Prove $\sin^2 a + \sin^2 b + \sin^2 c = 2(1-\cos a \cos b \cos c)$ when $a=b+c$ I want to prove this identity:

$$\sin^2 a + \sin^2 b + \sin^2 c = 2(1-\cos a \cos b \cos c) \qquad\text{when}\;a=b+c$$

Can somebody give a hint in the easiest way possible? I am debugging this for hours and can't get the left side to be the right.

If $a = b+ c$ then
$\cos a = \cos b\cos c - \sin b\sin c$
$\sin a = \sin b\cos c + \cos b\sin c$
So solving $\sin^2 a + \sin^2 b + \sin^2 c = 2(1-\cos a \cos b \cos c)$
involves solving
$\sin^2 b + \sin^2 c + \sin^2 b\cos^2 c + 2\cos b\sin b\cos c\sin c + \cos^2 b \sin^2 c = 2(1 - \cos b^2 \cos^2 c + \cos b\cos c\sin b\sin 2)$
$\sin^2 b + \sin^2 c + \sin^2 b\cos^2 c +\cos^2 b \sin^2 c +\cos^2 b \cos^2 c+\cos^2 b \cos^2 c= 2$
$\sin^2 b + \sin^2 c + \cos^2 c(\sin^2 b +\cos^2 b)+\cos^2 b(\sin^2 c + \cos^2 c) = 2$
$\sin^2 b  +\sin^2 c +\cos^2 c + \cos^2 b = 2$
which is always true.
So all $a = b+c$ will solve the equation.
...
So I have to wonder if the problem wasn't supposed to be not that we were told $a = b+c$ but we were suppose to solve the equation to find that $a = b+c$ is a solution.
Well, no point restarting from scratch.  If we *don't know that $a = b+c$ ...
Let $d = b+c$.
So 
$\sin^2 a + \sin^2 b + \sin^2 c = 2(1-\cos a \cos b \cos c)$
$\sin^2 a + \sin^2 d  + \sin^2 b + \sin^2 c = 2(1-\cos a \cos b \cos c) +\sin^2 d$
$\sin^2 a + 2(1 - \cos d\cos b \cos c) = 2(1-\cos a \cos b \cos c) +\sin^2 d$ (from above)
$1 - \cos^2 a - 1 + \cos^2 d = 2\cos b\cos c(\cos d - \cos a)$
$(\cos d - \cos a)(\cos d + \cos a) = 2\cos b\cos c(\cos d - \cos a)$.
So If $\cos d = \cos a$ then $a = \pm d = \pm (b + c)$ are one system of solutions.
Other wise 
$\cos d + \cos a = 2\cos b\cos c$
$\cos b\cos c - \sin b\sin c + \cos a = 2\cos b\cos c$
$\cos a = \cos b\cos c + \sin b\sin c = \cos (b-c)$ so 
$a = \pm (b-c)$.
So (modulo $2\pi$) all solution are $a = \pm b \pm c$.
