画像に写っている数学の問題のうち、問題6、7、8、9、10を解きます。

1. 問題の内容

画像に写っている数学の問題のうち、問題6、7、8、9、10を解きます。

2. 解き方の手順

**問題6**

(1) $\sin{\frac{5}{3}\pi}$

$\sin{\frac{5}{3}\pi} = \sin{(2\pi - \frac{1}{3}\pi)} = -\sin{\frac{\pi}{3}} = -\frac{\sqrt{3}}{2}$

(2) $\cos{\frac{5}{4}\pi}$

$\cos{\frac{5}{4}\pi} = \cos{(\pi + \frac{1}{4}\pi)} = -\cos{\frac{\pi}{4}} = -\frac{\sqrt{2}}{2}$

(3) $\tan{\frac{\pi}{6}}$

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

**問題7**

$0 \le \theta < 2\pi$ のとき

(1) $\sin{\theta} = \frac{\sqrt{2}}{2}$

$\theta = \frac{\pi}{4}, \frac{3\pi}{4}$

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

$\theta = \frac{\pi}{3}, \frac{5\pi}{3}$

(3) $\sin{\theta} \ge \frac{\sqrt{2}}{2}$

$\frac{\pi}{4} \le \theta \le \frac{3\pi}{4}$

(4) $\cos{\theta} < -\frac{1}{\sqrt{2}} = -\frac{\sqrt{2}}{2}$

$\frac{3\pi}{4} < \theta < \frac{5\pi}{4}$

**問題8**

$\alpha$が第2象限の角、$\beta$が第4象限の角で、$\sin{\alpha} = \frac{2}{3}, \cos{\beta} = \frac{1}{3}$であるとき

(1) $\sin{(\alpha + \beta)}$

$\cos{\alpha} = -\sqrt{1 - \sin^2{\alpha}} = -\sqrt{1 - (\frac{2}{3})^2} = -\sqrt{\frac{5}{9}} = -\frac{\sqrt{5}}{3}$

$\sin{\beta} = -\sqrt{1 - \cos^2{\beta}} = -\sqrt{1 - (\frac{1}{3})^2} = -\sqrt{\frac{8}{9}} = -\frac{2\sqrt{2}}{3}$

$\sin{(\alpha + \beta)} = \sin{\alpha}\cos{\beta} + \cos{\alpha}\sin{\beta} = (\frac{2}{3})(\frac{1}{3}) + (-\frac{\sqrt{5}}{3})(-\frac{2\sqrt{2}}{3}) = \frac{2}{9} + \frac{2\sqrt{10}}{9} = \frac{2 + 2\sqrt{10}}{9}$

(2) $\cos{(\alpha + \beta)}$

$\cos{(\alpha + \beta)} = \cos{\alpha}\cos{\beta} - \sin{\alpha}\sin{\beta} = (-\frac{\sqrt{5}}{3})(\frac{1}{3}) - (\frac{2}{3})(-\frac{2\sqrt{2}}{3}) = -\frac{\sqrt{5}}{9} + \frac{4\sqrt{2}}{9} = \frac{4\sqrt{2} - \sqrt{5}}{9}$

**問題9**

(1) $\cos{5\theta}\sin{2\theta}$

$\frac{1}{2}(\sin{(5\theta + 2\theta)} - \sin{(5\theta - 2\theta)}) = \frac{1}{2}(\sin{7\theta} - \sin{3\theta})$

(2) $\sin{3\theta}\sin{2\theta}$

$-\frac{1}{2}(\cos{(3\theta + 2\theta)} - \cos{(3\theta - 2\theta)}) = -\frac{1}{2}(\cos{5\theta} - \cos{\theta}) = \frac{1}{2}(\cos{\theta} - \cos{5\theta})$

(3) $\cos{4\theta}\cos{\theta}$

$\frac{1}{2}(\cos{(4\theta + \theta)} + \cos{(4\theta - \theta)}) = \frac{1}{2}(\cos{5\theta} + \cos{3\theta})$

(4) $\sin{3\theta}\cos{7\theta}$

$\frac{1}{2}(\sin{(3\theta + 7\theta)} + \sin{(3\theta - 7\theta)}) = \frac{1}{2}(\sin{10\theta} + \sin{(-4\theta)}) = \frac{1}{2}(\sin{10\theta} - \sin{4\theta})$

**問題10**

(1) $\cos{\frac{\pi}{12}}$ の値を求めよ。

$\cos{\frac{\pi}{12}} = \cos{(\frac{\pi}{3} - \frac{\pi}{4})} = \cos{\frac{\pi}{3}}\cos{\frac{\pi}{4}} + \sin{\frac{\pi}{3}}\sin{\frac{\pi}{4}} = (\frac{1}{2})(\frac{\sqrt{2}}{2}) + (\frac{\sqrt{3}}{2})(\frac{\sqrt{2}}{2}) = \frac{\sqrt{2} + \sqrt{6}}{4}$

(2) $y = 2\sin{x} - 2\cos{x}$ を一つの三角関数で表せ。

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

(3) $\sin{(\theta + \frac{\pi}{3})}, \cos{(\theta + \frac{\pi}{4})}$ を $\sin{\theta}$ と $\cos{\theta}$ で表せ。

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

$\cos{(\theta + \frac{\pi}{4})} = \cos{\theta}\cos{\frac{\pi}{4}} - \sin{\theta}\sin{\frac{\pi}{4}} = \frac{\sqrt{2}}{2}\cos{\theta} - \frac{\sqrt{2}}{2}\sin{\theta}$

(4) 関数 $y = \sqrt{3}\sin{x} + \cos{x}$ ($0 \le x \le 2\pi$) の最大値と最小値を求めよ。

$y = 2(\frac{\sqrt{3}}{2}\sin{x} + \frac{1}{2}\cos{x}) = 2(\sin{x}\cos{\frac{\pi}{6}} + \cos{x}\sin{\frac{\pi}{6}}) = 2\sin{(x + \frac{\pi}{6})}$

$0 \le x \le 2\pi$ より、$\frac{\pi}{6} \le x + \frac{\pi}{6} \le 2\pi + \frac{\pi}{6}$

よって、最大値は $2$ ($x = \frac{\pi}{3}$)、最小値は $-2$ ($x = \frac{4\pi}{3}$)

3. 最終的な答え

**問題6**

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

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

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

**問題7**

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

(2) $\theta = \frac{\pi}{3}, \frac{5\pi}{3}$

(3) $\frac{\pi}{4} \le \theta \le \frac{3\pi}{4}$

(4) $\frac{3\pi}{4} < \theta < \frac{5\pi}{4}$

**問題8**

(1) $\frac{2 + 2\sqrt{10}}{9}$

(2) $\frac{4\sqrt{2} - \sqrt{5}}{9}$

**問題9**

(1) $\frac{1}{2}(\sin{7\theta} - \sin{3\theta})$

(2) $\frac{1}{2}(\cos{\theta} - \cos{5\theta})$

(3) $\frac{1}{2}(\cos{5\theta} + \cos{3\theta})$

(4) $\frac{1}{2}(\sin{10\theta} - \sin{4\theta})$

**問題10**

(1) $\frac{\sqrt{2} + \sqrt{6}}{4}$

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

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

(4) 最大値：$2$、最小値：$-2$