Inverse trigonometric function

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In mathematics, the inverse trigonometric functions are a set of relationships closely related to the trigonometric functions. The principal inverses are listed in the following table.

Name	Usual notation	Definition	Range of x for real result	Range of usual principal value
arcsine	y = arcsin(x)	x = sin(y)	−1 to +1	−π/2 ≤ y ≤ π/2
arccosine	y = arccos(x)	x = cos(y)	−1 to +1	0 ≤ y ≤ π
arctangent	y = arctan(x)	x = tan(y)	all	−π/2 < y < π/2
arccotangent	y = arccot(x)	x = cot(y)	all	0 < y < π
arcsecant	y = arcsec(x)	x = sec(y)	−∞ to −1 or 1 to ∞	π/2 < y ≤ π or 0 ≤ y < π/2
arccosecant	y = arccsc(x)	x = cosec(y)	−∞ to −1 or 1 to ∞	−π/2 ≤ y < 0 or 0 < y ≤ π/2

The notations sin⁻¹, cos⁻¹, etc are often used for arcsin, arccos, etc, but this notation sometimes causes confusion between (e.g.) arcsin(x) and 1/sin(x).

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

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

In computer programming languages the functions arcsin, arccos, arctan, are usually called asin, acos, atan. Many programming languages also provide the two-argument atan2 function, which computes the arctangent of y/x given y and x, but with a range of [−π, π].

1 Relationships among the inverse trigonometric functions
2 Derivatives of inverse trigonometric functions
3 Expression as definite integrals
4 Infinite series
5 Continued fraction for arctangent
6 Indefinite integrals of inverse trigonometric functions
7 Recommended method of calculation
8 Two argument variant of arctangent
9 Logarithmic forms
- 9.1 Example proof
10 See also
11 External links

[edit] Relationships among the inverse trigonometric functions

Reciprocal arguments:

$\arccot x = \arctan \frac{1}{x},\$ if $\ x > 0$

$\arccot 0 = \frac{\pi}{2}$

$\arccot x = \pi + \arctan \frac{1}{x},\$ if $\ x < 0$

$\arcsec x = \arccos \frac{1}{x}$

$\arccsc x = \arcsin \frac{1}{x}$

Complementary angles:

$\arccos x = \frac{\pi}{2} - \arcsin x$

$\arccot x = \frac{\pi}{2} - \arctan x$

$\arccsc x = \frac{\pi}{2} - \arcsec x$

Negative arguments:

$\arcsin (-x) = - \arcsin x \!$

$\arccos (-x) = \pi - \arccos x \!$

$\arctan (-x) = - \arctan x \!$

$\arccot (-x) = \pi - \arccot x \!$

$\arcsec (-x) = \pi - \arcsec x \!$

$\arccsc (-x) = - \arccsc x \!$

If you only have a fragment of a sine table:

$\arccos x = \arcsin \sqrt{1-x^2}, \$ if $\ 0 \leq x \leq 1$

$\arctan x = \arcsin \frac{x}{\sqrt{x^2+1}}$

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

$\arcsin x = 2 \arctan \frac{x}{1+\sqrt{1-x^2}}$

$\arccos x = 2 \arctan \frac{\sqrt{1-x^2}}{1+x}, \$ if $\ -1 < x \leq +1$

$\arctan x = 2 \arctan \frac{x}{1+\sqrt{1+x^2}}$

[edit] Derivatives of inverse trigonometric functions

Simple derivatives for real values of $x$ are as follows:

$\begin{align} \frac{d}{dx} \arcsin x & {}= \frac{1}{\sqrt{1-x^2}}; \qquad |x| < 1\\ \frac{d}{dx} \arccos x & {}= \frac{-1}{\sqrt{1-x^2}}; \qquad |x| < 1\\ \frac{d}{dx} \arctan x & {}= \frac{1}{1+x^2}\\ \frac{d}{dx} \arccot x & {}= \frac{-1}{1+x^2}\\ \frac{d}{dx} \arcsec x & {}= \frac{1}{|x|\,\sqrt{x^2-1}}; \qquad |x| > 1\\ \frac{d}{dx} \arccsc x & {}= \frac{-1}{|x|\,\sqrt{x^2-1}}; \qquad |x| > 1 \end{align}$

