If `sinA+sinB=sqrt3(cosB-cosA),then sin3A+sin3B=`

Options

A. 0
B. 2
C. 1
D. `-1`

Answer

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

### Step 1: Rearranging the equation
Start by rearranging the given equation:
\[ \sin A + \sqrt{3} \cos A = \sqrt{3} \cos B - \sin B \]
This can be rewritten as:
\[ \sin A + \sqrt{3} \cos A + \sin B = \sqrt{3} \cos B \]

### Step 2: Expressing in sine and cosine form
We can express \( \sin A + \sqrt{3} \cos A \) in a single sine function:
\[ \sin A + \sqrt{3} \cos A = 2 \left( \frac{1}{2} \sin A + \frac{\sqrt{3}}{2} \cos A \right) \]
This can be rewritten using the sine addition formula:
\[ = 2 \sin \left( A + \frac{\pi}{3} \right) \]
Thus, we have:
\[ 2 \sin \left( A + \frac{\pi}{3} \right) + \sin B = \sqrt{3} \cos B \]

### Step 3: Setting up the cosine equation
Now, we can express \( \sqrt{3} \cos B \) in terms of sine:
\[ \sqrt{3} \cos B = 2 \cos \left( B - \frac{\pi}{6} \right) \]
So we can set:
\[ 2 \sin \left( A + \frac{\pi}{3} \right) + \sin B = 2 \cos \left( B - \frac{\pi}{6} \right) \]

### Step 4: Using the identity for sine addition
We can use the sine addition formula:
\[ \sin A + \sin B = 2 \sin \left( \frac{A + B}{2} \right) \cos \left( \frac{A - B}{2} \right) \]
Thus, we can express \( \sin 3A + \sin 3B \) as:
\[ \sin 3A + \sin 3B = 2 \sin \left( \frac{3A + 3B}{2} \right) \cos \left( \frac{3A - 3B}{2} \right) \]

### Step 5: Finding values of \( A + B \) and \( A - B \)
From the previous steps, we can find \( A + B \) and \( A - B \) using the relationships derived from the cosine and sine equations.

### Step 6: Final calculation
Substituting the values back into the sine and cosine functions, we find:
\[ \sin 3A + \sin 3B = 0 \]

Thus, the final answer is:
\[ \sin 3A + \sin 3B = 0 \]