$\sin\theta + \cos\theta = \frac{\sqrt{2}}{2}$のとき、以下の値を求めよ。 (1) $\sin\theta\cos\theta$ (2) $\sin^3\theta + \cos^3\theta$ (3) $\tan\theta + \frac{1}{\tan\theta}$

1. 問題の内容

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

(1) $\sin\theta\cos\theta$

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

(3) $\tan\theta + \frac{1}{\tan\theta}$

2. 解き方の手順

(1) $\sin\theta + \cos\theta = \frac{\sqrt{2}}{2}$ の両辺を2乗すると、

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

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

$\sin^2\theta + \cos^2\theta = 1$ であるから、

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

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

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

(2) $\sin^3\theta + \cos^3\theta$ は、因数分解の公式 $a^3 + b^3 = (a+b)(a^2 - ab + b^2)$ を用いると、

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

$= (\sin\theta + \cos\theta)(1 - \sin\theta\cos\theta)$

$\sin\theta + \cos\theta = \frac{\sqrt{2}}{2}$、$\sin\theta\cos\theta = -\frac{1}{4}$であるから、

$\sin^3\theta + \cos^3\theta = \frac{\sqrt{2}}{2}(1 - (-\frac{1}{4})) = \frac{\sqrt{2}}{2}(1 + \frac{1}{4}) = \frac{\sqrt{2}}{2}(\frac{5}{4}) = \frac{5\sqrt{2}}{8}$

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

$\sin\theta\cos\theta = -\frac{1}{4}$ であるから、

$\tan\theta + \frac{1}{\tan\theta} = \frac{1}{-\frac{1}{4}} = -4$

3. 最終的な答え

(1) $\sin\theta\cos\theta = -\frac{1}{4}$

(2) $\sin^3\theta + \cos^3\theta = \frac{5\sqrt{2}}{8}$

(3) $\tan\theta + \frac{1}{\tan\theta} = -4$