$\frac{\pi}{12} = \frac{\pi}{4} - \frac{\pi}{6}$ であることを用いて、$\sin \frac{\pi}{12}$ と $\cos \frac{\pi}{12}$ の値を求めよ。

1. 問題の内容

$\frac{\pi}{12} = \frac{\pi}{4} - \frac{\pi}{6}$ であることを用いて、$\sin \frac{\pi}{12}$ と $\cos \frac{\pi}{12}$ の値を求めよ。

2. 解き方の手順

$\sin (\alpha - \beta) = \sin \alpha \cos \beta - \cos \alpha \sin \beta$

$\cos (\alpha - \beta) = \cos \alpha \cos \beta + \sin \alpha \sin \beta$

である。

$\sin \frac{\pi}{12} = \sin (\frac{\pi}{4} - \frac{\pi}{6}) = \sin \frac{\pi}{4} \cos \frac{\pi}{6} - \cos \frac{\pi}{4} \sin \frac{\pi}{6}$

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

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

$\cos \frac{\pi}{4} = \frac{\sqrt{2}}{2}$

$\sin \frac{\pi}{6} = \frac{1}{2}$

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

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

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

3. 最終的な答え

$\sin \frac{\pi}{12} = \frac{\sqrt{6} - \sqrt{2}}{4}$

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