以下の三角関数の値を求める問題です。 (1) $\sin \frac{16}{3}\pi$ (2) $\cos \frac{7}{2}\pi$ (3) $\tan (-\frac{11}{6}\pi)$

1. 問題の内容

以下の三角関数の値を求める問題です。

(1) $\sin \frac{16}{3}\pi$

(2) $\cos \frac{7}{2}\pi$

(3) $\tan (-\frac{11}{6}\pi)$

2. 解き方の手順

(1) $\sin \frac{16}{3}\pi$

$\frac{16}{3}\pi = 5\pi + \frac{1}{3}\pi = 4\pi + \pi + \frac{1}{3}\pi$

$\sin \frac{16}{3}\pi = \sin (\pi + \frac{1}{3}\pi) = -\sin \frac{1}{3}\pi = -\sin 60^{\circ} = -\frac{\sqrt{3}}{2}$

(2) $\cos \frac{7}{2}\pi$

$\frac{7}{2}\pi = 3\pi + \frac{1}{2}\pi = 2\pi + \pi + \frac{1}{2}\pi$

$\cos \frac{7}{2}\pi = \cos (\pi + \frac{1}{2}\pi) = -\cos \frac{1}{2}\pi = -\cos 90^{\circ} = 0$

(3) $\tan (-\frac{11}{6}\pi)$

$-\frac{11}{6}\pi = -2\pi + \frac{1}{6}\pi$

$\tan (-\frac{11}{6}\pi) = \tan (\frac{1}{6}\pi) = \tan 30^{\circ} = \frac{1}{\sqrt{3}} = \frac{\sqrt{3}}{3}$

3. 最終的な答え

(1) $-\frac{\sqrt{3}}{2}$

(2) $0$

(3) $\frac{\sqrt{3}}{3}$