数学 – 円の中にひそむ関数 - 三角関数 – (三角関数の合成2)

数学読本〈2〉

学習環境

• 数式入力ソフト(TeX, MathML): MathType
• MathML対応ブラウザ: Firefox、Safari
• MathML非対応ブラウザ(Edge/Internet Explorer, Google Chrome...)用JavaScript Library: MathJax

数学読本〈2〉簡単な関数/平面図形と式/指数関数・対数関数/三角関数 (松坂 和夫(著)、岩波書店)の第8章(円の中にひそむ関数 - 三角関数)、8.2(加法定理)、三角関数の合成、問21.を取り組んでみる。

問21.

1. 
   $\begin{array}{l}\sin \theta -\cos \theta =0\\ \sqrt{2}\cos \frac{\pi }{4}\sin \theta -\sqrt{2}\sin \frac{\pi }{4}\cos \theta =0\\ \sin \theta \cos \frac{\pi }{4}-\cos \theta \sin \frac{\pi }{4}=0\\ \sin \left(\theta -\frac{\pi }{4}\right)=0\\ \theta -\frac{\pi }{4}=n\pi \\ \theta =\frac{\pi }{4}+n\pi \\ \theta =\frac{\pi }{4},\frac{5}{4}\pi \end{array}$
2. 
   $\begin{array}{l}\sqrt{2}\sin \theta \cos \frac{\pi }{4}+\sqrt{2}\cos \theta \sin \frac{\pi }{4}=-1\\ \sqrt{2}\sin \left(\theta +\frac{\pi }{4}\right)=-1\\ \sin \left(\theta +\frac{\pi }{4}\right)=-\frac{1}{\sqrt{2}}\\ \theta +\frac{\pi }{4}=\frac{5}{4}\pi +2n\pi ,\frac{7}{4}\pi +2n\pi \\ \theta =\pi +2n\pi ,\frac{3}{2}\pi +2n\pi \\ \theta =\pi ,\frac{3}{2}\pi \end{array}$
3. 
   $\begin{array}{l}2\sin \theta \cos \frac{\pi }{3}-2\cos \theta \sin \frac{\pi }{3}=0\\ \sin \theta \cos \frac{\pi }{3}-\cos \theta \sin \frac{\pi }{3}=0\\ \sin \left(\theta -\frac{\pi }{3}\right)=0\\ \theta -\frac{\pi }{3}=n\pi \\ \theta =\frac{\pi }{3}+n\pi \\ \theta =\frac{\pi }{3},\frac{4}{3}\pi \\ \end{array}$
4. 
   $\begin{array}{l}2\cos \theta \cos \frac{\pi }{3}-2\sin \theta \sin \frac{\pi }{3}=1\\ \cos \theta \cos \frac{\pi }{3}-\sin \theta \sin \frac{\pi }{3}=\frac{1}{2}\\ \cos \left(\theta +\frac{\pi }{3}\right)=\frac{1}{2}\\ \theta +\frac{\pi }{3}=\frac{\pi }{3}+2n\pi ,\frac{5}{3}\pi +2n\pi \\ \theta =2n\pi ,\frac{4}{3}\pi +2n\pi \\ \theta =0,\frac{4}{3}\pi \end{array}$