$\sin \theta + \cos \theta = \frac{1}{5}$ のとき、次の式の値を求めます。 (i) $\sin \theta \cos \theta$ (ii) $\sin^3 \theta + \cos^3 \theta$

1. 問題の内容

$\sin \theta + \cos \theta = \frac{1}{5}$ のとき、次の式の値を求めます。

(i) $\sin \theta \cos \theta$

(ii) $\sin^3 \theta + \cos^3 \theta$

2. 解き方の手順

(i) $\sin \theta + \cos \theta = \frac{1}{5}$ の両辺を2乗します。

$(\sin \theta + \cos \theta)^2 = (\frac{1}{5})^2$

$\sin^2 \theta + 2 \sin \theta \cos \theta + \cos^2 \theta = \frac{1}{25}$

$\sin^2 \theta + \cos^2 \theta = 1$ より、

$1 + 2 \sin \theta \cos \theta = \frac{1}{25}$

$2 \sin \theta \cos \theta = \frac{1}{25} - 1 = \frac{1}{25} - \frac{25}{25} = -\frac{24}{25}$

$\sin \theta \cos \theta = -\frac{12}{25}$

(ii) $\sin^3 \theta + \cos^3 \theta = (\sin \theta + \cos \theta)(\sin^2 \theta - \sin \theta \cos \theta + \cos^2 \theta)$

$\sin^3 \theta + \cos^3 \theta = (\sin \theta + \cos \theta)(\sin^2 \theta + \cos^2 \theta - \sin \theta \cos \theta)$

$\sin^3 \theta + \cos^3 \theta = (\frac{1}{5})(1 - (-\frac{12}{25})) = (\frac{1}{5})(1 + \frac{12}{25}) = (\frac{1}{5})(\frac{25}{25} + \frac{12}{25}) = (\frac{1}{5})(\frac{37}{25}) = \frac{37}{125}$

3. 最終的な答え

(i) $\sin \theta \cos \theta = -\frac{12}{25}$

(ii) $\sin^3 \theta + \cos^3 \theta = \frac{37}{125}$