1. ## trigonometry

If A+B+C=0 then prove that sin 2A + sin 2B + sin 2C = -4 sinA sinB sinC.

2. ## Re: trigonometry

sin 2A + sin 2B + sin 2C = 2 sin ( 2A + 2B )/2 cos ( 2A-2B)/2 + 2 sin C cos C
= 2 sin ( A + B ) cos ( A - B ) + 2 sin C cos C
= 2 sin ( - C ) cos ( A - B ) + 2 sin C cos C [ A+B+C= 0 ]
= - 2 sin C cos ( A - B ) + 2 sin C cos C
= - 2 sin C [ cos ( A - B ) - Cos C ]
= - 2 sin C [ - 2 sin ( A - B +C )/2 sin ( A - B - C )/2]
= - 4 sin C [ - sin ( - 2B )/2 sin ( 2 A ) / 2 ]
= - 4 sin C [ sin B sin A ]
= - 4 sin A sin B sin C = RHS

3. ## Re: trigonometry

$\displaystyle A+B+C = 0 \implies 2C = -2A -2B$

$\displaystyle \sin 2A + \sin 2B + \sin 2C$

$\displaystyle =\sin 2A + \sin 2B - \sin (2A + 2B)$

$\displaystyle =\sin 2A + \sin 2B - \sin 2A \cos 2B - \sin 2B \cos 2A$

$\displaystyle =\sin 2A (1 - \cos 2B) + \sin 2B (1 - \cos 2A)$

$\displaystyle =(2\sin A \cos A)(2 \sin^2 B) + (2\sin B \cos B)(2 \sin^2 A)$

$\displaystyle =4\sin A \sin B(\cos A\sin B + \cos B \sin A)$

$\displaystyle =4\sin A \sin B\sin (A + B)$

$\displaystyle =4\sin A \sin B\sin (-C)$

$\displaystyle =-4\sin A \sin B\sin C$