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Q. $\sin A+\sin B=\sqrt{3}(\cos B-\cos A)$
$\Rightarrow \sin 3 A+\sin 3 B$ is equal to

EAMCETEAMCET 2007

Solution:

Given, $\sin A+\sin B=\sqrt{3}(\cos B-\cos A)$
$\Rightarrow \sin A+\sqrt{3} \cos A=\sqrt{3} \cos B-\sin B$
$\Rightarrow \frac{1}{2} \sin A+\frac{\sqrt{3}}{2} \cos$
$A=\frac{\sqrt{3}}{2} \cos B-\frac{1}{2} \sin B$
$\Rightarrow \cos \frac{\pi}{3} \sin A+\sin \frac{\pi}{3} \cos A$
$=\sin \frac{\pi}{3} \cos B-\cos \frac{\pi}{3} \sin B$
$\Rightarrow \sin \left(A+\frac{\pi}{3}\right)=\sin \left(\frac{\pi}{3}-B\right)$
$\Rightarrow A+\frac{\pi}{3}=\frac{\pi}{3}-B$
$\Rightarrow A=-B$
Now, $\sin 3(A)+\sin 3 B$
$=\sin (-3 B)+\sin 3 B$
$=-\sin 3 B+\sin 3 B$
$=0$