Question

If $\sin x+\sin y=\sqrt{3}(\cos y-\cos x)$ then value of $\sin 3x+\sin 3y$ equals

• A. $3$
• B. $\sqrt{3}$
• C. $0$
• D. $1$

Answer 1

The correct option is C $0$

$\sin x+\sin y=\sqrt{3}(\cos y-\cos x)\Rightarrow 2\sin \left(\frac{x+y}{2}\right)\cos \left(\frac{x-y}{2}\right)=2\sqrt{3}\sin \left(\frac{x+y}{2}\right)\sin \left(\frac{x-y}{2}\right)\Rightarrow \tan \left(\frac{x-y}{2}\right)=\frac{1}{\sqrt{3}}$
$\Rightarrow x-y=\frac{\pi }{3}$ ...(1)
$\sin 3x+\sin 3y=2\sin \left[3\left(\frac{x+y}{2}\right)\right]\cos 3\left(\frac{x-y}{2}\right)$
$\Rightarrow \sin 3x+\sin 3y=2\sin \left[3\left(\frac{x+y}{2}\right)\right]\cos \frac{\pi }{2}$ ...{ from (1)}
$\therefore \sin 3x+\sin 3y=0$
Ans: C