(1) $0 < \alpha < \frac{\pi}{2}$, $\frac{\pi}{2} < \beta < \pi$ とする。$\sin \alpha = \frac{2}{3}$, $\sin \beta = \frac{3\sqrt{5}}{7}$ のとき、$\cos \alpha$, $\cos \beta$, $\sin(\alpha + \beta)$, $\cos 2\alpha$ の値を求める。 (2) $0 < \theta < \pi$ において $\cos \theta = -\frac{1}{4}$ のとき、$\sin \frac{\theta}{2}$, $\cos \frac{\theta}{2}$, $\cos 2\frac{\theta}{2}$ の値を求める。 (3) $\sqrt{3} \sin \theta - \cos \theta$ を $A \sin(\theta - B)$ の形に変形し、 $0 \le \theta < 2\pi$ のとき、$\sqrt{3} \sin \theta - \cos \theta = \sqrt{2}$ を満たす $\theta$ の値を求める。

1. 問題の内容

(1) $0 < \alpha < \frac{\pi}{2}$, $\frac{\pi}{2} < \beta < \pi$ とする。$\sin \alpha = \frac{2}{3}$, $\sin \beta = \frac{3\sqrt{5}}{7}$ のとき、$\cos \alpha$, $\cos \beta$, $\sin(\alpha + \beta)$, $\cos 2\alpha$ の値を求める。

(2) $0 < \theta < \pi$ において $\cos \theta = -\frac{1}{4}$ のとき、$\sin \frac{\theta}{2}$, $\cos \frac{\theta}{2}$, $\cos 2\frac{\theta}{2}$ の値を求める。

(3) $\sqrt{3} \sin \theta - \cos \theta$ を $A \sin(\theta - B)$ の形に変形し、 $0 \le \theta < 2\pi$ のとき、$\sqrt{3} \sin \theta - \cos \theta = \sqrt{2}$ を満たす $\theta$ の値を求める。

2. 解き方の手順

(1)

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

$0 < \alpha < \frac{\pi}{2}$ より $\cos \alpha > 0$ なので、

$\cos \alpha = \sqrt{\frac{5}{9}} = \frac{\sqrt{5}}{3}$

$\cos^2 \beta = 1 - \sin^2 \beta = 1 - (\frac{3\sqrt{5}}{7})^2 = 1 - \frac{45}{49} = \frac{4}{49}$

$\frac{\pi}{2} < \beta < \pi$ より $\cos \beta < 0$ なので、

$\cos \beta = -\sqrt{\frac{4}{49}} = -\frac{2}{7}$

$\sin(\alpha + \beta) = \sin \alpha \cos \beta + \cos \alpha \sin \beta = \frac{2}{3} \times (-\frac{2}{7}) + \frac{\sqrt{5}}{3} \times \frac{3\sqrt{5}}{7} = -\frac{4}{21} + \frac{15}{21} = \frac{11}{21}$

$\cos 2\alpha = 1 - 2\sin^2 \alpha = 1 - 2(\frac{2}{3})^2 = 1 - 2 \times \frac{4}{9} = 1 - \frac{8}{9} = \frac{1}{9}$

(2)

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

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

$\sin \frac{\theta}{2} = \sqrt{\frac{5}{8}} = \frac{\sqrt{10}}{4}$

$\cos^2 \frac{\theta}{2} = \frac{1 + \cos \theta}{2} = \frac{1 + (-\frac{1}{4})}{2} = \frac{1 - \frac{1}{4}}{2} = \frac{\frac{3}{4}}{2} = \frac{3}{8}$

$0 < \frac{\theta}{2} < \frac{\pi}{2}$ より $\cos \frac{\theta}{2} > 0$

$\cos \frac{\theta}{2} = \sqrt{\frac{3}{8}} = \frac{\sqrt{6}}{4}$

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

(3)

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

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

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

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

$0 \le \theta < 2\pi$ より $-\frac{\pi}{6} \le \theta - \frac{\pi}{6} < \frac{11\pi}{6}$

$\theta - \frac{\pi}{6} = \frac{\pi}{4}, \frac{3\pi}{4}$

$\theta = \frac{\pi}{4} + \frac{\pi}{6} = \frac{3\pi}{12} + \frac{2\pi}{12} = \frac{5\pi}{12}$

$\theta = \frac{3\pi}{4} + \frac{\pi}{6} = \frac{9\pi}{12} + \frac{2\pi}{12} = \frac{11\pi}{12}$

3. 最終的な答え

(1) $\cos \alpha = \frac{\sqrt{5}}{3}$, $\cos \beta = -\frac{2}{7}$, $\sin(\alpha + \beta) = \frac{11}{21}$, $\cos 2\alpha = \frac{1}{9}$

(2) $\sin \frac{\theta}{2} = \frac{\sqrt{10}}{4}$, $\cos \frac{\theta}{2} = \frac{\sqrt{6}}{4}$, $\cos \theta = -\frac{1}{4}$

(3) $2 \sin(\theta - \frac{\pi}{6})$, $\theta = \frac{5\pi}{12}, \frac{11\pi}{12}$