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 of θ radians will correspond to an arc whose length is rθ, where r is the radius of the circle. Thus in the unit 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 is x" is the same as "the angle whose cosine is x", 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 forms asin, acos, atan.^[6]

The notations sin⁻¹(x), cos⁻¹(x), tan⁻¹(x), etc., as introduced by John Herschel in 1813,^[7][8] are often used as well in English-language sources,^[1] much more than the also established sin^[−1](x), cos^[−1](x), tan^[−1](x) – conventions consistent with the notation of an inverse function, that is useful (for example) to define the multivalued version of each inverse trigonometric function: ${\displaystyle \tan ^{-1}(x)=\{\arctan(x)+\pi k\mid k\in \mathbb {Z} \}~.}$  However, this might appear to conflict logically with the common semantics for expressions such as sin²(x) (although only sin² x, without parentheses, is the really common use), which refer to numeric power rather than function composition, and therefore may result in confusion between notation for the reciprocal (multiplicative inverse) and inverse 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 an uppercase 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 the reciprocal, which should be represented by sin⁻¹(x), cos⁻¹(x), etc., or, better, by sin⁻¹ x, cos⁻¹ x, etc., it in turn creates yet another major source of ambiguity, especially since many popular high-level programming languages (e.g. Mathematica and MAGMA) 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, the ISO 80000-2 standard has specified solely the "arc" prefix for the inverse functions.

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 ${\displaystyle -1\leq x\leq 1,}$  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 (${\textstyle 0\leq y<{\frac {\pi }{2}}}$  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, ${\displaystyle \tan(\operatorname {arcsec}(x))={\sqrt {x^{2}-1}},}$  whereas with the range (${\textstyle 0\leq y<{\frac {\pi }{2}}}$  or ${\textstyle {\frac {\pi }{2}}<y\leq \pi }$ ), we would have to write ${\displaystyle \tan(\operatorname {arcsec}(x))=\pm {\sqrt {x^{2}-1}},}$  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 ${\textstyle 0<y\leq {\frac {\pi }{2}}).}$ 

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

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

The symbol ${\displaystyle \,\setminus \,}$  denotes set subtraction so that, for instance, ${\displaystyle \mathbb {R} \setminus (-1,1)=(-\infty ,-1]\cup [1,\infty )}$  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 ${\textstyle \pi \mathbb {Z} +(0,\pi )}$  and ${\displaystyle \pi \mathbb {Z} +{\bigl (}{-{\tfrac {\pi }{2}}},{\tfrac {\pi }{2}}{\bigr )}}$  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 ${\displaystyle \sin \theta \neq 0,}$  i.e. all real numbers that are not of the form ${\displaystyle \pi n}$  for some integer ${\displaystyle n,}$ 

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

Domain of tangent ${\displaystyle \tan }$  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 ${\displaystyle \cos \theta \neq 0,}$ 

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

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 ${\textstyle 2\pi k-{\frac {\pi }{2}}}$  (where ${\displaystyle k}$  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 ${\displaystyle 2\pi k,}$  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 ${\textstyle 2\pi k-{\frac {\pi }{2}},}$  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 ${\displaystyle 2\pi k,}$  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 Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "http://localhost:6011/en.wikipedia.org/v1/":): {\displaystyle \,\iff\,} 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^{[note 1]} 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 ${\textstyle \theta ={\frac {\pi }{2}}+\pi k=-{\frac {\pi }{2}}+\pi (k+1)}$  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 $${\displaystyle \operatorname {Parity} (h)={\begin{cases}0&{\text{if }}h{\text{ is even }}\\1&{\text{if }}h{\text{ is odd }}\\\end{cases}}}$$  it is possible to write a solution to ${\displaystyle \cos \theta =x}$  that doesn't involve the "plus or minus" ${\displaystyle \,\pm \,}$  symbol:

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

And similarly for the secant function,

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

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

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 ${\displaystyle -1\leq x\leq 1}$  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 $${\displaystyle \,\theta =\pm \arccos x+2\pi k\,\quad {\text{ for some }}k\in \mathbb {Z} }$$  which is a shorthand way of saying that (at least) one of the following statement is true:

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 ${\displaystyle x\neq -1}$  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 ${\displaystyle \,\arccos x\neq 0\,}$  and ${\displaystyle \,\arccos x\neq \pi ,\,}$  So assume this from now on. The solution to ${\displaystyle \cos \theta =x}$  is still $${\displaystyle \,\theta =\pm \arccos x+2\pi k\,\quad {\text{ for some }}k\in \mathbb {Z} }$$  which as before is shorthand for saying that one of statements (1) and (2) is true. However this time, because ${\displaystyle \,\arccos x\neq 0\,}$  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 ${\displaystyle \,-\pi \leq \theta \leq \pi \,}$  (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 ${\displaystyle \theta =-\pi /2}$ ).

