[ In computer programming languages, the inverse trigonometric functions are often called by the abbreviated forms , , .]
The notations , , , etc., as introduced by John Herschel in 1813,[ are often used as well in English-language sources,][ much more than the also established , , – conventions consistent with the notation of an inverse function, that is useful (for example) to define the multivalued version of each inverse trigonometric function: However, this might appear to conflict logically with the common semantics for expressions such as (although only , 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.]
The confusion is somewhat mitigated by the fact that each of the reciprocal trigonometric functions has its own name — for example, . Nevertheless, certain authors advise against using it, since it is ambiguous.[ Another precarious convention used by a small number of authors is to use an UPPERCASE first letter, along with a “” superscript: , , , etc.][ Although it is intended to avoid confusion with the reciprocal, which should be represented by , , etc., or, better, by , , 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 standard has specified solely the "arc" prefix for the inverse functions.
Basic concepts
Principal values
Since none of the six trigonometric functions are one-to-one, they must be restricted in order to have inverse functions. Therefore, the result ranges of the inverse functions are proper (i.e. strict) of the domains of the original functions.
For example, using in the sense of multivalued functions, just as the square root function could be defined from the function is defined so that For a given real number with there are multiple (in fact, countably infinitely many) numbers such that ; for example, but also etc. When only one value is desired, the function may be restricted to its principal branch. With this restriction, for each in the domain, the expression 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 or [For example: ] because the tangent function is nonnegative on this domain. This makes some computations more consistent. For example, using this range, whereas with the range or we would have to write since tangent is nonnegative on but nonpositive on For a similar reason, the same authors define the range of arccosecant to be or
Domains
If is allowed to be a complex number, then the range of applies only to its real part.
Solutions to elementary trigonometric equations
Each of the trigonometric functions is periodic in the real part of its argument, running through all its values twice in each interval of
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Sine and cosecant begin their period at (where is an integer), finish it at and then reverse themselves over to
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Cosine and secant begin their period at finish it at and then reverse themselves over to
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Tangent begins its period at finishes it at and then repeats it (forward) over to
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Cotangent begins its period at finishes it at and then repeats it (forward) over to
This periodicity is reflected in the general inverses, where 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 and all lie within appropriate ranges so that the relevant expressions below are well-defined.
Note that "for some " is just another way of saying "for some integer "
The symbol 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[The expression "LHS RHS" indicates that (a) the left hand side (i.e. LHS) and right hand side (i.e. RHS) are true, or else (b) the left hand side and right hand side are false; there is option (c) (e.g. it is possible for the LHS statement to be true and also simultaneously for the RHS statement to be false), because otherwise "LHS RHS" would not have been written. ]
To clarify, suppose that it is written "LHS RHS" where LHS (which abbreviates left hand side) and RHS are both statements that can individually be either be true or false. For example, if and are some given and fixed numbers and if the following is written:
then LHS is the statement "". Depending on what specific values and have, this LHS statement can either be true or false. For instance, LHS is true if and (because in this case ) but LHS is false if and (because in this case which is not equal to ); more generally, LHS is false if and Similarly, RHS is the statement " for some ". The RHS statement can also either true or false (as before, whether the RHS statement is true or false depends on what specific values and have). The logical equality symbol means that (a) if the LHS statement is true then the RHS statement is also true, and moreover (b) if the LHS statement is false then the RHS statement is also false. Similarly, means that (c) if the RHS statement is true then the LHS statement is also true, and moreover (d) if the RHS statement is false then the LHS statement is also false.
for more details and an example illustrating this concept).
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where the first four solutions can be written in expanded form as:
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For example, if then for some While if then for some where will be even if and it will be odd if The equations and have the same solutions as and respectively. In all equations above for those just solved (i.e. except for / and /), the integer in the solution's formula is uniquely determined by (for fixed and ).
With the help of integer parity
it is possible to write a solution to that doesn't involve the "plus or minus" symbol:
- if and only if for some
And similarly for the secant function,
- if and only if for some
where equals when the integer is even, and equals when it's odd.
Detailed example and explanation of the "plus or minus" symbol
The solutions to and involve the "plus or minus" symbol whose meaning is now clarified. Only the solution to will be discussed since the discussion for is the same.
We are given between and we know that there is an angle in some interval that satisfies We want to find this The table above indicates that the solution is
which is a shorthand way of saying that (at least) one of the following statement is true:
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for some integer
or
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for some integer
As mentioned above, if (which by definition only happens when ) then both statements (1) and (2) hold, although with different values for the integer : if is the integer from statement (1), meaning that holds, then the integer for statement (2) is (because ).
However, if then the integer is unique and completely determined by
If (which by definition only happens when ) then (because and so in both cases is equal to ) and so the statements (1) and (2) happen to be identical in this particular case (and so both hold).