For an example derivation, letting $\theta = \arcsin x \!$ , we get:

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

[edit] 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:

$\begin{align} \arcsin x &{}= \int_0^x \frac {1} {\sqrt{1 - z^2}}\,dz,\qquad |x| \leq 1\\ \arccos x &{}= \int_x^1 \frac {1} {\sqrt{1 - z^2}}\,dz,\qquad |x| \leq 1\\ \arctan x &{}= \int_0^x \frac 1 {z^2 + 1}\,dz,\\ \arccot x &{}= \int_x^\infty \frac {1} {z^2 + 1}\,dz,\\ \arcsec x &{}= \int_1^x \frac 1 {z \sqrt{z^2 - 1}}\,dz, \qquad x \geq 1\\ \arccsc x &{}= \int_x^\infty \frac {1} {z \sqrt{z^2 - 1}}\,dz, \qquad x \geq 1 \end{align}$

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

[edit] Infinite series

Like the sine and cosine functions, the inverse trigonometric functions can be calculated using infinite series, as follows:

$\begin{align} \arcsin z & {}= z + \left( \frac {1} {2} \right) \frac {z^3} {3} + \left( \frac {1 \cdot 3} {2 \cdot 4} \right) \frac {z^5} {5} + \left( \frac{1 \cdot 3 \cdot 5} {2 \cdot 4 \cdot 6 } \right) \frac{z^7} {7} + \cdots\\ & {}= \sum_{n=0}^\infty \left( \frac {(2n)!} {2^{2n}(n!)^2} \right) \frac {z^{2n+1}} {(2n+1)} ; \qquad | z | \le 1 \end{align}$

$\begin{align} \arccos z & {}= \frac {\pi} {2} - \arcsin z \\ & {}= \frac {\pi} {2} - (z + \left( \frac {1} {2} \right) \frac {z^3} {3} + \left( \frac {1 \cdot 3} {2 \cdot 4} \right) \frac {z^5} {5} + \left( \frac{1 \cdot 3 \cdot 5} {2 \cdot 4 \cdot 6 } \right) \frac{z^7} {7} + \cdots ) \\ & {}= \frac {\pi} {2} - \sum_{n=0}^\infty \left( \frac {(2n)!} {2^{2n}(n!)^2} \right) \frac {z^{2n+1}} {(2n+1)} ; \qquad | z | \le 1 \end{align}$

$\begin{align} \arctan z & {}= z - \frac {z^3} {3} +\frac {z^5} {5} -\frac {z^7} {7} +\cdots \\ & {}= \sum_{n=0}^\infty \frac {(-1)^n z^{2n+1}} {2n+1} ; \qquad | z | \le 1 \end{align}$

$\begin{align} \arccot z & {}= \frac {\pi} {2} - \arctan z \\ & {}= \frac {\pi} {2} - ( z - \frac {z^3} {3} +\frac {z^5} {5} -\frac {z^7} {7} +\cdots ) \\ & {}= \frac {\pi} {2} - \sum_{n=0}^\infty \frac {(-1)^n z^{2n+1}} {2n+1} ; \qquad | z | \le 1 \end{align}$

$\begin{align} \arcsec z & {}= \arccos\left(z^{-1}\right) \\ & {}= \frac {\pi} {2} - (z^{-1} + \left( \frac {1} {2} \right) \frac {z^{-3}} {3} + \left( \frac {1 \cdot 3} {2 \cdot 4} \right) \frac {z^{-5}} {5} + \left( \frac{1 \cdot 3 \cdot 5} {2 \cdot 4 \cdot 6 } \right) \frac{z^{-7}} {7} + \cdots ) \\ & {}= \frac {\pi} {2} - \sum_{n=0}^\infty \left( \frac {(2n)!} {2^{2n}(n!)^2} \right) \frac {z^{-(2n+1)}} {(2n+1)} ; \qquad \left| z \right| \ge 1 \end{align}$

