与えられた三角関数の値を計算する問題です。 (1) $\cos 390^{\circ}$, $\tan 765^{\circ}$, $\sin 1470^{\circ}$ (2) $\tan (-60^{\circ})$, $\sin (-120^{\circ})$, $\cos (-420^{\circ})$

1. 問題の内容

与えられた三角関数の値を計算する問題です。

(1) $\cos 390^{\circ}$, $\tan 765^{\circ}$, $\sin 1470^{\circ}$

(2) $\tan (-60^{\circ})$, $\sin (-120^{\circ})$, $\cos (-420^{\circ})$

2. 解き方の手順

(1)

$\cos 390^{\circ}$:

$390^{\circ} = 360^{\circ} + 30^{\circ}$なので、$\cos 390^{\circ} = \cos 30^{\circ}$。

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

$\tan 765^{\circ}$:

$765^{\circ} = 2 \times 360^{\circ} + 45^{\circ}$なので、$\tan 765^{\circ} = \tan 45^{\circ}$。

$\tan 45^{\circ} = 1$

$\sin 1470^{\circ}$:

$1470^{\circ} = 4 \times 360^{\circ} + 30^{\circ} + 30^{\circ}$なので、$1470 = 360 \times 4 + 30$なので$\sin 1470^{\circ} = \sin 30^{\circ}$

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

(2)

$\tan (-60^{\circ})$:

$\tan (-60^{\circ}) = -\tan 60^{\circ}$

$\tan 60^{\circ} = \sqrt{3}$なので、$\tan (-60^{\circ}) = -\sqrt{3}$

$\sin (-120^{\circ})$:

$\sin (-120^{\circ}) = -\sin 120^{\circ}$

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

したがって、$\sin (-120^{\circ}) = -\frac{\sqrt{3}}{2}$

$\cos (-420^{\circ})$:

$\cos (-420^{\circ}) = \cos (420^{\circ}) = \cos (360^{\circ} + 60^{\circ}) = \cos 60^{\circ}$

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

3. 最終的な答え

(1)

$\cos 390^{\circ} = \frac{\sqrt{3}}{2}$

$\tan 765^{\circ} = 1$

$\sin 1470^{\circ} = \frac{1}{2}$

(2)

$\tan (-60^{\circ}) = -\sqrt{3}$

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

$\cos (-420^{\circ}) = \frac{1}{2}$