2.8.6 $\arccos x$ 与 $\arccos y$ 的和与差

$$\arccos x+\arccos y$$$$\begin{array}{}\text{(2.157a)}& =\arccos \left(xy-\sqrt{1-x^2}\sqrt{1-y^2}\right)\left(x+y\ge 0\right)\end{array}$$$$\begin{array}{}\text{(2.157b)}& =2\pi -\arccos \left(xy-\sqrt{1-x^2}\sqrt{1-y^2}\right)\left(x+y<0\right).\end{array}$$$$\arccos x-\arccos y$$$$\begin{array}{}\text{(2.158a)}& =-\arccos \left(xy+\sqrt{1-x^2}\sqrt{1-y^2}\right)\left(x\ge y\right)\end{array}$$$$\begin{array}{}\text{(2.158b)}& =\arccos \left(xy+\sqrt{1-x^2}\sqrt{1-y^2}\right)\left(x<y\right).\end{array}$$