$\sin \alpha = \frac{1}{2}$ ($0 < \alpha < \frac{\pi}{2}$), $\sin \beta = \frac{1}{3}$ ($\frac{\pi}{2} < \beta < \pi$) のとき、$\sin(\alpha + \beta)$, $\cos(\alpha - \beta)$, $\tan(\alpha - \beta)$ の値を求める問題です。

1. 問題の内容

$\sin \alpha = \frac{1}{2}$ ($0 < \alpha < \frac{\pi}{2}$), $\sin \beta = \frac{1}{3}$ ($\frac{\pi}{2} < \beta < \pi$) のとき、$\sin(\alpha + \beta)$, $\cos(\alpha - \beta)$, $\tan(\alpha - \beta)$ の値を求める問題です。

2. 解き方の手順

まず、$\cos \alpha$ と $\cos \beta$ を求めます。

$\alpha$ は第1象限の角なので、$\cos \alpha > 0$ です。

$\beta$ は第2象限の角なので、$\cos \beta < 0$ です。

$\cos^2 \alpha + \sin^2 \alpha = 1$ より、

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

$\cos \alpha = \sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{2}$

$\cos^2 \beta + \sin^2 \beta = 1$ より、

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

$\cos \beta = -\sqrt{\frac{8}{9}} = -\frac{2\sqrt{2}}{3}$

次に、$\sin(\alpha + \beta)$, $\cos(\alpha - \beta)$, $\tan(\alpha - \beta)$ を加法定理を用いて計算します。

$\sin(\alpha + \beta) = \sin \alpha \cos \beta + \cos \alpha \sin \beta = (\frac{1}{2})(-\frac{2\sqrt{2}}{3}) + (\frac{\sqrt{3}}{2})(\frac{1}{3}) = -\frac{\sqrt{2}}{3} + \frac{\sqrt{3}}{6} = \frac{-\sqrt{2} \cdot 2 + \sqrt{3}}{6} = \frac{\sqrt{3} - 2\sqrt{2}}{6}$

$\cos(\alpha - \beta) = \cos \alpha \cos \beta + \sin \alpha \sin \beta = (\frac{\sqrt{3}}{2})(-\frac{2\sqrt{2}}{3}) + (\frac{1}{2})(\frac{1}{3}) = -\frac{\sqrt{6}}{3} + \frac{1}{6} = \frac{-2\sqrt{6} + 1}{6} = \frac{1 - 2\sqrt{6}}{6}$

$\tan(\alpha - \beta) = \frac{\tan \alpha - \tan \beta}{1 + \tan \alpha \tan \beta}$

$\tan \alpha = \frac{\sin \alpha}{\cos \alpha} = \frac{1/2}{\sqrt{3}/2} = \frac{1}{\sqrt{3}} = \frac{\sqrt{3}}{3}$

$\tan \beta = \frac{\sin \beta}{\cos \beta} = \frac{1/3}{-2\sqrt{2}/3} = -\frac{1}{2\sqrt{2}} = -\frac{\sqrt{2}}{4}$

$\tan(\alpha - \beta) = \frac{\frac{\sqrt{3}}{3} - (-\frac{\sqrt{2}}{4})}{1 + (\frac{\sqrt{3}}{3})(-\frac{\sqrt{2}}{4})} = \frac{\frac{4\sqrt{3} + 3\sqrt{2}}{12}}{1 - \frac{\sqrt{6}}{12}} = \frac{4\sqrt{3} + 3\sqrt{2}}{12 - \sqrt{6}} = \frac{(4\sqrt{3} + 3\sqrt{2})(12 + \sqrt{6})}{(12 - \sqrt{6})(12 + \sqrt{6})} = \frac{48\sqrt{3} + 12\sqrt{2} + 36\sqrt{2} + 6\sqrt{3}}{144 - 6} = \frac{54\sqrt{3} + 48\sqrt{2}}{138} = \frac{27\sqrt{3} + 24\sqrt{2}}{69}$

3. 最終的な答え

$\sin(\alpha + \beta) = \frac{\sqrt{3} - 2\sqrt{2}}{6}$

$\cos(\alpha - \beta) = \frac{1 - 2\sqrt{6}}{6}$

$\tan(\alpha - \beta) = \frac{27\sqrt{3} + 24\sqrt{2}}{69}$