Trigonometric Identities Question 143
Question: If $ \sin x+\sin y=3(\cos y-\cos x), $ then the value of $ \frac{\sin 3x}{\sin 3y} $ is
Options:
A) 1
B) - 1
C) 0
D) None of these
Show Answer
Answer:
Correct Answer: B
Solution:
We have $ \sin x+\sin y=3,(\cos y-\cos x) $
$ \Rightarrow ,\sin x+3\cos x=3\cos y-\sin y $ ?..(i)
$ \Rightarrow ,r\cos ,(x-\alpha )=r\cos ,(y+\alpha ), $ where $ r=\sqrt{10},,\tan \alpha =\frac{1}{3} $
$ \Rightarrow ,x-\alpha =\pm (y+\alpha ),\Rightarrow ,x=-y $ or $ x+y=2\alpha $ Clearly, $ x=-y $ satisfies (i);
$ \therefore \ \frac{\sin ,3x}{\sin ,3y}=\frac{-\sin ,3y}{\sin ,3y}=-1 $ .
BETA
