(1) 式 $(sin15^\circ + sin75^\circ)^2 + (cos75^\circ - cos15^\circ)^2$ の値を求めます。 (2) $sin\theta + cos\theta = \frac{3}{4}$ を満たすとき、$sin\theta cos\theta$ と $sin^2\theta cos\theta + sin\theta cos^2\theta$ の値を求めます。

1. 問題の内容

(1) 式 $(sin15^\circ + sin75^\circ)^2 + (cos75^\circ - cos15^\circ)^2$ の値を求めます。

(2) $sin\theta + cos\theta = \frac{3}{4}$ を満たすとき、$sin\theta cos\theta$ と $sin^2\theta cos\theta + sin\theta cos^2\theta$ の値を求めます。

2. 解き方の手順

(1)

まず、三角関数の和積の公式や、sin, cos の相互関係を利用して式を簡単にします。

$sin75^\circ = sin(45^\circ + 30^\circ) = sin45^\circ cos30^\circ + cos45^\circ sin30^\circ = \frac{\sqrt{2}}{2} \cdot \frac{\sqrt{3}}{2} + \frac{\sqrt{2}}{2} \cdot \frac{1}{2} = \frac{\sqrt{6} + \sqrt{2}}{4}$

$sin15^\circ = sin(45^\circ - 30^\circ) = sin45^\circ cos30^\circ - cos45^\circ sin30^\circ = \frac{\sqrt{2}}{2} \cdot \frac{\sqrt{3}}{2} - \frac{\sqrt{2}}{2} \cdot \frac{1}{2} = \frac{\sqrt{6} - \sqrt{2}}{4}$

$sin15^\circ + sin75^\circ = \frac{\sqrt{6} - \sqrt{2}}{4} + \frac{\sqrt{6} + \sqrt{2}}{4} = \frac{2\sqrt{6}}{4} = \frac{\sqrt{6}}{2}$

$cos75^\circ = cos(45^\circ + 30^\circ) = cos45^\circ cos30^\circ - sin45^\circ sin30^\circ = \frac{\sqrt{2}}{2} \cdot \frac{\sqrt{3}}{2} - \frac{\sqrt{2}}{2} \cdot \frac{1}{2} = \frac{\sqrt{6} - \sqrt{2}}{4}$

$cos15^\circ = cos(45^\circ - 30^\circ) = cos45^\circ cos30^\circ + sin45^\circ sin30^\circ = \frac{\sqrt{2}}{2} \cdot \frac{\sqrt{3}}{2} + \frac{\sqrt{2}}{2} \cdot \frac{1}{2} = \frac{\sqrt{6} + \sqrt{2}}{4}$

$cos75^\circ - cos15^\circ = \frac{\sqrt{6} - \sqrt{2}}{4} - \frac{\sqrt{6} + \sqrt{2}}{4} = \frac{-2\sqrt{2}}{4} = -\frac{\sqrt{2}}{2}$

したがって、

$(sin15^\circ + sin75^\circ)^2 + (cos75^\circ - cos15^\circ)^2 = (\frac{\sqrt{6}}{2})^2 + (-\frac{\sqrt{2}}{2})^2 = \frac{6}{4} + \frac{2}{4} = \frac{8}{4} = 2$

(2)

$sin\theta + cos\theta = \frac{3}{4}$ の両辺を2乗すると、

$(sin\theta + cos\theta)^2 = (\frac{3}{4})^2$

$sin^2\theta + 2sin\theta cos\theta + cos^2\theta = \frac{9}{16}$

$1 + 2sin\theta cos\theta = \frac{9}{16}$

$2sin\theta cos\theta = \frac{9}{16} - 1 = -\frac{7}{16}$

$sin\theta cos\theta = -\frac{7}{32}$

次に、$sin^2\theta cos\theta + sin\theta cos^2\theta$ を計算します。

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

$= (-\frac{7}{32}) (\frac{3}{4}) = -\frac{21}{128}$

3. 最終的な答え

(1) 2

(2) $sin\theta cos\theta = -\frac{7}{32}$

$sin^2\theta cos\theta + sin\theta cos^2\theta = -\frac{21}{128}$