(1) 次の三角関数を鋭角の三角関数で表す。 1. $\cos \frac{17}{5}\pi$ 2. $\cos \frac{5}{7}\pi$ 3. $\sin (-\frac{7}{8}\pi)$ 4. $\sin (-\frac{19}{8}\pi)$ (2) 次の値を求める。 1. $\tan (-\frac{\pi}{6})$ 2. $\tan (-\frac{\pi}{4})$

1. 問題の内容

(1) 次の三角関数を鋭角の三角関数で表す。

1. $\cos \frac{17}{5}\pi$

2. $\cos \frac{5}{7}\pi$

3. $\sin (-\frac{7}{8}\pi)$

4. $\sin (-\frac{19}{8}\pi)$

(2) 次の値を求める。

1. $\tan (-\frac{\pi}{6})$

2. $\tan (-\frac{\pi}{4})$

2. 解き方の手順

(1)

1. $\cos \frac{17}{5}\pi = \cos (\frac{17}{5}\pi - 2\pi \times 1) = \cos (\frac{17}{5}\pi - \frac{10}{5}\pi) = \cos \frac{7}{5}\pi = \cos (\pi + \frac{2}{5}\pi) = -\cos \frac{2}{5}\pi$

2. $\cos \frac{5}{7}\pi = \cos (\pi - \frac{2}{7}\pi) = -\cos \frac{2}{7}\pi$

3. $\sin (-\frac{7}{8}\pi) = -\sin \frac{7}{8}\pi = -\sin (\pi - \frac{1}{8}\pi) = -\sin \frac{1}{8}\pi$

4. $\sin (-\frac{19}{8}\pi) = -\sin \frac{19}{8}\pi = -\sin (2\pi + \frac{3}{8}\pi) = -\sin (\frac{3}{8}\pi) = -\sin (\frac{\pi}{2} + \frac{1}{8}\pi) = -\cos \frac{3}{8}\pi$

$\frac{3}{8}\pi > 0$

(2)

1. $\tan (-\frac{\pi}{6}) = -\tan \frac{\pi}{6} = -\frac{1}{\sqrt{3}} = -\frac{\sqrt{3}}{3}$

2. $\tan (-\frac{\pi}{4}) = -\tan \frac{\pi}{4} = -1$

3. 最終的な答え

(1)

1. $-\cos \frac{2}{5}\pi$

2. $-\cos \frac{2}{7}\pi$

3. $-\sin \frac{1}{8}\pi$

4. $-\cos \frac{3}{8}\pi$

(2)

1. $-\frac{\sqrt{3}}{3}$

2. $-1$