<s>[INST] 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.
 [/INST] 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$
$= 2 - 2 \cos a \cos b(\cos a\cos b + \sin a \sin b)$
$= 2 (1 - \cos a \cos b\cos(a-b)) = 2(1 - \cos a \cos b\cos c)$
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