Inverse trigonometric functions

Basic concepts

Principal values

Since none of the six trigonometric functions are one-to-one, they must be restricted in order to have inverse functions. Therefore, the result ranges of the inverse functions are proper (i.e. strict) subsets of the domains of the original functions.

For example, using function in the sense of multivalued functions, just as the square root function ${\displaystyle y=\sqrt{x}}$ could be defined from ${\displaystyle y^2=x,}$ the function ${\displaystyle y=\arcsin (x)}$ is defined so that ${\displaystyle \sin (y)=x.}$ For a given real number ${\displaystyle x,}$ with ${\displaystyle -1\le x\le 1,}$ there are multiple (in fact, countably infinitely many) numbers ${\displaystyle y}$ such that ${\displaystyle \sin (y)=x}$; for example, ${\displaystyle \sin (0)=0,}$ but also ${\displaystyle \sin (\pi )=0,}$ ${\displaystyle \sin (2\pi )=0,}$ etc. When only one value is desired, the function may be restricted to its principal branch. With this restriction, for each ${\displaystyle x}$ in the domain, the expression ${\displaystyle \arcsin (x)}$ will evaluate only to a single value, called its principal value. These properties apply to all the inverse trigonometric functions.

The principal inverses are listed in the following table.

Note: Some authors define the range of arcsecant to be ($0\le y<\frac{\pi }{2}$ or $\pi \le y<\frac{3\pi }{2}$ ), because the tangent function is nonnegative on this domain. This makes some computations more consistent. For example, using this range, ${\displaystyle \tan (\mathrm{arcsec}(x))=\sqrt{x^2-1},}$ whereas with the range ($0\le y<\frac{\pi }{2}$ or $\frac{\pi }{2}<y\le \pi$), we would have to write ${\displaystyle \tan (\mathrm{arcsec}(x))=\pm \sqrt{x^2-1},}$ since tangent is nonnegative on $0\le y<\frac{\pi }{2},$ but nonpositive on $\frac{\pi }{2}<y\le \pi .$ For a similar reason, the same authors define the range of arccosecant to be $(-\pi <y\le -\frac{\pi }{2}$ or $0<y\le \frac{\pi }{2}).$

Domains

If x is allowed to be a complex number, then the range of y applies only to its real part.

The table below displays names and domains of the inverse trigonometric functions along with the range of their usual principal values in radians.

The symbol ${\displaystyle \mathbb{R}=(-\mathrm{\infty },\mathrm{\infty })}$ denotes the set of all real numbers and ${\displaystyle \mathbb{Z}=\{\dots ,-2,-1,0,1,2,\dots \}}$ denotes the set of all integers. The set of all integer multiples of ${\displaystyle \pi }$ is denoted by

$${\displaystyle \pi \mathbb{Z}:=\{\pi n:n\in \mathbb{Z}\}=\{\dots ,-2\pi ,-\pi ,0,\pi ,2\pi ,\dots \}.}$$

The symbol ${\displaystyle \setminus }$ denotes set subtraction so that, for instance, ${\displaystyle \mathbb{R}\setminus (-1,1)=(-\mathrm{\infty },-1]\cup [1,\mathrm{\infty })}$ is the set of points in ${\displaystyle \mathbb{R}}$ (that is, real numbers) that are not in the interval ${\displaystyle (-1,1).}$

The Minkowski sum notation $\pi \mathbb{Z}+(0,\pi )$ and ${\displaystyle \pi \mathbb{Z}+(-\frac{\pi }{2},\frac{\pi }{2})}$ that is used above to concisely write the domains of ${\displaystyle \cot ,\csc ,\tan ,\text{ and }\sec }$ is now explained.

Domain of cotangent ${\displaystyle \cot }$ and cosecant ${\displaystyle \csc }$: The domains of ${\displaystyle \cot }$ and ${\displaystyle \csc }$ are the same. They are the set of all angles ${\displaystyle \theta }$ at which ${\displaystyle \sin \theta \ne 0,}$ i.e. all real numbers that are not of the form ${\displaystyle \pi n}$ for some integer ${\displaystyle n,}$

$${\displaystyle \begin{array}{cc}\pi \mathbb{Z}+(0,\pi )& =\cdots \cup (-2\pi ,-\pi )\cup (-\pi ,0)\cup (0,\pi )\cup (\pi ,2\pi )\cup \cdots \\ & =\mathbb{R}\setminus \pi \mathbb{Z}\end{array}}$$

Domain of tangent ${\displaystyle \tan }$ and secant ${\displaystyle \sec }$: The domains of ${\displaystyle \tan }$ and ${\displaystyle \sec }$ are the same. They are the set of all angles ${\displaystyle \theta }$ at which ${\displaystyle \cos \theta \ne 0,}$

$${\displaystyle \begin{array}{cc}\pi \mathbb{Z}+\left(-\frac{\pi }{2},\frac{\pi }{2}\right)& =\cdots \cup (-\frac{3\pi }{2},-\frac{\pi }{2})\cup (-\frac{\pi }{2},\frac{\pi }{2})\cup (\frac{\pi }{2},\frac{3\pi }{2})\cup \cdots \\ & =\mathbb{R}\setminus \left(\frac{\pi }{2}+\pi \mathbb{Z}\right)\end{array}}$$

Solutions to elementary trigonometric equations

Each of the trigonometric functions is periodic in the real part of its argument, running through all its values twice in each interval of ${\displaystyle 2\pi :}$

