Exercise 22 is asking us to solve a series of problems. First, we need to show that $\sin(5x) = 16\sin^5(x) - 20\sin^3(x) + 5\sin(x)$. Then, we need to solve the equation $\sin(5x) = 0$ for $x \in \mathbb{R}$ and verify that $\frac{\pi}{5}$ and $\frac{2\pi}{5}$ are solutions. Next, we must solve the equation $16X^5 - 20X^3 + 5X = 0$ for $X \in \mathbb{R}$. Finally, we have to deduce the values of $\sin(\frac{\pi}{5})$ and $\sin(\frac{2\pi}{5})$ from the previous questions.

1. Problem Description

Exercise 22 is asking us to solve a series of problems. First, we need to show that $\sin(5x) = 16\sin^5(x) - 20\sin^3(x) + 5\sin(x)$. Then, we need to solve the equation $\sin(5x) = 0$ for $x \in \mathbb{R}$ and verify that $\frac{\pi}{5}$ and $\frac{2\pi}{5}$ are solutions. Next, we must solve the equation $16X^5 - 20X^3 + 5X = 0$ for $X \in \mathbb{R}$. Finally, we have to deduce the values of $\sin(\frac{\pi}{5})$ and $\sin(\frac{2\pi}{5})$ from the previous questions.

2. Solution Steps

1. Showing the identity $\sin(5x) = 16\sin^5(x) - 20\sin^3(x) + 5\sin(x)$.

We know that $\sin(5x) = \sin(4x+x)$. Using the sine addition formula:

$\sin(5x) = \sin(4x)\cos(x) + \cos(4x)\sin(x)$.

We also know that

$\sin(4x) = 2\sin(2x)\cos(2x) = 2(2\sin(x)\cos(x))(\cos^2(x) - \sin^2(x)) = 4\sin(x)\cos(x)(\cos^2(x) - \sin^2(x))$.

$\cos(4x) = \cos^2(2x) - \sin^2(2x) = (\cos^2(x) - \sin^2(x))^2 - (2\sin(x)\cos(x))^2 = \cos^4(x) - 2\sin^2(x)\cos^2(x) + \sin^4(x) - 4\sin^2(x)\cos^2(x) = \cos^4(x) - 6\sin^2(x)\cos^2(x) + \sin^4(x)$.

So,

$\sin(5x) = 4\sin(x)\cos^2(x)(\cos^2(x) - \sin^2(x)) + (\cos^4(x) - 6\sin^2(x)\cos^2(x) + \sin^4(x))\sin(x)$

$\sin(5x) = 4\sin(x)\cos^4(x) - 4\sin^3(x)\cos^2(x) + \sin(x)\cos^4(x) - 6\sin^3(x)\cos^2(x) + \sin^5(x)$

$\sin(5x) = 5\sin(x)\cos^4(x) - 10\sin^3(x)\cos^2(x) + \sin^5(x)$

Since $\cos^2(x) = 1 - \sin^2(x)$:

$\sin(5x) = 5\sin(x)(1 - \sin^2(x))^2 - 10\sin^3(x)(1 - \sin^2(x)) + \sin^5(x)$

$\sin(5x) = 5\sin(x)(1 - 2\sin^2(x) + \sin^4(x)) - 10\sin^3(x) + 10\sin^5(x) + \sin^5(x)$

$\sin(5x) = 5\sin(x) - 10\sin^3(x) + 5\sin^5(x) - 10\sin^3(x) + 10\sin^5(x) + \sin^5(x)$

$\sin(5x) = 16\sin^5(x) - 20\sin^3(x) + 5\sin(x)$.

2. Solving $\sin(5x) = 0$.

$\sin(5x) = 0$ implies $5x = k\pi$ for $k \in \mathbb{Z}$. Therefore, $x = \frac{k\pi}{5}$ for $k \in \mathbb{Z}$.

For $x = \frac{\pi}{5}$, $\sin(5x) = \sin(\pi) = 0$.

For $x = \frac{2\pi}{5}$, $\sin(5x) = \sin(2\pi) = 0$.

So $\frac{\pi}{5}$ and $\frac{2\pi}{5}$ are solutions to $\sin(5x) = 0$.

3. Solving $16X^5 - 20X^3 + 5X = 0$.

We can factor out $X$ to get:

$X(16X^4 - 20X^2 + 5) = 0$.

Therefore, $X = 0$ is one solution.

Let $X = \sin(x)$. From part 1, we have $16\sin^5(x) - 20\sin^3(x) + 5\sin(x) = \sin(5x)$.

So, $16X^4 - 20X^2 + 5 = 0$ can be rewritten as $16\sin^4(x) - 20\sin^2(x) + 5 = 0$.

However, since $16X^5 - 20X^3 + 5X = X(16X^4 - 20X^2 + 5) = 0$, and $\sin(5x) = 16\sin^5(x) - 20\sin^3(x) + 5\sin(x)$,

the roots of $16X^5 - 20X^3 + 5X = 0$ correspond to solutions of $\sin(5x) = 0$.

Thus, $X = \sin(x)$ where $x = \frac{k\pi}{5}$ for $k \in \mathbb{Z}$.

The solutions for $X$ are $X = \sin(0) = 0$, $X = \sin(\frac{\pi}{5})$, $X = \sin(\frac{2\pi}{5})$, $X = \sin(\frac{3\pi}{5}) = \sin(\frac{2\pi}{5})$, $X = \sin(\frac{4\pi}{5}) = \sin(\frac{\pi}{5})$, $X = \sin(\pi) = 0$.

The distinct solutions are $0, \sin(\frac{\pi}{5}), \sin(\frac{2\pi}{5})$.

Therefore the solutions to $16X^5 - 20X^3 + 5X = 0$ are $X = 0, \sin(\frac{\pi}{5}), \sin(\frac{2\pi}{5}), -\sin(\frac{\pi}{5}), -\sin(\frac{2\pi}{5})$.

4. Deducing the values of $\sin(\frac{\pi}{5})$ and $\sin(\frac{2\pi}{5})$.

Let $X = \sin(\frac{\pi}{5})$. Then $X \ne 0$. Since $\sin(5x) = 0$, then $16\sin^5(x) - 20\sin^3(x) + 5\sin(x) = 0$, so $16X^4 - 20X^2 + 5 = 0$.

Let $Y = X^2$. Then $16Y^2 - 20Y + 5 = 0$.

Using the quadratic formula:

$Y = \frac{20 \pm \sqrt{20^2 - 4(16)(5)}}{2(16)} = \frac{20 \pm \sqrt{400 - 320}}{32} = \frac{20 \pm \sqrt{80}}{32} = \frac{20 \pm 4\sqrt{5}}{32} = \frac{5 \pm \sqrt{5}}{8}$.

Since $X = \sin(\frac{\pi}{5}) > 0$ and $\sin^2(x) = Y$, $X = \sqrt{Y}$.

We know that $\sin(\frac{\pi}{5}) < \sin(\frac{\pi}{2}) = 1$. Also, $\sin(\frac{2\pi}{5}) > 0$.

Therefore, $\sin(\frac{\pi}{5}) = \sqrt{\frac{5 - \sqrt{5}}{8}}$ and $\sin(\frac{2\pi}{5}) = \sqrt{\frac{5 + \sqrt{5}}{8}}$.

3. Final Answer

$\sin(\frac{\pi}{5}) = \sqrt{\frac{5 - \sqrt{5}}{8}}$

$\sin(\frac{2\pi}{5}) = \sqrt{\frac{5 + \sqrt{5}}{8}}$