Question

If $\sin x+\sin y=3\left( \cos y-\cos x \right)$, then the value of $\dfrac{\sin 3x}{\sin 3y}$
A. $1$
B. $-1$
C. $0$
D. None of these

Answer 1

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

Complete step by step answer:
Given that, $\sin x+\sin y=3\left( \cos y-\cos x \right)$
Apply distribution law of multiplication on the left-hand side of the above equation, then we will get
$\sin x+\sin y=3\cos y-3\cos x$
Rearrange the terms in the above equation so that all the $x$ terms are at one place and all the $y$ terms are at one place, then we will have
$\sin x+3\cos x=-\sin y+3\cos y$
Multiply the above equation with a constant $\dfrac{1}{\sqrt{{{1}^{2}}+{{3}^{2}}}}=\dfrac{1}{\sqrt{10}}$, then we will get
$\dfrac{1}{\sqrt{10}}\sin x+\dfrac{3}{\sqrt{10}}\cos x=-\dfrac{1}{\sqrt{10}}\sin y+\dfrac{3}{\sqrt{10}}\cos y$
Let us assume that $\sin \alpha =\dfrac{3}{\sqrt{10}}$, $\cos \alpha =\dfrac{1}{\sqrt{10}}$. Substituting these values in the above equation, then we will have
$\begin{align}
  & \cos \alpha \sin x+\sin \alpha \cos x=-\cos \alpha \sin y+\sin \alpha \cos y \\
 & \Rightarrow \sin x\cos \alpha +\sin \alpha \cos x=\sin \alpha \cos y-\cos \alpha \sin y \\
\end{align}$
Apply the trigonometric formulas $\sin \left( A+B \right)=\sin A\cos B+\sin B\cos A$, $\sin \left( A-B \right)=\sin A\cos B-\cos A\sin B$ in the above equation, then we will get
$\sin \left( x+\alpha \right)=\sin \left( \alpha -y \right)$
Equating on both sides of the above equation, then we will get
$x+\alpha =\alpha -y$
Cancelling the $\alpha $ which is on both sides of the above equation, then we will have
$x=-y$
Now considering the value $\dfrac{\sin 3x}{\sin 3y}$. Substituting $x=-y$ in the above value, then we will get
$\dfrac{\sin 3x}{\sin 3y}=\dfrac{\sin 3\left( -y \right)}{\sin 3y}$
We have the trigonometric formula $\sin \left( -\theta \right)=-\sin \theta $, applying this formula in the above equation, then we will have
$\begin{align}
  & \dfrac{\sin 3x}{\sin 3y}=\dfrac{-\sin 3y}{\sin 3y} \\
 & \therefore \dfrac{\sin 3x}{\sin 3y}=-1 \\
\end{align}$

So, the correct answer is “Option B”.

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