Question

If $A+B+C=\pi$, then the value of $\sin 2A+\sin 2B+\sin 2C$ is equal to

• A. $4\sin A\sin B\sin C$
• B. $4\sin A\sin B\sin C$
• C. $8\sin A\sin B\sin C$
• D. $2\sin A\sin B\sin C$

Answer 1

The correct option is A $4\sin A\sin B\sin C$

Given : $A+B+C=\pi$

We have to calculate , $\sin 2A+\sin 2B+\sin 2C$

As, $\sin x+\sin y=2\sin \left(\frac{x+y}{2}\right)\cos \left(\frac{x-y}{2}\right)$

$\sin 2A+\sin 2B+\sin 2C$$=2\sin \left(\frac{2A+2B}{2}\right)\cos \left(\frac{2A-2B}{2}\right)+\sin 2C$

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

We have, $A+B+C=\pi \Rightarrow A+B=\pi -C$ and also we know, $\sin 2x=2\sin x\cos x$

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

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

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

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

As, $\cos x-\cos y=2\sin \left(\frac{x+y}{2}\right)\sin \left(\frac{y-x}{2}\right)$

$\Rightarrow \sin 2A+\sin 2B+\sin 2C$$=2\sin C\left[2\sin \right(\frac{(A-B)+(A+B)}{2}\left)\sin \right(\frac{(A+B)-(A-B)}{2}\left)\right]$

$=2\sin C\left[2\sin A\sin B\right]$

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