加法定理を用いて、以下の三角関数の値を求める問題です。 (1) $\sin 195^\circ$, $\cos 195^\circ$, $\tan 195^\circ$ (2) $\sin 165^\circ$, $\cos 165^\circ$, $\tan 165^\circ$

1. 問題の内容

加法定理を用いて、以下の三角関数の値を求める問題です。

(1) $\sin 195^\circ$, $\cos 195^\circ$, $\tan 195^\circ$

(2) $\sin 165^\circ$, $\cos 165^\circ$, $\tan 165^\circ$

2. 解き方の手順

(1) $\sin 195^\circ$, $\cos 195^\circ$, $\tan 195^\circ$

まず、$195^\circ = 150^\circ + 45^\circ$と分解します。

$\sin(A+B) = \sin A \cos B + \cos A \sin B$

$\cos(A+B) = \cos A \cos B - \sin A \sin B$

$\tan(A+B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}$

$\sin 195^\circ = \sin (150^\circ + 45^\circ) = \sin 150^\circ \cos 45^\circ + \cos 150^\circ \sin 45^\circ = \frac{1}{2} \cdot \frac{\sqrt{2}}{2} + (-\frac{\sqrt{3}}{2}) \cdot \frac{\sqrt{2}}{2} = \frac{\sqrt{2} - \sqrt{6}}{4}$

$\cos 195^\circ = \cos (150^\circ + 45^\circ) = \cos 150^\circ \cos 45^\circ - \sin 150^\circ \sin 45^\circ = (-\frac{\sqrt{3}}{2}) \cdot \frac{\sqrt{2}}{2} - \frac{1}{2} \cdot \frac{\sqrt{2}}{2} = \frac{-\sqrt{6} - \sqrt{2}}{4} = -\frac{\sqrt{6} + \sqrt{2}}{4}$

$\tan 195^\circ = \frac{\sin 195^\circ}{\cos 195^\circ} = \frac{\frac{\sqrt{2} - \sqrt{6}}{4}}{\frac{-\sqrt{6} - \sqrt{2}}{4}} = \frac{\sqrt{2} - \sqrt{6}}{-\sqrt{6} - \sqrt{2}} = \frac{\sqrt{6} - \sqrt{2}}{\sqrt{6} + \sqrt{2}} = \frac{(\sqrt{6} - \sqrt{2})^2}{(\sqrt{6} + \sqrt{2})(\sqrt{6} - \sqrt{2})} = \frac{6 - 2\sqrt{12} + 2}{6 - 2} = \frac{8 - 4\sqrt{3}}{4} = 2 - \sqrt{3}$

(2) $\sin 165^\circ$, $\cos 165^\circ$, $\tan 165^\circ$

まず、$165^\circ = 120^\circ + 45^\circ$と分解します。

$\sin 165^\circ = \sin (120^\circ + 45^\circ) = \sin 120^\circ \cos 45^\circ + \cos 120^\circ \sin 45^\circ = \frac{\sqrt{3}}{2} \cdot \frac{\sqrt{2}}{2} + (-\frac{1}{2}) \cdot \frac{\sqrt{2}}{2} = \frac{\sqrt{6} - \sqrt{2}}{4}$

$\cos 165^\circ = \cos (120^\circ + 45^\circ) = \cos 120^\circ \cos 45^\circ - \sin 120^\circ \sin 45^\circ = (-\frac{1}{2}) \cdot \frac{\sqrt{2}}{2} - \frac{\sqrt{3}}{2} \cdot \frac{\sqrt{2}}{2} = \frac{-\sqrt{2} - \sqrt{6}}{4} = -\frac{\sqrt{2} + \sqrt{6}}{4}$

$\tan 165^\circ = \frac{\sin 165^\circ}{\cos 165^\circ} = \frac{\frac{\sqrt{6} - \sqrt{2}}{4}}{\frac{-\sqrt{2} - \sqrt{6}}{4}} = \frac{\sqrt{6} - \sqrt{2}}{-\sqrt{2} - \sqrt{6}} = \frac{\sqrt{6} - \sqrt{2}}{-(\sqrt{6} + \sqrt{2})} = - \frac{\sqrt{6} - \sqrt{2}}{\sqrt{6} + \sqrt{2}} = -(2 - \sqrt{3}) = -2 + \sqrt{3} = \sqrt{3} - 2$

3. 最終的な答え

(1)

$\sin 195^\circ = \frac{\sqrt{2} - \sqrt{6}}{4}$

$\cos 195^\circ = -\frac{\sqrt{6} + \sqrt{2}}{4}$

$\tan 195^\circ = 2 - \sqrt{3}$

(2)

$\sin 165^\circ = \frac{\sqrt{6} - \sqrt{2}}{4}$

$\cos 165^\circ = -\frac{\sqrt{2} + \sqrt{6}}{4}$

$\tan 165^\circ = \sqrt{3} - 2$