$\sin\theta + \cos\theta = \frac{1}{2}$ のとき、以下の値を求めます。 * $\sin\theta\cos\theta$ * $\tan\theta + \frac{1}{\tan\theta}$ * $\sin^3\theta + \cos^3\theta$

1. 問題の内容

$\sin\theta + \cos\theta = \frac{1}{2}$ のとき、以下の値を求めます。

* $\sin\theta\cos\theta$

* $\tan\theta + \frac{1}{\tan\theta}$

* $\sin^3\theta + \cos^3\theta$

2. 解き方の手順

(1) $\sin\theta\cos\theta$ の計算

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

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

$\sin^2\theta + 2\sin\theta\cos\theta + \cos^2\theta = \frac{1}{4}$

$\sin^2\theta + \cos^2\theta = 1$ なので、

$1 + 2\sin\theta\cos\theta = \frac{1}{4}$

$2\sin\theta\cos\theta = \frac{1}{4} - 1 = -\frac{3}{4}$

$\sin\theta\cos\theta = -\frac{3}{8}$

(2) $\tan\theta + \frac{1}{\tan\theta}$ の計算

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

$\sin\theta\cos\theta = -\frac{3}{8}$ より、

$\tan\theta + \frac{1}{\tan\theta} = \frac{1}{-\frac{3}{8}} = -\frac{8}{3}$

(3) $\sin^3\theta + \cos^3\theta$ の計算

$\sin^3\theta + \cos^3\theta = (\sin\theta + \cos\theta)(\sin^2\theta - \sin\theta\cos\theta + \cos^2\theta) = (\sin\theta + \cos\theta)(1 - \sin\theta\cos\theta)$

$\sin\theta + \cos\theta = \frac{1}{2}$ , $\sin\theta\cos\theta = -\frac{3}{8}$ より、

$\sin^3\theta + \cos^3\theta = \frac{1}{2}(1 - (-\frac{3}{8})) = \frac{1}{2}(1 + \frac{3}{8}) = \frac{1}{2}(\frac{8}{8} + \frac{3}{8}) = \frac{1}{2}(\frac{11}{8}) = \frac{11}{16}$

3. 最終的な答え

* $\sin\theta\cos\theta = -\frac{3}{8}$

* $\tan\theta + \frac{1}{\tan\theta} = -\frac{8}{3}$

* $\sin^3\theta + \cos^3\theta = \frac{11}{16}$