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Inverse trigonometric functions

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Inverse functions of sin, cos, tan, etc.

Trigonometry

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Calculus
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"Arctangent" redirects here. For other uses, seeArctangent (disambiguation).

Inmathematics, theinverse trigonometric functions (occasionally also calledantitrigonometric,^[1]cyclometric,^[2] orarcus functions^[3]) are theinverse functions of thetrigonometric functions, under suitably restricteddomains. Specifically, they are the inverses of thesine,cosine,tangent,cotangent,secant, andcosecant functions,^[4] and are used to obtain an angle from any of the angle's trigonometric ratios. Inverse trigonometric functions are widely used inengineering,navigation,physics, andgeometry.

Notation

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For a circle of radius 1, arcsin and arccos are the lengths of actual arcs determined by the quantities in question.

Several notations for the inverse trigonometric functions exist. The most common convention is to name inverse trigonometric functions using an arc- prefix:arcsin(x),arccos(x),arctan(x), etc.^[1] (This convention is used throughout this article.) This notation arises from the following geometric relationships:^{[citation needed]}when measuring in radians, an angle ofθ radians will correspond to anarc whose length isrθ, wherer is the radius of the circle. Thus in theunit circle, the cosine of x function is both the arc and the angle, because the arc of a circle of radius 1 is the same as the angle. Or, "the arc whose cosine isx" is the same as "the angle whose cosine isx", because the length of the arc of the circle in radii is the same as the measurement of the angle in radians.^[5] In computer programming languages, the inverse trigonometric functions are often called by the abbreviated formsasin,acos,atan.^[6]

The notationssin⁻¹(x),cos⁻¹(x),tan⁻¹(x), etc., as introduced byJohn Herschel in 1813,^[7]^[8] are often used as well in English-language sources,^[1] much more than the alsoestablishedsin^[−1](x),cos^[−1](x),tan^[−1](x) – conventions consistent with the notation of aninverse function, that is useful (for example) to define themultivalued version of each inverse trigonometric function: $\tan ^{-1}(x)=\{\arctan(x)+\pi k\mid k\in \mathbb {Z} \}~.$ However, this might appear to conflict logically with the common semantics for expressions such assin²(x) (although onlysin²x, without parentheses, is the really common use), which refer to numeric power rather thanfunction composition, and therefore may result in confusion between notation for thereciprocal (multiplicative inverse) andinverse function.^[9]

The confusion is somewhat mitigated by the fact that each of the reciprocal trigonometric functions has its own name — for example,(cos(x))⁻¹ = sec(x). Nevertheless, certain authors advise against using it, since it is ambiguous.^[1]^[10] Another precarious convention used by a small number of authors is to use anuppercase first letter, along with a “−1” superscript:Sin⁻¹(x),Cos⁻¹(x),Tan⁻¹(x), etc.^[11] Although it is intended to avoid confusion with thereciprocal, which should be represented bysin⁻¹(x),cos⁻¹(x), etc., or, better, bysin⁻¹x,cos⁻¹x, etc., it in turn creates yet another major source of ambiguity, especially since many popular high-level programming languages (e.g.Mathematica andMAGMA) use those very same capitalised representations for the standard trig functions, whereas others (Python,SymPy,NumPy,Matlab,MAPLE, etc.) use lower-case.

Hence, since 2009, theISO 80000-2 standard has specified solely the "arc" prefix for the inverse functions.

Basic concepts

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The points labelled1,Sec(θ),Csc(θ) represent the length of the line segment from the origin to that point.Sin(θ),Tan(θ), and1 are the heights to the line starting from thex-axis, whileCos(θ),1, andCot(θ) are lengths along thex-axis starting from the origin.

Principal values

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

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

The principal inverses are listed in the following table.

Name	Usual notation	Definition	Domain ofx for real result	Range of usual principal value (radians)	Range of usual principal value (degrees)
arcsine	y = arcsin(x)	x =sin(y)	−1 ≤x ≤ 1	−⁠π/2⁠ ≤y ≤⁠π/2⁠	−90° ≤y ≤ 90°
arccosine	y = arccos(x)	x =cos(y)	−1 ≤x ≤ 1	0 ≤y ≤ π	0° ≤y ≤ 180°
arctangent	y = arctan(x)	x =tan(y)	all real numbers	−⁠π/2⁠ <y <⁠π/2⁠	−90° <y < 90°
arccotangent	y = arccot(x)	x =cot(y)	all real numbers	0 <y < π	0° <y < 180°
arcsecant	y = arcsec(x)	x =sec(y)	\|x\| ≥ 1	0 ≤y <⁠π/2⁠ or⁠π/2⁠ <y ≤ π	0° ≤y < 90° or90° <y ≤ 180°
arccosecant	y = arccsc(x)	x =csc(y)	\|x\| ≥ 1	−⁠π/2⁠ ≤y < 0 or0 <y ≤⁠π/2⁠	−90° ≤y < 0 or0° <y ≤ 90°

Note: Some authors define the range of arcsecant to be( ${\textstyle 0\leq y<{\frac {\pi }{2}}}$ or ${\textstyle \pi \leq y<{\frac {3\pi }{2}}}$ ),^[12] because the tangent function is nonnegative on this domain. This makes some computations more consistent. For example, using this range, $\tan(\operatorname {arcsec}(x))={\sqrt {x^{2}-1}},$ whereas with the range( ${\textstyle 0\leq y<{\frac {\pi }{2}}}$ or ${\textstyle {\frac {\pi }{2}}<y\leq \pi }$ ), we would have to write $\tan(\operatorname {arcsec}(x))=\pm {\sqrt {x^{2}-1}},$ since tangent is nonnegative on ${\textstyle 0\leq y<{\frac {\pi }{2}},}$ but nonpositive on ${\textstyle {\frac {\pi }{2}}<y\leq \pi .}$ For a similar reason, the same authors define the range of arccosecant to be ${\textstyle (-\pi <y\leq -{\frac {\pi }{2}}}$ or ${\textstyle 0<y\leq {\frac {\pi }{2}}).}$

Domains

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Ifx is allowed to be acomplex number, then the range ofy applies only to its real part.

The table below displays names and domains of the inverse trigonometric functions along with therange of their usualprincipal values inradians.

