Consider a triangle ABC in which `cos A + cos B + cos C = sqrt(3) "sin"(pi)/(3)`
What is the value of `"sin"(A)/(2)"sin"(B)/(2)"sin"(C)/(2)`?

Options

A. `(1)/(2)`
B. `(1)/(4)`
C. `(1)/(8)`
D. `(1)/(16)`

Explanation

To solve the problem, we need to find the value of \( \frac{\sin A}{2} \cdot \frac{\sin B}{2} \cdot \frac{\sin C}{2} \) 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 \( \sin \left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2} \), so 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 use the identity for the sum of cosines in a triangle:
\[ \cos A + \cos B + \cos C = 1 + \frac{r}{R} \]
where \( r \) is the inradius and \( R \) is the circumradius. Setting this equal to \( \frac{3}{2} \):
\[ 1 + \frac{r}{R} = \frac{3}{2} \]
This implies:
\[ \frac{r}{R} = \frac{1}{2} \]

3. **Finding the Value of \( \sin A/2 \cdot \sin B/2 \cdot \sin C/2 \)**:
We can use the relationship between the circumradius \( R \) and the angles of the triangle:
\[ \sin A + \sin B + \sin C = \frac{r}{R} \cdot 2R = 2r \]
Since \( \frac{r}{R} = \frac{1}{2} \), we can use the formula:
\[ \sin A \cdot \sin B \cdot \sin C = \frac{r^3}{R^3} \]
We are interested in \( \sin \frac{A}{2} \cdot \sin \frac{B}{2} \cdot \sin \frac{C}{2} \). Using the half-angle identity:
\[ \sin \frac{A}{2} = \sqrt{\frac{1 - \cos A}{2}}, \quad \sin \frac{B}{2} = \sqrt{\frac{1 - \cos B}{2}}, \quad \sin \frac{C}{2} = \sqrt{\frac{1 - \cos C}{2}} \]
Therefore:
\[ \sin \frac{A}{2} \cdot \sin \frac{B}{2} \cdot \sin \frac{C}{2} = \sqrt{\frac{(1 - \cos A)(1 - \cos B)(1 - \cos C)}{8}} \]

4. **Calculating the Value**:
We know:
\[ 1 - \cos A = 1 - \cos B = 1 - \cos C = 1 - \frac{3}{2} + \frac{r}{R} \]
Using the previous results, we can simplify:
\[ \sin \frac{A}{2} \cdot \sin \frac{B}{2} \cdot \sin \frac{C}{2} = \sqrt{\frac{(1 - \frac{3}{2})(1 - \frac{3}{2})(1 - \frac{3}{2})}{8}} = \sqrt{\frac{(-\frac{1}{2})(-\frac{1}{2})(-\frac{1}{2})}{8}} = \frac{1}{8} \]

5. **Final Result**:
Thus, the value of \( \sin \frac{A}{2} \cdot \sin \frac{B}{2} \cdot \sin \frac{C}{2} \) is:
\[ \frac{1}{8} \]

### Conclusion:
The final answer is:
\[ \boxed{\frac{1}{8}} \]