• Sine and cosecant begin their period at $2\pi k-\frac{\pi }{2}$ (where ${\displaystyle k}$ is an integer), finish it at $2\pi k+\frac{\pi }{2},$ and then reverse themselves over $2\pi k+\frac{\pi }{2}$ to $2\pi k+\frac{3\pi }{2}.$
• Cosine and secant begin their period at ${\displaystyle 2\pi k,}$ finish it at ${\displaystyle 2\pi k+\pi .}$ and then reverse themselves over ${\displaystyle 2\pi k+\pi }$ to ${\displaystyle 2\pi k+2\pi .}$
• Tangent begins its period at $2\pi k-\frac{\pi }{2},$ finishes it at $2\pi k+\frac{\pi }{2},$ and then repeats it (forward) over $2\pi k+\frac{\pi }{2}$ to $2\pi k+\frac{3\pi }{2}.$
• Cotangent begins its period at ${\displaystyle 2\pi k,}$ finishes it at ${\displaystyle 2\pi k+\pi ,}$ and then repeats it (forward) over ${\displaystyle 2\pi k+\pi }$ to ${\displaystyle 2\pi k+2\pi .}$

This periodicity is reflected in the general inverses, where ${\displaystyle k}$ is some integer.

The following table shows how inverse trigonometric functions may be used to solve equalities involving the six standard trigonometric functions. It is assumed that the given values ${\displaystyle \theta ,}$ ${\displaystyle r,}$ ${\displaystyle s,}$ ${\displaystyle x,}$ and ${\displaystyle y}$ all lie within appropriate ranges so that the relevant expressions below are well-defined. Note that "for some ${\displaystyle k\in \mathbb{Z}}$" is just another way of saying "for some integer ${\displaystyle k.}$"

The symbol ${\displaystyle ⟺}$ is logical equality and indicates that if the left hand side is true then so is the right hand side and, conversely, if the right hand side is true then so is the left hand side (see this footnote for more details and an example illustrating this concept).

where the first four solutions can be written in expanded form as:

For example, if ${\displaystyle \cos \theta =-1}$ then ${\displaystyle \theta =\pi +2\pi k=-\pi +2\pi (1+k)}$ for some ${\displaystyle k\in \mathbb{Z}.}$ While if ${\displaystyle \sin \theta =\pm 1}$ then $\theta =\frac{\pi }{2}+\pi k=-\frac{\pi }{2}+\pi (k+1)$ for some ${\displaystyle k\in \mathbb{Z},}$ where ${\displaystyle k}$ will be even if ${\displaystyle \sin \theta =1}$ and it will be odd if ${\displaystyle \sin \theta =-1.}$ The equations ${\displaystyle \sec \theta =-1}$ and ${\displaystyle \csc \theta =\pm 1}$ have the same solutions as ${\displaystyle \cos \theta =-1}$ and ${\displaystyle \sin \theta =\pm 1,}$ respectively. In all equations above except for those just solved (i.e. except for ${\displaystyle \sin }$/${\displaystyle \csc \theta =\pm 1}$ and ${\displaystyle \cos }$/${\displaystyle \sec \theta =-1}$), the integer ${\displaystyle k}$ in the solution's formula is uniquely determined by ${\displaystyle \theta }$ (for fixed ${\displaystyle r,s,x,}$ and ${\displaystyle y}$).

With the help of integer parity $${\displaystyle \mathrm{Parity}(h)=\{\begin{array}{ll}0& \text{if }h\text{ is even }\\ 1& \text{if }h\text{ is odd }\end{array}}$$ it is possible to write a solution to ${\displaystyle \cos \theta =x}$ that doesn't involve the "plus or minus" ${\displaystyle \pm }$ symbol:

${\displaystyle cos\theta =x}$ if and only if ${\displaystyle \theta =(-1{)}^h\arccos (x)+\pi h+\pi \mathrm{Parity}(h)}$ for some ${\displaystyle h\in \mathbb{Z}.}$

And similarly for the secant function,

${\displaystyle sec\theta =r}$ if and only if ${\displaystyle \theta =(-1{)}^h\mathrm{arcsec}(r)+\pi h+\pi \mathrm{Parity}(h)}$ for some ${\displaystyle h\in \mathbb{Z},}$

where ${\displaystyle \pi h+\pi \mathrm{Parity}(h)}$ equals ${\displaystyle \pi h}$ when the integer ${\displaystyle h}$ is even, and equals ${\displaystyle \pi h+\pi }$ when it's odd.

Detailed example and explanation of the "plus or minus" symbol ±

The solutions to ${\displaystyle \cos \theta =x}$ and ${\displaystyle \sec \theta =x}$ involve the "plus or minus" symbol ${\displaystyle \pm ,}$ whose meaning is now clarified. Only the solution to ${\displaystyle \cos \theta =x}$ will be discussed since the discussion for ${\displaystyle \sec \theta =x}$ is the same. We are given ${\displaystyle x}$ between ${\displaystyle -1\le x\le 1}$ and we know that there is an angle ${\displaystyle \theta }$ in some interval that satisfies ${\displaystyle \cos \theta =x.}$ We want to find this ${\displaystyle \theta .}$ The table above indicates that the solution is $${\displaystyle \theta =\pm \arccos x+2\pi k\text{ for some }k\in \mathbb{Z}}$$ which is a shorthand way of saying that (at least) one of the following statement is true:

1. ${\displaystyle \theta =\arccos x+2\pi k}$ for some integer ${\displaystyle k,}$
   or
2. ${\displaystyle \theta =-\arccos x+2\pi k}$ for some integer ${\displaystyle k.}$

