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.
 A: dont know of an elegant method: 
on LHS substitute $c = a - b$
$LHS = \sin^2 a + \sin^2 b + (\sin a\cos b - \cos a \sin b)^2 $
$= \sin^2 a +  \sin^2 b + \sin^2a \cos^2b + \cos^2a \sin^2b - 2\cos a\cos b \sin a \sin b$
$ = \sin^2 a +  \sin^2 b + (1 - \cos^2a)\cos^2b + \cos^2a(1-\cos^2b)- 2\cos a\cos b \sin a \sin b $
$ = 2 - 2\cos^2a\cos^2b - 2\cos a\cos b \sin a \sin b$
$= 2 - 2 \cos a \cos b(\cos a\cos b + \sin a \sin b)$
$= 2 (1 - \cos a \cos b\cos(a-b)) = 2(1 - \cos a \cos b\cos c)$
A: 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 $a = \pm b \pm c$.
A: Let $\, X := e^{ia}, Y := e^{ib}, Z := e^{ic}. \,$ Now expand and factor
$$ W \!:=\! \sin^2 a \!+\! \sin^2 b \!+\! \sin^2 c \!-\! 2(1 \!-\! \cos a \cos b \cos c) \!=\! 
\frac{(X Y \!-\! Z) (Y Z \!-\! X) (Z X \!-\! Y) (X Y Z \!-\! 1)}{(2 X Y Z)^2}.\,$$
This leads to
  $$\, W = 4 \sin\frac{a+b-c}2 \sin\frac{a-b+c}2 \sin\frac{-a+b+c}2 \sin\frac{a+b+c}2 \,$$
which is zero iff at least one of
 $\, a+b-c,\, a-b+c,\, -a+b+c,\, a+b+c \,$ are multiples of $\,2\pi.\,$
A: 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. 
