$\theta$ が以下の値のとき、$\sin\theta$, $\cos\theta$, $\tan\theta$ の値をそれぞれ求めよ。 (1) $\theta = \frac{2}{3}\pi$ (2) $\theta = -\frac{\pi}{6}$ (3) $\theta = \frac{7}{4}\pi$

## 問題 5

1. 問題の内容

$\theta$ が以下の値のとき、$\sin\theta$, $\cos\theta$, $\tan\theta$ の値をそれぞれ求めよ。

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

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

(3) $\theta = \frac{7}{4}\pi$

2. 解き方の手順

(1) $\theta = \frac{2}{3}\pi$ のとき

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

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

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

(2) $\theta = -\frac{\pi}{6}$ のとき

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

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

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

(3) $\theta = \frac{7}{4}\pi$ のとき

* $\sin\left(\frac{7}{4}\pi\right) = \sin\left(2\pi - \frac{\pi}{4}\right) = -\sin\left(\frac{\pi}{4}\right) = -\frac{\sqrt{2}}{2}$

* $\cos\left(\frac{7}{4}\pi\right) = \cos\left(2\pi - \frac{\pi}{4}\right) = \cos\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2}$

* $\tan\left(\frac{7}{4}\pi\right) = \frac{\sin(\frac{7}{4}\pi)}{\cos(\frac{7}{4}\pi)} = \frac{-\frac{\sqrt{2}}{2}}{\frac{\sqrt{2}}{2}} = -1$

3. 最終的な答え

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

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

(3) $\sin\left(\frac{7}{4}\pi\right) = -\frac{\sqrt{2}}{2}$, $\cos\left(\frac{7}{4}\pi\right) = \frac{\sqrt{2}}{2}$, $\tan\left(\frac{7}{4}\pi\right) = -1$

## 問題 8

1. 問題の内容

$\sin\theta + \cos\theta = \frac{\sqrt{3}}{2}$ とする。次の式の値を求めよ。

(1) $\sin\theta\cos\theta$, $\sin^2\theta + \cos^2\theta$

(2) $\frac{\pi}{2} < \theta < \pi$ のとき、$\cos\theta - \sin\theta$

2. 解き方の手順

(1) $\sin\theta\cos\theta$, $\sin^2\theta + \cos^2\theta$ を求める。

* $\sin^2\theta + \cos^2\theta = 1$ (三角関数の基本公式)

* $(\sin\theta + \cos\theta)^2 = \sin^2\theta + 2\sin\theta\cos\theta + \cos^2\theta$

$\left(\frac{\sqrt{3}}{2}\right)^2 = 1 + 2\sin\theta\cos\theta$

$\frac{3}{4} = 1 + 2\sin\theta\cos\theta$

$2\sin\theta\cos\theta = \frac{3}{4} - 1 = -\frac{1}{4}$

$\sin\theta\cos\theta = -\frac{1}{8}$

(2) $\frac{\pi}{2} < \theta < \pi$ のとき、$\cos\theta - \sin\theta$ を求める。

$(\cos\theta - \sin\theta)^2 = \cos^2\theta - 2\sin\theta\cos\theta + \sin^2\theta = 1 - 2\sin\theta\cos\theta$

$(\cos\theta - \sin\theta)^2 = 1 - 2\left(-\frac{1}{8}\right) = 1 + \frac{1}{4} = \frac{5}{4}$

$\cos\theta - \sin\theta = \pm\sqrt{\frac{5}{4}} = \pm\frac{\sqrt{5}}{2}$

$\frac{\pi}{2} < \theta < \pi$ より、$\sin\theta > 0$ かつ $\cos\theta < 0$ なので、$\cos\theta - \sin\theta < 0$。

したがって、$\cos\theta - \sin\theta = -\frac{\sqrt{5}}{2}$

3. 最終的な答え

(1) $\sin\theta\cos\theta = -\frac{1}{8}$, $\sin^2\theta + \cos^2\theta = 1$

(2) $\cos\theta - \sin\theta = -\frac{\sqrt{5}}{2}$