(a+b)\cdot\sin(a-b)$$
Now use that sum-product relations:
$$\sin (a+b)+\sin (a-b)=2\sin (a)\cos (b)$$
$$2\sin(a+b)\cdot\sin(a-b)=\cos(2b)-\cos(2a)=2(\cos^2(b)-\cos^2(a))$$
So,
$$S=4(1-\cos^2(a))\cos^2(b)+2\cos^2(a)-2\cos^2(b)$$
$$S=4(1-\cos^2(a))\cos^2(b)+2\cos^2(a)-2\cos^2(b)$$
$$S=2[\cos^2(a)+\cos^2(b)]-4\cos^2(a)\cos^2(b)$$
and then the statement is wrong.