The table below shows how two angles ${\displaystyle \theta }$  and ${\displaystyle \varphi }$  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 ${\displaystyle \varphi }$  satisfy ${\displaystyle \left|\sin \theta \right|=\left|\sin \varphi \right|}$  if and only if they satisfy ${\displaystyle \left|\cos \theta \right|=\left|\cos \varphi \right|.}$ 

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:

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

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

$${\displaystyle {\begin{aligned}\sin \theta &=-\sin(-\theta )&&=-\sin(\pi +\theta )&&={\phantom {-}}\sin(\pi -\theta )\\&=-\cos \left({\frac {\pi }{2}}+\theta \right)&&={\phantom {-}}\cos \left({\frac {\pi }{2}}-\theta \right)&&=-\cos \left(-{\frac {\pi }{2}}-\theta \right)\\&={\phantom {-}}\cos \left(-{\frac {\pi }{2}}+\theta \right)&&=-\cos \left({\frac {3\pi }{2}}-\theta \right)&&=-\cos \left(-{\frac {3\pi }{2}}+\theta \right)\\[0.3ex]\cos \theta &={\phantom {-}}\cos(-\theta )&&=-\cos(\pi +\theta )&&=-\cos(\pi -\theta )\\&={\phantom {-}}\sin \left({\frac {\pi }{2}}+\theta \right)&&={\phantom {-}}\sin \left({\frac {\pi }{2}}-\theta \right)&&=-\sin \left(-{\frac {\pi }{2}}-\theta \right)\\&=-\sin \left(-{\frac {\pi }{2}}+\theta \right)&&=-\sin \left({\frac {3\pi }{2}}-\theta \right)&&={\phantom {-}}\sin \left(-{\frac {3\pi }{2}}+\theta \right)\\[0.3ex]\tan \theta &=-\tan(-\theta )&&={\phantom {-}}\tan(\pi +\theta )&&=-\tan(\pi -\theta )\\&=-\cot \left({\frac {\pi }{2}}+\theta \right)&&={\phantom {-}}\cot \left({\frac {\pi }{2}}-\theta \right)&&={\phantom {-}}\cot \left(-{\frac {\pi }{2}}-\theta \right)\\&=-\cot \left(-{\frac {\pi }{2}}+\theta \right)&&={\phantom {-}}\cot \left({\frac {3\pi }{2}}-\theta \right)&&=-\cot \left(-{\frac {3\pi }{2}}+\theta \right)\\[0.3ex]\end{aligned}}}$$ 

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 ${\textstyle \sin \left({\frac {\pi }{2}}-\theta \right)=\cos \theta ,}$  the equation ${\displaystyle \cos \theta =x}$  can be transformed into ${\textstyle \sin \left({\frac {\pi }{2}}-\theta \right)=x,}$  which allows for the solution to the equation ${\displaystyle \;\sin \varphi =x\;}$  (where ${\textstyle \varphi :={\frac {\pi }{2}}-\theta }$ ) to be used; that solution being: ${\displaystyle \varphi =(-1)^{k}\arcsin(x)+\pi k\;{\text{ for some }}k\in \mathbb {Z} ,}$  which becomes: $${\displaystyle {\frac {\pi }{2}}-\theta ~=~(-1)^{k}\arcsin(x)+\pi k\quad {\text{ for some }}k\in \mathbb {Z} }$$  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: $${\displaystyle \theta ~=~(-1)^{h+1}\arcsin(x)+\pi h+{\frac {\pi }{2}}\quad {\text{ for some }}h\in \mathbb {Z} .}$$  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 ${\displaystyle \;\arcsin x.\;}$