$\theta$ 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。具体的には、(1) $\theta = \frac{19}{6}\pi$, (2) $\theta = 3\pi$, (3) $\theta = -\frac{\pi}{6}$, (4) $\theta = -\frac{3}{4}\pi$ のそれぞれについて、$\sin\theta$, $\cos\theta$, $\tan\theta$ を計算します。

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。具体的には、(1) $\theta = \frac{19}{6}\pi$, (2) $\theta = 3\pi$, (3) $\theta = -\frac{\pi}{6}$, (4) $\theta = -\frac{3}{4}\pi$ のそれぞれについて、$\sin\theta$, $\cos\theta$, $\tan\theta$ を計算します。

2. 解き方の手順

(1) $\theta = \frac{19}{6}\pi$

まず、$\theta$ が一周を超える角度なので、基準となる角度を求めます。

$\frac{19}{6}\pi = \frac{12}{6}\pi + \frac{7}{6}\pi = 2\pi + \frac{7}{6}\pi$

したがって、$\sin(\frac{19}{6}\pi) = \sin(\frac{7}{6}\pi)$, $\cos(\frac{19}{6}\pi) = \cos(\frac{7}{6}\pi)$, $\tan(\frac{19}{6}\pi) = \tan(\frac{7}{6}\pi)$ となります。

$\frac{7}{6}\pi = \pi + \frac{\pi}{6}$ であるので、

$\sin(\frac{7}{6}\pi) = -\sin(\frac{\pi}{6}) = -\frac{1}{2}$

$\cos(\frac{7}{6}\pi) = -\cos(\frac{\pi}{6}) = -\frac{\sqrt{3}}{2}$

$\tan(\frac{7}{6}\pi) = \frac{\sin(\frac{7}{6}\pi)}{\cos(\frac{7}{6}\pi)} = \frac{-\frac{1}{2}}{-\frac{\sqrt{3}}{2}} = \frac{1}{\sqrt{3}} = \frac{\sqrt{3}}{3}$

(2) $\theta = 3\pi$

$3\pi = 2\pi + \pi$ であるので、

$\sin(3\pi) = \sin(\pi) = 0$

$\cos(3\pi) = \cos(\pi) = -1$

$\tan(3\pi) = \frac{\sin(3\pi)}{\cos(3\pi)} = \frac{0}{-1} = 0$

(3) $\theta = -\frac{\pi}{6}$

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

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

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

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

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

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

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

3. 最終的な答え

(1) $\sin(\frac{19}{6}\pi) = -\frac{1}{2}$, $\cos(\frac{19}{6}\pi) = -\frac{\sqrt{3}}{2}$, $\tan(\frac{19}{6}\pi) = \frac{\sqrt{3}}{3}$

(2) $\sin(3\pi) = 0$, $\cos(3\pi) = -1$, $\tan(3\pi) = 0$

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

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