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?
 A: On the LHS, you have $$2s_a^2c_b^2+2c_a^2s_b^2$$ after grouping.
On the RHS,
$$1-(c_a^2-s_a^2)(c_b^2-s_b^2)=(c_a^2+s_a^2)(c_b^2+s_b^2)-(c_a^2-s_a^2)(c_b^2-s_b^2)=2s_a^2c_b^2+2c_a^2s_b^2.$$
A: On RHS we have,
$$
1-\cos(2a)\cos(2b) = 1-\cos^2a\cos^2b+\cos^2a\sin^2b+\sin^2a\cos^2b-\sin^2a\sin^2b
= (\cos^2a\sin^2b+\sin^2a\cos^2b)+1-\cos^2a+\sin^b\cos^2a-\sin^2a\sin^2b
= (\cos^2a\sin^2b+\sin^2a\cos^2b)+\sin^2b\cos^2a+\sin^2a-\sin^2a\sin^2b
= 2(\cos^2a\sin^2b+\sin^2a\cos^2b)
= LHS
$$ 
Hence proved
A: We have $$
\begin{align}
\sin^2(a+b)+\sin^2(a-b)&=\frac{1-\cos(2(a+b))}{2}+\frac{1-\cos(2(a-b))}{2}\\
&=1-\frac{\color{red}{\cos(2a+2b)}+\color{blue}{\cos(2a-2b)}}{2}\\
&=1-\frac{\color{red}{\cos 2a\cos 2b-\sin 2a\sin 2b}+ \color{blue}{\cos 2a\cos 2b+\sin 2a\sin 2b}}{2}\\
&=1-\frac{2\cos 2a\cos 2b}{2}\\
&=1-\cos 2a\cos 2b.
\end{align}
$$
A: HINT:
Use Prove that $\cos (A + B)\cos (A - B) = {\cos ^2}A - {\sin ^2}B$ on
$$\sin^2(A+B)+\sin^2(A-B)=1-\{\cos^2(A+B)-\sin^2(A-B)\}$$
A: Other approach: let $u=a+b,v=a-b$.
Then
$$\begin{align}\cos(u+v)\cos(u-v)&=(\cos u\cos v-\sin u\sin v)\cdot(\cos u\cos v+\sin u\sin v)\\
&=\cos^2u\cos^2v-\sin^2u\sin^2v\\
&=(1-\sin^2u)(1-\sin^2v)-\sin^2u\sin^2v\\
&=1-\sin^2u-\sin^2v.\end{align}$$
A: Given that
$$ \cos(u) = \frac{e^{iu} + e^{-iu}}{2} , \sin(u) = \frac{e^{iu} - e^{-iu}}{2i} \text{ , and } e^{2u}+ e^{-2u} = \left(e^{u} - e^{-u}\right)^2 + 2,$$
we prove the identity by
$$\begin{align} 1 - \cos(2a) \cos(2b) & = 1 - \frac{ \left( e^{2ia} + e^{-2ia} \right) \left(e^{2ib} + e^{-2ib} \right)}{4}\\
& = \frac{e^{2i(a + b)} + e^{-2i(a + b)} + e^{2i(a - b)} + e^{-2i(a - b)} - 4}{-4}\\
& = \frac{ \left( e^{i(a + b)} - e^{-i(a + b)} \right)^2  + 2 + \left( e^{i(a - b)} - e^{-i(a - b)} \right)^2 + 2 - 4}{4i^2} \\
& = \sin^2(a+b) + \sin^2(a - b)\end{align} $$
A: 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
$$
\begin{align}
\sin^2(a+b)+\sin^2(a-b)&=\frac{1-\cos(2a+2b)}{2}+\frac{1-\cos(2a-2b)}{2}\\
&=1-\frac{1}{2}\bigg[\cos(2a+2b)+\cos(2a-2b)\bigg]\\
&=1-\frac{1}{2}\bigg[(\cos 2a\cos 2b-\sin 2a\sin 2b)\\
&\qquad\qquad\qquad +(\cos 2a\cos 2b+\sin 2a\sin 2b)\bigg]\\
&=1-\frac{1}{2}\bigg[2\cos 2a\cos 2b\bigg]\\
&=1-\cos 2a\cos 2b.
\end{align}
$$
A: $\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.
