The problem consists of five sub-problems: a) Place the points A, B, C, D, E, F, G, H, J, and I on the trigonometric circle, corresponding to the real numbers $\frac{\pi}{4}$, $-\frac{\pi}{3}$, $\frac{5\pi}{6}$, $\frac{2\pi}{3}$, $-\frac{3\pi}{4}$, $\frac{\pi}{6}$, $\frac{\pi}{3}$, $\pi$, $\frac{\pi}{2}$, and $0$ respectively. b) Deduce in a table the sine and cosine values of $0$, $\frac{\pi}{6}$, $\frac{\pi}{4}$, $\frac{\pi}{3}$, $\frac{\pi}{2}$, and $\pi$. c) Let $m \in \mathbb{R}$. For which value of $m$ does there exist an angle $x$ such that $\sin x = -2m + 3$? d) Given $\cos(\frac{7\pi}{8}) = -\frac{\sqrt{2+\sqrt{2}}}{2}$, calculate $\cos(\frac{\pi}{8})$ and $\sin(\frac{3\pi}{8})$. e) Simplify the expressions for $A$, $B$, and $C$, where: $A = \cos(-x) + \sin(-x) + \sin(\pi - x) + \cos(\pi - x)$ $B = \sin(x) + \sin(\frac{\pi}{2} - x) + \cos(\frac{\pi}{2} - x) - \cos(x - \frac{\pi}{2})$ $C = \cos(\frac{\pi}{10}) + \cos(\frac{9\pi}{10}) + \cos(\frac{6\pi}{10}) + \cos(\frac{4\pi}{10})$

1. Problem Description

The problem consists of five sub-problems:

a) Place the points A, B, C, D, E, F, G, H, J, and I on the trigonometric circle, corresponding to the real numbers $\frac{\pi}{4}$, $-\frac{\pi}{3}$, $\frac{5\pi}{6}$, $\frac{2\pi}{3}$, $-\frac{3\pi}{4}$, $\frac{\pi}{6}$, $\frac{\pi}{3}$, $\pi$, $\frac{\pi}{2}$, and $0$ respectively.

b) Deduce in a table the sine and cosine values of $0$, $\frac{\pi}{6}$, $\frac{\pi}{4}$, $\frac{\pi}{3}$, $\frac{\pi}{2}$, and $\pi$.

c) Let $m \in \mathbb{R}$. For which value of $m$ does there exist an angle $x$ such that $\sin x = -2m + 3$?

d) Given $\cos(\frac{7\pi}{8}) = -\frac{\sqrt{2+\sqrt{2}}}{2}$, calculate $\cos(\frac{\pi}{8})$ and $\sin(\frac{3\pi}{8})$.

e) Simplify the expressions for $A$, $B$, and $C$, where:

$A = \cos(-x) + \sin(-x) + \sin(\pi - x) + \cos(\pi - x)$

$B = \sin(x) + \sin(\frac{\pi}{2} - x) + \cos(\frac{\pi}{2} - x) - \cos(x - \frac{\pi}{2})$

$C = \cos(\frac{\pi}{10}) + \cos(\frac{9\pi}{10}) + \cos(\frac{6\pi}{10}) + \cos(\frac{4\pi}{10})$

2. Solution Steps

a) Placing the points on the trigonometric circle is a visual task that cannot be performed here. The angles must be converted to degrees to accurately place each point.

b) We can calculate the sine and cosine values for the given angles:

$\cos(0) = 1, \sin(0) = 0$

$\cos(\frac{\pi}{6}) = \frac{\sqrt{3}}{2}, \sin(\frac{\pi}{6}) = \frac{1}{2}$

$\cos(\frac{\pi}{4}) = \frac{\sqrt{2}}{2}, \sin(\frac{\pi}{4}) = \frac{\sqrt{2}}{2}$

$\cos(\frac{\pi}{3}) = \frac{1}{2}, \sin(\frac{\pi}{3}) = \frac{\sqrt{3}}{2}$

$\cos(\frac{\pi}{2}) = 0, \sin(\frac{\pi}{2}) = 1$

$\cos(\pi) = -1, \sin(\pi) = 0$

c) Since $-1 \le \sin x \le 1$, we have $-1 \le -2m + 3 \le 1$.

Subtracting 3 from all sides: $-4 \le -2m \le -2$.

Dividing by -2 and reversing the inequalities: $1 \le m \le 2$.

