$\theta$が次の値のとき、$\sin\theta$, $\cos\theta$, $\tan\theta$の値を求める。 (1) $\theta = \frac{13}{4}\pi$ (2) $\theta = -\frac{19}{6}\pi$ (3) $\theta = -5\pi$

1. 問題の内容

$\theta$が次の値のとき、$\sin\theta$, $\cos\theta$, $\tan\theta$の値を求める。

(1) $\theta = \frac{13}{4}\pi$

(2) $\theta = -\frac{19}{6}\pi$

(3) $\theta = -5\pi$

2. 解き方の手順

(1) $\theta = \frac{13}{4}\pi$ の場合：

$\frac{13}{4}\pi = \frac{16-3}{4}\pi = 4\pi - \frac{3}{4}\pi = 2(2\pi) - \frac{3}{4}\pi$

なので、$\frac{13}{4}\pi$ は $-\frac{3}{4}\pi$ と同じ位置にある。

$-\frac{3}{4}\pi = -\frac{3}{4}(180^\circ) = -135^\circ$

$\sin\left(\frac{13}{4}\pi\right) = \sin\left(-\frac{3}{4}\pi\right) = -\frac{\sqrt{2}}{2}$

$\cos\left(\frac{13}{4}\pi\right) = \cos\left(-\frac{3}{4}\pi\right) = -\frac{\sqrt{2}}{2}$

$\tan\left(\frac{13}{4}\pi\right) = \tan\left(-\frac{3}{4}\pi\right) = 1$

(2) $\theta = -\frac{19}{6}\pi$ の場合：

$-\frac{19}{6}\pi = -\frac{18+1}{6}\pi = -3\pi - \frac{1}{6}\pi$

$-3\pi = -1\pi - 2\pi$

$-3\pi - \frac{1}{6}\pi = -1\pi - \frac{1}{6}\pi - 2\pi = -\frac{7}{6}\pi - 2\pi$

$-\frac{7}{6}\pi = -180^\circ - 30^\circ = -210^\circ$

$\sin\left(-\frac{19}{6}\pi\right) = \sin\left(-\frac{7}{6}\pi\right) = -\frac{1}{2}$

$\cos\left(-\frac{19}{6}\pi\right) = \cos\left(-\frac{7}{6}\pi\right) = -\frac{\sqrt{3}}{2}$

$\tan\left(-\frac{19}{6}\pi\right) = \tan\left(-\frac{7}{6}\pi\right) = \frac{1}{\sqrt{3}} = \frac{\sqrt{3}}{3}$

(3) $\theta = -5\pi$ の場合：

$-5\pi = -2(2\pi) - \pi$

$-5\pi$ は $-\pi$ と同じ位置にある。

$\sin(-5\pi) = \sin(-\pi) = 0$

$\cos(-5\pi) = \cos(-\pi) = -1$

$\tan(-5\pi) = \tan(-\pi) = 0$

3. 最終的な答え

(1) $\sin\left(\frac{13}{4}\pi\right) = -\frac{\sqrt{2}}{2}$, $\cos\left(\frac{13}{4}\pi\right) = -\frac{\sqrt{2}}{2}$, $\tan\left(\frac{13}{4}\pi\right) = 1$

(2) $\sin\left(-\frac{19}{6}\pi\right) = -\frac{1}{2}$, $\cos\left(-\frac{19}{6}\pi\right) = -\frac{\sqrt{3}}{2}$, $\tan\left(-\frac{19}{6}\pi\right) = \frac{\sqrt{3}}{3}$

(3) $\sin(-5\pi) = 0$, $\cos(-5\pi) = -1$, $\tan(-5\pi) = 0$