Question

If $sin\,x+sin\,y=3\left(cos\,y-cos\,x\right)$, then the value of $\frac{sin\,3x}{sin\,3y}$ 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).
$\frac{\sin \,3 x}{\sin \,3 y}=\frac{-\sin\, 3 y}{\sin \,3 y}=1$