As mentioned above, if ${\displaystyle \arccos x=\pi }$ (which by definition only happens when ${\displaystyle x=\cos \pi =-1}$) then both statements (1) and (2) hold, although with different values for the integer ${\displaystyle k}$: if ${\displaystyle K}$ is the integer from statement (1), meaning that ${\displaystyle \theta =\pi +2\pi K}$ holds, then the integer ${\displaystyle k}$ for statement (2) is ${\displaystyle K+1}$ (because ${\displaystyle \theta =-\pi +2\pi (1+K)}$). However, if ${\displaystyle x\ne -1}$ then the integer ${\displaystyle k}$ is unique and completely determined by ${\displaystyle \theta .}$ If ${\displaystyle \arccos x=0}$ (which by definition only happens when ${\displaystyle x=\cos 0=1}$) then ${\displaystyle \pm \arccos x=0}$ (because ${\displaystyle +\arccos x=+0=0}$ and ${\displaystyle -\arccos x=-0=0}$ so in both cases ${\displaystyle \pm \arccos x}$ is equal to ${\displaystyle 0}$) and so the statements (1) and (2) happen to be identical in this particular case (and so both hold). Having considered the cases ${\displaystyle \arccos x=0}$ and ${\displaystyle \arccos x=\pi ,}$ we now focus on the case where ${\displaystyle \arccos x\ne 0}$ and ${\displaystyle \arccos x\ne \pi ,}$ So assume this from now on. The solution to ${\displaystyle \cos \theta =x}$ is still $${\displaystyle \theta =\pm \arccos x+2\pi k\text{ for some }k\in \mathbb{Z}}$$ which as before is shorthand for saying that one of statements (1) and (2) is true. However this time, because ${\displaystyle \arccos x\ne 0}$ and ${\displaystyle 0<\arccos x<\pi ,}$ statements (1) and (2) are different and furthermore, exactly one of the two equalities holds (not both). Additional information about ${\displaystyle \theta }$ is needed to determine which one holds. For example, suppose that ${\displaystyle x=0}$ and that all that is known about ${\displaystyle \theta }$ is that ${\displaystyle -\pi \le \theta \le \pi }$ (and nothing more is known). Then $${\displaystyle \arccos x=\arccos 0=\frac{\pi }{2}}$$ and moreover, in this particular case ${\displaystyle k=0}$ (for both the ${\displaystyle +}$ case and the ${\displaystyle -}$ case) and so consequently, $${\displaystyle \theta =\pm \arccos x+2\pi k=\pm \left(\frac{\pi }{2}\right)+2\pi (0)=\pm \frac{\pi }{2}.}$$ This means that ${\displaystyle \theta }$ could be either ${\displaystyle \pi /2}$ or ${\displaystyle -\pi /2.}$ Without additional information it is not possible to determine which of these values ${\displaystyle \theta }$ has. An example of some additional information that could determine the value of ${\displaystyle \theta }$ would be knowing that the angle is above the ${\displaystyle x}$-axis (in which case ${\displaystyle \theta =\pi /2}$) or alternatively, knowing that it is below the ${\displaystyle x}$-axis (in which case ${\displaystyle \theta =-\pi /2}$).

Equal identical trigonometric functions

The table below shows how two angles ${\displaystyle \theta }$ and ${\displaystyle \phi }$ must be related if their values under a given trigonometric function are equal or negatives of each other.

The vertical double arrow ${\displaystyle \Updownarrow }$ in the last row indicates that ${\displaystyle \theta }$ and ${\displaystyle \phi }$ satisfy ${\displaystyle \left|\sin \theta \right|=\left|\sin \phi \right|}$ if and only if they satisfy ${\displaystyle \left|\cos \theta \right|=\left|\cos \phi \right|.}$

Set of all solutions to elementary trigonometric equations

Thus given a single solution ${\displaystyle \theta }$ to an elementary trigonometric equation (${\displaystyle \sin \theta =y}$ is such an equation, for instance, and because ${\displaystyle \sin (\arcsin y)=y}$ always holds, ${\displaystyle \theta :=\arcsin y}$ is always a solution), the set of all solutions to it are:

Transforming equations

The equations above can be transformed by using the reflection and shift identities:

These formulas imply, in particular, that the following hold:

$${\displaystyle \begin{array}{cccccc}\sin \theta & =-\sin (-\theta )& & =-\sin (\pi +\theta )& & =\phantom{-}\sin (\pi -\theta )\\ & =-\cos \left(\frac{\pi }{2}+\theta \right)& & =\phantom{-}\cos \left(\frac{\pi }{2}-\theta \right)& & =-\cos \left(-\frac{\pi }{2}-\theta \right)\\ & =\phantom{-}\cos \left(-\frac{\pi }{2}+\theta \right)& & =-\cos \left(\frac{3\pi }{2}-\theta \right)& & =-\cos \left(-\frac{3\pi }{2}+\theta \right)\\ \cos \theta & =\phantom{-}\cos (-\theta )& & =-\cos (\pi +\theta )& & =-\cos (\pi -\theta )\\ & =\phantom{-}\sin \left(\frac{\pi }{2}+\theta \right)& & =\phantom{-}\sin \left(\frac{\pi }{2}-\theta \right)& & =-\sin \left(-\frac{\pi }{2}-\theta \right)\\ & =-\sin \left(-\frac{\pi }{2}+\theta \right)& & =-\sin \left(\frac{3\pi }{2}-\theta \right)& & =\phantom{-}\sin \left(-\frac{3\pi }{2}+\theta \right)\\ \tan \theta & =-\tan (-\theta )& & =\phantom{-}\tan (\pi +\theta )& & =-\tan (\pi -\theta )\\ & =-\cot \left(\frac{\pi }{2}+\theta \right)& & =\phantom{-}\cot \left(\frac{\pi }{2}-\theta \right)& & =\phantom{-}\cot \left(-\frac{\pi }{2}-\theta \right)\\ & =-\cot \left(-\frac{\pi }{2}+\theta \right)& & =\phantom{-}\cot \left(\frac{3\pi }{2}-\theta \right)& & =-\cot \left(-\frac{3\pi }{2}+\theta \right)\end{array}}$$