Thus, for any $m$ in the interval $[1, 2]$, there exists an angle $x$ such that $\sin x = -2m + 3$. The question asks for what value of m will there exist a solution. We need the range where -2m+3 is in [-1,1] or -1<=-2m+3<=1 or -4<=-2m<=-

2. Dividing everything by -2 and switching the signs, 1<=m<=

2. Any m in this interval will have a solution for x. We want one value, and any value in this interval will do. Let's choose the midpoint: $m = \frac{1+2}{2} = \frac{3}{2}$.

d) We are given $\cos(\frac{7\pi}{8}) = -\frac{\sqrt{2+\sqrt{2}}}{2}$.

We want to find $\cos(\frac{\pi}{8})$ and $\sin(\frac{3\pi}{8})$.

Using the half-angle formula, $\cos(\frac{x}{2}) = \pm\sqrt{\frac{1+\cos(x)}{2}}$.

$\cos(\frac{\pi}{8}) = \cos(\frac{\pi/4}{2}) = \sqrt{\frac{1+\cos(\frac{\pi}{4})}{2}} = \sqrt{\frac{1+\frac{\sqrt{2}}{2}}{2}} = \sqrt{\frac{2+\sqrt{2}}{4}} = \frac{\sqrt{2+\sqrt{2}}}{2}$. Since $\frac{\pi}{8}$ is in the first quadrant, the cosine is positive.

$\sin(\frac{3\pi}{8}) = \cos(\frac{\pi}{2} - \frac{3\pi}{8}) = \cos(\frac{4\pi}{8} - \frac{3\pi}{8}) = \cos(\frac{\pi}{8}) = \frac{\sqrt{2+\sqrt{2}}}{2}$.

e) Simplifying the expressions:

$A = \cos(-x) + \sin(-x) + \sin(\pi - x) + \cos(\pi - x) = \cos(x) - \sin(x) + \sin(x) - \cos(x) = 0$.

$B = \sin(x) + \sin(\frac{\pi}{2} - x) + \cos(\frac{\pi}{2} - x) - \cos(x - \frac{\pi}{2}) = \sin(x) + \cos(x) + \sin(x) - \sin(x) = \sin(x) + \cos(x)$.

$C = \cos(\frac{\pi}{10}) + \cos(\frac{9\pi}{10}) + \cos(\frac{6\pi}{10}) + \cos(\frac{4\pi}{10}) = \cos(\frac{\pi}{10}) + \cos(\pi - \frac{\pi}{10}) + \cos(\frac{3\pi}{5}) + \cos(\frac{2\pi}{5}) = \cos(\frac{\pi}{10}) - \cos(\frac{\pi}{10}) + \cos(\frac{3\pi}{5}) + \cos(\frac{2\pi}{5}) = \cos(\frac{3\pi}{5}) + \cos(\frac{2\pi}{5})$.

$C = \cos(\frac{3\pi}{5}) + \cos(\frac{2\pi}{5}) = \cos(108^\circ) + \cos(72^\circ) = -\sin(18^\circ) + \sin(18^\circ + 36^\circ) = -\frac{\sqrt{5}-1}{4} + \frac{\sqrt{5}+1}{4} = \frac{2}{4} = \frac{1}{2}$

3. Final Answer

a) Not possible to provide exact positions without a graphical tool.

b)

| Angle | $0$ | $\frac{\pi}{6}$ | $\frac{\pi}{4}$ | $\frac{\pi}{3}$ | $\frac{\pi}{2}$ | $\pi$ |

|-----------|-----|-----------------|-----------------|-----------------|-----------------|-------|

| Sine | $0$ | $\frac{1}{2}$ | $\frac{\sqrt{2}}{2}$ | $\frac{\sqrt{3}}{2}$ | $1$ | $0$ |

| Cosine | $1$ | $\frac{\sqrt{3}}{2}$ | $\frac{\sqrt{2}}{2}$ | $\frac{1}{2}$ | $0$ | $-1$ |

c) $m = \frac{3}{2}$ (or any value in $[1, 2]$)

d) $\cos(\frac{\pi}{8}) = \frac{\sqrt{2+\sqrt{2}}}{2}$, $\sin(\frac{3\pi}{8}) = \frac{\sqrt{2+\sqrt{2}}}{2}$

e) $A = 0$, $B = \sin(x) + \cos(x)$, $C = \frac{1}{2}$