Question

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

• A. $1$
• B. $-1$
• C. $0$
• D. None of these

Answer 1

Answer: (B)

$-1$

$sinx+siny=3cosy-3cosx$

$\frac{1}{\sqrt{10}}sinx+\frac{3}{\sqrt{10}}cosx=\frac{-1}{\sqrt{10}}siny+\frac{3}{\sqrt{10}}cosy$

$sin\alpha =3/\sqrt{10}$

$cos\alpha =1/\sqrt{10}$

$sinxcos\alpha +sin\alpha cosx=sin\alpha cosy-cos\alpha siny$

$sin(x+\alpha )=sin(\alpha -y)$

$\Rightarrow x+\alpha =\alpha -y\Rightarrow x=-y$

$\frac{sin3x}{sin3y}=\frac{sin(-3y)}{sin3y}=\frac{-sin3y}{sin3y}$

$=-1$