`sin^(2) A + sin ^(2) B +sin^(2) C =`

Options

A. `2+2 cos A cos B cos C`
B. `2+2 sin A sin B sin C`
C. `1+ 2 cos A cos B cos C`
D. `1-2 sin A sin B sin C`

Answer

To solve the equation \( \sin^2 A + \sin^2 B + \sin^2 C \), we can use the identity for sine squared. The identity states:

\[ \sin^2 \theta = \frac{1 - \cos 2\theta}{2} \]

### Step-by-Step Solution:

1. **Apply the identity for each sine squared term:**
\[ \sin^2 A = \frac{1 - \cos 2A}{2}, \quad \sin^2 B = \frac{1 - \cos 2B}{2}, \quad \sin^2 C = \frac{1 - \cos 2C}{2} \]

2. **Substitute these identities into the equation:**
\[ \sin^2 A + \sin^2 B + \sin^2 C = \frac{1 - \cos 2A}{2} + \frac{1 - \cos 2B}{2} + \frac{1 - \cos 2C}{2} \]

3. **Combine the fractions:**
\[ = \frac{(1 - \cos 2A) + (1 - \cos 2B) + (1 - \cos 2C)}{2} \]
\[ = \frac{3 - (\cos 2A + \cos 2B + \cos 2C)}{2} \]

4. **Final expression:**
Thus, we have:
\[ \sin^2 A + \sin^2 B + \sin^2 C = \frac{3 - (\cos 2A + \cos 2B + \cos 2C)}{2} \]