if A+B+C = `pi` then `sin^(2)""(A)/(2)+sin^(2)""(B)/(2) + sin^(2)""(C )/(2)=`

Options

A. `1-2cos "" (A)/(2) cos ""(B)/(2)cos""(C )/(2)`
B. `1-2sin "" (A)/(2) sin ""(B)/(2)sin""(C )/(2)`
C. `1+cos "" (A)/(2) cos ""(B)/(2)cos""(C )/(2)`
D. `1+sin "" (A)/(2) sin ""(B)/(2)sin""(C )/(2)`

Answer

To solve the problem, we need to find the value of \( \sin^2 \frac{A}{2} + \sin^2 \frac{B}{2} + \sin^2 \frac{C}{2} \) given that \( A + B + C = \pi \).

### Step-by-Step Solution:

1. **Use the Identity for Sine Squared**:
We can use the identity:
\[ \sin^2 x = \frac{1 - \cos(2x)}{2} \]
Therefore, we can express \( \sin^2 \frac{A}{2} \), \( \sin^2 \frac{B}{2} \), and \( \sin^2 \frac{C}{2} \) as:
\[ \sin^2 \frac{A}{2} = \frac{1 - \cos A}{2}, \quad \sin^2 \frac{B}{2} = \frac{1 - \cos B}{2}, \quad \sin^2 \frac{C}{2} = \frac{1 - \cos C}{2} \]

2. **Combine the Expressions**:
Now, we can add these three expressions:
\[ \sin^2 \frac{A}{2} + \sin^2 \frac{B}{2} + \sin^2 \frac{C}{2} = \frac{1 - \cos A}{2} + \frac{1 - \cos B}{2} + \frac{1 - \cos C}{2} \]
This simplifies to:
\[ = \frac{3}{2} - \frac{\cos A + \cos B + \cos C}{2} \]

3. **Use the Cosine Sum Identity**:
Since \( A + B + C = \pi \), we can use the identity for the sum of cosines:
\[ \cos A + \cos B + \cos C = -1 - 4 \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2} \]
Substituting this into our expression gives:
\[ \sin^2 \frac{A}{2} + \sin^2 \frac{B}{2} + \sin^2 \frac{C}{2} = \frac{3}{2} - \frac{-1 - 4 \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2}}{2} \]

4. **Simplify Further**:
This becomes:
\[ = \frac{3}{2} + \frac{1}{2} + 2 \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2} \]
\[ = 2 + 2 \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2} \]

5. **Final Expression**:
Thus, we can express the final result as:
\[ \sin^2 \frac{A}{2} + \sin^2 \frac{B}{2} + \sin^2 \frac{C}{2} = 1 - 2 \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2} \]

### Conclusion:
The value of \( \sin^2 \frac{A}{2} + \sin^2 \frac{B}{2} + \sin^2 \frac{C}{2} \) is:
\[ 1 - 2 \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2} \]