Inverse trigonometric functions

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

For example, using function in the sense of multivalued functions, just as the square root function y = √x could be defined from y² = x, the function y = arcsin(x) is defined so that sin(y) = x. For a given real number x, with −1 ≤ x ≤ 1, there are multiple (in fact, countably infinite) numbers y such that sin(y) = x; for example, sin(0) = 0, but also sin(π) = 0, sin(2π) = 0, etc. When only one value is desired, the function may be restricted to its principal branch. With this restriction, for each x in the domain, the expression 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 ≤ y < π/2 or π ≤ y < 3π/2 ), because the tangent function is nonnegative on this domain. This makes some computations more consistent. For example, using this range, tan(arcsec(x)) = √x² − 1, whereas with the range ( 0 ≤ y < π/2 or π/2 < y ≤ π ), we would have to write tan(arcsec(x)) = ±√x² − 1, since tangent is nonnegative on 0 ≤ y < π/2, but nonpositive on π/2 < y ≤ π. For a similar reason, the same authors define the range of arccosecant to be −π < y ≤ −π/2 or 0 < y ≤ π/2.)

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

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

• Sine and cosecant begin their period at 2πk − π/2 (where k is an integer), finish it at 2πk + π/2, and then reverse themselves over 2πk + π/2 to 2πk + 3π/2.
• Cosine and secant begin their period at 2πk, finish it at 2πk + π, and then reverse themselves over 2πk + π to 2πk + 2π.
• Tangent begins its period at 2πk − π/2, finishes it at 2πk + π/2, and then repeats it (forward) over 2πk + π/2 to 2πk + 3π/2.
• Cotangent begins its period at 2πk, finishes it at 2πk + π, and then repeats it (forward) over 2πk + π to 2πk + 2π.

This periodicity is reflected in the general inverses, where k is some integer.

The following table shows how inverse trigonometric functions may be used to solve equalities involving the six standard trigonometric functions, where it is assumed that r, s, x, and y all lie within the appropriate range.

The symbol ⇔ is logical equality. The expression "LHS ⇔ RHS" indicates that either (a) the left hand side (i.e. LHS) and right hand side (i.e. RHS) are both true, or else (b) the left hand side and right hand side are both false; there is no option (c) (e.g. it is not possible for the LHS statement to be true and also simultaneously for the RHS statement to false), because otherwise "LHS ⇔ RHS" would not have been written (see this footnote[note 1] for an example illustrating this concept).

In the table below, we show how two angles θ and φ must be related, if their values under a given trigonometric function are equal or negatives of each other.

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, then applying the Pythagorean theorem and definitions of the trigonometric ratios. Purely algebraic derivations are longer. It's worth noting that for arcsecant and arccosecant, the diagram assumes that x is positive, and thus the result has to be corrected through the use of absolute values and the signum (sgn) operation.

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

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:

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

Negative arguments:

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

Reciprocal arguments:

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

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

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

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

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

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, ${\displaystyle \tan \left({\tfrac {\theta }{2}}\right)={\tfrac {\sin(\theta )}{1+\cos(\theta )}}}$, we get:

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

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

This is derived from the tangent addition formula

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

by letting

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

Main article: Differentiation of trigonometric functions

The derivatives for complex values of z are as follows:

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

Only for real values of x:

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

For a sample derivation: if ${\displaystyle \theta =\arcsin(x)}$, we get:

${\displaystyle {\frac {d\arcsin(x)}{dx}}={\frac {d\theta }{d\sin(\theta )}}={\frac {d\theta }{\cos(\theta )\,d\theta }}={\frac {1}{\cos(\theta )}}={\frac {1}{\sqrt {1-\sin ^{2}(\theta )}}}={\frac {1}{\sqrt {1-x^{2}}}}}$

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

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

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

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, ${\textstyle {\tfrac {1}{\sqrt {1-z^{2}}}}}$, 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 ${\textstyle {\frac {1}{1+z^{2}}}}$ in a geometric series, and applying the integral definition above (see Leibniz series).

