In a `!ABC cos^(2)A/2+cos^(2)B/2+cos^(2)C/2=`

Options

A. `2-r/R`
B. `2-r/(2R)`
C. `2+r/(2R)`
D. none of these

Answer

To solve the problem, we need to find the value of \( \cos^2 \frac{A}{2} + \cos^2 \frac{B}{2} + \cos^2 \frac{C}{2} \) in terms of the angles \( A, B, C \) of triangle \( ABC \).

### Step-by-Step Solution:

1. **Use the Half-Angle Identity**:
We know that:
\[ \cos^2 \frac{A}{2} = \frac{1 + \cos A}{2} \]
Similarly, we can express \( \cos^2 \frac{B}{2} \) and \( \cos^2 \frac{C}{2} \):
\[ \cos^2 \frac{B}{2} = \frac{1 + \cos B}{2} \]
\[ \cos^2 \frac{C}{2} = \frac{1 + \cos C}{2} \]

2. **Combine the Expressions**:
Now, we can combine these expressions:
\[ \cos^2 \frac{A}{2} + \cos^2 \frac{B}{2} + \cos^2 \frac{C}{2} = \frac{1 + \cos A}{2} + \frac{1 + \cos B}{2} + \frac{1 + \cos C}{2} \]
Simplifying this, we get:
\[ = \frac{3}{2} + \frac{\cos A + \cos B + \cos C}{2} \]

3. **Use the Cosine Sum Formula**:
We know from triangle properties that:
\[ \cos A + \cos B + \cos C = 1 + 4 \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2} \]
Substituting this into our expression gives:
\[ \cos^2 \frac{A}{2} + \cos^2 \frac{B}{2} + \cos^2 \frac{C}{2} = \frac{3}{2} + \frac{1 + 4 \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2}}{2} \]

4. **Final Simplification**:
Now we can simplify further:
\[ = \frac{3}{2} + \frac{1}{2} + 2 \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2} \]
\[ = 2 + 2 \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2} \]

5. **Conclusion**:
Therefore, the final result is:
\[ \cos^2 \frac{A}{2} + \cos^2 \frac{B}{2} + \cos^2 \frac{C}{2} = 2 + 2 \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2} \]