If `sin x + sin y = 3 ( cos y - cos x )`, then the value of `sin 3x + sin 3y` is

Options

A. 1
B. `-1`
C. 0
D. none of these

Answer

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

### Step 1: Rearranging the Equation
Starting with the given equation:
\[ \sin x + \sin y = 3 (\cos y - \cos x) \]
We can rewrite this as:
\[ \sin x + \sin y = 3 \cos y - 3 \cos x \]
This gives us:
\[ \sin x + 3 \cos x = 3 \cos y - \sin y \]

### Step 2: Dividing by \(\sqrt{10}\)
Next, we divide both sides by \(\sqrt{10}\):
\[ \frac{1}{\sqrt{10}} \sin x + \frac{3}{\sqrt{10}} \cos x = \frac{3}{\sqrt{10}} \cos y - \frac{1}{\sqrt{10}} \sin y \]

### Step 3: Identifying Sine and Cosine Values
We can let:
\[ \sin \alpha = \frac{3}{\sqrt{10}}, \quad \cos \alpha = \frac{1}{\sqrt{10}} \]
This means we can express the left side as:
\[ \sin x \cos \alpha + \cos x \sin \alpha = \sin(x + \alpha) \]
And the right side becomes:
\[ \sin \alpha \cos y - \cos \alpha \sin y = \sin(\alpha - y) \]

### Step 4: Setting the Sine Functions Equal
Now we have:
\[ \sin(x + \alpha) = \sin(\alpha - y) \]
From the property of sine, we can equate the angles:
\[ x + \alpha = \alpha - y \quad \text{or} \quad x + \alpha = \pi - (\alpha - y) \]

### Step 5: Solving for \(y\)
From the first equation:
\[ x + y = 0 \quad \Rightarrow \quad y = -x \]
From the second equation:
\[ x + y = \pi - 2\alpha \]

### Step 6: Finding \( \sin 3x + \sin 3y \)
Using \( y = -x \):
\[ \sin 3x + \sin 3y = \sin 3x + \sin(-3x) = \sin 3x - \sin 3x = 0 \]

### Final Answer
Thus, the value of \( \sin 3x + \sin 3y \) is:
\[ \boxed{0} \]