# Trigonometric functions/R/Inverse functions/Analytic properties/Section

## Corollary

The real sine function induces a bijective, strictly increasing function

${\displaystyle [-\pi /2,\pi /2]\longrightarrow [-1,1],}$

and the real cosine function induces a bijective, strictly decreasing function

${\displaystyle [0,\pi ]\longrightarrow [-1,1].}$

### Proof

${\displaystyle \Box }$

## Corollary

The real tangent function induces a bijective, strictly increasing function

${\displaystyle ]-\pi /2,\pi /2[\longrightarrow \mathbb {R} ,}$

and the real cotangent function induces a bijective strictly decreasing function

${\displaystyle [0,\pi ]\longrightarrow \mathbb {R} .}$

### Proof

${\displaystyle \Box }$

Due to the bijectivity of sine, cosine, tangent and cotangent on suitable interval, there exist the following inverse functions.

## Definition

The inverse function of the real sine function is

${\displaystyle [-1,1]\longrightarrow [-{\frac {\pi }{2}},{\frac {\pi }{2}}],x\longmapsto \arcsin x,}$
and is called arcsine.

## Definition

The inverse function of the real cosine function is

${\displaystyle [-1,1]\longrightarrow [0,\pi ],x\longmapsto \arccos x,}$
and is called arccosine.

## Definition

The inverse function of the real tangent function is

${\displaystyle \mathbb {R} \longrightarrow ]-{\frac {\pi }{2}},{\frac {\pi }{2}}[,x\longmapsto \arctan x,}$
and is called arctangent.

## Definition

The inverse function of the real cotangent function is

${\displaystyle \mathbb {R} \longrightarrow ]0,\pi [,x\longmapsto \operatorname {arccot} x,}$
and is called arccotangent.

## Theorem

The inverse trigonometric functions have the following

derivatives.
1. ${\displaystyle {}{\left(\arcsin x\right)}'={\frac {1}{\sqrt {1-x^{2}}}}\,.}$
2. ${\displaystyle {}{\left(\arccos x\right)}'=-{\frac {1}{\sqrt {1-x^{2}}}}\,.}$
3. ${\displaystyle {}{\left(\arctan x\right)}'={\frac {1}{1+x^{2}}}\,.}$
4. ${\displaystyle {}{\left(\operatorname {arccot} x\right)}'=-{\frac {1}{1+x^{2}}}\,.}$

### Proof

For example, for the arctangent, we have, due to fact,

{\displaystyle {}{\begin{aligned}(\arctan x)^{\prime }&={\frac {1}{\frac {1}{\cos ^{2}(\arctan x)}}}\\&={\frac {1}{\frac {\cos ^{2}(\arctan x)+\sin ^{2}(\arctan x)}{\cos ^{2}(\arctan x)}}}\\&={\frac {1}{1+\tan ^{2}(\arctan x)}}\\&={\frac {1}{1+x^{2}}}.\end{aligned}}}
${\displaystyle \Box }$