If `sinA+sinB=C,cosA+cosB=D`, then the value of `sin(A+B)`=

Options

A. CD
B. `(CD)/(C^(2)+D^(2))`
C. `(C^(2)+D^(2))/(2CD)`
D. `(2CD)/(C^(2)+D^(2))`

Explanation

To solve the problem where we need to find the value of \( \sin(A+B) \) given that \( \sin A + \sin B = C \) and \( \cos A + \cos B = D \), we can follow these steps:

### Step-by-Step Solution:

1. **Use the sum-to-product identities**:
We know that:
\[ \sin A + \sin B = 2 \sin\left(\frac{A+B}{2}\right) \cos\left(\frac{A-B}{2}\right) \]
and
\[ \cos A + \cos B = 2 \cos\left(\frac{A+B}{2}\right) \cos\left(\frac{A-B}{2}\right) \]

2. **Set up the equations**:
From the given conditions, we can write:
\[ 2 \sin\left(\frac{A+B}{2}\right) \cos\left(\frac{A-B}{2}\right) = C \]
\[ 2 \cos\left(\frac{A+B}{2}\right) \cos\left(\frac{A-B}{2}\right) = D \]

3. **Divide the two equations**:
Dividing the first equation by the second gives:
\[ \frac{2 \sin\left(\frac{A+B}{2}\right) \cos\left(\frac{A-B}{2}\right)}{2 \cos\left(\frac{A+B}{2}\right) \cos\left(\frac{A-B}{2}\right)} = \frac{C}{D} \]
This simplifies to:
\[ \tan\left(\frac{A+B}{2}\right) = \frac{C}{D} \]

4. **Use the double angle identity for sine**:
We know that:
\[ \sin(A+B) = 2 \sin\left(\frac{A+B}{2}\right) \cos\left(\frac{A+B}{2}\right) \]
We can express \( \sin\left(\frac{A+B}{2}\right) \) and \( \cos\left(\frac{A+B}{2}\right) \) in terms of \( \tan\left(\frac{A+B}{2}\right) \).

5. **Express sine and cosine in terms of \( C \) and \( D \)**:
Let \( \tan\left(\frac{A+B}{2}\right) = \frac{C}{D} \). Then:
\[ \sin\left(\frac{A+B}{2}\right) = \frac{C}{\sqrt{C^2 + D^2}}, \quad \cos\left(\frac{A+B}{2}\right) = \frac{D}{\sqrt{C^2 + D^2}} \]

6. **Substitute back into the sine formula**:
Now substituting these into the sine double angle formula:
\[ \sin(A+B) = 2 \cdot \frac{C}{\sqrt{C^2 + D^2}} \cdot \frac{D}{\sqrt{C^2 + D^2}} = \frac{2CD}{C^2 + D^2} \]

### Final Answer:
Thus, the value of \( \sin(A+B) \) is:
\[ \sin(A+B) = \frac{2CD}{C^2 + D^2} \]