以下の2つの式の値を求める問題です。 (1) $\sqrt{3} \sin{\frac{\pi}{12}} + \cos{\frac{\pi}{12}}$ (2) $\sin{\frac{5\pi}{12}} - \cos{\frac{5\pi}{12}}$

1. 問題の内容

以下の2つの式の値を求める問題です。

(1) $\sqrt{3} \sin{\frac{\pi}{12}} + \cos{\frac{\pi}{12}}$

(2) $\sin{\frac{5\pi}{12}} - \cos{\frac{5\pi}{12}}$

2. 解き方の手順

(1) $\sin$ と $\cos$ の合成公式を利用します。

$a \sin{\theta} + b \cos{\theta} = \sqrt{a^2 + b^2} \sin{(\theta + \alpha)}$

ただし、$\cos{\alpha} = \frac{a}{\sqrt{a^2 + b^2}}$, $\sin{\alpha} = \frac{b}{\sqrt{a^2 + b^2}}$

今回は、$a = \sqrt{3}$, $b = 1$, $\theta = \frac{\pi}{12}$ です。

$\sqrt{a^2 + b^2} = \sqrt{(\sqrt{3})^2 + 1^2} = \sqrt{3 + 1} = \sqrt{4} = 2$

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

よって、$\alpha = \frac{\pi}{6}$ です。

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

(2) $\sin{\frac{5\pi}{12}} - \cos{\frac{5\pi}{12}}$ を変形します。

$\sin{\frac{5\pi}{12}} - \cos{\frac{5\pi}{12}} = \sqrt{2} (\frac{1}{\sqrt{2}} \sin{\frac{5\pi}{12}} - \frac{1}{\sqrt{2}} \cos{\frac{5\pi}{12}}) = \sqrt{2} (\cos{\frac{\pi}{4}} \sin{\frac{5\pi}{12}} - \sin{\frac{\pi}{4}} \cos{\frac{5\pi}{12}})$

$\sin{(\alpha - \beta)} = \sin{\alpha} \cos{\beta} - \cos{\alpha} \sin{\beta}$ の公式を用いると、

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

$\sin{(-x)} = - \sin{x}$ であるので、

$-\sqrt{2} \sin{(-\frac{\pi}{6})} = \sqrt{2} \sin{\frac{\pi}{6}} = \sqrt{2} \cdot \frac{1}{2} = \frac{\sqrt{2}}{2}$

3. 最終的な答え

(1) $\sqrt{2}$

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