Question

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

• A. 0
• B. 2
• C. 1
• D. -1

Answer 1

Answer: (A)

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, $\sin 3(A)+\sin 3B$
$=\sin (-3B)+\sin 3B$
$=-\sin 3B+\sin 3B$
$=0$