2.8.8 arcsin x,arccos x 及 arctan x 间的特殊关系

$$\begin{array}{}\text{(2.161a)}& 2\arcsin x=\arcsin \left(2x\sqrt{1-x^2}\right)\left(\left|x\right|\le \frac{1}{\sqrt{2}}\right)\end{array}$$$$\begin{array}{}\text{(2.161b)}& =\pi -\arcsin \left(2x\sqrt{1-x^2}\right)\left(\frac{1}{\sqrt{2}}<x\le 1\right)\end{array}$$$$\begin{array}{}\text{(2.161c)}& =-\pi -\arcsin \left(2x\sqrt{1-x^2}\right)\left(-1\le x<-\frac{1}{\sqrt{2}}\right)\end{array}$$$$\begin{array}{}\text{(2.162a)}& 2\arccos x=\arccos \left(2x^2-1\right)\left(0\le x\le 1\right)\end{array}$$$$\begin{array}{}\text{(2.162b)}& =2\pi -\arccos \left(2x^2-1\right)\left(-1\le x<0\right).\end{array}$$$$\begin{array}{}\text{(2.163a)}& 2\arctan x=\arctan \frac{2x}{1-x^2}\left(\left|x\right|<1\right)\end{array}$$$$\begin{array}{}\text{(2.163b)}& =\pi +\arctan \frac{2x}{1-x^2}\left(x>1\right)\end{array}$$$$\begin{array}{}\text{(2.163c)}& =-\pi +\arctan \frac{2x}{1-x^2}\left(x<-1\right).\end{array}$$$$\begin{array}{}\text{(2.164)}& \cos \left(n\arccos x\right)=T_n\left(x\right)\left(n\ge 1\right),\end{array}$$

其中的 $n\ge 1$ 也可能为分数,

$$\begin{array}{}\text{(2.165)}& T_n\left(x\right)=\frac{{(x+\sqrt{x^2-1})}^n+{(x-\sqrt{x^2-1})}^n}{2}.\end{array}$$

对任意整数 $n,T_n\left(x\right)$ 是关于 $x$ 的多项式 (切比雪夫多项式). 关于切比雪夫多项式的性质参见第 1283 页 19.6.3.