In a `triangle 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 \( \cos^2 \frac{A}{2} + \cos^2 \frac{B}{2} + \cos^2 \frac{C}{2} \) in triangle \( ABC \), we can follow these steps:

### Step-by-Step Solution:

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

2. **Substitute into the Expression**:
Substitute these identities into the original expression:
\[ \cos^2 \frac{A}{2} + \cos^2 \frac{B}{2} + \cos^2 \frac{C}{2} = \frac{1 + \cos A}{2} + \frac{1 + \cos B}{2} + \frac{1 + \cos C}{2} \]

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

4. **Use the Cosine Sum Identity**:
From the properties of triangles, we know that \( \cos A + \cos B + \cos C = 1 + \frac{r}{R} \), where \( r \) is the inradius and \( R \) is the circumradius of triangle \( ABC \).

5. **Substitute the Cosine Sum**:
Substitute this identity back into the expression:
\[ = \frac{3 + \left(1 + \frac{r}{R}\right)}{2} = \frac{4 + \frac{r}{R}}{2} \]

6. **Simplify the Expression**:
Simplify the expression:
\[ = 2 + \frac{r}{2R} \]

### Final Result:
Thus, we have:
\[ \cos^2 \frac{A}{2} + \cos^2 \frac{B}{2} + \cos^2 \frac{C}{2} = 2 + \frac{r}{2R} \]