Name	Symbol		Domain		Image/Range	Inverse function		Domain		Image of principal values
sine	$\sin$	$: {\displaystyle :}$	$\mathbb {R}$	$\to$	$[-1,1]$	$\arcsin$	$: {\displaystyle :}$	$[-1,1]$	$\to$	$\left[-{\tfrac {\pi }{2}},{\tfrac {\pi }{2}}\right]$
cosine	$\cos$	$: {\displaystyle :}$	$\mathbb {R}$	$\to$	$[-1,1]$	$\arccos$	$: {\displaystyle :}$	$[-1,1]$	$\to$	$[0,\pi ]$
tangent	$\tan$	$: {\displaystyle :}$	$\pi \mathbb {Z} +\left(-{\tfrac {\pi }{2}},{\tfrac {\pi }{2}}\right)$	$\to$	$\mathbb {R}$	$\arctan$	$: {\displaystyle :}$	$\mathbb {R}$	$\to$	$\left(-{\tfrac {\pi }{2}},{\tfrac {\pi }{2}}\right)$
cotangent	$\cot$	$: {\displaystyle :}$	$\pi \mathbb {Z} +(0,\pi )$	$\to$	$\mathbb {R}$	$\operatorname {arccot}$	$: {\displaystyle :}$	$\mathbb {R}$	$\to$	$(0,\pi )$
secant	$\sec$	$: {\displaystyle :}$	$\pi \mathbb {Z} +\left(-{\tfrac {\pi }{2}},{\tfrac {\pi }{2}}\right)$	$\to$	$\mathbb {R} \setminus (-1,1)$	$\operatorname {arcsec}$	$: {\displaystyle :}$	$\mathbb {R} \setminus (-1,1)$	$\to$	$[\,0,\;\pi \,]\;\;\;\setminus \left\{{\tfrac {\pi }{2}}\right\}$
cosecant	$\csc$	$: {\displaystyle :}$	$\pi \mathbb {Z} +(0,\pi )$	$\to$	$\mathbb {R} \setminus (-1,1)$	$\operatorname {arccsc}$	$: {\displaystyle :}$	$\mathbb {R} \setminus (-1,1)$	$\to$	$\left[-{\tfrac {\pi }{2}},{\tfrac {\pi }{2}}\right]\setminus \{0\}$

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

$\pi \mathbb {Z} ~:=~\{\pi n\;:\;n\in \mathbb {Z} \}~=~\{\ldots ,\,-2\pi ,\,-\pi ,\,0,\,\pi ,\,2\pi ,\,\ldots \}.$

The symbol $\,\setminus \,$ denotesset subtraction so that, for instance, $\mathbb {R} \setminus (-1,1)=(-\infty ,-1]\cup [1,\infty )$ is the set of points in $\mathbb {R}$ (that is, real numbers) that arenot in the interval $(-1,1).$

TheMinkowski sum notation ${\textstyle \pi \mathbb {Z} +(0,\pi )}$ and $\pi \mathbb {Z} +{\bigl (}{-{\tfrac {\pi }{2}}},{\tfrac {\pi }{2}}{\bigr )}$ that is used above to concisely write the domains of $\cot ,\csc ,\tan ,{\text{ and }}\sec$ is now explained.

Domain of cotangent $\cot$ and cosecant $\csc$ : The domains of $\,\cot \,$ and $\,\csc \,$ are the same. They are the set of all angles $\theta$ at which $\sin \theta \neq 0,$ i.e. all real numbers that arenot of the form $\pi n$ for some integer $n, {\displaystyle n,}$

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

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

${\begin{aligned}\pi \mathbb {Z} +\left(-{\tfrac {\pi }{2}},{\tfrac {\pi }{2}}\right)&=\cdots \cup {\bigl (}{-{\tfrac {3\pi }{2}}},{-{\tfrac {\pi }{2}}}{\bigr )}\cup {\bigl (}{-{\tfrac {\pi }{2}}},{\tfrac {\pi }{2}}{\bigr )}\cup {\bigl (}{\tfrac {\pi }{2}},{\tfrac {3\pi }{2}}{\bigr )}\cup \cdots \\&=\mathbb {R} \setminus \left({\tfrac {\pi }{2}}+\pi \mathbb {Z} \right)\\\end{aligned}}$

Solutions to elementary trigonometric equations

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

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

This periodicity is reflected in the general inverses, where $k {\displaystyle k}$ 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 $\theta ,$ $r, {\displaystyle r,}$ $s, {\displaystyle s,}$ $x, {\displaystyle x,}$ and $y {\displaystyle y}$ all lie within appropriate ranges so that the relevant expressions below arewell-defined. Note that "for some $k\in \mathbb {Z}$ " is just another way of saying "for someinteger $k . {\displaystyle k.}$ "

The symbol $\,\iff \,$ islogical 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^{[note 1]} for more details and an example illustrating this concept).


Equation	if and only if	Solution
$\sin \theta =y$	$\iff$	$\theta =\,$	$(-1)^{k}$	$\arcsin(y)$	$+$		$\pi k$	for some $k\in \mathbb {Z}$
$\csc \theta =r$	$\iff$	$\theta =\,$	$(-1)^{k}$	$\operatorname {arccsc}(r)$	$+$		$\pi k$	for some $k\in \mathbb {Z}$
$\cos \theta =x$	$\iff$	$\theta =\,$	$\pm \,$	$\arccos(x)$	$+$	$2 {\displaystyle 2}$	$\pi k$	for some $k\in \mathbb {Z}$
$\sec \theta =r$	$\iff$	$\theta =\,$	$\pm \,$	$\operatorname {arcsec}(r)$	$+$	$2 {\displaystyle 2}$	$\pi k$	for some $k\in \mathbb {Z}$
$\tan \theta =s$	$\iff$	$\theta =\,$		$\arctan(s)$	$+$		$\pi k$	for some $k\in \mathbb {Z}$
$\cot \theta =r$	$\iff$	$\theta =\,$		$\operatorname {arccot}(r)$	$+$		$\pi k$	for some $k\in \mathbb {Z}$

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


Equation	if and only if	Solution
$\sin \theta =y$	$\iff$	$\theta =\;\;\;\,\arcsin(y)+2\pi k$ or $\theta =-\arcsin(y)+2\pi k+\pi$	for some $k\in \mathbb {Z}$
$\csc \theta =r$	$\iff$	$\theta =\;\;\;\,\operatorname {arccsc}(r)+2\pi k$ or $\theta =-\operatorname {arccsc}(r)+2\pi k+\pi$	for some $k\in \mathbb {Z}$
$\cos \theta =x$	$\iff$	$\theta =\;\;\;\,\arccos(x)+2\pi k$ or $\theta =-\arccos(x)+2\pi k$	for some $k\in \mathbb {Z}$
$\sec \theta =r$	$\iff$	$\theta =\;\;\;\,\operatorname {arcsec}(r)+2\pi k$ or $\theta =-\operatorname {arcsec}(r)+2\pi k$	for some $k\in \mathbb {Z}$

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

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

cos\;\theta =x\quad

if and only if

\quad \theta =(-1)^{h}\arccos(x)+\pi h+\pi \operatorname {Parity} (h)\quad

for some

h\in \mathbb {Z} .

