与えられた三角関数の式を、$\sin \theta$ と $\cos \theta$ を用いて分解（展開）し、式を完成させる問題です。具体的には、以下の式を完成させます。 $\sqrt{2} \sin(\theta + \frac{\pi}{4}) = \frac{◻ \sin \theta + ◻ \cos \theta}{◻}$ $\frac{4}{3} \sin(\theta + \frac{3\pi}{4}) + \frac{1}{\sqrt{2}} \cos(\theta + \frac{\pi}{4}) = \frac{◻ \sin \theta + ◻ \cos \theta}{◻}$

1. 問題の内容

与えられた三角関数の式を、$\sin \theta$ と $\cos \theta$ を用いて分解（展開）し、式を完成させる問題です。具体的には、以下の式を完成させます。

$\sqrt{2} \sin(\theta + \frac{\pi}{4}) = \frac{◻ \sin \theta + ◻ \cos \theta}{◻}$

$\frac{4}{3} \sin(\theta + \frac{3\pi}{4}) + \frac{1}{\sqrt{2}} \cos(\theta + \frac{\pi}{4}) = \frac{◻ \sin \theta + ◻ \cos \theta}{◻}$

2. 解き方の手順

(1) $\sqrt{2} \sin(\theta + \frac{\pi}{4})$ を加法定理を用いて展開します。

$\sin(\theta + \frac{\pi}{4}) = \sin \theta \cos \frac{\pi}{4} + \cos \theta \sin \frac{\pi}{4}$

$\cos \frac{\pi}{4} = \sin \frac{\pi}{4} = \frac{1}{\sqrt{2}}$ なので、

$\sin(\theta + \frac{\pi}{4}) = \frac{1}{\sqrt{2}} \sin \theta + \frac{1}{\sqrt{2}} \cos \theta$

よって、$\sqrt{2} \sin(\theta + \frac{\pi}{4}) = \sqrt{2} (\frac{1}{\sqrt{2}} \sin \theta + \frac{1}{\sqrt{2}} \cos \theta) = \sin \theta + \cos \theta$

したがって、$\sqrt{2} \sin(\theta + \frac{\pi}{4}) = \frac{\sqrt{2} \sin \theta + \sqrt{2} \cos \theta}{\sqrt{2}} = \sin \theta + \cos \theta$

(2) $\sin(\theta + \frac{3\pi}{4})$ と $\cos(\theta + \frac{\pi}{4})$ を加法定理を用いて展開します。

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

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

$\cos \frac{3\pi}{4} = - \frac{1}{\sqrt{2}}, \sin \frac{3\pi}{4} = \frac{1}{\sqrt{2}}, \cos \frac{\pi}{4} = \sin \frac{\pi}{4} = \frac{1}{\sqrt{2}}$ なので、

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

$\cos(\theta + \frac{\pi}{4}) = \frac{1}{\sqrt{2}} \cos \theta - \frac{1}{\sqrt{2}} \sin \theta$

したがって、

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

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

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

$= (-\frac{4\sqrt{2}}{6} - \frac{3}{6}) \sin \theta + (\frac{4\sqrt{2}}{6} + \frac{3}{6}) \cos \theta$

$= \frac{(-4\sqrt{2} - 3) \sin \theta + (4\sqrt{2} + 3) \cos \theta}{6}$

$= \frac{(-4\sqrt{2} - 3) \sin \theta + (4\sqrt{2} + 3) \cos \theta}{6}$

3. 最終的な答え

$\sqrt{2} \sin(\theta + \frac{\pi}{4}) = \sin \theta + \cos \theta$

$\frac{4}{3} \sin(\theta + \frac{3\pi}{4}) + \frac{1}{\sqrt{2}} \cos(\theta + \frac{\pi}{4}) = \frac{(-4\sqrt{2} - 3) \sin \theta + (4\sqrt{2} + 3) \cos \theta}{6}$