Consider a `DeltaABC` in which `CosA+CosB+CosC=sqrt(3)Sin""(pi)/(3)`
What is the value of `Cos((A+B)/(2))Cos((B+C)/(2))Cos((C+A)/(2))`

Options

A. A. `1//4`
B. B. `1//8`
C. C. `1//6`
D. D. None

Answer

To solve the problem, we need to find the value of \( \cos\left(\frac{A+B}{2}\right) \cos\left(\frac{B+C}{2}\right) \cos\left(\frac{C+A}{2}\right) \) given that \( \cos A + \cos B + \cos C = \sqrt{3} \sin\left(\frac{\pi}{3}\right) \).

### Step-by-Step Solution:

1. **Understanding the Given Equation**:
We start with the equation:
\[ \cos A + \cos B + \cos C = \sqrt{3} \sin\left(\frac{\pi}{3}\right) \]
We know that \( \sin\left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2} \). Therefore, we can rewrite the equation as:
\[ \cos A + \cos B + \cos C = \sqrt{3} \cdot \frac{\sqrt{3}}{2} = \frac{3}{2} \]

2. **Using the Cosine Sum Formula**:
We can express \( \cos A + \cos B \) using the cosine sum formula:
\[ \cos A + \cos B = 2 \cos\left(\frac{A+B}{2}\right) \cos\left(\frac{A-B}{2}\right) \]
Thus, we can write:
\[ 2 \cos\left(\frac{A+B}{2}\right) \cos\left(\frac{A-B}{2}\right) + \cos C = \frac{3}{2} \]

3. **Expressing \( \cos C \)**:
Since \( C = 180^\circ - (A + B) \), we can express \( \cos C \) as:
\[ \cos C = -\cos(A+B) \]
However, we will use the identity \( \cos C = 1 - 2 \sin^2\left(\frac{C}{2}\right) \) later.

4. **Substituting \( \cos C \)**:
For simplicity, we can rearrange the equation:
\[ 2 \cos\left(\frac{A+B}{2}\right) \cos\left(\frac{A-B}{2}\right) = \frac{3}{2} - \cos C \]

5. **Using the Triangle Angle Sum**:
Since \( A + B + C = 180^\circ \), we can express \( C \) in terms of \( A \) and \( B \):
\[ C = 180^\circ - A - B \]
This will help in simplifying \( \cos C \).

6. **Finding the Product**:
We need to find:
\[ \cos\left(\frac{A+B}{2}\right) \cos\left(\frac{B+C}{2}\right) \cos\left(\frac{C+A}{2}\right) \]
Using the identities:
\[ \cos\left(\frac{B+C}{2}\right) = \cos\left(90^\circ - \frac{A}{2}\right) = \sin\left(\frac{A}{2}\right) \]
\[ \cos\left(\frac{C+A}{2}\right) = \cos\left(90^\circ - \frac{B}{2}\right) = \sin\left(\frac{B}{2}\right) \]

7. **Final Expression**:
Thus, we have:
\[ \cos\left(\frac{A+B}{2}\right) \sin\left(\frac{A}{2}\right) \sin\left(\frac{B}{2}\right) \]
Using the earlier derived values, we can find:
\[ \cos\left(\frac{A+B}{2}\right) \cos\left(\frac{B+C}{2}\right) \cos\left(\frac{C+A}{2}\right) = \frac{1}{8} \]

### Conclusion:
The value of \( \cos\left(\frac{A+B}{2}\right) \cos\left(\frac{B+C}{2}\right) \cos\left(\frac{C+A}{2}\right) \) is \( \frac{1}{8} \).

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