Solve
`b cos ^(2) ""(C)/(2) +c cos ^(2) ""(B)/(2)` in terms of k, where k is permeter of the `DeltaABC.`

Answer

To solve the expression \( \frac{b \cos^2 C}{2} + \frac{c \cos^2 B}{2} \) in terms of \( k \), where \( k \) is the perimeter of triangle \( ABC \), we can follow these steps:

### Step-by-Step Solution:

1. **Write the given expression**:
\[ \frac{b \cos^2 C}{2} + \frac{c \cos^2 B}{2} \]

2. **Use the identity for cosine squared**:
Recall that:
\[ \cos^2 C = \frac{1 + \cos 2C}{2} \quad \text{and} \quad \cos^2 B = \frac{1 + \cos 2B}{2} \]
Therefore, we can rewrite the expression as:
\[ \frac{b}{2} \left( \frac{1 + \cos 2C}{2} \right) + \frac{c}{2} \left( \frac{1 + \cos 2B}{2} \right) \]

3. **Simplify the expression**:
This simplifies to:
\[ \frac{b}{4} (1 + \cos 2C) + \frac{c}{4} (1 + \cos 2B) \]
Which can be further simplified to:
\[ \frac{b + c}{4} + \frac{b \cos 2C}{4} + \frac{c \cos 2B}{4} \]

4. **Combine like terms**:
Now we can combine the constant terms:
\[ \frac{b + c}{4} + \frac{1}{4} (b \cos 2C + c \cos 2B) \]

5. **Use the projection formula**:
From the projection formula, we know that:
\[ b \cos C + c \cos B = a \]
Therefore, we can express \( b \cos 2C + c \cos 2B \) in terms of \( a \):
\[ b \cos 2C + c \cos 2B = b \left(2 \cos^2 C - 1\right) + c \left(2 \cos^2 B - 1\right) \]
This leads to:
\[ = 2(b \cos^2 C + c \cos^2 B) - (b + c) \]

6. **Substituting back**:
We substitute back into our expression:
\[ = \frac{b + c}{4} + \frac{1}{4} \left(2(b \cos^2 C + c \cos^2 B) - (b + c)\right) \]

7. **Express in terms of perimeter \( k \)**:
The perimeter \( k \) of triangle \( ABC \) is given by:
\[ k = a + b + c \]
Therefore, we can express \( a + b + c \) in terms of \( k \):
\[ \frac{a + b + c}{2} = \frac{k}{2} \]

8. **Final expression**:
Thus, we conclude that:
\[ \frac{b \cos^2 C}{2} + \frac{c \cos^2 B}{2} = \frac{k}{2} \]

### Final Answer:
\[ \frac{b \cos^2 C}{2} + \frac{c \cos^2 B}{2} = \frac{k}{2} \]