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

Options

A. `(1)/(4)`
B. `(1)/(2)`
C. `(1)/(16)`
D. None of these

Explanation

To solve the problem, we start with the given equation for triangle ABC:

\[ \cos A + \cos B + \cos C = \sqrt{3} \sin\left(\frac{\pi}{3}\right) \]

Since \(\sin\left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2}\), we can rewrite the equation as:

\[ \cos A + \cos B + \cos C = \sqrt{3} \cdot \frac{\sqrt{3}}{2} = \frac{3}{2} \]

Next, we will use the identity for the sum of cosines in a triangle. The sum of cosines can be expressed in terms of the angles of the triangle:

\[ \cos A + \cos B + \cos C = 1 + \frac{r}{R} \]

where \(r\) is the inradius and \(R\) is the circumradius of triangle ABC. Setting this equal to \(\frac{3}{2}\):

\[ 1 + \frac{r}{R} = \frac{3}{2} \]

This leads to:

\[ \frac{r}{R} = \frac{3}{2} - 1 = \frac{1}{2} \]

From this, we can conclude that:

\[ r = \frac{1}{2} R \]

Next, we need to find the value of:

\[ \cos\left(\frac{A+B}{2}\right) \cos\left(\frac{B+C}{2}\right) \cos\left(\frac{C+A}{2}\right) \]

Using the identity for the cosine of half-angles, we can express each cosine term in terms of the angles of the triangle.

Recall that in a triangle \(A + B + C = \pi\), so:

\[ \frac{A+B}{2} = \frac{\pi - C}{2} = \frac{\pi}{2} - \frac{C}{2} \]

Thus:

\[ \cos\left(\frac{A+B}{2}\right) = \sin\left(\frac{C}{2}\right) \]

Similarly, we find:

\[ \cos\left(\frac{B+C}{2}\right) = \sin\left(\frac{A}{2}\right) \]
\[ \cos\left(\frac{C+A}{2}\right) = \sin\left(\frac{B}{2}\right) \]

Therefore, we can rewrite the expression we need to evaluate as:

\[ \cos\left(\frac{A+B}{2}\right) \cos\left(\frac{B+C}{2}\right) \cos\left(\frac{C+A}{2}\right) = \sin\left(\frac{C}{2}\right) \sin\left(\frac{A}{2}\right) \sin\left(\frac{B}{2}\right) \]

Using the identity for the product of sines, we can relate this to the area of the triangle or the circumradius, but for simplicity, we can use the known result for a triangle where \( \cos A + \cos B + \cos C = \frac{3}{2} \).

In such a case, it is known that:

\[ \sin\left(\frac{A}{2}\right) \sin\left(\frac{B}{2}\right) \sin\left(\frac{C}{2}\right) = \frac{1}{8} \]

Thus, we have:

\[ \cos\left(\frac{A+B}{2}\right) \cos\left(\frac{B+C}{2}\right) \cos\left(\frac{C+A}{2}\right) = \frac{1}{8} \]

So the final answer is:

\[ \cos\left(\frac{A+B}{2}\right) \cos\left(\frac{B+C}{2}\right) \cos\left(\frac{C+A}{2}\right) = \frac{1}{8} \]