Having considered the cases and we now focus on the case where and So assume this from now on. The solution to is still
which as before is shorthand for saying that one of statements (1) and (2) is true. However this time, because and statements (1) and (2) are different and furthermore, exactly one of the two equalities holds (not both). Additional information about is needed to determine which one holds. For example, suppose that and that that is known about is that (and nothing more is known). Then
and moreover, in this particular case (for both the case and the case) and so consequently,
This means that could be either or Without additional information it is not possible to determine which of these values has.
An example of some additional information that could determine the value of would be knowing that the angle is above the -axis (in which case ) or alternatively, knowing that it is below the -axis (in which case ).
Equal identical trigonometric functions
- Set of all solutions to elementary trigonometric equations
Thus given a single solution to an elementary trigonometric equation ( is such an equation, for instance, and because always holds, is always a solution), the set of all solutions to it are:
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Transforming equations
The equations above can be transformed by using the reflection and shift identities:
+ Transforming equations by shifts and reflections |
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These formulas imply, in particular, that the following hold:
where swapping swapping and swapping gives the analogous equations for respectively.
So for example, by using the equality the equation can be transformed into which allows for the solution to the equation (where ) to be used; that solution being:
which becomes:
where using the fact that and substituting proves that another solution to is:
The substitution may be used express the right hand side of the above formula in terms of instead of
Relationships between trigonometric functions and inverse trigonometric functions
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 then applying the Pythagorean theorem and definitions of the trigonometric ratios. It is worth noting that for arcsecant and arccosecant, the diagram assumes that is positive, and thus the result has to be corrected through the use of and the Sign function (sgn) operation.
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Relationships among the inverse trigonometric functions
Complementary angles:
\arccos(x) &= \frac{\pi}{2} - \arcsin(x) \\0.5em
\arccot(x) &= \frac{\pi}{2} - \arctan(x) \\0.5em
\arccsc(x) &= \frac{\pi}{2} - \arcsec(x)
\end{align}
Negative arguments:
\arcsin(-x) &= -\arcsin(x) \\
\arccsc(-x) &= -\arccsc(x) \\
\arccos(-x) &= \pi -\arccos(x) \\
\arcsec(-x) &= \pi -\arcsec(x) \\
\arctan(-x) &= -\arctan(x) \\
\arccot(-x) &= \pi -\arccot(x)
\end{align}
Reciprocal arguments:
\arcsin\left(\frac{1}{x}\right) &= \arccsc(x) & \\0.3em
\arccsc\left(\frac{1}{x}\right) &= \arcsin(x) & \\0.3em
\arccos\left(\frac{1}{x}\right) &= \arcsec(x) & \\0.3em
\arcsec\left(\frac{1}{x}\right) &= \arccos(x) & \\0.3em
\arctan\left(\frac{1}{x}\right) &= \arccot(x) &= \frac{\pi}{2} - \arctan(x) \, , \text{ if } x > 0 \\0.3em
\arctan\left(\frac{1}{x}\right) &= \arccot(x) - \pi &= -\frac{\pi}{2} - \arctan(x) \, , \text{ if } x < 0 \\0.3em
\arccot\left(\frac{1}{x}\right) &= \arctan(x) &= \frac{\pi}{2} - \arccot(x) \, , \text{ if } x > 0 \\0.3em
\arccot\left(\frac{1}{x}\right) &= \arctan(x) + \pi &= \frac{3\pi}{2} - \arccot(x) \, , \text{ if } x < 0
\end{align}
The identities above can be used with (and derived from) the fact that and are reciprocals (i.e. ), as are and and and
Useful identities if one only has a fragment of a sine table:
\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}-\sgn(x)\arcsin(x) \\
\arctan(x) &= \arcsin\left(\frac{x}{\sqrt{1 + x^2}}\right) \\
\arccot(x) &= \arccos\left(\frac{x}{\sqrt{1 + x^2}}\right)
\end{align}
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
- .
It is obtained by recognizing that .
