2.8.3 主值间的关系

$$\arcsin x=\frac{\pi }{2}-\arccos x=\arctan \frac{x}{\sqrt{1-x^2}}$$$$\begin{array}{}\text{(2.145)}& =\{\begin{array}{ll}-\arccos \sqrt{1-x^2}& \left(-1\le x\le 0\right),\\ \arccos \sqrt{1-x^2}& \left(0\le x\le 1\right),\end{array}\end{array}$$$$\arccos x=\frac{\pi }{2}-\arcsin x=\mathrm{arccot}\frac{x}{\sqrt{1-x^2}}$$$$\begin{array}{}\text{(2.146)}& =\{\begin{array}{ll}\pi -\arcsin \sqrt{1-x^2}& \left(\pi -1\le x\le 0\right),\\ \arcsin \sqrt{1-x^2}& \left(0\le x\le 1\right),\end{array}\end{array}$$$$\begin{array}{}\text{(2.147)}& \arctan x=\frac{\pi }{2}-\mathrm{arccot}x=\arcsin \frac{x}{\sqrt{1+x^2}},\end{array}$$$$\arctan x=\{\begin{array}{ll}\mathrm{arccot}\frac{1}{x}-\pi & \left(x<0\right)\\ \mathrm{arccot}\frac{1}{x}& \left(x>0\right)\end{array}=\{\begin{array}{ll}-\arccos \frac{1}{\sqrt{1+x^2}}& \left(x\le 0\right),\\ \arccos \frac{1}{\sqrt{1+x^2}}& \left(x\ge 0\right),\end{array}$$

(2.148)

$$\begin{array}{}\text{(2.149)}& \mathrm{arccot}x=\frac{\pi }{2}-\arctan x=\arccos \frac{x}{\sqrt{1+x^2}},\end{array}$$$$\mathrm{arccot}x=\{\begin{array}{ll}\arctan \frac{1}{x}+\pi & \left(x<0\right)\\ \arctan \frac{1}{x}& \left(x>0\right)\end{array}=\{\begin{array}{ll}\pi -\arcsin \frac{1}{\sqrt{1+x^2}}& \left(x\le 0\right),\\ \arcsin \frac{1}{\sqrt{1+x^2}}& \left(x\ge 0\right).\end{array}$$

(2.150)