Question

In $\Delta ABC$ prove: $\sin 2A+\sin 2B+\sin 2C=4\sin A\sin B\sin C$.

Answer 1

Given:

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

$\Rightarrow$$A+B+C=180$

Formula:

$1]\sin C+\sin D=2\sin \frac{C+D}{2}\cos \frac{C-D}{2}$ and

$2]sin2A=2sinA.cosA$

$L.H.S.=\sin 2A+\sin 2B+\sin 2C$

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

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

$=2\sin A[\cos A+\cos (B-C\left)\right]$

$=2\sin A\left[\cos \right(B-C)-\cos (B+C\left)\right]$

$[\because \cos A=-\cos (B+C\left)\right]$

$=2\sin A\cdot 2\sin B\sin C$

$=4\sin A\sin B\sin C$.