From the half-angle formula, , we get:
\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{align}
Arctangent addition formula
This is derived from the tangent addition formula
by letting
In calculus
Derivatives of inverse trigonometric functions
The for complex values of z are as follows:
\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} \arccot(z) &{} = -\frac{1}{1+z^2} \; ; &z &{}\neq -i, +i \\
\frac{d}{dz} \arcsec(z) &{} = \frac{1}{z^2 \sqrt{1 - \frac{1}{z^{2}}}} \; ; &z &{}\neq -1, 0, +1 \\
\frac{d}{dz} \arccsc(z) &{} = -\frac{1}{z^2 \sqrt{1 - \frac{1}{z^{2}}}} \; ; &z &{}\neq -1, 0, +1
\end{align}
Only for real values of x:
\frac{d}{dx} \arcsec(x) &{} = \frac{1} 1\\>> 1
\end{align}
These formulas can be derived in terms of the derivatives of trigonometric functions. For example, if , then so
Expression as definite integrals
Integrating the derivative and fixing the value at one point gives an expression for the inverse trigonometric function as a definite integral:
\arcsin(x) &{}= \int_0^x \frac{1}{\sqrt{1 - z^2}} \, dz \; , & | &{} \leq 1\\
\arccos(x) &{}= \int_x^1 \frac{1}{\sqrt{1 - z^2}} \, dz \; , & | &{} \leq 1\\
\arctan(x) &{}= \int_0^x \frac{1}{z^2 + 1} \, dz \; ,\\
\arccot(x) &{}= \int_x^\infty \frac{1}{z^2 + 1} \, dz \; ,\\
\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\\
\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{align}
When x equals 1, the integrals with limited domains are improper integrals, but still well-defined.
Infinite series
Similar to the sine and cosine functions, the inverse trigonometric functions can also be calculated using power series, as follows. For arcsine, the series can be derived by expanding its derivative, , 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 in a geometric series, and applying the integral definition above (see Leibniz series).
\begin{align}
\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 | \le 1
\end{align}
= 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 | \le 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, , , and so on. Another series is given by:
Leonhard Euler found a series for the arctangent that converges more quickly than its Taylor series:
(The term in the sum for n = 0 is the empty product, so is 1.)
Alternatively, this can be expressed as
Another series for the arctangent function is given by
where is the imaginary unit.
Continued fractions for arctangent
Two alternatives to the power series for arctangent are these generalized continued fractions:
\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}}}}} =
The second of these is valid in the cut complex plane. There are two cuts, from − i to the point at infinity, going down the imaginary axis, and from i to the point at infinity, going up the same axis. It works best for real numbers running from −1 to 1. The partial denominators are the odd natural numbers, and the partial numerators (after the first) are just ( nz)2, with each perfect square appearing once. The first was developed by Leonhard Euler; the second by Carl Friedrich Gauss utilizing the Gaussian hypergeometric series.
Indefinite integrals of inverse trigonometric functions
For real and complex values of z:
\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 \arccot(z) \, dz &{}= z \, \arccot(z) + \frac{1}{2} \ln \left( 1 + z^2 \right) + C\\
\int \arcsec(z) \, dz &{}= z \, \arcsec(z) - \ln \left + C\\
\int \arccsc(z) \, dz &{}= z \, \arccsc(z) + \ln \left + C
\end{align}
For real x ≥ 1:
\int \arcsec(x) \, dx &{}= x \, \arcsec(x) - \ln \left( x + \sqrt{x^2-1} \right) + C\\
\int \arccsc(x) \, dx &{}= x \, \arccsc(x) + \ln \left( x + \sqrt{x^2-1} \right) + C
\end{align}
For all real x not between -1 and 1:
\int \arcsec(x) \, dx &{}= x \, \arcsec(x) - \sgn(x) \ln\left | + C\\
\int \arccsc(x) \, dx &{}= x \, \arccsc(x) + \sgn(x) \ln\left | + C
\end{align}
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:
\int \arcsec(x) \, dx &{}= x \, \arcsec(x) - \operatorname{arcosh}( | ) + C\\
\int \arccsc(x) \, dx &{}= x \, \arccsc(x) + \operatorname{arcosh}( | ) + C\\
\end{align}
The absolute value in the argument of the arcosh function creates a negative half of its graph, making it identical to the signum logarithmic function shown above.
All of these antiderivatives can be derived using integration by parts and the simple derivative forms shown above.
Example
Using (i.e. integration by parts), set
u &= \arcsin(x) & dv &= dx \\
du &= \frac{dx}{\sqrt{1-x^2}} & v &= x
\end{align}
Then
which by the simple substitution yields the final result:
Extension to the complex plane
Since the inverse trigonometric functions are analytic functions, they can be extended from the Number line to the complex plane. This results in functions with multiple sheets and . One possible way of defining the extension is:
where the part of the imaginary axis which does not lie strictly between the branch points (−i and +i) is the branch cut between the principal sheet and other sheets. The path of the integral must not cross a branch cut. For z not on a branch cut, a straight line path from 0 to z is such a path. For z on a branch cut, the path must approach from for the upper branch cut and from for the lower branch cut.
The arcsine function may then be defined as:
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;
which has the same cut as arcsin;
which has the same cut as arctan;
where the part of the real axis between −1 and +1 inclusive is the cut between the principal sheet of arcsec and other sheets;
which has the same cut as arcsec.
Logarithmic forms
These functions may also be expressed using complex logarithms. This extends their domains to the complex plane in a natural fashion. The following identities for principal values of the functions hold everywhere that they are defined, even on their branch cuts.
