(1) $\cos A = \frac{1}{5}$のとき、$\sin A$, $\tan A$の値を求め、空欄を埋める問題。 (2) $\sin A = \frac{3}{5}$のとき、$\cos A$, $\tan A$の値を求め、空欄を埋める問題。

1. 問題の内容

(1) $\cos A = \frac{1}{5}$のとき、$\sin A$, $\tan A$の値を求め、空欄を埋める問題。

(2) $\sin A = \frac{3}{5}$のとき、$\cos A$, $\tan A$の値を求め、空欄を埋める問題。

2. 解き方の手順

(1)

$\sin^2 A + \cos^2 A = 1$ より、

$\sin^2 A = 1 - \cos^2 A$

$\sin^2 A = 1 - (\frac{1}{5})^2 = 1 - \frac{1}{25} = \frac{24}{25}$

$\sin A > 0$ より、

$\sin A = \sqrt{\frac{24}{25}} = \frac{\sqrt{24}}{5} = \frac{2\sqrt{6}}{5}$

$\tan A = \frac{\sin A}{\cos A} = \sin A \div \cos A = \frac{2\sqrt{6}}{5} \div \frac{1}{5} = \frac{2\sqrt{6}}{5} \times 5 = 2\sqrt{6}$

(2)

$\sin^2 A + \cos^2 A = 1$ より、

$\cos^2 A = 1 - \sin^2 A$

$\cos^2 A = 1 - (\frac{3}{5})^2 = 1 - \frac{9}{25} = \frac{16}{25}$

$\cos A > 0$ より、

$\cos A = \sqrt{\frac{16}{25}} = \frac{4}{5}$

$\tan A = \frac{\sin A}{\cos A} = \sin A \div \cos A = \frac{3}{5} \div \frac{4}{5} = \frac{3}{5} \times \frac{5}{4} = \frac{3}{4}$

3. 最終的な答え

(1)

$\sin^2 A = 1 - (\frac{1}{5})^2 = \frac{24}{25}$

$\sin A = \frac{2\sqrt{6}}{5}$

$\tan A = 2\sqrt{6}$

(2)

$\cos^2 A = 1 - (\frac{3}{5})^2 = \frac{16}{25}$

$\cos A = \frac{4}{5}$

$\tan A = \frac{3}{4}$