If `cosA+cosB=m` and `sinA+sinB=n` then `sin(A+B)=`

Options

A. `(mn)/(m^(2)+n^(2))`
B. `(2mn)/(m^(2)+n^(2))`
C. `(m^(2)+n^(2))/(2mn)`
D. `(mn)/(m+n)`

Answer

To solve the problem, we need to find the value of \( \sin(A + B) \) given that \( \cos A + \cos B = m \) and \( \sin A + \sin B = n \).

### Step-by-Step Solution:

1. **Use the sum-to-product identities**:
We know the following identities:
\[ \cos A + \cos B = 2 \cos\left(\frac{A + B}{2}\right) \cos\left(\frac{A - B}{2}\right) \]
\[ \sin A + \sin B = 2 \sin\left(\frac{A + B}{2}\right) \cos\left(\frac{A - B}{2}\right) \]
From the problem, we have:
\[ 2 \cos\left(\frac{A + B}{2}\right) \cos\left(\frac{A - B}{2}\right) = m \quad \text{(1)} \]
\[ 2 \sin\left(\frac{A + B}{2}\right) \cos\left(\frac{A - B}{2}\right) = n \quad \text{(2)} \]

2. **Divide equation (2) by equation (1)**:
\[ \frac{2 \sin\left(\frac{A + B}{2}\right) \cos\left(\frac{A - B}{2}\right)}{2 \cos\left(\frac{A + B}{2}\right) \cos\left(\frac{A - B}{2}\right)} = \frac{n}{m} \]
This simplifies to:
\[ \frac{\sin\left(\frac{A + B}{2}\right)}{\cos\left(\frac{A + B}{2}\right)} = \frac{n}{m} \]
Thus, we have:
\[ \tan\left(\frac{A + B}{2}\right) = \frac{n}{m} \quad \text{(3)} \]

3. **Use the double angle formula 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 \( \tan\left(\frac{A + B}{2}\right) \):
\[ \sin\left(\frac{A + B}{2}\right) = \frac{n}{\sqrt{m^2 + n^2}}, \quad \cos\left(\frac{A + B}{2}\right) = \frac{m}{\sqrt{m^2 + n^2}} \]

4. **Substitute back into the sine double angle formula**:
\[ \sin(A + B) = 2 \cdot \frac{n}{\sqrt{m^2 + n^2}} \cdot \frac{m}{\sqrt{m^2 + n^2}} = \frac{2mn}{m^2 + n^2} \]

### Final Result:
Thus, the value of \( \sin(A + B) \) is:
\[ \sin(A + B) = \frac{2mn}{m^2 + n^2} \]