And similarly for the secant function,

sec\;\theta =r\quad

if and only if

\quad \theta =(-1)^{h}\operatorname {arcsec}(r)+\pi h+\pi \operatorname {Parity} (h)\quad

for some

h\in \mathbb {Z} ,

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

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

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

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

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

Equal identical trigonometric functions

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


Equation	if and only if	Solution (for some $k\in \mathbb {Z}$ )	Also a solution to
${\phantom {-}}\sin \theta =\sin \varphi$	$\iff$	$\theta ={\phantom {\quad }}(-1)^{k}\varphi +{\phantom {2}}\pi k{\phantom {+\pi }}$	${\phantom {-}}\csc \theta =\csc \varphi$
${\phantom {-}}\cos \theta =\cos \varphi$	$\iff$	$\theta ={\phantom {-1\quad }}\pm \varphi +2\pi k{\phantom {+\pi }}$	${\phantom {-}}\sec \theta =\sec \varphi$
${\phantom {-}}\tan \theta =\tan \varphi$	$\iff$	$\theta ={\phantom {(-1)^{k+1}}}\varphi +{\phantom {2}}\pi k{\phantom {+\pi }}$	${\phantom {-}}\cot \theta =\cot \varphi$
$-\sin \theta =\sin \varphi$	$\iff$	$\theta =(-1)^{k+1}\varphi +{\phantom {2}}\pi k{\phantom {+\pi }}$	$-\csc \theta =\csc \varphi$
$-\cos \theta =\cos \varphi$	$\iff$	$\theta ={\phantom {-1\quad }}\pm \varphi +2\pi k+\pi {\phantom {+\pi }}$	$-\sec \theta =\sec \varphi$
$-\tan \theta =\tan \varphi$	$\iff$	$\theta ={\phantom {-1\quad }}-\varphi +{\phantom {2}}\pi k{\phantom {+\pi }}$	$-\cot \theta =\cot \varphi$
${\begin{aligned}{\phantom {-}}\left\|\sin \theta \right\|&=\left\|\sin \varphi \right\|\\&\Updownarrow \\{\phantom {-}}\left\|\cos \theta \right\|&=\left\|\cos \varphi \right\|\end{aligned}}$	$\iff$	$\theta ={\phantom {-1\quad }}\pm \varphi +{\phantom {2}}\pi k{\phantom {+\pi }}$	${\begin{aligned}{\phantom {-}}\left\|\tan \theta \right\|&=\left\|\tan \varphi \right\|\\\left\|\csc \theta \right\|&=\left\|\csc \varphi \right\|\\\left\|\sec \theta \right\|&=\left\|\sec \varphi \right\|\\\left\|\cot \theta \right\|&=\left\|\cot \varphi \right\|\end{aligned}}$

Set of all solutions to elementary trigonometric equations

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


If $\theta$ solves	then	Set of all solutions (in terms of $\theta$ )
$\;\sin \theta =y$	then	$\{\varphi :\sin \varphi =y\}=\,$	$(\theta$	$\,+\,2$	$\pi \mathbb {Z} )$	$\,\cup \,(-\theta$	$-\pi$	$+2\pi \mathbb {Z} )$
$\;\csc \theta =r$	then	$\{\varphi :\csc \varphi =r\}=\,$	$(\theta$	$\,+\,2$	$\pi \mathbb {Z} )$	$\,\cup \,(-\theta$	$-\pi$	$+2\pi \mathbb {Z} )$
$\;\cos \theta =x$	then	$\{\varphi :\cos \varphi =x\}=\,$	$(\theta$	$\,+\,2$	$\pi \mathbb {Z} )$	$\,\cup \,(-\theta$		$+2\pi \mathbb {Z} )$
$\;\sec \theta =r$	then	$\{\varphi :\sec \varphi =r\}=\,$	$(\theta$	$\,+\,2$	$\pi \mathbb {Z} )$	$\,\cup \,(-\theta$		$+2\pi \mathbb {Z} )$
$\;\tan \theta =s$	then	$\{\varphi :\tan \varphi =s\}=\,$	$\theta$	$\,+\,$	$\pi \mathbb {Z}$
$\;\cot \theta =r$	then	$\{\varphi :\cot \varphi =r\}=\,$	$\theta$	$\,+\,$	$\pi \mathbb {Z}$

Transforming equations

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The equations above can be transformed by using the reflection and shift identities:^[13]

Transforming equations by shifts and reflections
Argument: ${\underline {\;~~~~~~\;}}=$	$-\theta$	${\frac {\pi }{2}}\pm \theta$	$\pi \pm \theta$	${\frac {3\pi }{2}}\pm \theta$	$2k\pi \pm \theta ,$ $(k\in \mathbb {Z} )$
$\sin {\underline {\;~~~~~~~~~~~~~~\;}}=$	$-\sin \theta$	${\phantom {-}}\cos \theta$	$\mp \sin \theta$	$-\cos \theta$	$\pm \sin \theta$
$\csc {\underline {\;~~~~~~~~~~~~~~\;}}=$	$-\csc \theta$	${\phantom {-}}\sec \theta$	$\mp \csc \theta$	$-\sec \theta$	$\pm \csc \theta$
$\cos {\underline {\;~~~~~~~~~~~~~~\;}}=$	${\phantom {-}}\cos \theta$	$\mp \sin \theta$	$-\cos \theta$	$\pm \sin \theta$	${\phantom {-}}\cos \theta$
$\sec {\underline {\;~~~~~~~~~~~~~~\;}}=$	${\phantom {-}}\sec \theta$	$\mp \csc \theta$	$-\sec \theta$	$\pm \csc \theta$	${\phantom {-}}\sec \theta$
$\tan {\underline {\;~~~~~~~~~~~~~~\;}}=$	$-\tan \theta$	$\mp \cot \theta$	$\pm \tan \theta$	$\mp \cot \theta$	$\pm \tan \theta$
$\cot {\underline {\;~~~~~~~~~~~~~~\;}}=$	$-\cot \theta$	$\mp \tan \theta$	$\pm \cot \theta$	$\mp \tan \theta$	$\pm \cot \theta$

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

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

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

Relationships between trigonometric functions and inverse trigonometric functions

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Trigonometric functions of inverse trigonometric functions are tabulated below. A quick way to derive them is by considering the geometry of a right-angled triangle, with one side of length 1 and another side of length $x, {\displaystyle x,}$ then applying thePythagorean theorem and definitions of the trigonometric ratios. It is worth noting that for arcsecant and arccosecant, the diagram assumes that $x {\displaystyle x}$ is positive, and thus the result has to be corrected through the use ofabsolute values and thesignum (sgn) operation.

