$\sin \theta + \cos \theta = \frac{1}{2}$ ($ \frac{\pi}{2} \le \theta \le \pi $) のとき、$\sin 2\theta$, $\cos \theta - \sin \theta$, $\cos 2\theta$, $\tan 2\theta$, $\sin(\theta + \frac{3}{4}\pi)$ の値を求める問題です。

1. 問題の内容

$\sin \theta + \cos \theta = \frac{1}{2}$ ($\frac{\pi}{2} \le \theta \le \pi$) のとき、$\sin 2\theta$, $\cos \theta - \sin \theta$, $\cos 2\theta$, $\tan 2\theta$, $\sin(\theta + \frac{3}{4}\pi)$ の値を求める問題です。

2. 解き方の手順

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

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

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

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

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

よって、$\sin 2\theta = -\frac{3}{4}$

(2) $\sin \theta + \cos \theta = \frac{1}{2}$ より、$\cos \theta = \frac{1}{2} - \sin \theta$ となります。これを $\sin^2 \theta + \cos^2 \theta = 1$ に代入します。

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

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

$2 \sin^2 \theta - \sin \theta - \frac{3}{4} = 0$

$8 \sin^2 \theta - 4 \sin \theta - 3 = 0$

$\sin \theta = \frac{4 \pm \sqrt{16 - 4(8)(-3)}}{16} = \frac{4 \pm \sqrt{16 + 96}}{16} = \frac{4 \pm \sqrt{112}}{16} = \frac{4 \pm 4\sqrt{7}}{16} = \frac{1 \pm \sqrt{7}}{4}$

$\frac{\pi}{2} \le \theta \le \pi$ なので、$\sin \theta > 0$ です。

よって、$\sin \theta = \frac{1 + \sqrt{7}}{4}$

$\cos \theta = \frac{1}{2} - \sin \theta = \frac{1}{2} - \frac{1 + \sqrt{7}}{4} = \frac{2 - 1 - \sqrt{7}}{4} = \frac{1 - \sqrt{7}}{4}$

$\cos \theta - \sin \theta = \frac{1 - \sqrt{7}}{4} - \frac{1 + \sqrt{7}}{4} = \frac{1 - \sqrt{7} - 1 - \sqrt{7}}{4} = \frac{-2\sqrt{7}}{4} = -\frac{\sqrt{7}}{2}$

(3) $\cos 2\theta = 1 - 2\sin^2 \theta = 1 - 2 (\frac{1 + \sqrt{7}}{4})^2 = 1 - 2(\frac{1 + 2\sqrt{7} + 7}{16}) = 1 - \frac{8 + 2\sqrt{7}}{8} = 1 - (1 + \frac{\sqrt{7}}{4}) = -\frac{\sqrt{7}}{4}$

(4) $\tan 2\theta = \frac{\sin 2\theta}{\cos 2\theta} = \frac{-\frac{3}{4}}{-\frac{\sqrt{7}}{4}} = \frac{3}{\sqrt{7}} = \frac{3\sqrt{7}}{7}$

(5) $\sin(\theta + \frac{3}{4}\pi) = \sin \theta \cos \frac{3}{4}\pi + \cos \theta \sin \frac{3}{4}\pi = \sin \theta (-\frac{\sqrt{2}}{2}) + \cos \theta (\frac{\sqrt{2}}{2}) = \frac{\sqrt{2}}{2}(\cos \theta - \sin \theta) = \frac{\sqrt{2}}{2}(-\frac{\sqrt{7}}{2}) = -\frac{\sqrt{14}}{4}$

3. 最終的な答え

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

$\cos \theta - \sin \theta = -\frac{\sqrt{7}}{2}$

$\cos 2\theta = -\frac{\sqrt{7}}{4}$

$\tan 2\theta = \frac{3\sqrt{7}}{7}$

$\sin(\theta + \frac{3}{4}\pi) = -\frac{\sqrt{14}}{4}$