${\displaystyle {\begin{aligned}\arcsin(z)&=z+\left({\frac {1}{2}}\right){\frac {z^{3}}{3}}+\left({\frac {1\cdot 3}{2\cdot 4}}\right){\frac {z^{5}}{5}}+\left({\frac {1\cdot 3\cdot 5}{2\cdot 4\cdot 6}}\right){\frac {z^{7}}{7}}+\cdots \\[5pt]&=\sum _{n=0}^{\infty }{\frac {(2n-1)!!}{(2n)!!}}{\frac {z^{2n+1}}{2n+1}}\\[5pt]&=\sum _{n=0}^{\infty }{\frac {(2n)!}{(2^{n}n!)^{2}}}{\frac {z^{2n+1}}{2n+1}}\,;\qquad |z|\leq 1\end{aligned}}}$

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

Series for the other inverse trigonometric functions can be given in terms of these according to the relationships given above. For example, ${\displaystyle \arccos(x)=\pi /2-\arcsin(x)}$, ${\displaystyle \operatorname {arccsc}(x)=\arcsin(1/x)}$, and so on. Another series is given by:[18]

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

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

${\displaystyle \arctan(z)={\frac {z}{1+z^{2}}}\sum _{n=0}^{\infty }\prod _{k=1}^{n}{\frac {2kz^{2}}{(2k+1)(1+z^{2})}}.}$[19]

(The term in the sum for n = 0 is the empty product, so is 1.)

Alternatively, this can be expressed as

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

Another series for the arctangent function is given by

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

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

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

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

The second of these is valid in the cut complex plane. There are two cuts, from −i to 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.

For real and complex values of z:

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

For real x ≥ 1:

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

For all real x not between -1 and 1:

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

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 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:

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

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 using integration by parts and the simple derivative forms shown above.

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

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

Then

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

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

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

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.

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

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:

${\displaystyle ce^{\theta i}=c\cos(\theta )+ci\sin(\theta )}$

or

${\displaystyle ce^{\theta i}=a+bi}$

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 ${\displaystyle \theta }$.

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

or

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

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 above and simplifying.

${\displaystyle {\begin{aligned}&a&&b&&c&&\theta &&\theta _{\text{simplified}}&&\theta _{a,b\in \mathbb {R} }\\\arcsin(z)\ \ &{\sqrt {1-z^{2}}}&&z&&1&&-i\ln \left({\frac {{\sqrt {1-z^{2}}}+zi}{1}}\right)&&=-i\ln \left({\sqrt {1-z^{2}}}+zi\right)&&\operatorname {Im} \left(\ln \left({\sqrt {1-z^{2}}}+zi\right)\right)\\\arccos(z)\ \ &z&&{\sqrt {1-z^{2}}}&&1&&-i\ln \left({\frac {z+i{\sqrt {1-z^{2}}}}{1}}\right)&&=-i\ln \left(z+{\sqrt {z^{2}-1}}\right)&&\operatorname {Im} \left(\ln \left(z+{\sqrt {z^{2}-1}}\right)\right)\\\arctan(z)\ \ &1&&z&&{\sqrt {1+z^{2}}}&&i\ln \left({\frac {\sqrt {1+z^{2}}}{1+zi}}\right)&&={\frac {i}{2}}\ln \left({\frac {i+z}{i-z}}\right)&&\operatorname {Im} \left(\ln \left(1+zi\right)\right)\\\operatorname {arccot}(z)\ \ &z&&1&&{\sqrt {z^{2}+1}}&&i\ln \left({\frac {\sqrt {z^{2}+1}}{z+i}}\right)&&={\frac {i}{2}}\ln \left({\frac {z-i}{z+i}}\right)&&\operatorname {Im} \left(\ln \left(z+i\right)\right)\\\operatorname {arcsec}(z)\ \ &1&&{\sqrt {z^{2}-1}}&&z&&-i\ln \left({\frac {1+i{\sqrt {z^{2}-1}}}{z}}\right)&&=-i\ln \left({\frac {1}{z}}+{\sqrt {{\frac {1}{z^{2}}}-1}}\right)&&\operatorname {Im} \left(\ln \left(1+{\sqrt {1-z^{2}}}\right)\right)\\\operatorname {arccsc}(z)\ \ &{\sqrt {z^{2}-1}}&&1&&z&&-i\ln \left({\frac {{\sqrt {z^{2}-1}}+i}{z}}\right)&&=-i\ln \left({\sqrt {1-{\frac {1}{z^{2}}}}}+{\frac {i}{z}}\right)&&\operatorname {Im} \left(\ln \left({\sqrt {z^{2}-1}}+i\right)\right)\\\end{aligned}}}$

