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:
\[ \sin x + \sin y = 3(\cos y - \cos x) \]
This can be rewritten as:
\[ \sin x + \sin y = 3\cos y - 3\cos x \]

### Step 2: Rearranging terms
Next, we rearrange the equation to group similar terms:
\[ \sin x + 3\cos x = 3\cos y - \sin y \]

### Step 3: Introduce parameters
We can express the coefficients in terms of \( r \) and \( \theta \):
Let \( r = \sqrt{1^2 + 3^2} = \sqrt{10} \) and \( \tan \theta = \frac{3}{1} \) (so \( \theta = \tan^{-1}(3) \)).
Then, we can write:
\[ \sin x = r \sin \theta \cos x + r \cos \theta \sin x \]
This gives us:
\[ \sqrt{10}(\cos \theta \sin x + \sin \theta \cos x) = \sqrt{10}(\sin \theta \cos y - \cos \theta \sin y) \]

### Step 4: Using sine addition and subtraction formulas
Recognizing the left-hand side as \( \sin(x + \theta) \) and the right-hand side as \( \sin(\theta - y) \):
\[ \sin(x + \theta) = \sin(\theta - y) \]

### Step 5: Setting the angles equal
Since the sine function is periodic, we can equate the angles:
\[ x + \theta = \theta - y \quad \text{or} \quad x + \theta = \pi - (\theta - y) \]
From the first equation:
\[ x + y = 0 \implies x = -y \]

### Step 6: Finding \( \frac{\sin 3x}{\sin 3y} \)
Now, substituting \( x = -y \):
\[ \frac{\sin 3x}{\sin 3y} = \frac{\sin(3(-y))}{\sin(3y)} = \frac{-\sin(3y)}{\sin(3y)} = -1 \]

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