The problem asks us to compute the value of $A = \cos(\frac{\pi}{12})\cos(\frac{5\pi}{12}) + \sin(\frac{\pi}{12})\sin(\frac{5\pi}{12})$ and $B = \cos(\frac{5\pi}{12})\cos(\frac{\pi}{12}) - \sin(\frac{5\pi}{12})\sin(\frac{\pi}{12})$. Then, we are asked to deduce that $\sin(\frac{\pi}{12})\sin(\frac{5\pi}{12}) = \frac{1}{4}$.

1. Problem Description

The problem asks us to compute the value of $A = \cos(\frac{\pi}{12})\cos(\frac{5\pi}{12}) + \sin(\frac{\pi}{12})\sin(\frac{5\pi}{12})$ and $B = \cos(\frac{5\pi}{12})\cos(\frac{\pi}{12}) - \sin(\frac{5\pi}{12})\sin(\frac{\pi}{12})$. Then, we are asked to deduce that $\sin(\frac{\pi}{12})\sin(\frac{5\pi}{12}) = \frac{1}{4}$.

2. Solution Steps

First, we calculate A:

Using the trigonometric identity $\cos(a-b) = \cos(a)\cos(b) + \sin(a)\sin(b)$, we have:

$A = \cos(\frac{\pi}{12})\cos(\frac{5\pi}{12}) + \sin(\frac{\pi}{12})\sin(\frac{5\pi}{12}) = \cos(\frac{5\pi}{12} - \frac{\pi}{12}) = \cos(\frac{4\pi}{12}) = \cos(\frac{\pi}{3}) = \frac{1}{2}$.

Next, we calculate B:

Using the trigonometric identity $\cos(a+b) = \cos(a)\cos(b) - \sin(a)\sin(b)$, we have:

$B = \cos(\frac{5\pi}{12})\cos(\frac{\pi}{12}) - \sin(\frac{5\pi}{12})\sin(\frac{\pi}{12}) = \cos(\frac{5\pi}{12} + \frac{\pi}{12}) = \cos(\frac{6\pi}{12}) = \cos(\frac{\pi}{2}) = 0$.

We are given that $\cos(\frac{\pi}{12})\cos(\frac{5\pi}{12}) = \frac{1}{4}$.

From the calculation of $B$, we have

$\cos(\frac{5\pi}{12})\cos(\frac{\pi}{12}) - \sin(\frac{5\pi}{12})\sin(\frac{\pi}{12}) = 0$

$\cos(\frac{5\pi}{12})\cos(\frac{\pi}{12}) = \sin(\frac{5\pi}{12})\sin(\frac{\pi}{12})$

$\sin(\frac{5\pi}{12})\sin(\frac{\pi}{12}) = \frac{1}{4}$.

3. Final Answer

$\sin(\frac{\pi}{12})\sin(\frac{5\pi}{12}) = \frac{1}{4}$