# If $A+B+C=\pi$ then $\sin 2A+\sin 2B+\sin 2C$ is equal to

$\begin{array}{1 1}(a)\;4\sin A\sin B\sin C&(b)\;\sin A\sin B\sin C\\(c)\;2\sin A\sin B\sin C&(d)\;None\;of\;these\end{array}$

We have :-
$\sin 2A+\sin 2B+\sin 2C$
$\Rightarrow (\sin 2A+\sin 2B)+\sin 2C$
$\Rightarrow 2\sin\big(\large\frac{2A+2B}{2}\big)$$\cos\big(\large\frac{2A-2B}{2}\big)$$+\sin 2C$
$\Rightarrow 2\sin(A+B)\cos(A-B)+\sin 2C$
$\Rightarrow 2\sin C\cos(A-B)+2\sin C\cos C$
We have $A+B+C=\pi$
$A+B=\pi-C$
$\Rightarrow 2\sin C[\cos(A-B)+\cos C]$
$C=\pi-(A+B)$
$\Rightarrow 2\sin C[\cos(A-B)-\cos(A+B)]$
$\Rightarrow 2\sin C(2\sin A\sin B)$
$\Rightarrow 4\sin A\sin B\sin C$
Hence (a) is the correct option.
answered Oct 15, 2013