In this sense, all of the inverse trig functions can be thought of as specific cases of the complex-valued log function. Since this definition works for any complex-valued ${\displaystyle z}$, this definition allows for hyperbolic angles as outputs and can be used to further define the inverse hyperbolic functions. Elementary proofs of the relations may also proceed via expansion to exponential forms of the trigonometric functions.

${\displaystyle \sin(\phi )=z}$

${\displaystyle \phi =\arcsin(z)}$

Using the exponential definition of sine, one obtains

${\displaystyle z={\frac {e^{\phi i}-e^{-\phi i}}{2i}}}$

Let

${\displaystyle \xi =e^{\phi i}}$

Solving for ${\displaystyle \phi }$

${\displaystyle z={\frac {\xi -{\frac {1}{\xi }}}{2i}}}$

${\displaystyle 2iz={\xi -{\frac {1}{\xi }}}}$

${\displaystyle {\xi -2iz-{\frac {1}{\xi }}}=0}$

${\displaystyle \xi ^{2}-2i\xi z-1\,=\,0}$

${\displaystyle \xi =iz\pm {\sqrt {1-z^{2}}}=e^{\phi i}}$

${\displaystyle \phi i=\ln \left(iz\pm {\sqrt {1-z^{2}}}\right)}$

${\displaystyle \phi =-i\ln \left(iz\pm {\sqrt {1-z^{2}}}\right)}$

(the positive branch is chosen)

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

Color wheel graphs of inverse trigonometric functions in the complex plane

${\displaystyle \arcsin(z)}$	${\displaystyle \arccos(z)}$	${\displaystyle \arctan(z)}$	${\displaystyle \operatorname {arccot}(z)}$	${\displaystyle \operatorname {arcsec}(z)}$	${\displaystyle \operatorname {arccsc}(z)}$

A right triangle.

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

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

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

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

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

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

The two-argument atan2 function computes the arctangent of y / x given y and x, but with a range of (−π, π]. In other words, atan2(y, x) is the angle between the positive x-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 standard arctan function, that is with range of (−π/2, π/2), it can be expressed as follows:

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

It also equals the principal value of the argument of the complex number x + iy.

This function may also be defined using the tangent half-angle formulae as follows:

${\displaystyle \operatorname {atan2} (y,x)=2\arctan \left({\frac {y}{{\sqrt {x^{2}+y^{2}}}+x}}\right)}$

provided that either x > 0 or y ≠ 0. However this fails if given x ≤ 0 and y = 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 in ISO standards such as the C programming language, but a few authors may use the opposite convention (x, y) so some caution is warranted. These variations are detailed at atan2.

In many applications[20] the solution ${\displaystyle y}$ of the equation ${\displaystyle x=\tan(y)}$ is to come as close as possible to a given value ${\displaystyle -\infty <\eta <\infty }$. The adequate solution is produced by the parameter modified arctangent function

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

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

For angles near 0 and π, arccosine is ill-conditioned and will thus calculate the angle with reduced accuracy in a computer implementation (due to the limited number of digits).[21] Similarly, arcsine is inaccurate for angles near −π/2 and π/2.