与えられた角度 $\alpha$ に対する、$\sin \alpha$, $\cos \alpha$, $\tan \alpha$ の値を表に埋めて完成させる問題です。角度 $\alpha$ は、0°, 30°, 45°, 60°, 90°, 120°, 135°, 150°, 180° です。

1. 問題の内容

与えられた角度 $\alpha$ に対する、$\sin \alpha$, $\cos \alpha$, $\tan \alpha$ の値を表に埋めて完成させる問題です。角度 $\alpha$ は、0°, 30°, 45°, 60°, 90°, 120°, 135°, 150°, 180° です。

2. 解き方の手順

各角度に対する三角関数の値を求めます。単位円をイメージするとわかりやすいでしょう。

* $\alpha = 0^\circ$:

* $\sin 0^\circ = 0$

* $\cos 0^\circ = 1$

* $\tan 0^\circ = \frac{\sin 0^\circ}{\cos 0^\circ} = \frac{0}{1} = 0$

* $\alpha = 30^\circ$:

* $\sin 30^\circ = \frac{1}{2}$

* $\cos 30^\circ = \frac{\sqrt{3}}{2}$

* $\tan 30^\circ = \frac{\sin 30^\circ}{\cos 30^\circ} = \frac{1/2}{\sqrt{3}/2} = \frac{1}{\sqrt{3}} = \frac{\sqrt{3}}{3}$

* $\alpha = 45^\circ$:

* $\sin 45^\circ = \frac{\sqrt{2}}{2}$

* $\cos 45^\circ = \frac{\sqrt{2}}{2}$

* $\tan 45^\circ = \frac{\sin 45^\circ}{\cos 45^\circ} = \frac{\sqrt{2}/2}{\sqrt{2}/2} = 1$

* $\alpha = 60^\circ$:

* $\sin 60^\circ = \frac{\sqrt{3}}{2}$

* $\cos 60^\circ = \frac{1}{2}$

* $\tan 60^\circ = \frac{\sin 60^\circ}{\cos 60^\circ} = \frac{\sqrt{3}/2}{1/2} = \sqrt{3}$

* $\alpha = 90^\circ$:

* $\sin 90^\circ = 1$

* $\cos 90^\circ = 0$

* $\tan 90^\circ = \frac{\sin 90^\circ}{\cos 90^\circ} = \frac{1}{0} = \text{undefined (定義されない)}$

* $\alpha = 120^\circ = 180^\circ - 60^\circ$:

* $\sin 120^\circ = \sin (180^\circ - 60^\circ) = \sin 60^\circ = \frac{\sqrt{3}}{2}$

* $\cos 120^\circ = -\cos (180^\circ - 120^\circ) = -\cos 60^\circ = -\frac{1}{2}$

* $\tan 120^\circ = \frac{\sin 120^\circ}{\cos 120^\circ} = \frac{\sqrt{3}/2}{-1/2} = -\sqrt{3}$

* $\alpha = 135^\circ = 180^\circ - 45^\circ$:

* $\sin 135^\circ = \sin (180^\circ - 45^\circ) = \sin 45^\circ = \frac{\sqrt{2}}{2}$

* $\cos 135^\circ = -\cos (180^\circ - 135^\circ) = -\cos 45^\circ = -\frac{\sqrt{2}}{2}$

* $\tan 135^\circ = \frac{\sin 135^\circ}{\cos 135^\circ} = \frac{\sqrt{2}/2}{-\sqrt{2}/2} = -1$

* $\alpha = 150^\circ = 180^\circ - 30^\circ$:

* $\sin 150^\circ = \sin (180^\circ - 30^\circ) = \sin 30^\circ = \frac{1}{2}$

* $\cos 150^\circ = -\cos (180^\circ - 150^\circ) = -\cos 30^\circ = -\frac{\sqrt{3}}{2}$

* $\tan 150^\circ = \frac{\sin 150^\circ}{\cos 150^\circ} = \frac{1/2}{-\sqrt{3}/2} = -\frac{1}{\sqrt{3}} = -\frac{\sqrt{3}}{3}$

* $\alpha = 180^\circ$:

* $\sin 180^\circ = 0$

* $\cos 180^\circ = -1$

* $\tan 180^\circ = \frac{\sin 180^\circ}{\cos 180^\circ} = \frac{0}{-1} = 0$

3. 最終的な答え

| $\alpha$ | 0° | 30° | 45° | 60° | 90° | 120° | 135° | 150° | 180° |

|---|---|---|---|---|---|---|---|---|---|

| $\sin \alpha$ | 0 | $\frac{1}{2}$ | $\frac{\sqrt{2}}{2}$ | $\frac{\sqrt{3}}{2}$ | 1 | $\frac{\sqrt{3}}{2}$ | $\frac{\sqrt{2}}{2}$ | $\frac{1}{2}$ | 0 |

| $\cos \alpha$ | 1 | $\frac{\sqrt{3}}{2}$ | $\frac{\sqrt{2}}{2}$ | $\frac{1}{2}$ | 0 | $-\frac{1}{2}$ | $-\frac{\sqrt{2}}{2}$ | $-\frac{\sqrt{3}}{2}$ | -1 |

| $\tan \alpha$ | 0 | $\frac{\sqrt{3}}{3}$ | 1 | $\sqrt{3}$ | undefined | $-\sqrt{3}$ | -1 | $-\frac{\sqrt{3}}{3}$ | 0 |