If `sinA+sinB=C,&cosA+cosB=D`, then `sin(A+B)=?`

Options

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

Answer

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

### Step-by-Step Solution:

1. **Square Both Sides**:
Start by squaring both equations:
\[ (\sin A + \sin B)^2 = C^2 \quad \text{and} \quad (\cos A + \cos B)^2 = D^2 \]

2. **Expand the Squares**:
Expand both squared equations:
\[ \sin^2 A + \sin^2 B + 2 \sin A \sin B = C^2 \]
\[ \cos^2 A + \cos^2 B + 2 \cos A \cos B = D^2 \]

3. **Use the Pythagorean Identity**:
Recall the identity \( \sin^2 A + \cos^2 A = 1 \) and \( \sin^2 B + \cos^2 B = 1 \):
\[ (\sin^2 A + \cos^2 A) + (\sin^2 B + \cos^2 B) = 2 \]
This implies:
\[ \sin^2 A + \sin^2 B + \cos^2 A + \cos^2 B = 2 \]

4. **Combine the Equations**:
Adding the two expanded equations:
\[ C^2 + D^2 = (\sin^2 A + \sin^2 B) + (\cos^2 A + \cos^2 B) + 2(\sin A \sin B + \cos A \cos B) \]
Thus, we can write:
\[ C^2 + D^2 = 2 + 2(\sin A \sin B + \cos A \cos B) \]

5. **Simplify**:
Rearranging gives:
\[ C^2 + D^2 - 2 = 2(\sin A \sin B + \cos A \cos B) \]
Therefore:
\[ \sin A \sin B + \cos A \cos B = \frac{C^2 + D^2 - 2}{2} \]

6. **Use the Sine Addition Formula**:
Recall the sine addition formula:
\[ \sin(A + B) = \sin A \cos B + \cos A \sin B \]
We need to express this in terms of \( C \) and \( D \).

7. **Express \( \sin(A + B) \)**:
From the earlier steps, we can express:
\[ \sin(A + B) = \sin A \cos B + \cos A \sin B \]
Using the values of \( C \) and \( D \):
\[ \sin(A + B) = \frac{C D}{2} \]

### Final Result:
Thus, we conclude that:
\[ \sin(A + B) = \frac{C D}{2} \]