与えられた三角関数の値を求めます。具体的には、以下の値を計算します。 (1) $\sin \frac{10}{3}\pi$ (2) $\cos(-\frac{4}{3}\pi)$ (3) $\tan \frac{13}{4}\pi$ (4) $\sin \frac{17}{18}\pi + \cos \frac{13}{18}\pi + \sin \frac{7}{9}\pi - \sin \frac{\pi}{18}$

## 解答

1. 問題の内容

与えられた三角関数の値を求めます。具体的には、以下の値を計算します。

(1) $\sin \frac{10}{3}\pi$

(2) $\cos(-\frac{4}{3}\pi)$

(3) $\tan \frac{13}{4}\pi$

(4) $\sin \frac{17}{18}\pi + \cos \frac{13}{18}\pi + \sin \frac{7}{9}\pi - \sin \frac{\pi}{18}$

2. 解き方の手順

(1) $\sin \frac{10}{3}\pi$の計算

$\sin \frac{10}{3}\pi = \sin (\frac{4}{3}\pi + 2\pi) = \sin \frac{4}{3}\pi = \sin (\pi + \frac{\pi}{3})$

$\sin (\pi + \frac{\pi}{3}) = -\sin \frac{\pi}{3} = -\frac{\sqrt{3}}{2}$

(2) $\cos(-\frac{4}{3}\pi)$の計算

$\cos(-\frac{4}{3}\pi) = \cos(\frac{4}{3}\pi) = \cos(\pi + \frac{\pi}{3})$

$\cos(\pi + \frac{\pi}{3}) = -\cos \frac{\pi}{3} = -\frac{1}{2}$

(3) $\tan \frac{13}{4}\pi$の計算

$\tan \frac{13}{4}\pi = \tan (\frac{13}{4}\pi - 2\pi) = \tan (\frac{5}{4}\pi) = \tan (\pi + \frac{\pi}{4})$

$\tan (\pi + \frac{\pi}{4}) = \tan \frac{\pi}{4} = 1$

(4) $\sin \frac{17}{18}\pi + \cos \frac{13}{18}\pi + \sin \frac{7}{9}\pi - \sin \frac{\pi}{18}$の計算

$\sin \frac{17}{18}\pi = \sin (\pi - \frac{\pi}{18}) = \sin \frac{\pi}{18}$

$\sin \frac{7}{9}\pi = \sin (\frac{14}{18}\pi) = \sin (\pi - \frac{4}{18}\pi) = \sin (\frac{4}{18}\pi) = \sin (\frac{2}{9}\pi)$

$\cos \frac{13}{18}\pi = \cos (\frac{\pi}{2} + \frac{4}{18}\pi) = -\sin \frac{4}{18}\pi = -\sin \frac{2}{9}\pi$

よって、

$\sin \frac{17}{18}\pi + \cos \frac{13}{18}\pi + \sin \frac{7}{9}\pi - \sin \frac{\pi}{18} = \sin \frac{\pi}{18} - \sin \frac{2}{9}\pi + \sin \frac{2}{9}\pi - \sin \frac{\pi}{18} = 0$

3. 最終的な答え

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

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

(3) $1$

(4) $0$