where swapping ${\displaystyle \sin \leftrightarrow \csc ,}$ swapping ${\displaystyle \cos \leftrightarrow \sec ,}$ and swapping ${\displaystyle \tan \leftrightarrow \cot }$ gives the analogous equations for ${\displaystyle \csc ,\sec ,\text{ and }\cot ,}$ respectively.

So for example, by using the equality $\sin \left(\frac{\pi }{2}-\theta \right)=\cos \theta ,$ the equation ${\displaystyle \cos \theta =x}$ can be transformed into $\sin \left(\frac{\pi }{2}-\theta \right)=x,$ which allows for the solution to the equation ${\displaystyle \sin \phi =x}$ (where $\phi :=\frac{\pi }{2}-\theta$) to be used; that solution being: ${\displaystyle \phi =(-1{)}^k\arcsin (x)+\pi k\text{ for some }k\in \mathbb{Z},}$ which becomes: $${\displaystyle \frac{\pi }{2}-\theta =(-1{)}^k\arcsin (x)+\pi k\text{ for some }k\in \mathbb{Z}}$$ where using the fact that ${\displaystyle (-1{)}^k=(-1{)}^{-k}}$ and substituting ${\displaystyle h:=-k}$ proves that another solution to ${\displaystyle \cos \theta =x}$ is: $${\displaystyle \theta =(-1{)}^{h+1}\arcsin (x)+\pi h+\frac{\pi }{2}\text{ for some }h\in \mathbb{Z}.}$$ The substitution ${\displaystyle \arcsin x=\frac{\pi }{2}-\arccos x}$ may be used express the right hand side of the above formula in terms of ${\displaystyle \arccos x}$ instead of ${\displaystyle \arcsin x.}$

Relationships between trigonometric functions and inverse trigonometric functions

Trigonometric functions of inverse trigonometric functions are tabulated below. They may be derived from the Pythagorean identities. Another way is by considering the geometry of a right-angled triangle, with one side of length 1 and another side of length ${\displaystyle x,}$ then applying the Pythagorean theorem and definitions of the trigonometric ratios. It is worth noting that for arcsecant and arccosecant, the diagram assumes that ${\displaystyle x}$ is positive, and thus the result has to be corrected through the use of absolute values and the signum (sgn) operation.

Relationships among the inverse trigonometric functions

Complementary angles:

${\displaystyle \begin{array}{cc}\arccos (x)& =\frac{\pi }{2}-\arcsin (x)\\ \mathrm{arccot}(x)& =\frac{\pi }{2}-\arctan (x)\\ \mathrm{arccsc}(x)& =\frac{\pi }{2}-\mathrm{arcsec}(x)\end{array}}$

Negative arguments:

${\displaystyle \begin{array}{cc}\arcsin (-x)& =-\arcsin (x)\\ \mathrm{arccsc}(-x)& =-\mathrm{arccsc}(x)\\ \arccos (-x)& =\pi -\arccos (x)\\ \mathrm{arcsec}(-x)& =\pi -\mathrm{arcsec}(x)\\ \arctan (-x)& =-\arctan (x)\\ \mathrm{arccot}(-x)& =\pi -\mathrm{arccot}(x)\end{array}}$

Reciprocal arguments:

${\displaystyle \begin{array}{ccc}\arcsin \left(\frac{1}{x}\right)& =\mathrm{arccsc}(x)& \\ \mathrm{arccsc}\left(\frac{1}{x}\right)& =\arcsin (x)& \\ \arccos \left(\frac{1}{x}\right)& =\mathrm{arcsec}(x)& \\ \mathrm{arcsec}\left(\frac{1}{x}\right)& =\arccos (x)& \\ \arctan \left(\frac{1}{x}\right)& =\mathrm{arccot}(x)& =\frac{\pi }{2}-\arctan (x),\text{ if }x>0\\ \arctan \left(\frac{1}{x}\right)& =\mathrm{arccot}(x)-\pi & =-\frac{\pi }{2}-\arctan (x),\text{ if }x<0\\ \mathrm{arccot}\left(\frac{1}{x}\right)& =\arctan (x)& =\frac{\pi }{2}-\mathrm{arccot}(x),\text{ if }x>0\\ \mathrm{arccot}\left(\frac{1}{x}\right)& =\arctan (x)+\pi & =\frac{3\pi }{2}-\mathrm{arccot}(x),\text{ if }x<0\end{array}}$

The identities above can be used with (and derived from) the fact that ${\displaystyle \sin }$ and ${\displaystyle \csc }$ are reciprocals (i.e. ${\displaystyle \csc =\frac{1}{\sin }}$), as are ${\displaystyle \cos }$ and ${\displaystyle \sec ,}$ and ${\displaystyle \tan }$ and ${\displaystyle \cot .}$

