# Definition:Inverse Trigonometric Function

## Definition

As there are six basic trigonometric functions, so each of these has its inverse functions.

As follows:

### Real Numbers

Arcsine Function

From Shape of Sine Function, we have that $\sin x$ is continuous and strictly increasing on the interval $\closedint {-\dfrac \pi 2} {\dfrac \pi 2}$.

$\map \sin {-\dfrac {\pi} 2} = -1$

and:

$\sin \dfrac {\pi} 2 = 1$

Therefore, let $g: \closedint {-\dfrac \pi 2} {\dfrac \pi 2} \to \closedint {-1} 1$ be the restriction of $\sin x$ to $\closedint {-\dfrac \pi 2} {\dfrac \pi 2}$.

Thus from Inverse of Strictly Monotone Function, $g \paren x$ admits an inverse function, which will be continuous and strictly increasing on $\closedint {-1} 1$.

This function is called arcsine of $x$ and is written $\arcsin x$.

Thus:

The domain of $\arcsin x$ is $\closedint {-1} 1$
The image of $\arcsin x$ is $\closedint {-\dfrac \pi 2} {\dfrac \pi 2}$.

### Complex Plane

The principal branch of the complex inverse sine function is defined as:

$\map \arcsin z = \dfrac 1 i \, \map \Ln {i z + \sqrt {1 - z^2} }$

where:

$\Ln$ denotes the principal branch of the complex natural logarithm
$\sqrt {1 - z^2}$ denotes the principal square root of $1 - z^2$.

### Real Numbers

Arccosine Function

From Shape of Cosine Function, we have that $\cos x$ is continuous and strictly decreasing on the interval $\left[{0 \,.\,.\, \pi}\right]$.

From Cosine of Multiple of Pi, $\cos \pi = -1$ and $\cos 0 = 1$.

Therefore, let $g: \left[{0 \,.\,.\, \pi}\right] \to \left[{-1 \,.\,.\, 1}\right]$ be the restriction of $\cos x$ to $\left[{0 \,.\,.\, \pi}\right]$.

Thus from Inverse of Strictly Monotone Function, $g \left({x}\right)$ admits an inverse function, which will be continuous and strictly decreasing on $\left[{-1 \,.\,.\, 1}\right]$.

This function is called arccosine of $x$ and is written $\arccos x$.

Thus:

• The domain of $\arccos x$ is $\left[{-1 \,.\,.\, 1}\right]$
• The image of $\arccos x$ is $\left[{0 \,.\,.\, \pi}\right]$.

### Complex Plane

The principal branch of the complex inverse cosine function is defined as:

$\map \arccos z = \dfrac 1 i \, \map \Ln {z + \sqrt {z^2 - 1} }$

where:

$\Ln$ denotes the principal branch of the complex natural logarithm
$\sqrt {z^2 - 1}$ denotes the principal square root of $z^2 - 1$.

### Real Numbers

Arctangent Function

From Shape of Tangent Function, we have that $\tan x$ is continuous and strictly increasing on the interval $\left({-\dfrac \pi 2 \,.\,.\, \dfrac \pi 2}\right)$.

From the same source, we also have that:

• $\tan x \to + \infty$ as $x \to \dfrac \pi 2 ^-$
• $\tan x \to - \infty$ as $x \to -\dfrac \pi 2 ^+$

Let $g: \left({-\dfrac \pi 2 \,.\,.\, \dfrac \pi 2}\right) \to \R$ be the restriction of $\tan x$ to $\left({-\dfrac \pi 2 \,.\,.\, \dfrac \pi 2}\right)$.

Thus from Inverse of Strictly Monotone Function, $g \left({x}\right)$ admits an inverse function, which will be continuous and strictly increasing on $\R$.

This function is called arctangent of $x$ and is written $\arctan x$.

Thus:

• The domain of $\arctan x$ is $\R$
• The image of $\arctan x$ is $\left({-\dfrac \pi 2 \,.\,.\, \dfrac \pi 2}\right)$.

### Complex Plane

The principal branch of the complex inverse tangent function is defined as:

$\map \arctan z := \dfrac 1 {2 i} \, \map \Ln {\dfrac {i - z} {i + z} }$

where $\Ln$ denotes the principal branch of the complex natural logarithm.

### Real Numbers

Arccotangent Function

From Shape of Cotangent Function, we have that $\cot x$ is continuous and strictly decreasing on the interval $\openint 0 \pi$.

From the same source, we also have that:

$\cot x \to + \infty$ as $x \to 0^+$
$\cot x \to - \infty$ as $x \to \pi^-$

Let $g: \openint 0 \pi \to \R$ be the restriction of $\cot x$ to $\openint 0 \pi$.

Thus from Inverse of Strictly Monotone Function, $\map g x$ admits an inverse function, which will be continuous and strictly decreasing on $\R$.

This function is called arccotangent of $x$ and is written $\arccot x$.

Thus:

The domain of $\arccot x$ is $\R$
The image of $\arccot x$ is $\openint 0 \pi$.

### Complex Plane

The principal branch of the complex inverse cotangent function is defined as:

$\map \arccot z := \dfrac 1 {2 i} \, \map \Ln {\dfrac {z + i} {z - i} }$

where $\Ln$ denotes the principal branch of the complex natural logarithm.

