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.

Ignoring the condition $a=b+c$, we can see that the equation can be manipulated into an interesting form:
$$\begin{align}
\sin^2 a + \sin^2 b + \sin^2 c &= 2 ( 1 - \cos a\cos b\cos c ) \\
( 1 - \cos^2 a ) + ( 1 - \cos^2 b ) + ( 1 - \cos^2 c ) &=
2 - 2 \cos a\cos b\cos c \\
1 - \cos^2 b - \cos^2 c &=
\cos^2 a - 2 \cos a\cos b\cos c \\
\qquad 1- \cos^2 b - \cos^2 c + \cos^2 b\cos^2 c &= \cos^2 a - 2 \cos a \cos b\cos c + \cos^2 b \cos^2 c \\
(1-\cos^2b)(1-\cos^2 c) &= ( \cos a - \cos b \cos c )^2 \\
\sin^2 b\sin^2 c &= ( \cos a - \cos b \cos c )^2 \\
0 &= ( \cos a - \cos b \cos c )^2 -\sin^2 b\sin^2 c \\
0 &= (\cos a - \cos b \cos c + \sin b \sin c) \\
&\phantom{=}\cdot( \cos a - \cos b \cos c - \sin b\sin c) \\
0 &= (\cos a - \cos(b+c))( \cos a - \cos(b-c))
\end{align}$$
Thus, $\cos a = \cos(b\pm c)$, so that

$$a = \pm b \pm c + 2\pi k \quad\text{(with independent $\pm$s)}$$ 

In particular, $a=b+c$ is one of those cases.

It's perhaps worth noting that the original equation reduces even further using the "difference-to-product" identity:
$$\cos\theta - \cos\phi = - 2 \sin\frac{\theta+\phi}{2}\cdot\sin\frac{\theta-\phi}{2}$$
to
$$\sin\frac{a+b+c}{2}\cdot\sin\frac{-a+b+c}{2}\cdot\sin\frac{a-b+c}{2}\cdot\sin\frac{a+b-c}{2} = 0$$
which bears an interesting resemblance to Heron's formula for the area of a triangle. Moreover, had the sign of the $2\cos a\cos b\cos c$ term been reversed, then we'd have instead
$$(\cos a + \cos(b+c))( \cos a + \cos(b-c)) = 0$$
which gives rise to another Heron-like form:
$$\cos\frac{a+b+c}{2}\cdot\cos\frac{-a+b+c}{2}\cdot\cos\frac{a-b+c}{2}\cdot\cos\frac{a+b-c}{2} = 0$$
This relation is satisfied, for instance, when $a+b+c=180^\circ$; that is, when $a$, $b$, $c$ are the angles of a triangle.
I guess my point is that the equation in question isn't entirely random. 