Useful identities if one only has a fragment of a sine table:

${\displaystyle \begin{array}{cc}\arcsin (x)& =\frac{1}{2}\arccos \left(1-2x^2\right),\text{ if }0\le x\le 1\\ \arcsin (x)& =\arctan \left(\frac{x}{\sqrt{1-x^2}}\right)\\ \arccos (x)& =\frac{1}{2}\arccos \left(2x^2-1\right),\text{ if }0\le x\le 1\\ \arccos (x)& =\arctan \left(\frac{\sqrt{1-x^2}}{x}\right)\\ \arccos (x)& =\arcsin \left(\sqrt{1-x^2}\right),\text{ if }0\le x\le 1\text{ , from which you get }\\ \arccos & \left(\frac{1-x^2}{1+x^2}\right)=\arcsin \left(\frac{2x}{1+x^2}\right),\text{ if }0\le x\le 1\\ \arcsin & \left(\sqrt{1-x^2}\right)=\frac{\pi }{2}-\mathrm{sgn}(x)\arcsin (x)\\ \arctan (x)& =\arcsin \left(\frac{x}{\sqrt{1+x^2}}\right)\\ \mathrm{arccot}(x)& =\arccos \left(\frac{x}{\sqrt{1+x^2}}\right)\end{array}}$

Whenever the square root of a complex number is used here, we choose the root with the positive real part (or positive imaginary part if the square was negative real).

A useful form that follows directly from the table above is

${\displaystyle \arctan (x)=\arccos \left(\sqrt{\frac{1}{1+x^2}}\right),\text{ if }x\ge 0}$.

It is obtained by recognizing that ${\displaystyle \cos \left(\arctan \left(x\right)\right)=\sqrt{\frac{1}{1+x^2}}=\cos \left(\arccos \left(\sqrt{\frac{1}{1+x^2}}\right)\right)}$.

From the half-angle formula, ${\displaystyle \tan \left(\frac{\theta }{2}\right)=\frac{\sin (\theta )}{1+\cos (\theta )}}$, we get:

${\displaystyle \begin{array}{cc}\arcsin (x)& =2\arctan \left(\frac{x}{1+\sqrt{1-x^2}}\right)\\ \arccos (x)& =2\arctan \left(\frac{\sqrt{1-x^2}}{1+x}\right),\text{ if }-1<x\le 1\\ \arctan (x)& =2\arctan \left(\frac{x}{1+\sqrt{1+x^2}}\right)\end{array}}$

Arctangent addition formula

${\displaystyle \arctan (u)\pm \arctan (v)=\arctan \left(\frac{u\pm v}{1\mp uv}\right)(\mathrm{mod}\pi ),uv\ne 1.}$

This is derived from the tangent addition formula

${\displaystyle \tan (\alpha \pm \beta )=\frac{\tan (\alpha )\pm \tan (\beta )}{1\mp \tan (\alpha )\tan (\beta )},}$

by letting

${\displaystyle \alpha =\arctan (u),\beta =\arctan (v).}$

In calculus

Derivatives of inverse trigonometric functions

The derivatives for complex values of z are as follows:

${\displaystyle \begin{array}{cccc}\frac{d}{dz}\arcsin (z)& =\frac{1}{\sqrt{1-z^2}};& z& \ne -1,+1\\ \frac{d}{dz}\arccos (z)& =-\frac{1}{\sqrt{1-z^2}};& z& \ne -1,+1\\ \frac{d}{dz}\arctan (z)& =\frac{1}{1+z^2};& z& \ne -i,+i\\ \frac{d}{dz}\mathrm{arccot}(z)& =-\frac{1}{1+z^2};& z& \ne -i,+i\\ \frac{d}{dz}\mathrm{arcsec}(z)& =\frac{1}{z^2\sqrt{1-\frac{1}{z^2}}};& z& \ne -1,0,+1\\ \frac{d}{dz}\mathrm{arccsc}(z)& =-\frac{1}{z^2\sqrt{1-\frac{1}{z^2}}};& z& \ne -1,0,+1\end{array}}$

Only for real values of x:

${\displaystyle \begin{array}{ccc}\frac{d}{dx}\mathrm{arcsec}(x)& =\frac{1}{|x|\sqrt{x^2-1}};& |x|>1\\ \frac{d}{dx}\mathrm{arccsc}(x)& =-\frac{1}{|x|\sqrt{x^2-1}};& |x|>1\end{array}}$

These formulas can be derived in terms of the derivatives of trigonometric functions. For example, if ${\displaystyle x=\sin \theta }$, then $dx/d\theta =\cos \theta =\sqrt{1-x^2},$ so

${\displaystyle \frac{d}{dx}\arcsin (x)=\frac{d\theta }{dx}=\frac{1}{dx/d\theta }=\frac{1}{\sqrt{1-x^2}}.}$

Expression as definite integrals

Integrating the derivative and fixing the value at one point gives an expression for the inverse trigonometric function as a definite integral:

${\displaystyle \begin{array}{cccc}\arcsin (x)& ={\int }_0^x\frac{1}{\sqrt{1-z^2}}dz,& |x|& \le 1\\ \arccos (x)& ={\int }_x^1\frac{1}{\sqrt{1-z^2}}dz,& |x|& \le 1\\ \arctan (x)& ={\int }_0^x\frac{1}{z^2+1}dz,\\ \mathrm{arccot}(x)& ={\int }_x^{\mathrm{\infty }}\frac{1}{z^2+1}dz,\\ \mathrm{arcsec}(x)& ={\int }_1^x\frac{1}{z\sqrt{z^2-1}}dz=\pi +{\int }_{-x}^{-1}\frac{1}{z\sqrt{z^2-1}}dz,& x& \ge 1\\ \mathrm{arccsc}(x)& ={\int }_x^{\mathrm{\infty }}\frac{1}{z\sqrt{z^2-1}}dz={\int }_{-\mathrm{\infty }}^{-x}\frac{1}{z\sqrt{z^2-1}}dz,& x& \ge 1\end{array}}$

When x equals 1, the integrals with limited domains are improper integrals, but still well-defined.