$\theta$	$\sin(\theta )$	$\cos(\theta )$	$\tan(\theta )$
$\arcsin(x)$	$\sin(\arcsin(x))=x$	$\cos(\arcsin(x))={\sqrt {1-x^{2}}}$	$\tan(\arcsin(x))={\frac {x}{\sqrt {1-x^{2}}}}$
$\arccos(x)$	$\sin(\arccos(x))={\sqrt {1-x^{2}}}$	$\cos(\arccos(x))=x$	$\tan(\arccos(x))={\frac {\sqrt {1-x^{2}}}{x}}$
$\arctan(x)$	$\sin(\arctan(x))={\frac {x}{\sqrt {1+x^{2}}}}$	$\cos(\arctan(x))={\frac {1}{\sqrt {1+x^{2}}}}$	$\tan(\arctan(x))=x$
$\operatorname {arccot}(x)$	$\sin(\operatorname {arccot}(x))={\frac {1}{\sqrt {1+x^{2}}}}$	$\cos(\operatorname {arccot}(x))={\frac {x}{\sqrt {1+x^{2}}}}$	$\tan(\operatorname {arccot}(x))={\frac {1}{x}}$
$\operatorname {arcsec}(x)$	$\sin(\operatorname {arcsec}(x))={\frac {\sqrt {x^{2}-1}}{\|x\|}}$	$\cos(\operatorname {arcsec}(x))={\frac {1}{x}}$	$\tan(\operatorname {arcsec}(x))=\operatorname {sgn}(x){\sqrt {x^{2}-1}}$
$\operatorname {arccsc}(x)$	$\sin(\operatorname {arccsc}(x))={\frac {1}{x}}$	$\cos(\operatorname {arccsc}(x))={\frac {\sqrt {x^{2}-1}}{\|x\|}}$	$\tan(\operatorname {arccsc}(x))={\frac {\operatorname {sgn}(x)}{\sqrt {x^{2}-1}}}$

Relationships among the inverse trigonometric functions

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The usual principal values of the arcsin(x) (red) and arccos(x) (blue) functions graphed on the cartesian plane.

The usual principal values of the arctan(x) and arccot(x) functions graphed on the cartesian plane.

Principal values of the arcsec(x) and arccsc(x) functions graphed on the cartesian plane.

Complementary angles:

{\begin{aligned}\arccos(x)&={\frac {\pi }{2}}-\arcsin(x)\\[0.5em]\operatorname {arccot}(x)&={\frac {\pi }{2}}-\arctan(x)\\[0.5em]\operatorname {arccsc}(x)&={\frac {\pi }{2}}-\operatorname {arcsec}(x)\end{aligned}}

Negative arguments:

{\begin{aligned}\arcsin(-x)&=-\arcsin(x)\\\operatorname {arccsc}(-x)&=-\operatorname {arccsc}(x)\\\arccos(-x)&=\pi -\arccos(x)\\\operatorname {arcsec}(-x)&=\pi -\operatorname {arcsec}(x)\\\arctan(-x)&=-\arctan(x)\\\operatorname {arccot}(-x)&=\pi -\operatorname {arccot}(x)\end{aligned}}

Reciprocal arguments:

{\begin{aligned}\arcsin \left({\frac {1}{x}}\right)&=\operatorname {arccsc}(x)&\\[0.3em]\operatorname {arccsc} \left({\frac {1}{x}}\right)&=\arcsin(x)&\\[0.3em]\arccos \left({\frac {1}{x}}\right)&=\operatorname {arcsec}(x)&\\[0.3em]\operatorname {arcsec} \left({\frac {1}{x}}\right)&=\arccos(x)&\\[0.3em]\arctan \left({\frac {1}{x}}\right)&=\operatorname {arccot}(x)&={\frac {\pi }{2}}-\arctan(x)\,,{\text{ if }}x>0\\[0.3em]\arctan \left({\frac {1}{x}}\right)&=\operatorname {arccot}(x)-\pi &=-{\frac {\pi }{2}}-\arctan(x)\,,{\text{ if }}x<0\\[0.3em]\operatorname {arccot} \left({\frac {1}{x}}\right)&=\arctan(x)&={\frac {\pi }{2}}-\operatorname {arccot}(x)\,,{\text{ if }}x>0\\[0.3em]\operatorname {arccot} \left({\frac {1}{x}}\right)&=\arctan(x)+\pi &={\frac {3\pi }{2}}-\operatorname {arccot}(x)\,,{\text{ if }}x<0\end{aligned}}

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

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

{\begin{aligned}\arcsin(x)&={\frac {1}{2}}\arccos \left(1-2x^{2}\right)\,,{\text{ if }}0\leq x\leq 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\leq x\leq 1\\\arccos(x)&=\arctan \left({\frac {\sqrt {1-x^{2}}}{x}}\right)\\\arccos(x)&=\arcsin \left({\sqrt {1-x^{2}}}\right)\,,{\text{ if }}0\leq x\leq 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\leq x\leq 1\\\arcsin &\left({\sqrt {1-x^{2}}}\right)={\frac {\pi }{2}}-\operatorname {sgn}(x)\arcsin(x)\\\arctan(x)&=\arcsin \left({\frac {x}{\sqrt {1+x^{2}}}}\right)\\\operatorname {arccot}(x)&=\arccos \left({\frac {x}{\sqrt {1+x^{2}}}}\right)\end{aligned}}

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

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

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

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

{\begin{aligned}\arcsin(x)&=2\arctan \left({\frac {x}{1+{\sqrt {1-x^{2}}}}}\right)\\[0.5em]\arccos(x)&=2\arctan \left({\frac {\sqrt {1-x^{2}}}{1+x}}\right)\,,{\text{ if }}-1<x\leq 1\\[0.5em]\arctan(x)&=2\arctan \left({\frac {x}{1+{\sqrt {1+x^{2}}}}}\right)\end{aligned}}

Arctangent addition formula

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\arctan(u)\pm \arctan(v)=\arctan \left({\frac {u\pm v}{1\mp uv}}\right){\pmod {\pi }}\,,\quad uv\neq 1\,.

