If `sin x+siny=3(cosy-cosx),`then the value of `(sin3x)/(sin3y),` is

Options

A. 1
B. `-1`
C. 0
D. `+-1`

Answer

To solve the equation \( \sin x + \sin y = 3(\cos y - \cos x) \) and find the value of \( \frac{\sin 3x}{\sin 3y} \), we can follow these steps:

### Step 1: Rewrite the Given Equation
We start with the equation:
\[ \sin x + \sin y = 3(\cos y - \cos x) \]

### Step 2: Rearranging the Equation
Rearranging gives us:
\[ \sin x + \sin y + 3\cos x - 3\cos y = 0 \]

### Step 3: Use the Sine Addition Formula
We can express \( \sin x + \sin y \) using the sine addition formula:
\[ \sin x + \sin y = 2 \sin\left(\frac{x+y}{2}\right) \cos\left(\frac{x-y}{2}\right) \]
And for \( \cos y - \cos x \):
\[ \cos y - \cos x = -2 \sin\left(\frac{y+x}{2}\right) \sin\left(\frac{y-x}{2}\right) \]

### Step 4: Substitute Back into the Equation
Substituting these into our rearranged equation gives:
\[ 2 \sin\left(\frac{x+y}{2}\right) \cos\left(\frac{x-y}{2}\right) + 6 \sin\left(\frac{x+y}{2}\right) \sin\left(\frac{y-x}{2}\right) = 0 \]

### Step 5: Factor Out Common Terms
Factoring out \( \sin\left(\frac{x+y}{2}\right) \):
\[ \sin\left(\frac{x+y}{2}\right) \left( 2 \cos\left(\frac{x-y}{2}\right) + 6 \sin\left(\frac{y-x}{2}\right) \right) = 0 \]

### Step 6: Solve for \( x \) and \( y \)
This gives us two cases:
1. \( \sin\left(\frac{x+y}{2}\right) = 0 \) which implies \( x + y = n\pi \) for some integer \( n \).
2. \( 2 \cos\left(\frac{x-y}{2}\right) + 6 \sin\left(\frac{y-x}{2}\right) = 0 \).

### Step 7: Find Relationship Between \( x \) and \( y \)
From the first case, we can express \( y \) in terms of \( x \):
\[ y = n\pi - x \]

### Step 8: Substitute into \( \frac{\sin 3x}{\sin 3y} \)
We need to find:
\[ \frac{\sin 3x}{\sin 3y} = \frac{\sin 3x}{\sin(3(n\pi - x))} \]
Using the property \( \sin(\pi - \theta) = \sin \theta \):
\[ \sin(3(n\pi - x)) = \sin(3n\pi - 3x) = (-1)^{3n} \sin(3x) \]

### Step 9: Simplify the Expression
Thus, we have:
\[ \frac{\sin 3x}{\sin 3y} = \frac{\sin 3x}{(-1)^{3n} \sin 3x} = -1 \]

### Final Answer
Therefore, the value of \( \frac{\sin 3x}{\sin 3y} \) is:
\[ \boxed{-1} \]