Infinite series

Similar to the sine and cosine functions, the inverse trigonometric functions can also be calculated using power series, as follows. For arcsine, the series can be derived by expanding its derivative, $\frac{1}{\sqrt{1-z^2}}$, as a binomial series, and integrating term by term (using the integral definition as above). The series for arctangent can similarly be derived by expanding its derivative $\frac{1}{1+z^2}$ in a geometric series, and applying the integral definition above (see Leibniz series).

$${\displaystyle \begin{array}{cc}\arcsin (z)& =z+\left(\frac{1}{2}\right)\frac{z^3}{3}+\left(\frac{1\cdot 3}{2\cdot 4}\right)\frac{z^5}{5}+\left(\frac{1\cdot 3\cdot 5}{2\cdot 4\cdot 6}\right)\frac{z^7}{7}+\cdots \\ & =\sum _{n=0}^{\mathrm{\infty }}\frac{(2n-1)!!}{(2n)!!}\frac{z^{2n+1}}{2n+1}\\ & =\sum _{n=0}^{\mathrm{\infty }}\frac{(2n)!}{(2^{n}n!{)}^2}\frac{z^{2n+1}}{2n+1};|z|\le 1\end{array}}$$ The Taylor series for arctangent is also known as the arctangent series or Gregory's series. $${\displaystyle \arctan (z)=z-\frac{z^3}{3}+\frac{z^5}{5}-\frac{z^7}{7}+\cdots =\sum _{n=0}^{\mathrm{\infty }}\frac{(-1{)}^n{z}^{2n+1}}{2n+1};|z|\le 1z\ne i,-i}$$

Series for the other inverse trigonometric functions can be given in terms of these according to the relationships given above. For example, ${\displaystyle \arccos (x)=\pi /2-\arcsin (x)}$, ${\displaystyle \mathrm{arccsc}(x)=\arcsin (1/x)}$, and so on. Another series is given by:

$${\displaystyle 2{\left(\arcsin \left(\frac{x}{2}\right)\right)}^2=\sum _{n=1}^{\mathrm{\infty }}\frac{x^{2n}}{n^2(\genfrac{}{}{0ex}{}{2n}{n})}.}$$

Leonhard Euler found a series for the arctangent that converges more quickly than its Taylor series:

$${\displaystyle \arctan (z)=\frac{z}{1+z^2}\sum _{n=0}^{\mathrm{\infty }}\prod _{k=1}^n\frac{2k{z}^2}{(2k+1)(1+z^2)}.}$$ (The term in the sum for n = 0 is the empty product, so is 1.)

Alternatively, this can be expressed as

$${\displaystyle \arctan (z)=\sum _{n=0}^{\mathrm{\infty }}\frac{2^{2n}(n!{)}^2}{(2n+1)!}\frac{z^{2n+1}}{(1+z^2{)}^{n+1}}.}$$

Another series for the arctangent function is given by

$${\displaystyle \arctan (z)=i\sum _{n=1}^{\mathrm{\infty }}\frac{1}{2n-1}\left(\frac{1}{(1+2i/z{)}^{2n-1}}-\frac{1}{(1-2i/z{)}^{2n-1}}\right),}$$

where ${\displaystyle i=\sqrt{-1}}$ is the imaginary unit.

Continued fractions for arctangent

Two alternatives to the power series for arctangent are these generalized continued fractions:

$${\displaystyle \arctan (z)=\frac{z}{1+\frac{(1z{)}^2}{3-1z^2+\frac{(3z{)}^2}{5-3z^2+\frac{(5z{)}^2}{7-5z^2+\frac{(7z{)}^2}{9-7z^2+\ddots }}}}}=\frac{z}{1+\frac{(1z{)}^2}{3+\frac{(2z{)}^2}{5+\frac{(3z{)}^2}{7+\frac{(4z{)}^2}{9+\ddots }}}}}}$$

The second of these is valid in the cut complex plane. There are two cuts, from −i to the point at infinity, going down the imaginary axis, and from i to the point at infinity, going up the same axis. It works best for real numbers running from −1 to 1. The partial denominators are the odd natural numbers, and the partial numerators (after the first) are just (nz)², with each perfect square appearing once. The first was developed by Leonhard Euler, the second by Carl Friedrich Gauss utilizing the Gaussian hypergeometric series.