This is derived from the tangentaddition formula

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

by letting

\alpha =\arctan(u)\,,\quad \beta =\arctan(v)\,.

In calculus

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Derivatives of inverse trigonometric functions

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Main article:Differentiation of trigonometric functions

Thederivatives for complex values ofz are as follows:

{\begin{aligned}{\frac {d}{dz}}\arcsin(z)&{}={\frac {1}{\sqrt {1-z^{2}}}}\;;&z&{}\neq -1,+1\\{\frac {d}{dz}}\arccos(z)&{}=-{\frac {1}{\sqrt {1-z^{2}}}}\;;&z&{}\neq -1,+1\\{\frac {d}{dz}}\arctan(z)&{}={\frac {1}{1+z^{2}}}\;;&z&{}\neq -i,+i\\{\frac {d}{dz}}\operatorname {arccot}(z)&{}=-{\frac {1}{1+z^{2}}}\;;&z&{}\neq -i,+i\\{\frac {d}{dz}}\operatorname {arcsec}(z)&{}={\frac {1}{z^{2}{\sqrt {1-{\frac {1}{z^{2}}}}}}}\;;&z&{}\neq -1,0,+1\\{\frac {d}{dz}}\operatorname {arccsc}(z)&{}=-{\frac {1}{z^{2}{\sqrt {1-{\frac {1}{z^{2}}}}}}}\;;&z&{}\neq -1,0,+1\end{aligned}}

Only for real values ofx:

{\begin{aligned}{\frac {d}{dx}}\operatorname {arcsec}(x)&{}={\frac {1}{|x|{\sqrt {x^{2}-1}}}}\;;&|x|>1\\{\frac {d}{dx}}\operatorname {arccsc}(x)&{}=-{\frac {1}{|x|{\sqrt {x^{2}-1}}}}\;;&|x|>1\end{aligned}}

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

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

Expression as definite integrals

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

{\begin{aligned}\arcsin(x)&{}=\int _{0}^{x}{\frac {1}{\sqrt {1-z^{2}}}}\,dz\;,&|x|&{}\leq 1\\\arccos(x)&{}=\int _{x}^{1}{\frac {1}{\sqrt {1-z^{2}}}}\,dz\;,&|x|&{}\leq 1\\\arctan(x)&{}=\int _{0}^{x}{\frac {1}{z^{2}+1}}\,dz\;,\\\operatorname {arccot}(x)&{}=\int _{x}^{\infty }{\frac {1}{z^{2}+1}}\,dz\;,\\\operatorname {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&{}\geq 1\\\operatorname {arccsc}(x)&{}=\int _{x}^{\infty }{\frac {1}{z{\sqrt {z^{2}-1}}}}\,dz=\int _{-\infty }^{-x}{\frac {1}{z{\sqrt {z^{2}-1}}}}\,dz\;,&x&{}\geq 1\\\end{aligned}}

Whenx equals 1, the integrals with limited domains areimproper integrals, but still well-defined.

Infinite series

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

{\begin{aligned}\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 \\[5pt]&=\sum _{n=0}^{\infty }{\frac {(2n-1)!!}{(2n)!!}}{\frac {z^{2n+1}}{2n+1}}\\[5pt]&=\sum _{n=0}^{\infty }{\frac {(2n)!}{(2^{n}n!)^{2}}}{\frac {z^{2n+1}}{2n+1}}\,;\qquad |z|\leq 1\end{aligned}}

\arctan(z)=z-{\frac {z^{3}}{3}}+{\frac {z^{5}}{5}}-{\frac {z^{7}}{7}}+\cdots =\sum _{n=0}^{\infty }{\frac {(-1)^{n}z^{2n+1}}{2n+1}}\,;\qquad |z|\leq 1\qquad z\neq i,-i

Series for the other inverse trigonometric functions can be given in terms of these according to the relationships given above. For example, $\arccos(x)=\pi /2-\arcsin(x)$ , $\operatorname {arccsc}(x)=\arcsin(1/x)$ , and so on. Another series is given by:^[14]

2\left(\arcsin \left({\frac {x}{2}}\right)\right)^{2}=\sum _{n=1}^{\infty }{\frac {x^{2n}}{n^{2}{\binom {2n}{n}}}}.

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

\arctan(z)={\frac {z}{1+z^{2}}}\sum _{n=0}^{\infty }\prod _{k=1}^{n}{\frac {2kz^{2}}{(2k+1)(1+z^{2})}}.

^[15]

(The term in the sum forn = 0 is theempty product, so is 1.)

Alternatively, this can be expressed as

\arctan(z)=\sum _{n=0}^{\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

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

where $i={\sqrt {-1}}$ is theimaginary unit.^[16]

Continued fractions for arctangent

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

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

The second of these is valid in the cut complex plane. There are two cuts, from −i to thepoint at infinity, going down the imaginary axis, and fromi 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 byLeonhard Euler; the second byCarl Friedrich Gauss utilizing theGaussian hypergeometric series.

