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

If sin⁡x+sin⁡y=3(cos⁡y−cos⁡x)\sin x+\sin y=3\left( \cos y-\cos x \right), then the value of sin⁡3xsin⁡3y\dfrac{\sin 3x}{\sin 3y}
A. 11
B. −1-1
C. 00
D. None of these

Explanation

Solution

In this problem we need to calculate the value of sin⁡3xsin⁡3y\dfrac{\sin 3x}{\sin 3y} where we have given sin⁡x+sin⁡y=3(cos⁡y−cos⁡x)\sin x+\sin y=3\left( \cos y-\cos x \right). Consider the given equation and rearrange the terms such that all the xx terms are at one place and all the yy terms are at one place. Now multiply the equation with a constant and use the trigonometric formulas sin⁡(A+B)=sin⁡Acos⁡B+sin⁡Bcos⁡A\sin \left( A+B \right)=\sin A\cos B+\sin B\cos A, sin⁡(A−B)=sin⁡Acos⁡B−cos⁡Asin⁡B\sin \left( A-B \right)=\sin A\cos B-\cos A\sin B. On equating the equation, the trigonometric ratio sin⁡\sin is cancelled and we will get only angles. From this equation we can find the relation between the terms xx and yy. Now use the obtained relation to calculate the required value.

Complete step by step answer:
Given that, sin⁡x+sin⁡y=3(cos⁡y−cos⁡x)\sin x+\sin y=3\left( \cos y-\cos x \right)
Apply distribution law of multiplication on the left-hand side of the above equation, then we will get
sin⁡x+sin⁡y=3cos⁡y−3cos⁡x\sin x+\sin y=3\cos y-3\cos x
Rearrange the terms in the above equation so that all the xx terms are at one place and all the yy terms are at one place, then we will have
sin⁡x+3cos⁡x=−sin⁡y+3cos⁡y\sin x+3\cos x=-\sin y+3\cos y
Multiply the above equation with a constant 112+32=110\dfrac{1}{\sqrt{{{1}^{2}}+{{3}^{2}}}}=\dfrac{1}{\sqrt{10}}, then we will get
110sin⁡x+310cos⁡x=−110sin⁡y+310cos⁡y\dfrac{1}{\sqrt{10}}\sin x+\dfrac{3}{\sqrt{10}}\cos x=-\dfrac{1}{\sqrt{10}}\sin y+\dfrac{3}{\sqrt{10}}\cos y
Let us assume that sin⁡α=310\sin \alpha =\dfrac{3}{\sqrt{10}}, cos⁡α=110\cos \alpha =\dfrac{1}{\sqrt{10}}. Substituting these values in the above equation, then we will have
cos⁡αsin⁡x+sin⁡αcos⁡x=−cos⁡αsin⁡y+sin⁡αcos⁡y ⇒sin⁡xcos⁡α+sin⁡αcos⁡x=sin⁡αcos⁡y−cos⁡αsin⁡y \begin{aligned} & \cos \alpha \sin x+\sin \alpha \cos x=-\cos \alpha \sin y+\sin \alpha \cos y \\\ & \Rightarrow \sin x\cos \alpha +\sin \alpha \cos x=\sin \alpha \cos y-\cos \alpha \sin y \\\ \end{aligned}
Apply the trigonometric formulas sin⁡(A+B)=sin⁡Acos⁡B+sin⁡Bcos⁡A\sin \left( A+B \right)=\sin A\cos B+\sin B\cos A, sin⁡(A−B)=sin⁡Acos⁡B−cos⁡Asin⁡B\sin \left( A-B \right)=\sin A\cos B-\cos A\sin B in the above equation, then we will get
sin⁡(x+α)=sin⁡(α−y)\sin \left( x+\alpha \right)=\sin \left( \alpha -y \right)
Equating on both sides of the above equation, then we will get
x+α=α−yx+\alpha =\alpha -y
Cancelling the α\alpha which is on both sides of the above equation, then we will have
x=−yx=-y
Now considering the value sin⁡3xsin⁡3y\dfrac{\sin 3x}{\sin 3y}. Substituting x=−yx=-y in the above value, then we will get
sin⁡3xsin⁡3y=sin⁡3(−y)sin⁡3y\dfrac{\sin 3x}{\sin 3y}=\dfrac{\sin 3\left( -y \right)}{\sin 3y}
We have the trigonometric formula sin⁡(−θ)=−sin⁡θ\sin \left( -\theta \right)=-\sin \theta , applying this formula in the above equation, then we will have
sin⁡3xsin⁡3y=−sin⁡3ysin⁡3y ∴sin⁡3xsin⁡3y=−1 \begin{aligned} & \dfrac{\sin 3x}{\sin 3y}=\dfrac{-\sin 3y}{\sin 3y} \\\ & \therefore \dfrac{\sin 3x}{\sin 3y}=-1 \\\ \end{aligned}

So, the correct answer is “Option B”.

Note: In this problem we have only trigonometric ratio sin⁡\sin , so we have used the formulas sin⁡(A+B)=sin⁡Acos⁡B+sin⁡Bcos⁡A\sin \left( A+B \right)=\sin A\cos B+\sin B\cos A, sin⁡(A−B)=sin⁡Acos⁡B−cos⁡Asin⁡B\sin \left( A-B \right)=\sin A\cos B-\cos A\sin B and sin⁡(−θ)=−sin⁡θ\sin \left( -\theta \right)=-\sin \theta . This type of problems may also come with the trigonometric ratio cos⁡\cos , then we need to use the formulas cos⁡(A+B)=cos⁡Acos⁡B−sin⁡Asin⁡B\cos \left( A+B \right)=\cos A\cos B-\sin A\sin B, cos⁡(A−B)=cos⁡Acos⁡B+sin⁡Asin⁡B\cos \left( A-B \right)=\cos A\cos B+\sin A\sin B, cos⁡(−θ)=cos⁡θ\cos \left( -\theta \right)=\cos \theta .