Indefinite integrals of inverse trigonometric functions

For real and complex values of z:

${\displaystyle \begin{array}{cc}\int \arcsin (z)dz& =z\arcsin (z)+\sqrt{1-z^2}+C\\ \int \arccos (z)dz& =z\arccos (z)-\sqrt{1-z^2}+C\\ \int \arctan (z)dz& =z\arctan (z)-\frac{1}{2}\ln \left(1+z^2\right)+C\\ \int \mathrm{arccot}(z)dz& =z\mathrm{arccot}(z)+\frac{1}{2}\ln \left(1+z^2\right)+C\\ \int \mathrm{arcsec}(z)dz& =z\mathrm{arcsec}(z)-\ln \left[z\left(1+\sqrt{\frac{z^2-1}{z^2}}\right)\right]+C\\ \int \mathrm{arccsc}(z)dz& =z\mathrm{arccsc}(z)+\ln \left[z\left(1+\sqrt{\frac{z^2-1}{z^2}}\right)\right]+C\end{array}}$

For real x ≥ 1:

${\displaystyle \begin{array}{cc}\int \mathrm{arcsec}(x)dx& =x\mathrm{arcsec}(x)-\ln \left(x+\sqrt{x^2-1}\right)+C\\ \int \mathrm{arccsc}(x)dx& =x\mathrm{arccsc}(x)+\ln \left(x+\sqrt{x^2-1}\right)+C\end{array}}$

For all real x not between -1 and 1:

${\displaystyle \begin{array}{cc}\int \mathrm{arcsec}(x)dx& =x\mathrm{arcsec}(x)-\mathrm{sgn}(x)\ln \left|x+\sqrt{x^2-1}\right|+C\\ \int \mathrm{arccsc}(x)dx& =x\mathrm{arccsc}(x)+\mathrm{sgn}(x)\ln \left|x+\sqrt{x^2-1}\right|+C\end{array}}$

The absolute value is necessary to compensate for both negative and positive values of the arcsecant and arccosecant functions. The signum function is also necessary due to the absolute values in the derivatives of the two functions, which create two different solutions for positive and negative values of x. These can be further simplified using the logarithmic definitions of the inverse hyperbolic functions:

${\displaystyle \begin{array}{cc}\int \mathrm{arcsec}(x)dx& =x\mathrm{arcsec}(x)-\mathrm{arcosh}(|x|)+C\\ \int \mathrm{arccsc}(x)dx& =x\mathrm{arccsc}(x)+\mathrm{arcosh}(|x|)+C\end{array}}$

The absolute value in the argument of the arcosh function creates a negative half of its graph, making it identical to the signum logarithmic function shown above.

All of these antiderivatives can be derived using integration by parts and the simple derivative forms shown above.

Example

Using ${\displaystyle \int udv=uv-\int vdu}$ (i.e. integration by parts), set

${\displaystyle \begin{array}{cccc}u& =\arcsin (x)& dv& =dx\\ du& =\frac{dx}{\sqrt{1-x^2}}& v& =x\end{array}}$

Then

${\displaystyle \int \arcsin (x)dx=x\arcsin (x)-\int \frac{x}{\sqrt{1-x^2}}dx,}$

which by the simple substitution ${\displaystyle w=1-x^2,dw=-2xdx}$ yields the final result:

${\displaystyle \int \arcsin (x)dx=x\arcsin (x)+\sqrt{1-x^2}+C}$

Extension to the complex plane

Since the inverse trigonometric functions are analytic functions, they can be extended from the real line to the complex plane. This results in functions with multiple sheets and branch points. One possible way of defining the extension is:

${\displaystyle \arctan (z)={\int }_0^z\frac{dx}{1+x^2}z\ne -i,+i}$

where the part of the imaginary axis which does not lie strictly between the branch points (−i and +i) is the branch cut between the principal sheet and other sheets. The path of the integral must not cross a branch cut. For z not on a branch cut, a straight line path from 0 to z is such a path. For z on a branch cut, the path must approach from Re[x] > 0 for the upper branch cut and from Re[x] < 0 for the lower branch cut.

The arcsine function may then be defined as:

${\displaystyle \arcsin (z)=\arctan \left(\frac{z}{\sqrt{1-z^2}}\right)z\ne -1,+1}$

where (the square-root function has its cut along the negative real axis and) the part of the real axis which does not lie strictly between −1 and +1 is the branch cut between the principal sheet of arcsin and other sheets;

${\displaystyle \arccos (z)=\frac{\pi }{2}-\arcsin (z)z\ne -1,+1}$

which has the same cut as arcsin;

${\displaystyle \mathrm{arccot}(z)=\frac{\pi }{2}-\arctan (z)z\ne -i,i}$

which has the same cut as arctan;

${\displaystyle \mathrm{arcsec}(z)=\arccos \left(\frac{1}{z}\right)z\ne -1,0,+1}$

where the part of the real axis between −1 and +1 inclusive is the cut between the principal sheet of arcsec and other sheets;

${\displaystyle \mathrm{arccsc}(z)=\arcsin \left(\frac{1}{z}\right)z\ne -1,0,+1}$

which has the same cut as arcsec.

Logarithmic forms

These functions may also be expressed using complex logarithms. This extends their domains to the complex plane in a natural fashion. The following identities for principal values of the functions hold everywhere that they are defined, even on their branch cuts.