\arcsin(z) &{}= -i \ln \left( \sqrt{1-z^2} + iz \right) = i \ln \left( \sqrt{1-z^2} - iz \right) &{}= \arccsc\left(\frac{1}{z}\right) \\10pt
\arccos(z) &{}= -i \ln \left( i \sqrt {1-z^2} + z \right) = \frac{\pi}{2} - \arcsin(z) &{}= \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) &{}= \arccot\left(\frac{1}{z}\right) \\10pt
\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
\arcsec(z) &{}= -i \ln \left( i \sqrt{1 - \frac{1}{z^2}} + \frac{1}{z} \right) = \frac{\pi}{2} - \arccsc(z) &{}= \arccos\left(\frac{1}{z}\right) \\10pt
\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{align}
Generalization
Because all of the inverse trigonometric functions output an angle of a right triangle, they can be generalized by using Euler's formula to form a right triangle in the complex plane. Algebraically, this gives us:
or
where is the adjacent side, is the opposite side, and is the hypotenuse. From here, we can solve for .
e^{\ln(c) + i\theta} & = a + ib \\
\ln c + i\theta & = \ln(a + ib) \\
\theta & = \operatorname{Im}\left( \ln(a + ib) \right)
\end{align}
or
Simply taking the imaginary part works for any real-valued and , but if or 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 also removes 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 , 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
The table below shows the values of a, b, and c for each of the inverse trig functions and the equivalent expressions for that result from plugging the values into the equations above and simplifying.
& a & & b & & c && -i\ln\left(\frac{a+ib}{c}\right) && \theta && \theta_{a,b\in\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) \\
\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) \\
\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) \\
\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{align}
The particular form of the simplified expression can cause the output to differ from the usual principal branch of each of the inverse trig functions. The formulations given will output the usual principal branch when using the and principal branch for every function except arccotangent in the column. Arccotangent in the column will output on its usual principal branch by using the and 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 , the definitions allow for as outputs and can be used to further define the inverse 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
\sin(\phi) &= z \\
\phi &= \arcsin(z)
\end{align}
Using the exponential definition of sine, and letting
z &= \frac{e^{i \phi} - e^{-i \phi}}{2i} \\10mu
2iz &= \xi - \frac{1}{\xi} \\5mu
0 &= \xi^2 - 2i z \xi - 1 \\5mu
\xi &= iz \pm \sqrt{1 - z^2} \\5mu
\phi &= -i \ln \left(iz \pm \sqrt{1 - z^2}\right)
\end{align}
(the positive branch is chosen)
{ | style="text-align:center;"
| Domain coloring]] of '''inverse trigonometric functions in the [[complex plane]]''' | Arcsine of z in the complex plane.]] | Arccosine of z in the complex plane.]] | Arctangent of z in the complex plane.]] |
| | | |
>+ | Arccosecant of z in the complex plane.]] | Arcsecant of z in the complex plane.]] | Arccotangent of z in the complex plane.]] |
| | | |
Applications
Finding the angle of 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
Often, the hypotenuse is unknown and would need to be calculated before using arcsine or arccosine using the Pythagorean Theorem: where is the length of the hypotenuse. Arctangent comes in handy in this situation, as the length of the hypotenuse is not needed.
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:
= \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
Two-argument variant of arctangent
The two-argument function computes the arctangent of given and , but with a range of . In other words, is the angle between the positive -axis of a plane and the point on it, with positive sign for counter-clockwise angles (upper half-plane, ), and negative sign for clockwise angles (lower half-plane, ). 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 , it can be expressed as follows:
It also equals the principal value of the argument of the complex number .
This limited version of the function above may also be defined using the tangent half-angle formulae as follows:
provided that either or . However this fails if given and so the expression is unsuitable for computational use.
The above argument order (, ) seems to be the most common, and in particular is used in such as the C programming language, but a few authors may use the opposite convention (, ) so some caution is warranted.
Arctangent function with location parameter
In many applications[when a time varying angle crossing should be mapped by a smooth line instead of a saw toothed one (robotics, astronomy, angular movement in general)] the solution of the equation is to come as close as possible to a given value . 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 rounds to the nearest integer.
Numerical accuracy
For angles near 0 and , arccosine is ill-conditioned, and similarly with arcsine for angles near −/2 and /2. Computer applications thus need to consider the stability of inputs to these functions and the sensitivity of their calculations, or use alternate methods.
See also
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Arcsine distribution
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Inverse exsecant
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Inverse versine
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Inverse hyperbolic functions
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List of integrals of inverse trigonometric functions
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List of trigonometric identities
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Trigonometric function
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Trigonometric functions of matrices
Notes
External links