这是一份uwa西澳大学MATH1021的成功案例

The function Arccos $\mathrm{x}$. When the graph of $y=\cos x$ is considered, we see that it is decreasing in the (maximum) interval $[0, \pi]$, so the inverse function exists for this particular restriction.
It follows that the function
$$
y=\operatorname{Arccos} x \in[0, \pi], \quad x \in[-1,1],
$$
is only defined by reflecting the graph of $y=\cos x, x \in[0, \pi]$, and by the relation
$$
\cos (\operatorname{Arccos} x)=x, \quad x \in[-1,1] .
$$
Using the method of differentiation of an inverse function, given in Section 2.2, we obtain
$$
\frac{d}{d x} \operatorname{Arccos} x=\frac{1}{-\sin (\operatorname{Arccos} x)}=-\frac{1}{\sqrt{1-\cos ^{2}(\operatorname{Arccos} x)}}=-\frac{1}{\sqrt{1-x^{2}}}
$$
Summary.
$\begin{cases}y=\operatorname{Arccos} x \in[0, \pi], & \text { for } x \in[-1,1], \ \cos (\operatorname{Arccos} x)=x, & \text { for } x \in[-1,1], \ \frac{d}{d x} \operatorname{Arccos} x=-\frac{1}{\sqrt{1-x^{2}}}, & \text { for } x \in]-1,1[.\end{cases}$

MATH1021 COURSE NOTES ：

By combining the derivatives we get
$$
\left.\frac{d}{d x}{\operatorname{Arccos} x+\operatorname{Arcsin} x}=-\frac{1}{\sqrt{1-x^{2}}}+\frac{1}{\sqrt{1-x^{2}}}=0 \quad \text { for } x \in\right]-1,1[,
$$
and
$$
\frac{d}{d x}{\operatorname{Arctan} x+\operatorname{Arccot} x}=\frac{1}{1+x^{2}}-\frac{1}{1+x^{2}}=0, \quad \text { for } x \in \mathbb{R} .
$$
Since the functions are of class $C^{1}$ with the derivative 0 , they must be constants. These constants are found by choosing e.g. $x=0$. Hence, by a trivial continuous extension in the first case,
$\operatorname{Arccos} x+\operatorname{Arcsin} x=\frac{\pi}{2} \quad$ for $x \in[-1,1]$,
and
$\operatorname{Arctan} x+\operatorname{Arccot} x=\frac{\pi}{2} \quad$ for $x \in \mathbb{R}$.