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

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

A

1

B

– 1

C

0

D

None of these

Answer

– 1

Explanation

Solution

We have sin⁡x+sin⁡y=3(cos⁡y−cos⁡x)\sin x + \sin y = 3(\cos y - \cos x)

⇒sin⁡x+3cos⁡x=3cos⁡y−sin⁡y\Rightarrow \sin x + 3\cos x = 3\cos y - \sin y …..(i)

⇒rcos⁡(x−α)=rcos⁡(y+α),\Rightarrow r\cos(x - \alpha) = r\cos(y + \alpha),

where r=10,tan⁡α=13r = \sqrt{10},\tan\alpha = \frac{1}{3}

⇒x−α=±(y+α)⇒x=−y\Rightarrow x - \alpha = \pm (y + \alpha) \Rightarrow x = - y or x+y=2αx + y = 2\alpha

Clearly, x=−yx = - ysatisfies (i); ∴ 6musin⁡3xsin⁡3y=−sin⁡3ysin⁡3y=−1\therefore\mspace{6mu}\frac{\sin 3x}{\sin 3y} = \frac{- \sin 3y}{\sin 3y} = - 1.