数学 – 円の中にひそむ関数 - 三角関数 – (三角関数の諸公式(半角、2倍角、3倍角・・・))

数学読本〈2〉

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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)、三角関数の諸公式、問28.を取り組んでみる。

問28.

1. 
   $\begin{array}{l}{\left(\frac{1-\tan \theta }{1+\tan \theta }\right)}^2\\ ={\left(\frac{1-\frac{\sin \theta }{\cos \theta }}{1+\frac{\sin \theta }{\cos \theta }}\right)}^2\\ ={\left(\frac{\cos \theta -\sin \theta }{\cos \theta +\sin \theta }\right)}^2\\ =\frac{1-2\sin \theta \cos \theta }{1+2\sin \theta \cos \theta }\\ =\frac{1-\sin 2\theta }{1+\sin 2\theta }\end{array}$
2. 
   $\begin{array}{l}\cos 3\theta +\sin 3\theta \\ =\cos 2\theta \cos \theta -\sin 2\theta \sin \theta +\sin 2\theta \cos \theta +\cos 2\theta \sin \theta \\ =\left({\cos }^2\theta -{\sin }^2\theta \right)\cos \theta -2{\sin }^2\theta \cos \theta +2\sin \theta {\cos }^2\theta +\left({\cos }^2\theta -{\sin }^2\theta \right)\sin \theta \\ =\left(2{\cos }^2\theta -1\right)\cos \theta -2\left(1-{\cos }^2\theta \right)\cos \theta +2\sin \theta \left(1-{\sin }^2\theta \right)+\left(1-2{\sin }^2\theta \right)\sin \theta \\ =2{\cos }^3\theta -\cos \theta -2\cos \theta +2{\cos }^3\theta +2\sin \theta -2{\sin }^3\theta +\sin \theta -2{\sin }^3\theta \\ =4\left({\cos }^3\theta -{\sin }^3\theta \right)-3\left(\cos \theta +\sin \theta \right)\\ =\left(\cos \theta -\sin \theta \right)\left(4\left({\cos }^2\theta +\sin \theta \cos \theta +{\sin }^2\theta \right)-3\right)\\ =\left(\cos \theta -\sin \theta \right)\left(4\sin \theta \cos \theta +1\right)\\ =\left(\cos \theta -\sin \theta \right)\left(1+2\sin 2\theta \right)\end{array}$
3. 
   $\begin{array}{l}\tan 3\theta \\ =\frac{\tan 2\theta +\tan \theta }{1-\tan 2\theta \tan \theta }\\ =\frac{\frac{2\tan \theta }{1-{\tan }^2\theta }+\tan \theta }{1-\frac{2\tan \theta }{1-{\tan }^2\theta }·\tan \theta }\\ =\frac{2\tan \theta +\tan \theta -{\tan }^3\theta }{1-{\tan }^2\theta -2{\tan }^2\theta }\\ =\frac{3\tan \theta -{\tan }^3\theta }{1-3{\tan }^2\theta }\end{array}$
4. 
   $\begin{array}{l}\frac{1+\sin \theta -\cos \theta }{1+\sin \theta +\cos \theta }\\ =\frac{1+2\sin \frac{\theta }{2}\cos \frac{\theta }{2}-\left({\cos }^2\frac{\theta }{2}-{\sin }^2\frac{\theta }{2}\right)}{1+2\sin \frac{\theta }{2}\cos \frac{\theta }{2}+\left({\cos }^2\frac{\theta }{2}-{\sin }^2\frac{\theta }{2}\right)}\\ =\frac{1+2\sin \frac{\theta }{2}\cos \frac{\theta }{2}-\left(1-2{\sin }^2\frac{\theta }{2}\right)}{1+2\sin \frac{\theta }{2}cos\frac{\theta }{2}+\left(2{\cos }^2\frac{\theta }{2}-1\right)}\\ =\frac{2\sin \frac{\theta }{2}\cos \frac{\theta }{2}+2{\sin }^2\frac{\theta }{2}}{2\sin \frac{\theta }{2}\cos \frac{\theta }{2}+2{\cos }^2\frac{\theta }{2}}\\ =\frac{2\sin \frac{\theta }{2}\left(\cos \frac{\theta }{2}+\sin \frac{\theta }{2}\right)}{2\cos \frac{\theta }{2}\left(\sin \frac{\theta }{2}+\cos \frac{\theta }{2}\right)}\\ =\tan \frac{\theta }{2}\end{array}$
5. 
   $\begin{array}{l}\tan \frac{\theta }{2}=\frac{\tan \frac{\theta }{3}+\tan \frac{\theta }{6}}{1-\tan \frac{\theta }{3}\tan \frac{\theta }{6}}\\ \tan \frac{\theta }{2}\left(1-\tan \frac{\theta }{3}\tan \frac{\theta }{6}\right)=\tan \frac{\theta }{3}+\tan \frac{\theta }{6}\\ \tan \frac{\theta }{2}-\tan \frac{\theta }{3}-\tan \frac{\theta }{6}=\tan \frac{\theta }{2}\tan \frac{\theta }{3}\tan \frac{\theta }{6}\end{array}$