三角関数の値を求めます。 (1) $\sin \frac{13}{6}\pi$ (2) $\cos \frac{8}{3}\pi$ (3) $\tan \frac{9}{4}\pi$

## 問題 7

1. 問題の内容

三角関数の値を求めます。

(1) $\sin \frac{13}{6}\pi$

(2) $\cos \frac{8}{3}\pi$

(3) $\tan \frac{9}{4}\pi$

2. 解き方の手順

(1) $\sin \frac{13}{6}\pi = \sin (\frac{12}{6}\pi + \frac{1}{6}\pi) = \sin (2\pi + \frac{1}{6}\pi) = \sin \frac{\pi}{6} = \frac{1}{2}$

(2) $\cos \frac{8}{3}\pi = \cos (\frac{6}{3}\pi + \frac{2}{3}\pi) = \cos (2\pi + \frac{2}{3}\pi) = \cos \frac{2}{3}\pi = -\frac{1}{2}$

(3) $\tan \frac{9}{4}\pi = \tan (\frac{8}{4}\pi + \frac{1}{4}\pi) = \tan (2\pi + \frac{1}{4}\pi) = \tan \frac{\pi}{4} = 1$

3. 最終的な答え

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

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

(3) $1$

## 問題 8

1. 問題の内容

三角関数の値を求めます。

(1) $\sin \frac{4}{3}\pi$

(2) $\cos \frac{11}{6}\pi$

(3) $\tan \frac{5}{4}\pi$

2. 解き方の手順

(1) $\sin \frac{4}{3}\pi = \sin (\pi + \frac{1}{3}\pi) = -\sin \frac{\pi}{3} = -\frac{\sqrt{3}}{2}$

(2) $\cos \frac{11}{6}\pi = \cos (2\pi - \frac{1}{6}\pi) = \cos (-\frac{\pi}{6}) = \cos \frac{\pi}{6} = \frac{\sqrt{3}}{2}$

(3) $\tan \frac{5}{4}\pi = \tan (\pi + \frac{\pi}{4}) = \tan \frac{\pi}{4} = 1$

3. 最終的な答え

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

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

(3) $1$

## 問題 9

1. 問題の内容

与えられた角 $\theta$ に対して、$\sin \theta, \cos \theta, \tan \theta$ の値を求めます。

(1) $\theta = \frac{9}{4}\pi$

(2) $\theta = -\frac{8}{3}\pi$

2. 解き方の手順

(1) $\theta = \frac{9}{4}\pi$ のとき

$\sin \frac{9}{4}\pi = \sin (2\pi + \frac{\pi}{4}) = \sin \frac{\pi}{4} = \frac{\sqrt{2}}{2}$

$\cos \frac{9}{4}\pi = \cos (2\pi + \frac{\pi}{4}) = \cos \frac{\pi}{4} = \frac{\sqrt{2}}{2}$

$\tan \frac{9}{4}\pi = \tan (2\pi + \frac{\pi}{4}) = \tan \frac{\pi}{4} = 1$

(2) $\theta = -\frac{8}{3}\pi$ のとき

$\sin (-\frac{8}{3}\pi) = \sin (-2\pi - \frac{2}{3}\pi) = \sin (-\frac{2}{3}\pi) = -\sin \frac{2}{3}\pi = -\frac{\sqrt{3}}{2}$

$\cos (-\frac{8}{3}\pi) = \cos (-2\pi - \frac{2}{3}\pi) = \cos (-\frac{2}{3}\pi) = \cos \frac{2}{3}\pi = -\frac{1}{2}$

$\tan (-\frac{8}{3}\pi) = \tan (-2\pi - \frac{2}{3}\pi) = \tan (-\frac{2}{3}\pi) = \tan \frac{\pi}{3} = \sqrt{3}$

3. 最終的な答え

(1) $\sin \frac{9}{4}\pi = \frac{\sqrt{2}}{2}$, $\cos \frac{9}{4}\pi = \frac{\sqrt{2}}{2}$, $\tan \frac{9}{4}\pi = 1$

(2) $\sin (-\frac{8}{3}\pi) = -\frac{\sqrt{3}}{2}$, $\cos (-\frac{8}{3}\pi) = -\frac{1}{2}$, $\tan (-\frac{8}{3}\pi) = \sqrt{3}$