126

计算$\operatorname{sen} 15^\circ$、$\cos 15^\circ$和$\operatorname{tg} 15^\circ$。

查看分辨率

$\operatorname{sen} 15^\circ = \operatorname{sen}\dfrac{\pi}{12} = \dfrac{\ell_{12}}{2}$。让我们计算$\ell_{12}$：设$Q$为内接正六边形的顶点，$P$为六边形一侧弧的中点，则$PQ = \ell_{12}$。半径 $OP$ 在六边形的中点 $R$ 处垂直切割六边形的边。

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在三角形$PQR$中，矩形在$R$中：

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

（因为 $OR$ 是 等边三角形 与 1 之间的高度）。然后

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

*注意：*在第六章中我们会看到这些值也写为$\operatorname{sen} 15^\circ = \dfrac{\sqrt{6} - \sqrt{2}}{4}$、$\cos 15^\circ = \dfrac{\sqrt{6} + \sqrt{2}}{4}$和$\operatorname{tg} 15^\circ = 2 - \sqrt{3}$。

127

计算$\operatorname{sen}\dfrac{\pi}{8}$、$\cos\dfrac{\pi}{8}$和$\operatorname{tg}\dfrac{\pi}{8}$。

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

确定集合 $A = \left\{x = \operatorname{tg}\dfrac{k\pi}{3} \;\middle|\; k \in \mathbb{Z}\right\}$ 的元素。

查看分辨率

给$k$赋值：

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

然后$A = \{-\sqrt{3}, 0, \sqrt{3}\}$。

129

已知$A = \left\{x = \operatorname{sen}\dfrac{k\pi}{6} \;\middle|\; k \in \mathbb{Z}\right\}$和$B = \left\{x = \cos\dfrac{k\pi}{4} \;\middle|\; k \in \mathbb{Z}\right\}$，确定$A \cap B$。

Ver resposta

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

130

(MAPOFEI-74) 计算$930^\circ$弧的所有三角函数。

Ver resposta

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