126

Calculate $\operatorname{sen} 15^\circ$, $\cos 15^\circ$ and $\operatorname{tg} 15^\circ$.

See the resolution

$\operatorname{sen} 15^\circ = \operatorname{sen}\dfrac{\pi}{12} = \dfrac{\ell_{12}}{2}$. Let us calculate $\ell_{12}$: let $Q$ be a vertex of the inscribed regular hexagon and $P$ be the midpoint of the arc on one side of the hexagon, so that $PQ = \ell_{12}$. The radius $OP$ cuts the side of the hexagon at its midpoint $R$, perpendicularly.

Ative o JavaScript para ver esta figura interativa.

In triangle $PQR$, rectangle in $R$:

$$PQ = \ell_{12} \qquad QR = \frac{\ell_6}{2} = \frac{1}{2} \qquad RP = OP - OR = 1 - \frac{\sqrt{3}}{2}$$

(since $OR$ is the height of equilateral triangle beside 1). Then

$$\ell_{12}^2 = \left(\frac{1}{2}\right)^2 + \left(1 - \frac{\sqrt{3}}{2}\right)^2 = \frac{1}{4} + 1 - \sqrt{3} + \frac{3}{4} = 2 - \sqrt{3}$$$$\operatorname{sen}\frac{\pi}{12} = \frac{\ell_{12}}{2} = \frac{\sqrt{2 - \sqrt{3}}}{2} \qquad \cos\frac{\pi}{12} = \sqrt{1 - \frac{2 - \sqrt{3}}{4}} = \frac{\sqrt{2 + \sqrt{3}}}{2} \qquad \operatorname{tg}\frac{\pi}{12} = \frac{\sqrt{2 - \sqrt{3}}}{\sqrt{2 + \sqrt{3}}}.$$

Note: in chapter VI we will see that these values ​​are also written $\operatorname{sen} 15^\circ = \dfrac{\sqrt{6} - \sqrt{2}}{4}$, $\cos 15^\circ = \dfrac{\sqrt{6} + \sqrt{2}}{4}$ and $\operatorname{tg} 15^\circ = 2 - \sqrt{3}$.

127

Calculate $\operatorname{sen}\dfrac{\pi}{8}$, $\cos\dfrac{\pi}{8}$ and $\operatorname{tg}\dfrac{\pi}{8}$.

Ver resposta

$\operatorname{sen}\dfrac{\pi}{8} = \dfrac{\sqrt{2 - \sqrt{2}}}{2}$, $\cos\dfrac{\pi}{8} = \dfrac{\sqrt{2 + \sqrt{2}}}{2}$, $\operatorname{tg}\dfrac{\pi}{8} = \dfrac{\sqrt{2 - \sqrt{2}}}{\sqrt{2 + \sqrt{2}}}$

128

Determine the elements of the set $A = \left\{x = \operatorname{tg}\dfrac{k\pi}{3} \;\middle|\; k \in \mathbb{Z}\right\}$.

See the resolution

Giving values ​​to $k$:

$k$ $0$ $1$ $2$ $3$ $4$ $5$ $6$
$x$ $\operatorname{tg} 0 = 0$ $\operatorname{tg}\frac{\pi}{3} = \sqrt{3}$ $\operatorname{tg}\frac{2\pi}{3} = -\sqrt{3}$ $\operatorname{tg}\pi = 0$ $\operatorname{tg}\frac{4\pi}{3} = \sqrt{3}$ $\operatorname{tg}\frac{5\pi}{3} = -\sqrt{3}$ $\operatorname{tg} 2\pi = 0$ (repeat)

Then $A = \{-\sqrt{3}, 0, \sqrt{3}\}$.

129

Determine $A \cap B$ knowing that $A = \left\{x = \operatorname{sen}\dfrac{k\pi}{6} \;\middle|\; k \in \mathbb{Z}\right\}$ and $B = \left\{x = \cos\dfrac{k\pi}{4} \;\middle|\; k \in \mathbb{Z}\right\}$.

Ver resposta

$A \cap B = \{-1, 0, 1\}$

130

(MAPOFEI-74) Calculate all trigonometric functions of an arc of $930^\circ$.

Ver resposta

$\operatorname{sen} 930^\circ = -\dfrac{1}{2}$, $\cos 930^\circ = -\dfrac{\sqrt{3}}{2}$, $\operatorname{tg} 930^\circ = \dfrac{\sqrt{3}}{3}$, $\operatorname{cotg} 930^\circ = \sqrt{3}$, $\sec 930^\circ = -\dfrac{2\sqrt{3}}{3}$, $\operatorname{cossec} 930^\circ = -2$