<s>[INST] $\sin 2a + \sin 2b + \sin 2c - \sin 2(a+b+c) = 4 \sin (a+b) \sin (b+c) \sin (c+a)$ I am clueless about this homework question. Looking at it, I see I could use the compound angle and and sum formulas, and tried using them. Unfortunately, couldn't proceed beyond that. Help?
$$\sin 2a + \sin 2b + \sin 2c - \sin 2(a+b+c) = 4 \sin (a+b) \sin (b+c) \sin (c+a)$$
 [/INST] Thanks to @labbhattacharjee. Here's a step by step version of their solution:
$$\sin 2a + \sin 2b + \sin 2c - \sin 2(a+b+c) = 4 \sin (a+b) \sin (b+c) \sin (c+a)$$
LHS:
$ = 2 \sin (a+b) \cos (a-b) + \sin 2c - \sin((2a+2b)+2c) $
$ = 2 \sin (a+b) \cos (a-b) + 2 \cos (a+b+2c) \sin (-a-b) $
$ = 2 \sin (a+b) \cos (a-b) + 2 \cos (a+b+2c) \sin (-a-b) $ 
$ = 2 \sin (a+b) \cos (a-b) - 2 \cos (a+b+2c) \sin (a+b) $
$ = 2 \sin (a+b) (\cos (a-b) - \cos (a+b+2c)) $
$ = 2 \sin (a+b) (-2 \sin (c+a) \sin (-b-c)) $
$ = 2 \sin (a+b) (2 \sin (c+a) \sin (b+c)) $
$ = 4 \sin (a+b) \sin (b+c) \sin (c+a) $
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