与えられた表の空欄に、三角関数 ($\sin \theta$, $\cos \theta$, $\tan \theta$) の値、角度の弧度法表示、度数法表示を埋める問題です。

1. 問題の内容

与えられた表の空欄に、三角関数 ($\sin \theta$, $\cos \theta$, $\tan \theta$) の値、角度の弧度法表示、度数法表示を埋める問題です。

2. 解き方の手順

まず、角度と三角関数の関係を確認します。

* $\sin \theta$: 単位円上の点のy座標

* $\cos \theta$: 単位円上の点のx座標

* $\tan \theta = \frac{\sin \theta}{\cos \theta}$

与えられた情報から、以下の手順で表を埋めます。

1. -45°:

* $\sin(-45^\circ) = -\frac{1}{\sqrt{2}}$

* $\cos(-45^\circ) = \frac{1}{\sqrt{2}}$

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

2. -30°:

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

3. 0°:

* $\sin(0^\circ) = 0$

* $\cos(0^\circ) = 1$

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

4. $\frac{\pi}{6}$: (30°)

* $\cos(\frac{\pi}{6}) = \frac{\sqrt{3}}{2}$

* $\sin(\frac{\pi}{6}) = \frac{1}{2}$

* $\tan(\frac{\pi}{6}) = \frac{\sin(\frac{\pi}{6})}{\cos(\frac{\pi}{6})} = \frac{\frac{1}{2}}{\frac{\sqrt{3}}{2}} = \frac{1}{\sqrt{3}}$

5. $\frac{\pi}{4}$: (45°)

* $\sin(\frac{\pi}{4}) = \frac{1}{\sqrt{2}}$

* $\cos(\frac{\pi}{4}) = \frac{1}{\sqrt{2}}$

* $\tan(\frac{\pi}{4}) = \frac{\sin(\frac{\pi}{4})}{\cos(\frac{\pi}{4})} = \frac{\frac{1}{\sqrt{2}}}{\frac{1}{\sqrt{2}}} = 1$

6. $\frac{\pi}{2}$: (90°)

* $\sin(\frac{\pi}{2}) = 1$

* $\cos(\frac{\pi}{2}) = 0$

* $\tan(\frac{\pi}{2})$ は定義されない (X)

7. $\frac{2\pi}{3}$: (120°)

* $\sin(\frac{2\pi}{3}) = \frac{\sqrt{3}}{2}$

* $\cos(\frac{2\pi}{3}) = -\frac{1}{2}$

* $\tan(\frac{2\pi}{3}) = \frac{\sin(\frac{2\pi}{3})}{\cos(\frac{2\pi}{3})} = \frac{\frac{\sqrt{3}}{2}}{-\frac{1}{2}} = -\sqrt{3}$

8. $\frac{3\pi}{4}$: (135°)

* $\sin(\frac{3\pi}{4}) = \frac{1}{\sqrt{2}}$

* $\cos(\frac{3\pi}{4}) = -\frac{1}{\sqrt{2}}$

* $\tan(\frac{3\pi}{4}) = \frac{\sin(\frac{3\pi}{4})}{\cos(\frac{3\pi}{4})} = \frac{\frac{1}{\sqrt{2}}}{-\frac{1}{\sqrt{2}}} = -1$

9. $\pi$: (180°)

* $\sin(\pi) = 0$

* $\cos(\pi) = -1$

* $\tan(\pi) = \frac{\sin(\pi)}{\cos(\pi)} = \frac{0}{-1} = 0$

最終的な表は以下のようになります。

| 角度 (度数法) | 角度 (弧度法) | $\sin \theta$ | $\cos \theta$ | $\tan \theta$ |

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

| -45° | $-\frac{\pi}{4}$ | $-\frac{1}{\sqrt{2}}$ | $\frac{1}{\sqrt{2}}$ | -1 |

| -30° | $-\frac{\pi}{6}$ | $-\frac{1}{2}$ | $\frac{\sqrt{3}}{2}$ | $-\frac{1}{\sqrt{3}}$ |

| 0° | 0 | 0 | 1 | 0 |

| 30° | $\frac{\pi}{6}$ | $\frac{1}{2}$ | $\frac{\sqrt{3}}{2}$ | $\frac{1}{\sqrt{3}}$ |

| 45° | $\frac{\pi}{4}$ | $\frac{1}{\sqrt{2}}$ | $\frac{1}{\sqrt{2}}$ | 1 |

| 90° | $\frac{\pi}{2}$ | 1 | 0 | X |

| 120° | $\frac{2\pi}{3}$ | $\frac{\sqrt{3}}{2}$ | $-\frac{1}{2}$ | $-\sqrt{3}$ |

| 135° | $\frac{3\pi}{4}$ | $\frac{1}{\sqrt{2}}$ | $-\frac{1}{\sqrt{2}}$ | -1 |

| 180° | $\pi$ | 0 | -1 | 0 |

3. 最終的な答え

上記の表をご覧ください。