Indefinite integrals of inverse trigonometric functions

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For real and complex values ofz:

{\begin{aligned}\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 \operatorname {arccot}(z)\,dz&{}=z\,\operatorname {arccot}(z)+{\frac {1}{2}}\ln \left(1+z^{2}\right)+C\\\int \operatorname {arcsec}(z)\,dz&{}=z\,\operatorname {arcsec}(z)-\ln \left[z\left(1+{\sqrt {\frac {z^{2}-1}{z^{2}}}}\right)\right]+C\\\int \operatorname {arccsc}(z)\,dz&{}=z\,\operatorname {arccsc}(z)+\ln \left[z\left(1+{\sqrt {\frac {z^{2}-1}{z^{2}}}}\right)\right]+C\end{aligned}}

For realx ≥ 1:

{\begin{aligned}\int \operatorname {arcsec}(x)\,dx&{}=x\,\operatorname {arcsec}(x)-\ln \left(x+{\sqrt {x^{2}-1}}\right)+C\\\int \operatorname {arccsc}(x)\,dx&{}=x\,\operatorname {arccsc}(x)+\ln \left(x+{\sqrt {x^{2}-1}}\right)+C\end{aligned}}

For all realx not between -1 and 1:

{\begin{aligned}\int \operatorname {arcsec}(x)\,dx&{}=x\,\operatorname {arcsec}(x)-\operatorname {sgn}(x)\ln \left|x+{\sqrt {x^{2}-1}}\right|+C\\\int \operatorname {arccsc}(x)\,dx&{}=x\,\operatorname {arccsc}(x)+\operatorname {sgn}(x)\ln \left|x+{\sqrt {x^{2}-1}}\right|+C\end{aligned}}

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 thederivatives 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 theinverse hyperbolic functions:

{\begin{aligned}\int \operatorname {arcsec}(x)\,dx&{}=x\,\operatorname {arcsec}(x)-\operatorname {arcosh} (|x|)+C\\\int \operatorname {arccsc}(x)\,dx&{}=x\,\operatorname {arccsc}(x)+\operatorname {arcosh} (|x|)+C\\\end{aligned}}

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 usingintegration by parts and the simple derivative forms shown above.

Example

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Using $\int u\,dv=uv-\int v\,du$ (i.e.integration by parts), set

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

Then

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

which by the simplesubstitution $w=1-x^{2},\ dw=-2x\,dx$ yields the final result:

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

Extension to the complex plane

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ARiemann surface for the argument of the relationtanz =x. The orange sheet in the middle is the principal sheet representingarctanx. The blue sheet above and green sheet below are displaced by2π and−2π respectively.

Since the inverse trigonometric functions areanalytic functions, they can be extended from thereal line to the complex plane. This results in functions with multiple sheets andbranch points. One possible way of defining the extension is:

\arctan(z)=\int _{0}^{z}{\frac {dx}{1+x^{2}}}\quad z\neq -i,+i

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

The arcsine function may then be defined as:

\arcsin(z)=\arctan \left({\frac {z}{\sqrt {1-z^{2}}}}\right)\quad z\neq -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;

\arccos(z)={\frac {\pi }{2}}-\arcsin(z)\quad z\neq -1,+1

which has the same cut as arcsin;

\operatorname {arccot}(z)={\frac {\pi }{2}}-\arctan(z)\quad z\neq -i,i

which has the same cut as arctan;

\operatorname {arcsec}(z)=\arccos \left({\frac {1}{z}}\right)\quad z\neq -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;

\operatorname {arccsc}(z)=\arcsin \left({\frac {1}{z}}\right)\quad z\neq -1,0,+1

which has the same cut as arcsec.

Logarithmic forms

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These functions may also be expressed usingcomplex logarithms. This extends theirdomains to thecomplex 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.

{\begin{aligned}\arcsin(z)&{}=-i\ln \left({\sqrt {1-z^{2}}}+iz\right)=i\ln \left({\sqrt {1-z^{2}}}-iz\right)&{}=\operatorname {arccsc} \left({\frac {1}{z}}\right)\\[10pt]\arccos(z)&{}=-i\ln \left(i{\sqrt {1-z^{2}}}+z\right)={\frac {\pi }{2}}-\arcsin(z)&{}=\operatorname {arcsec} \left({\frac {1}{z}}\right)\\[10pt]\arctan(z)&{}=-{\frac {i}{2}}\ln \left({\frac {i-z}{i+z}}\right)=-{\frac {i}{2}}\ln \left({\frac {1+iz}{1-iz}}\right)&{}=\operatorname {arccot} \left({\frac {1}{z}}\right)\\[10pt]\operatorname {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)\\[10pt]\operatorname {arcsec}(z)&{}=-i\ln \left(i{\sqrt {1-{\frac {1}{z^{2}}}}}+{\frac {1}{z}}\right)={\frac {\pi }{2}}-\operatorname {arccsc}(z)&{}=\arccos \left({\frac {1}{z}}\right)\\[10pt]\operatorname {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{aligned}}

Generalization

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

ce^{i\theta }=c\cos(\theta )+ic\sin(\theta )

ce^{i\theta }=a+ib

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

{\begin{aligned}e^{\ln(c)+i\theta }&=a+ib\\\ln c+i\theta &=\ln(a+ib)\\\theta &=\operatorname {Im} \left(\ln(a+ib)\right)\end{aligned}}

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

Simply taking the imaginary part works for any real-valued $a {\displaystyle a}$ and $b {\displaystyle b}$ , but if $a {\displaystyle a}$ or $b {\displaystyle b}$ 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 $\ln(a+bi)$ also removes $c {\displaystyle c}$ 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 $z {\displaystyle z}$ , 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 thePythagorean Theorem relation

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 $\theta$ that result from plugging the values into the equations $\theta =-i\ln \left({\tfrac {a+ib}{c}}\right)$ above and simplifying.

{\begin{aligned}&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)&&\operatorname {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)&&\operatorname {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)&&\operatorname {Im} \left(\ln \left(1+iz\right)\right)\\\operatorname {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)&&\operatorname {Im} \left(\ln \left(z+i\right)\right)\\\operatorname {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)&&\operatorname {Im} \left(\ln \left({\frac {1}{z}}+{\sqrt {{\frac {1}{z^{2}}}-1}}\right)\right)\\\operatorname {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)&&\operatorname {Im} \left(\ln \left({\sqrt {1-{\frac {1}{z^{2}}}}}+{\frac {i}{z}}\right)\right)\\\end{aligned}}

The particular form of the simplified expression can cause the output to differ from theusual principal branch of each of the inverse trig functions. The formulations given will output the usual principal branch when using the $\operatorname {Im} \left(\ln z\right)\in (-\pi ,\pi ]$ and $\operatorname {Re} \left({\sqrt {z}}\right)\geq 0$ principal branch for every function except arccotangent in the $\theta$ column. Arccotangent in the $\theta$ column will output on its usual principal branch by using the $\operatorname {Im} \left(\ln z\right)\in [0,2\pi )$ and $\operatorname {Im} \left({\sqrt {z}}\right)\geq 0$ convention.

