## 2019年11月27日水曜日

### 数学 - 円の中にひそむ関数 - 三角関数 - 加法定理 - 三角関数の諸公式の正接 - 正弦と余弦と正接、等式の証明、3倍角、半角、式の変形

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

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

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

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

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