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?

Well, let's the start the manipulation at the left hand side. Using the identity
$$\sin^2\theta=\frac{1-\cos 2\theta}{2}$$
we get
$$
\begin{align}
\sin^2(a+b)+\sin^2(a-b)&=\frac{1-\cos(2a+2b)}{2}+\frac{1-\cos(2a-2b)}{2}\\
&=1-\frac{1}{2}\bigg[\cos(2a+2b)+\cos(2a-2b)\bigg]\\
&=1-\frac{1}{2}\bigg[(\cos 2a\cos 2b-\sin 2a\sin 2b)\\
&\qquad\qquad\qquad +(\cos 2a\cos 2b+\sin 2a\sin 2b)\bigg]\\
&=1-\frac{1}{2}\bigg[2\cos 2a\cos 2b\bigg]\\
&=1-\cos 2a\cos 2b.
\end{align}
$$