In this sense, all of the inverse trig functions can be thought of as specific cases of the complex-valued log function. Since these definition work for any complex-valued $z {\displaystyle z}$ , the definitions allow forhyperbolic angles as outputs and can be used to further define theinverse hyperbolic functions. It's possible to algebraically prove these relations by starting with the exponential forms of the trigonometric functions and solving for the inverse function.

Example proof

[edit]

{\begin{aligned}\sin(\phi )&=z\\\phi &=\arcsin(z)\end{aligned}}

Using theexponential definition of sine, and letting $\xi =e^{i\phi },$

{\begin{aligned}z&={\frac {e^{i\phi }-e^{-i\phi }}{2i}}\\[10mu]2iz&=\xi -{\frac {1}{\xi }}\\[5mu]0&=\xi ^{2}-2iz\xi -1\\[5mu]\xi &=iz\pm {\sqrt {1-z^{2}}}\\[5mu]\phi &=-i\ln \left(iz\pm {\sqrt {1-z^{2}}}\right)\end{aligned}}

(the positive branch is chosen)

\phi =\arcsin(z)=-i\ln \left(iz+{\sqrt {1-z^{2}}}\right)

Color wheel graphs ofinverse trigonometric functions in thecomplex plane

$\arcsin(z)$	$\arccos(z)$	$\arctan(z)$



$\operatorname {arccsc}(z)$	$\operatorname {arcsec}(z)$	$\operatorname {arccot}(z)$

Applications

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Finding the angle of a right triangle

[edit]

Aright triangle with sides relative to an angle at the $A {\displaystyle A}$ point.

Aright triangle with sides relative to an angle at the $A {\displaystyle A}$ point.

Inverse trigonometric functions are useful when trying to determine the remaining two angles of aright triangle when the lengths of the sides of the triangle are known. Recalling the right-triangle definitions of sine and cosine, it follows that

\theta =\arcsin \left({\frac {\text{opposite}}{\text{hypotenuse}}}\right)=\arccos \left({\frac {\text{adjacent}}{\text{hypotenuse}}}\right).

Often, the hypotenuse is unknown and would need to be calculated before using arcsine or arccosine using thePythagorean Theorem: $a^{2}+b^{2}=h^{2}$ where $h {\displaystyle h}$ is the length of the hypotenuse. Arctangent comes in handy in this situation, as the length of the hypotenuse is not needed.

\theta =\arctan \left({\frac {\text{opposite}}{\text{adjacent}}}\right)\,.

For example, suppose a roof drops 8 feet as it runs out 20 feet. The roof makes an angleθ with the horizontal, whereθ may be computed as follows:

\theta =\arctan \left({\frac {\text{opposite}}{\text{adjacent}}}\right)=\arctan \left({\frac {\text{rise}}{\text{run}}}\right)=\arctan \left({\frac {8}{20}}\right)\approx 21.8^{\circ }\,.

In computer science and engineering

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Two-argument variant of arctangent

[edit]

Main article:atan2

The two-argumentatan2 function computes the arctangent ofy/x giveny andx, but with a range of(−π, π]. In other words,atan2(y,x) is the angle between the positivex-axis of a plane and the point(x,y) on it, with positive sign for counter-clockwise angles (upper half-plane,y > 0), and negative sign for clockwise angles (lower half-plane,y < 0). It was first introduced in many computer programming languages, but it is now also common in other fields of science and engineering.

In terms of the standardarctan function, that is with range of(−π/2, π/2), it can be expressed as follows:

$\operatorname {atan2} (y,x)={\begin{cases}\arctan \left({\frac {y}{x}}\right)&\quad x>0\\\arctan \left({\frac {y}{x}}\right)+\pi &\quad y\geq 0,\;x<0\\\arctan \left({\frac {y}{x}}\right)-\pi &\quad y<0,\;x<0\\{\frac {\pi }{2}}&\quad y>0,\;x=0\\-{\frac {\pi }{2}}&\quad y<0,\;x=0\\{\text{undefined}}&\quad y=0,\;x=0\end{cases}}$

It also equals theprincipal value of theargument of thecomplex numberx +iy.

This limited version of the function above may also be defined using thetangent half-angle formulae as follows: $\operatorname {atan2} (y,x)=2\arctan \left({\frac {y}{{\sqrt {x^{2}+y^{2}}}+x}}\right)$ provided that eitherx > 0 ory ≠ 0. However this fails if givenx ≤ 0 andy = 0 so the expression is unsuitable for computational use.

The above argument order (y,x) seems to be the most common, and in particular is used inISO standards such as theC programming language, but a few authors may use the opposite convention (x,y) so some caution is warranted.(See variations atatan2 § Realizations of the function in common computer languages.)

Arctangent function with location parameter

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In many applications^[17] the solution $y {\displaystyle y}$ of the equation $x=\tan(y)$ is to come as close as possible to a given value $-\infty <\eta <\infty$ . The adequate solution is produced by the parameter modified arctangent function

y=\arctan _{\eta }(x):=\arctan(x)+\pi \,\operatorname {rni} \left({\frac {\eta -\arctan(x)}{\pi }}\right)\,.

The function $\operatorname {rni}$ rounds to the nearest integer.

Numerical accuracy

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For angles near 0 andπ, arccosine isill-conditioned, and similarly with arcsine for angles near −π/2 andπ/2. Computer applications thus need to consider the stability of inputs to these functions and the sensitivity of their calculations, or use alternate methods.^[18]

Notes

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^The expression "LHS $\,\iff \,$ RHS" indicates thateither (a) the left hand side (i.e. LHS) and right hand side (i.e. RHS) areboth true, or else (b) the left hand side and right hand side areboth false; there isno option (c) (e.g. it isnot possible for the LHS statement to be true and also simultaneously for the RHS statement to be false), because otherwise "LHS $\,\iff \,$ RHS" would not have been written.
To clarify, suppose that it is written "LHS $\,\iff \,$ RHS" where LHS (which abbreviatesleft hand side) and RHS are both statements that can individually be either be true or false. For example, if $\theta$ and $s {\displaystyle s}$ are some given and fixed numbers and if the following is written: $\tan \theta =s\,\iff \,\theta =\arctan(s)+\pi k\quad {\text{ for some }}k\in \mathbb {Z}$ then LHS is the statement " $\tan \theta =s$ ". Depending on what specific values $\theta$ and $s {\displaystyle s}$ have, this LHS statement can either be true or false. For instance, LHS is true if $\theta =0$ and $s=0$ (because in this case $\tan \theta =\tan 0=s$ ) but LHS is false if $\theta =0$ and $s=2$ (because in this case $\tan \theta =\tan 0=s$ which is not equal to $s=2$ ); more generally, LHS is false if $\theta =0$ and $s\neq 0.$ Similarly, RHS is the statement " $\theta =\arctan(s)+\pi k$ for some $k\in \mathbb {Z}$ ". The RHS statement can also either true or false (as before, whether the RHS statement is true or false depends on what specific values $\theta$ and $s {\displaystyle s}$ have). The logical equality symbol $\,\iff \,$ means that (a) if the LHS statement is true then the RHS statement is alsonecessarily true, and moreover (b) if the LHS statement is false then the RHS statement is alsonecessarily false. Similarly, $\,\iff \,$ also means that (c) if the RHS statement is true then the LHS statement is alsonecessarily true, and moreover (d) if the RHS statement is false then the LHS statement is alsonecessarily false.

