If $A + B + C = 180^{0}$ , then
$sin 2 A + sin 2 B + sin 2 C = 4 sin A sin B sin C$

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Solution

Given:
$A + B + C = π$
To Prove:
$sin 2 A + sin 2 B + sin 2 C = 4 sin A sin B sin C$

We have that:
$A + B = π - C$
$sin (A + B) = sin (π - C) = sin C$
$sin 2 A + sin 2 B = 2 sin (A + B) cos (A - B)$
$sin 2 A + sin 2 B = 2 sin C cos (A - B)$ $(1)$

$cos (A + B) = cos (π - C) = - cos C$
$sin 2 C = 2 sin C cos C$
$sin 2 C = - 2 sin C cos (A + B)$ $(2)$

Adding $(1)$ and $2)$ :
$sin 2 A + sin 2 B + sin 2 C = 2 sin C [cos (A - B) - cos (A + B)]$
$sin 2 A + sin 2 B + sin 2 C = 2 sin C [cos A cos B + sin A sin B - (cos A cos B - sin A sin B)]$
$sin 2 A + sin 2 B + sin 2 C = 2 sin C [2 sin A sin B]$
$sin 2 A + sin 2 B + sin 2 C = 4 sin A sin B sin C$
$∴$ Hence proved

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