If $A + B + C = 0$ , (Note $0$ , $\neq π$ ) then prove that $sin 2 A + sin 2 B + sin 2 C = 2 (sin A + sin B + sin C) (1 + cos A + cos B + cos C)$

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Solution

By multiplication $(3 \times 4 = 12$ terms)
R.H.S. $= 2 [\sum sin A 3 + \sum sin A cos A 3 + \sum sin (A + B) 6]$
Now $sin (A + B) = sin (0 - C) = - sin C$
$= 0 + \sum sin 2 A =$ L.H.S.

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