(1) $\frac{cos\alpha}{1 - sin\alpha} - tan\alpha$ を簡単にせよ。 (2) $sin\theta + cos\theta = \sqrt{2}$ のとき $sin\theta cos\theta$ の値を求めよ。 (3) $(1 - sin\theta)(1 + sin\theta) - \frac{1}{1 + tan^2\theta}$ の値を求めよ。 (4) $sin\theta - cos\theta = \frac{1}{2}$ のとき $sin^3\theta - cos^3\theta$ の値を求めよ。

1. 問題の内容

(1) $\frac{cos\alpha}{1 - sin\alpha} - tan\alpha$ を簡単にせよ。

(2) $sin\theta + cos\theta = \sqrt{2}$ のとき $sin\theta cos\theta$ の値を求めよ。

(3) $(1 - sin\theta)(1 + sin\theta) - \frac{1}{1 + tan^2\theta}$ の値を求めよ。

(4) $sin\theta - cos\theta = \frac{1}{2}$ のとき $sin^3\theta - cos^3\theta$ の値を求めよ。

2. 解き方の手順

(1)

$\frac{cos\alpha}{1 - sin\alpha} - tan\alpha = \frac{cos\alpha}{1 - sin\alpha} - \frac{sin\alpha}{cos\alpha}$

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

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

(2)

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

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

$sin^2\theta + 2sin\theta cos\theta + cos^2\theta = 2$

$1 + 2sin\theta cos\theta = 2$

$2sin\theta cos\theta = 1$

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

(3)

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

$= cos^2\theta - \frac{1}{\frac{cos^2\theta + sin^2\theta}{cos^2\theta}} = cos^2\theta - \frac{1}{\frac{1}{cos^2\theta}} = cos^2\theta - cos^2\theta = 0$

(4)

$sin\theta - cos\theta = \frac{1}{2}$ の両辺を2乗すると、

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

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

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

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

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

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

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

3. 最終的な答え

(1) $\frac{1}{cos\alpha}$

(2) $\frac{1}{2}$

(3) $0$

(4) $\frac{11}{16}$