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

1. 問題の内容

以下の三角関数の値を求めよ。

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

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

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

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

2. 解き方の手順

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

$\sin{\frac{\pi}{6}} = \frac{1}{2}$、$\cos{\frac{\pi}{4}} = \frac{\sqrt{2}}{2}$、$\cos{\frac{\pi}{6}} = \frac{\sqrt{3}}{2}$、$\sin{\frac{\pi}{4}} = \frac{\sqrt{2}}{2}$なので、

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

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

$\cos{\frac{\pi}{3}} = \frac{1}{2}$、$\cos{\frac{\pi}{4}} = \frac{\sqrt{2}}{2}$、$\sin{\frac{\pi}{3}} = \frac{\sqrt{3}}{2}$、$\sin{\frac{\pi}{4}} = \frac{\sqrt{2}}{2}$なので、

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

(3) $\tan{\frac{13}{12}\pi} = \tan{(\pi + \frac{\pi}{12})} = \tan{\frac{\pi}{12}} = \tan{(\frac{\pi}{3} - \frac{\pi}{4})} = \frac{\tan{\frac{\pi}{3}} - \tan{\frac{\pi}{4}}}{1 + \tan{\frac{\pi}{3}}\tan{\frac{\pi}{4}}}$

$\tan{\frac{\pi}{3}} = \sqrt{3}$、$\tan{\frac{\pi}{4}} = 1$なので、

$\tan{\frac{13}{12}\pi} = \frac{\sqrt{3} - 1}{1 + \sqrt{3}} = \frac{(\sqrt{3}-1)(1-\sqrt{3})}{(1+\sqrt{3})(1-\sqrt{3})} = \frac{\sqrt{3} - 3 - 1 + \sqrt{3}}{1-3} = \frac{2\sqrt{3} - 4}{-2} = 2 - \sqrt{3}$

(4) 半角の公式を用いる。

$\sin^2{\frac{\theta}{2}} = \frac{1 - \cos{\theta}}{2}$なので、

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

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