$\theta$ が与えられたときに、$\sin \theta$, $\cos \theta$, $\tan \theta$ の値をそれぞれ求める問題です。 (1) $\theta = \frac{5}{4}\pi$ (2) $\theta = \frac{11}{6}\pi$ (3) $\theta = -\frac{\pi}{3}$

1. 問題の内容

$\theta$ が与えられたときに、$\sin \theta$, $\cos \theta$, $\tan \theta$ の値をそれぞれ求める問題です。

(1) $\theta = \frac{5}{4}\pi$

(2) $\theta = \frac{11}{6}\pi$

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

2. 解き方の手順

(1) $\theta = \frac{5}{4}\pi$ の場合:

$\frac{5}{4}\pi$ は第3象限の角です。基準となる角は $\frac{5}{4}\pi - \pi = \frac{\pi}{4}$ です。

$\sin \frac{5}{4}\pi = -\sin \frac{\pi}{4} = -\frac{\sqrt{2}}{2}$

$\cos \frac{5}{4}\pi = -\cos \frac{\pi}{4} = -\frac{\sqrt{2}}{2}$

$\tan \frac{5}{4}\pi = \frac{\sin \frac{5}{4}\pi}{\cos \frac{5}{4}\pi} = \frac{-\frac{\sqrt{2}}{2}}{-\frac{\sqrt{2}}{2}} = 1$

(2) $\theta = \frac{11}{6}\pi$ の場合:

$\frac{11}{6}\pi$ は第4象限の角です。基準となる角は $2\pi - \frac{11}{6}\pi = \frac{\pi}{6}$ です。

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

$\cos \frac{11}{6}\pi = \cos \frac{\pi}{6} = \frac{\sqrt{3}}{2}$

$\tan \frac{11}{6}\pi = \frac{\sin \frac{11}{6}\pi}{\cos \frac{11}{6}\pi} = \frac{-\frac{1}{2}}{\frac{\sqrt{3}}{2}} = -\frac{1}{\sqrt{3}} = -\frac{\sqrt{3}}{3}$

(3) $\theta = -\frac{\pi}{3}$ の場合:

$-\frac{\pi}{3}$ は第4象限の角です。

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

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

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

3. 最終的な答え

(1) $\theta = \frac{5}{4}\pi$:

$\sin \theta = -\frac{\sqrt{2}}{2}$

$\cos \theta = -\frac{\sqrt{2}}{2}$

$\tan \theta = 1$

(2) $\theta = \frac{11}{6}\pi$:

$\sin \theta = -\frac{1}{2}$

$\cos \theta = \frac{\sqrt{3}}{2}$

$\tan \theta = -\frac{\sqrt{3}}{3}$

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

$\sin \theta = -\frac{\sqrt{3}}{2}$

$\cos \theta = \frac{1}{2}$

$\tan \theta = -\sqrt{3}$