How can I prove this trigonometric equation with squares of sines? Here is the equation:
$$\sin^2(a+b)+\sin^2(a-b)=1-\cos(2a)\cos(2b)$$ 
Following from comment help,
$${\left(\sin a \cos b + \cos a \sin b\right)}^2 + {\left(\sin a \cos b - \cos a \sin b\right)}^2$$ 
$$=\sin^2 a \cos^2b + \cos^2 a \sin^2 b + \sin^2 a \cos^2 b + \cos^2 a \sin^2 b$$
I am stuck here, how do I proceed from here?
Edit: from answers I understand how to prove,but how to prove from where I am stuck?

$\sin^2(a+b)+\sin^2(a−b)
$
$=\sin^2a\cos^2b+\cos^2a\sin^2b+\sin^2a\cos^2b+\cos^2a\sin^2b
$
$=2(\sin^2a\cos^2b+\cos^2a\sin^2b)
$
$=2\left ({\left (\dfrac{(1-\cos(2a)}{2}\right )(\cos^2b)+\left (\dfrac{\cos(2a)+1}{2}\right )(\sin^2b)}\right )
$
$=1+\cos(2a)\sin^2b-\cos(2a)\cos^2b
$
$=1-\cos(2a)\cos(2b)
$
Hence proved.
