The problem consists of several exercises. Exercise 5 asks us to consider two functions, $f(x) = 2\cos(x - \frac{\pi}{6}) - \sin x$ and $g(x) = 2\sin(x + \frac{\pi}{4}) - \sqrt{2} \sin x$. We need to show that there exists a constant $K$ such that $f(x) = K \cdot g(x)$ for all $x$. Then, using $x = \frac{\pi}{6}$, we need to calculate $\sin(\frac{5\pi}{12})$.

1. Problem Description

The problem consists of several exercises. Exercise 5 asks us to consider two functions, $f(x) = 2\cos(x - \frac{\pi}{6}) - \sin x$ and $g(x) = 2\sin(x + \frac{\pi}{4}) - \sqrt{2} \sin x$. We need to show that there exists a constant $K$ such that $f(x) = K \cdot g(x)$ for all $x$. Then, using $x = \frac{\pi}{6}$, we need to calculate $\sin(\frac{5\pi}{12})$.

2. Solution Steps

First, let's find the value of K such that $f(x) = K \cdot g(x)$.

Let's set $x = 0$. Then we have:

$f(0) = 2\cos(-\frac{\pi}{6}) - \sin(0) = 2\cos(\frac{\pi}{6}) - 0 = 2 \cdot \frac{\sqrt{3}}{2} = \sqrt{3}$

$g(0) = 2\sin(\frac{\pi}{4}) - \sqrt{2} \sin(0) = 2 \cdot \frac{\sqrt{2}}{2} - 0 = \sqrt{2}$

Then, $f(0) = K \cdot g(0)$, so $\sqrt{3} = K \cdot \sqrt{2}$, which implies $K = \frac{\sqrt{3}}{\sqrt{2}} = \frac{\sqrt{6}}{2}$.

Now, using $x = \frac{\pi}{6}$, we have:

$f(\frac{\pi}{6}) = 2\cos(\frac{\pi}{6} - \frac{\pi}{6}) - \sin(\frac{\pi}{6}) = 2\cos(0) - \sin(\frac{\pi}{6}) = 2 \cdot 1 - \frac{1}{2} = \frac{3}{2}$

$g(\frac{\pi}{6}) = 2\sin(\frac{\pi}{6} + \frac{\pi}{4}) - \sqrt{2} \sin(\frac{\pi}{6}) = 2\sin(\frac{2\pi + 3\pi}{12}) - \sqrt{2} \cdot \frac{1}{2} = 2\sin(\frac{5\pi}{12}) - \frac{\sqrt{2}}{2}$

We have $f(\frac{\pi}{6}) = K \cdot g(\frac{\pi}{6})$. Substituting the values, we get:

$\frac{3}{2} = \frac{\sqrt{6}}{2} \cdot (2\sin(\frac{5\pi}{12}) - \frac{\sqrt{2}}{2})$

$\frac{3}{2} = \sqrt{6} \sin(\frac{5\pi}{12}) - \frac{\sqrt{12}}{4}$

$\frac{3}{2} = \sqrt{6} \sin(\frac{5\pi}{12}) - \frac{2\sqrt{3}}{4} = \sqrt{6} \sin(\frac{5\pi}{12}) - \frac{\sqrt{3}}{2}$

$\sqrt{6} \sin(\frac{5\pi}{12}) = \frac{3}{2} + \frac{\sqrt{3}}{2} = \frac{3+\sqrt{3}}{2}$

$\sin(\frac{5\pi}{12}) = \frac{3+\sqrt{3}}{2\sqrt{6}} = \frac{3+\sqrt{3}}{2\sqrt{6}} \cdot \frac{\sqrt{6}}{\sqrt{6}} = \frac{3\sqrt{6}+3\sqrt{2}}{12} = \frac{\sqrt{6}+\sqrt{2}}{4}$

Alternatively, $\sin(\frac{5\pi}{12}) = \sin(\frac{2\pi}{12}+\frac{3\pi}{12}) = \sin(\frac{\pi}{6}+\frac{\pi}{4}) = \sin(\frac{\pi}{6})\cos(\frac{\pi}{4}) + \cos(\frac{\pi}{6})\sin(\frac{\pi}{4}) = \frac{1}{2} \cdot \frac{\sqrt{2}}{2} + \frac{\sqrt{3}}{2} \cdot \frac{\sqrt{2}}{2} = \frac{\sqrt{2} + \sqrt{6}}{4}$.

3. Final Answer

$\sin(\frac{5\pi}{12}) = \frac{\sqrt{6} + \sqrt{2}}{4}$