References

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Abramowitz, Milton;Stegun, Irene A., eds. (1972).Handbook of Mathematical Functions with Formulas, Graphs, and Mathematical Tables. New York:Dover Publications.ISBN 978-0-486-61272-0.

^^a ^b ^c ^dHall, Arthur Graham; Frink, Fred Goodrich (Jan 1909)."Chapter II. The Acute Angle [14] Inverse trigonometric functions". Written at Ann Arbor, Michigan, USA.Trigonometry. Vol. Part I: Plane Trigonometry. New York, USA:Henry Holt and Company / Norwood Press / J. S. Cushing Co. - Berwick & Smith Co., Norwood, Massachusetts, USA. p. 15. Retrieved2017-08-12.[…]α = arcsin m: It is frequently read "arc-sinem" or "anti-sinem," since two mutually inverse functions are said each to be theanti-function of the other. […] A similar symbolic relation holds for the othertrigonometric functions. […] This notation is universally used in Europe and is fast gaining ground in this country. A less desirable symbol,α = sin^-1m, is still found in English and American texts. The notationα = inv sinm is perhaps better still on account of its general applicability. […]
^Klein, Felix (1924) [1902].Elementarmathematik vom höheren Standpunkt aus: Arithmetik, Algebra, Analysis (in German). Vol. 1 (3rd ed.). Berlin: J. Springer. Translated asElementary Mathematics from an Advanced Standpoint: Arithmetic, Algebra, Analysis. Translated by Hedrick, E. R.; Noble, C. A. Macmillan. 1932.ISBN 978-0-486-43480-3.{{cite book}}:ISBN / Date incompatibility (help)
^Hazewinkel, Michiel (1994) [1987].Encyclopaedia of Mathematics (unabridged reprint ed.).Kluwer Academic Publishers /Springer Science & Business Media.ISBN 978-155608010-4.
Bronshtein, I. N.; Semendyayev, K. A.; Musiol, Gerhard; Mühlig, Heiner. "Cyclometric or Inverse Trigonometric Functions".Handbook of Mathematics (6th ed.). Berlin: Springer.§ 2.8, pp. 85–89.doi:10.1007/978-3-663-46221-8 (inactive 1 Jul 2025).{{cite book}}: CS1 maint: DOI inactive as of July 2025 (link)
However, the term "arcus function" can also refer to the function giving theargument of a complex number, sometimes called thearcus.
^Weisstein, Eric W."Inverse Trigonometric Functions".mathworld.wolfram.com. Retrieved2020-08-29.
^Beach, Frederick Converse; Rines, George Edwin, eds. (1912). "Inverse trigonometric functions".The Americana: a universal reference library. Vol. 21.
^Cook, John D. (11 Feb 2021)."Trig functions across programming languages".johndcook.com (blog). Retrieved2021-03-10.
^Cajori, Florian (1919).A History of Mathematics (2 ed.). New York, NY:The Macmillan Company. p. 272.
^Herschel, John Frederick William (1813)."On a remarkable Application of Cotes's Theorem".Philosophical Transactions.103 (1). Royal Society, London: 8.doi:10.1098/rstl.1813.0005.
^"Inverse trigonometric functions". Wiki.Brilliant Math & Science (brilliant.org). Retrieved2020-08-29.
^Korn, Grandino Arthur;Korn, Theresa M. (2000) [1961]. "21.2.-4. Inverse Trigonometric Functions".Mathematical handbook for scientists and engineers: Definitions, theorems, and formulars for reference and review (3 ed.). Mineola, New York, USA:Dover Publications, Inc. p. 811.ISBN 978-0-486-41147-7.
^Bhatti, Sanaullah; Nawab-ud-Din; Ahmed, Bashir; Yousuf, S. M.; Taheem, Allah Bukhsh (1999). "Differentiation of Trigonometric, Logarithmic and Exponential Functions". In Ellahi, Mohammad Maqbool; Dar, Karamat Hussain; Hussain, Faheem (eds.).Calculus and Analytic Geometry (1 ed.).Lahore: Punjab Textbook Board. p. 140.
^For example:
Stewart, James; Clegg, Daniel; Watson, Saleem (2021). "Inverse Functions and Logarithms".Calculus: Early Transcendentals (9th ed.). Cengage Learning. § 1.5, p. 64.ISBN 978-1-337-61392-7.
^Abramowitz & Stegun 1972, p. 73, 4.3.44
^Borwein, Jonathan; Bailey, David; Gingersohn, Roland (2004).Experimentation in Mathematics: Computational Paths to Discovery (1 ed.). Wellesley, MA, USA:A. K. Peters. p. 51.ISBN 978-1-56881-136-9.
^Hwang Chien-Lih (2005), "An elementary derivation of Euler's series for the arctangent function",The Mathematical Gazette,89 (516):469–470,doi:10.1017/S0025557200178404,S2CID 123395287
^S. M. Abrarov and B. M. Quine (2018), "A formula for pi involving nested radicals",The Ramanujan Journal,46 (3):657–665,arXiv:1610.07713,doi:10.1007/s11139-018-9996-8,S2CID 119150623
^when a time varying angle crossing $\pm \pi /2$ should be mapped by a smooth line instead of a saw toothed one (robotics, astronomy, angular movement in general)^{[citation needed]}
^Gade, Kenneth (2010)."A non-singular horizontal position representation"(PDF).The Journal of Navigation.63 (3).Cambridge University Press:395–417.Bibcode:2010JNav...63..395G.doi:10.1017/S0373463309990415.