Question

If $sinx+siny=3\left(cosy-cosx\right)$, then the value of $\frac{sin3x}{sin3y}$ is

• A. $1$
• B. $-1$
• C. $0$
• D. $3$

Answer 1

Answer: (B)

We have, $\sin x+\sin y=3(\cos y-\cos x)$
$\Rightarrow \sin x+3\cos x=3\cos y-\sin y\dots (i)$
$\Rightarrow r\cos (x-\alpha )=r\cos (y+\alpha )$,
where $r=\sqrt{10}$ and $\tan \alpha =\frac{1}{3}$
$\Rightarrow x-\alpha =\pm (y+\alpha )$
$\Rightarrow x=-y$ or $x-y=2\alpha$
Thus, $x=-y$ satisfies the relation (i).
$\frac{\sin 3x}{\sin 3y}=\frac{-\sin 3y}{\sin 3y}=1$