### Real Numbers

Arcsecant Function

From Shape of Secant Function, we have that $\sec x$ is continuous and strictly increasing on the intervals $\left[{0 \,.\,.\, \dfrac \pi 2}\right)$ and $\left({\dfrac \pi 2 \,.\,.\, \pi}\right]$.

From the same source, we also have that:

• $\sec x \to + \infty$ as $x \to \dfrac \pi 2^-$
• $\sec x \to - \infty$ as $x \to \dfrac \pi 2^+$

Let $g: \left[{0 \,.\,.\, \dfrac \pi 2}\right) \to \left[{1 \,.\,.\, \infty}\right)$ be the restriction of $\sec x$ to $\left[{0 \,.\,.\, \dfrac \pi 2}\right)$.

Let $h: \left({\dfrac \pi 2 \,.\,.\, \pi}\right] \to \left({-\infty \,.\,.\, -1}\right]$ be the restriction of $\sec x$ to $\left({\dfrac \pi 2 \,.\,.\, \pi}\right]$.

Let $f: \left[{0 \,.\,.\, \pi}\right] \setminus \dfrac \pi 2 \to \R \setminus \left({-1 \,.\,.\, 1}\right)$:

$f\left({x}\right) = \begin{cases} g\left({x}\right) & : 0 \le x < \dfrac \pi 2 \\ h\left({x}\right) & : \dfrac \pi 2 < x \le \pi \end{cases}$

From Inverse of Strictly Monotone Function, $g \left({x}\right)$ admits an inverse function, which will be continuous and strictly increasing on $\left[{1 \,.\,.\, \infty}\right)$.

From Inverse of Strictly Monotone Function, $h \left({x}\right)$ admits an inverse function, which will be continuous and strictly increasing on $\left({-\infty \,.\,.\, -1}\right]$.

As both the domain and range of $g$ and $h$ are disjoint, it follows that:

$f^{-1}\left({x}\right) = \begin{cases} g^{-1}\left({x}\right) & : x \ge 1 \\ h^{-1}\left({x}\right) & : x \le -1 \end{cases}$

This function $f^{-1} \left({x}\right)$ is called arcsecant of $x$ and is written $\operatorname{arcsec} x$.

Thus:

• The domain of $\operatorname{arcsec} x$ is $\R \setminus \left({-1 \,.\,.\, 1}\right)$
• The image of $\operatorname{arcsec} x$ is $\left[{0 \,.\,.\, \pi}\right] \setminus \dfrac \pi 2$.

### Complex Plane

The principal branch of the complex inverse secant function is defined as:

$\forall z \in \C_{\ne 0}: \map \arcsec z := \dfrac 1 i \, \map \Ln {\dfrac {1 + \sqrt {1 - z^2} } z}$

where:

$\Ln$ denotes the principal branch of the complex natural logarithm
$\sqrt {1 - z^2}$ denotes the principal square root of $1 - z^2$.

### Real Numbers

Arccosecant Function

From Shape of Cosecant Function, we have that $\csc x$ is continuous and strictly decreasing on the intervals $\hointr {-\dfrac \pi 2} 0$ and $\hointl 0 {\dfrac \pi 2}$.

From the same source, we also have that:

$\csc x \to + \infty$ as $x \to 0^+$
$\csc x \to - \infty$ as $x \to 0^-$

Let $g: \hointr {-\dfrac \pi 2} 0 \to \hointl {-\infty} {-1}$ be the restriction of $\csc x$ to $\hointr {-\dfrac \pi 2} 0$.

Let $h: \hointl 0 {\dfrac \pi 2} \to \hointr 1 \infty$ be the restriction of $\csc x$ to $\hointl 0 {\dfrac \pi 2}$.

Let $f: \closedint {-\dfrac \pi 2} {\dfrac \pi 2} \setminus \set 0 \to \R \setminus \openint {-1} 1$:

$\map f x = \begin{cases} \map g x & : -\dfrac \pi 2 \le x < 0 \\ \map h x & : 0 < x \le \dfrac \pi 2 \end{cases}$

From Inverse of Strictly Monotone Function, $\map g x$ admits an inverse function, which will be continuous and strictly decreasing on $\hointl {-\infty} {-1}$.

From Inverse of Strictly Monotone Function, $\map h x$ admits an inverse function, which will be continuous and strictly decreasing on $\hointr 1 \infty$.

As both the domain and range of $g$ and $h$ are disjoint, it follows that:

$\map {f^{-1} } x = \begin{cases} \map {g^{-1} } x & : x \le -1 \\ \map {h^{-1} } x & : x \ge 1 \end{cases}$

This function $\map {f^{-1} } x$ is called arccosecant of $x$ and is written $\arccsc x$.

Thus:

The domain of $\arccsc x$ is $\R \setminus \openint {-1} 1$
The image of $\arccsc x$ is $\closedint {-\dfrac \pi 2} {\dfrac \pi 2} \setminus \set 0$.

### Complex Plane

The principal branch of the complex inverse cosecant function is defined as:

$\forall z \in \C_{\ne 0}: \map \arccsc z := \dfrac 1 i \, \map \Ln {\dfrac {i + \sqrt {z^2 - 1} } z}$

where:

$\Ln$ denotes the principal branch of the complex natural logarithm
$\sqrt {z^2 - 1}$ denotes the principal square root of $z^2 - 1$.

## Also see

• Results about inverse trigonometric functions can be found here.