## 2016年5月18日水曜日

### 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

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

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

3. $\begin{array}{l}\mathrm{tan}3\theta \\ =\frac{\mathrm{tan}2\theta +\mathrm{tan}\theta }{1-\mathrm{tan}2\theta \mathrm{tan}\theta }\\ =\frac{\frac{2\mathrm{tan}\theta }{1-{\mathrm{tan}}^{2}\theta }+\mathrm{tan}\theta }{1-\frac{2\mathrm{tan}\theta }{1-{\mathrm{tan}}^{2}\theta }·\mathrm{tan}\theta }\\ =\frac{2\mathrm{tan}\theta +\mathrm{tan}\theta -{\mathrm{tan}}^{3}\theta }{1-{\mathrm{tan}}^{2}\theta -2{\mathrm{tan}}^{2}\theta }\\ =\frac{3\mathrm{tan}\theta -{\mathrm{tan}}^{3}\theta }{1-3{\mathrm{tan}}^{2}\theta }\end{array}$

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

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