If sin x + sin y =

Solution:

We have,

$\sin x+\sin y=\sqrt{3}(\cos y-\cos x)$

$\Rightarrow 2 \sin \left(\frac{x+y}{2}\right) \cos \left(\frac{x-y}{2}\right)=2 \sqrt{3} \sin \left(\frac{x+y}{2}\right) \sin \left(\frac{x-y}{2}\right)$

$\Rightarrow 2 \sin \left(\frac{x+y}{2}\right) \cos \left(\frac{x-y}{2}\right)-2 \sqrt{3} \sin \left(\frac{x+y}{2}\right) \sin \left(\frac{x-y}{2}\right)=0$

$\Rightarrow 2 \sin \left(\frac{x+y}{2}\right)\left[\cos \left(\frac{x-y}{2}\right)-\sqrt{3} \sin \frac{x-y}{2}\right]=0$

$\Rightarrow \sin \left(\frac{x+y}{2}\right)\left[\cos \left(\frac{x-y}{2}\right)-\sqrt{3} \sin \frac{x-y}{2}\right]=0$

$\Rightarrow \sin \frac{x+y}{2}=0$ or, $\cos \left(\frac{x-y}{2}\right)-\sqrt{3} \sin \left(\frac{x-y}{2}\right)=0$

$\Rightarrow \frac{x+y}{2}=0 \quad$ or, $\tan \left(\frac{x-y}{2}\right)=\frac{1}{\sqrt{3}}=\tan \frac{\pi}{6}$

$\Rightarrow x=-y \quad$ or, $\frac{x-y}{2}=\frac{\pi}{6}$

$\Rightarrow x=-y \quad$ or, $x-y=\frac{\pi}{3}$

Case - I

When $x=-y$

In this case

$\sin 3 x+\sin 3 y=\sin (-3 y)+\sin 3 y=-\sin 3 y+\sin 3 y=0$

Case - II

when $x-y=\frac{\pi}{3}$

or, $3 x=\pi+3 y$

So, $\sin 3 \mathrm{x}+\sin 3 \mathrm{y}=\sin (\pi+3 y)+\sin 3 y$

$=-\sin 3 y+\sin 3 y$

$=0$