以下の三角関数の値を求める問題です。 (1) $\sin{\frac{5}{12}\pi}$ (2) $\cos{\frac{\pi}{12}}$ (3) $\tan{\frac{13}{12}\pi}$ (4) $\sin{\frac{7}{8}\pi}$

1. 問題の内容

以下の三角関数の値を求める問題です。

(1) $\sin{\frac{5}{12}\pi}$

(2) $\cos{\frac{\pi}{12}}$

(3) $\tan{\frac{13}{12}\pi}$

(4) $\sin{\frac{7}{8}\pi}$

2. 解き方の手順

(1) $\sin{\frac{5}{12}\pi}$

$\frac{5}{12}\pi = \frac{\pi}{6} + \frac{\pi}{4}$ より、

$\sin{\frac{5}{12}\pi} = \sin(\frac{\pi}{6} + \frac{\pi}{4}) = \sin{\frac{\pi}{6}}\cos{\frac{\pi}{4}} + \cos{\frac{\pi}{6}}\sin{\frac{\pi}{4}} = \frac{1}{2} \cdot \frac{\sqrt{2}}{2} + \frac{\sqrt{3}}{2} \cdot \frac{\sqrt{2}}{2} = \frac{\sqrt{2} + \sqrt{6}}{4}$

(2) $\cos{\frac{\pi}{12}}$

$\frac{\pi}{12} = \frac{\pi}{3} - \frac{\pi}{4}$ より、

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

(3) $\tan{\frac{13}{12}\pi}$

$\frac{13}{12}\pi = \pi + \frac{\pi}{12}$ より、

$\tan{\frac{13}{12}\pi} = \tan{(\pi + \frac{\pi}{12})} = \tan{\frac{\pi}{12}}$

$\frac{\pi}{12} = \frac{\pi}{3} - \frac{\pi}{4}$ より、

$\tan{\frac{\pi}{12}} = \tan(\frac{\pi}{3} - \frac{\pi}{4}) = \frac{\tan{\frac{\pi}{3}} - \tan{\frac{\pi}{4}}}{1 + \tan{\frac{\pi}{3}}\tan{\frac{\pi}{4}}} = \frac{\sqrt{3} - 1}{1 + \sqrt{3}} = \frac{(\sqrt{3} - 1)^2}{3 - 1} = \frac{3 - 2\sqrt{3} + 1}{2} = \frac{4 - 2\sqrt{3}}{2} = 2 - \sqrt{3}$

(4) $\sin{\frac{7}{8}\pi}$

$\frac{7}{8}\pi = \pi - \frac{\pi}{8}$ より、

$\sin{\frac{7}{8}\pi} = \sin{(\pi - \frac{\pi}{8})} = \sin{\frac{\pi}{8}}$

$\sin^2{\frac{\pi}{8}} = \frac{1 - \cos{\frac{\pi}{4}}}{2} = \frac{1 - \frac{\sqrt{2}}{2}}{2} = \frac{2 - \sqrt{2}}{4}$

$\sin{\frac{\pi}{8}} = \sqrt{\frac{2 - \sqrt{2}}{4}} = \frac{\sqrt{2 - \sqrt{2}}}{2}$

3. 最終的な答え

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

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

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

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