三角関数の値を求める問題です。具体的には、以下の値を求めます。 (1) $\sin 195^\circ$ (2) $\cos 195^\circ$ (3) $\tan 105^\circ$ (4) $\sin \frac{11}{12}\pi$ (5) $\cos \frac{11}{12}\pi$ (6) $\tan \frac{13}{12}\pi$

1. 問題の内容

三角関数の値を求める問題です。具体的には、以下の値を求めます。

(1) $\sin 195^\circ$

(2) $\cos 195^\circ$

(3) $\tan 105^\circ$

(4) $\sin \frac{11}{12}\pi$

(5) $\cos \frac{11}{12}\pi$

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

2. 解き方の手順

(1) $\sin 195^\circ$

$195^\circ = 180^\circ + 15^\circ$ なので、

$\sin 195^\circ = \sin(180^\circ + 15^\circ) = -\sin 15^\circ$

$\sin 15^\circ = \sin(45^\circ - 30^\circ) = \sin 45^\circ \cos 30^\circ - \cos 45^\circ \sin 30^\circ = \frac{\sqrt{2}}{2} \cdot \frac{\sqrt{3}}{2} - \frac{\sqrt{2}}{2} \cdot \frac{1}{2} = \frac{\sqrt{6} - \sqrt{2}}{4}$

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

(2) $\cos 195^\circ$

$195^\circ = 180^\circ + 15^\circ$ なので、

$\cos 195^\circ = \cos(180^\circ + 15^\circ) = -\cos 15^\circ$

$\cos 15^\circ = \cos(45^\circ - 30^\circ) = \cos 45^\circ \cos 30^\circ + \sin 45^\circ \sin 30^\circ = \frac{\sqrt{2}}{2} \cdot \frac{\sqrt{3}}{2} + \frac{\sqrt{2}}{2} \cdot \frac{1}{2} = \frac{\sqrt{6} + \sqrt{2}}{4}$

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

(3) $\tan 105^\circ$

$\tan 105^\circ = \tan(60^\circ + 45^\circ) = \frac{\tan 60^\circ + \tan 45^\circ}{1 - \tan 60^\circ \tan 45^\circ} = \frac{\sqrt{3} + 1}{1 - \sqrt{3}} = \frac{(\sqrt{3} + 1)(1 + \sqrt{3})}{(1 - \sqrt{3})(1 + \sqrt{3})} = \frac{3 + 2\sqrt{3} + 1}{1 - 3} = \frac{4 + 2\sqrt{3}}{-2} = -2 - \sqrt{3}$

(4) $\sin \frac{11}{12}\pi$

$\sin \frac{11}{12}\pi = \sin (\pi - \frac{\pi}{12}) = \sin \frac{\pi}{12} = \sin 15^\circ = \frac{\sqrt{6} - \sqrt{2}}{4}$

(5) $\cos \frac{11}{12}\pi$

$\cos \frac{11}{12}\pi = \cos(\pi - \frac{\pi}{12}) = - \cos \frac{\pi}{12} = - \cos 15^\circ = - \frac{\sqrt{6} + \sqrt{2}}{4}$

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

$\tan \frac{13}{12}\pi = \tan(\pi + \frac{\pi}{12}) = \tan \frac{\pi}{12} = \tan 15^\circ = \frac{\sin 15^\circ}{\cos 15^\circ} = \frac{\frac{\sqrt{6} - \sqrt{2}}{4}}{\frac{\sqrt{6} + \sqrt{2}}{4}} = \frac{\sqrt{6} - \sqrt{2}}{\sqrt{6} + \sqrt{2}} = \frac{(\sqrt{6} - \sqrt{2})^2}{6 - 2} = \frac{6 - 2\sqrt{12} + 2}{4} = \frac{8 - 4\sqrt{3}}{4} = 2 - \sqrt{3}$

3. 最終的な答え

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

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

(3) $\tan 105^\circ = -2 - \sqrt{3}$

(4) $\sin \frac{11}{12}\pi = \frac{\sqrt{6} - \sqrt{2}}{4}$

(5) $\cos \frac{11}{12}\pi = - \frac{\sqrt{6} + \sqrt{2}}{4}$

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