${\displaystyle \begin{array}{ccc}\arcsin (z)& =-i\ln \left(\sqrt{1-z^2}+iz\right)=i\ln \left(\sqrt{1-z^2}-iz\right)& =\mathrm{arccsc}\left(\frac{1}{z}\right)\\ \arccos (z)& =-i\ln \left(z+i\sqrt{1-z^2}\right)=i\ln \left(z-i\sqrt{1-z^2}\right)=\frac{\pi }{2}-\arcsin (z)& =\mathrm{arcsec}\left(\frac{1}{z}\right)\\ \arctan (z)& =-\frac{i}{2}\ln \left(\frac{i-z}{i+z}\right)=-\frac{i}{2}\ln \left(\frac{1+iz}{1-iz}\right)& =\mathrm{arccot}\left(\frac{1}{z}\right)\\ \mathrm{arccot}(z)& =-\frac{i}{2}\ln \left(\frac{z+i}{z-i}\right)=-\frac{i}{2}\ln \left(\frac{iz-1}{iz+1}\right)& =\arctan \left(\frac{1}{z}\right)\\ \mathrm{arcsec}(z)& =-i\ln \left(\frac{1}{z}+i\sqrt{1-\frac{1}{z^2}}\right)=i\ln \left(\frac{1}{z}-i\sqrt{1-\frac{1}{z^2}}\right)=\frac{\pi }{2}-\mathrm{arccsc}(z)& =\arccos \left(\frac{1}{z}\right)\\ \mathrm{arccsc}(z)& =-i\ln \left(\sqrt{1-\frac{1}{z^2}}+\frac{i}{z}\right)=i\ln \left(\sqrt{1-\frac{1}{z^2}}-\frac{i}{z}\right)& =\arcsin \left(\frac{1}{z}\right)\end{array}}$

Generalization

Because all of the inverse trigonometric functions output an angle of a right triangle, they can be generalized by using Euler's formula to form a right triangle in the complex plane. Algebraically, this gives us:

${\displaystyle c{e}^{i\theta }=c\cos (\theta )+ic\sin (\theta )}$

or

${\displaystyle c{e}^{i\theta }=a+ib}$

where ${\displaystyle a}$ is the adjacent side, ${\displaystyle b}$ is the opposite side, and ${\displaystyle c}$ is the hypotenuse. From here, we can solve for ${\displaystyle \theta }$.

${\displaystyle \begin{array}{cc}{e}^{\ln (c)+i\theta }& =a+ib\\ \ln c+i\theta & =\ln (a+ib)\\ \theta & =\mathrm{Im}\left(\ln (a+ib)\right)\end{array}}$

or

${\displaystyle \theta =-i\ln \left(\frac{a+ib}{c}\right)}$

Simply taking the imaginary part works for any real-valued ${\displaystyle a}$ and ${\displaystyle b}$, but if ${\displaystyle a}$ or ${\displaystyle b}$ is complex-valued, we have to use the final equation so that the real part of the result isn't excluded. Since the length of the hypotenuse doesn't change the angle, ignoring the real part of ${\displaystyle \ln (a+bi)}$ also removes ${\displaystyle c}$ from the equation. In the final equation, we see that the angle of the triangle in the complex plane can be found by inputting the lengths of each side. By setting one of the three sides equal to 1 and one of the remaining sides equal to our input ${\displaystyle z}$, we obtain a formula for one of the inverse trig functions, for a total of six equations. Because the inverse trig functions require only one input, we must put the final side of the triangle in terms of the other two using the Pythagorean theorem relation

${\displaystyle a^2+b^2=c^2}$

The table below shows the values of a, b, and c for each of the inverse trig functions and the equivalent expressions for ${\displaystyle \theta }$ that result from plugging the values into the equations ${\displaystyle \theta =-i\ln \left(\frac{a+ib}{c}\right)}$ above and simplifying.

${\displaystyle \begin{array}{cccccccccccc}& a& & b& & c& & -i\ln \left(\frac{a+ib}{c}\right)& & \theta & & {\theta }_{a,b\in \mathbb{R}}\\ \arcsin (z)& \sqrt{1-z^2}& & z& & 1& & -i\ln \left(\frac{\sqrt{1-z^2}+iz}{1}\right)& & =-i\ln \left(\sqrt{1-z^2}+iz\right)& & \mathrm{Im}\left(\ln \left(\sqrt{1-z^2}+iz\right)\right)\\ \arccos (z)& z& & \sqrt{1-z^2}& & 1& & -i\ln \left(\frac{z+i\sqrt{1-z^2}}{1}\right)& & =-i\ln \left(z+\sqrt{z^2-1}\right)& & \mathrm{Im}\left(\ln \left(z+\sqrt{z^2-1}\right)\right)\\ \arctan (z)& 1& & z& & \sqrt{1+z^2}& & -i\ln \left(\frac{1+iz}{\sqrt{1+z^2}}\right)& & =-\frac{i}{2}\ln \left(\frac{i-z}{i+z}\right)& & \mathrm{Im}\left(\ln \left(1+iz\right)\right)\\ \mathrm{arccot}(z)& z& & 1& & \sqrt{z^2+1}& & -i\ln \left(\frac{z+i}{\sqrt{z^2+1}}\right)& & =-\frac{i}{2}\ln \left(\frac{z+i}{z-i}\right)& & \mathrm{Im}\left(\ln \left(z+i\right)\right)\\ \mathrm{arcsec}(z)& 1& & \sqrt{z^2-1}& & z& & -i\ln \left(\frac{1+i\sqrt{z^2-1}}{z}\right)& & =-i\ln \left(\frac{1}{z}+\sqrt{\frac{1}{z^2}-1}\right)& & \mathrm{Im}\left(\ln \left(\frac{1}{z}+\sqrt{\frac{1}{z^2}-1}\right)\right)\\ \mathrm{arccsc}(z)& \sqrt{z^2-1}& & 1& & z& & -i\ln \left(\frac{\sqrt{z^2-1}+i}{z}\right)& & =-i\ln \left(\sqrt{1-\frac{1}{z^2}}+\frac{i}{z}\right)& & \mathrm{Im}\left(\ln \left(\sqrt{1-\frac{1}{z^2}}+\frac{i}{z}\right)\right)\end{array}}$