加法定理を用いて、以下の三角関数の値を求めます。 (1) $\sin 105^\circ$ (2) $\cos 105^\circ$ (3) $\tan 165^\circ$ (4) $\sin 15^\circ$ (5) $\cos \frac{5}{12}\pi$ (6) $\tan \frac{7}{12}\pi$

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ます。

(1) $\sin 105^\circ$

(2) $\cos 105^\circ$

(3) $\tan 165^\circ$

(4) $\sin 15^\circ$

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

(6) $\tan \frac{7}{12}\pi$

2. 解き方の手順

(1) $\sin 105^\circ$

$105^\circ = 60^\circ + 45^\circ$ なので、加法定理 $\sin(A+B) = \sin A \cos B + \cos A \sin B$ を用いると、

$\sin 105^\circ = \sin (60^\circ + 45^\circ) = \sin 60^\circ \cos 45^\circ + \cos 60^\circ \sin 45^\circ = \frac{\sqrt{3}}{2} \cdot \frac{\sqrt{2}}{2} + \frac{1}{2} \cdot \frac{\sqrt{2}}{2} = \frac{\sqrt{6} + \sqrt{2}}{4}$

(2) $\cos 105^\circ$

$105^\circ = 60^\circ + 45^\circ$ なので、加法定理 $\cos(A+B) = \cos A \cos B - \sin A \sin B$ を用いると、

$\cos 105^\circ = \cos (60^\circ + 45^\circ) = \cos 60^\circ \cos 45^\circ - \sin 60^\circ \sin 45^\circ = \frac{1}{2} \cdot \frac{\sqrt{2}}{2} - \frac{\sqrt{3}}{2} \cdot \frac{\sqrt{2}}{2} = \frac{\sqrt{2} - \sqrt{6}}{4}$

(3) $\tan 165^\circ$

$165^\circ = 120^\circ + 45^\circ$ なので、加法定理 $\tan(A+B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}$ を用いると、

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

(4) $\sin 15^\circ$

$15^\circ = 45^\circ - 30^\circ$ なので、加法定理 $\sin(A-B) = \sin A \cos B - \cos A \sin B$ を用いると、

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

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

$\frac{5}{12}\pi = \frac{2}{12}\pi + \frac{3}{12}\pi = \frac{\pi}{6} + \frac{\pi}{4}$ なので、加法定理 $\cos(A+B) = \cos A \cos B - \sin A \sin B$ を用いると、

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

(6) $\tan \frac{7}{12}\pi$

$\frac{7}{12}\pi = \frac{3}{12}\pi + \frac{4}{12}\pi = \frac{\pi}{4} + \frac{\pi}{3}$ なので、加法定理 $\tan(A+B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}$ を用いると、

$\tan \frac{7}{12}\pi = \tan (\frac{\pi}{4} + \frac{\pi}{3}) = \frac{\tan \frac{\pi}{4} + \tan \frac{\pi}{3}}{1 - \tan \frac{\pi}{4} \tan \frac{\pi}{3}} = \frac{1 + \sqrt{3}}{1 - 1 \cdot \sqrt{3}} = \frac{1 + \sqrt{3}}{1 - \sqrt{3}} = \frac{(1+\sqrt{3})(1+\sqrt{3})}{(1-\sqrt{3})(1+\sqrt{3})} = \frac{1 + 2\sqrt{3} + 3}{1 - 3} = \frac{4 + 2\sqrt{3}}{-2} = -2 - \sqrt{3}$

3. 最終的な答え

(1) $\frac{\sqrt{6} + \sqrt{2}}{4}$

(2) $\frac{\sqrt{2} - \sqrt{6}}{4}$

(3) $-2 + \sqrt{3}$

(4) $\frac{\sqrt{6} - \sqrt{2}}{4}$

(5) $\frac{\sqrt{6} - \sqrt{2}}{4}$

(6) $-2 - \sqrt{3}$