$\begin{align} \arccsc z & {}= \arcsin\left(z^{-1}\right) \\ & {}= z^{-1} + \left( \frac {1} {2} \right) \frac {z^{-3}} {3} + \left( \frac {1 \cdot 3} {2 \cdot 4 } \right) \frac {z^{-5}} {5} + \left( \frac {1 \cdot 3 \cdot 5} {2 \cdot 4 \cdot 6} \right) \frac {z^{-7}} {7} +\cdots \\ & {}= \sum_{n=0}^\infty \left( \frac {(2n)!} {2^{2n}(n!)^2} \right) \frac {z^{-(2n+1)}} {2n+1} ; \qquad \left| z \right| \ge 1 \end{align}$

Leonhard Euler found a more efficient series for the arctangent, which is:

$\arctan x = \frac{x}{1+x^2} \sum_{n=0}^\infty \prod_{k=1}^n \frac{2k x^2}{(2k+1)(1+x^2)}.$

[edit] Continued fraction for arctangent

A faster alternative to the power series for arctangent is its generalized continued fraction:

$\arctan(z)=\cfrac{z}{1 + \cfrac{z^2}{3 + \cfrac{4 z^2}{5 + \cfrac{9 z^2}{7 + \cfrac{16 z^2}{9 + \cfrac{25 z^2}{\ddots\,}}}}}}\,$

This 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. It was developed by Carl Friedrich Gauss, utilizing the hypergeometric series.

[edit] Indefinite integrals of inverse trigonometric functions

$\begin{align} \int \arcsin x\,dx &{}= x\,\arcsin x + \sqrt{1-x^2} + C\\ \int \arccos x\,dx &{}= x\,\arccos x - \sqrt{1-x^2} + C\\ \int \arctan x\,dx &{}= x\,\arctan x - \frac{1}{2}\ln\left(1+x^2\right) + C\\ \int \arccot x\,dx &{}= x\,\arccot x + \frac{1}{2}\ln\left(1+x^2\right) + C\\ \int \arcsec x\,dx &{}= x\,\arcsec x - \ln\left(x+\sqrt{x^2-1}\right) + C\\ \int \arccsc x\,dx &{}= x\,\arccsc x + \ln\left(x+\sqrt{x^2-1}\right) + C \end{align}$

These are all easily derived using integration by parts and the simple derivative forms shown above.

[edit] Recommended method of calculation

To calculate arcsine, use:

$\arcsin x = 2 \arctan \frac{x}{1+\sqrt{1-x^2}}.$

To calculate arccosine, use:

$\arccos x = \frac{\pi}{2} - \arcsin x.$

To calculate arctangent for x near zero, use the continued fraction above. To calculate arctangent for other values of x, use:

$\arctan x = 2 \arctan \frac{x}{1+\sqrt{1+x^2}}.$

To calculate arccotangent, use:

$\arccot x = \frac{\pi}{2} - \arctan x.$

To calculate arcsecant, use:

$\arcsec x = \frac{\pi}{2} - \arcsin \frac{1}{x}.$

To calculate arccosecant, use:

$\arccsc x = \arcsin \frac{1}{x}.$

[edit] Two argument variant of arctangent

Many computer programming languages also provide the two-argument atan2 function, which computes the arctangent of y/x given y and x, but with a range of [−π, π]. This function may be computed using the half-angle formula as follows:

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

provided that either x>0 or y≠0.

[edit] Logarithmic forms

These functions may also be expressed using natural logarithms. This extends in a natural fashion their domain to the whole of the complex plane.

$\begin{align} \arcsin x &{}= -i\,\ln\left(i\,x+\sqrt{1-x^2}\right) &{}= \arccsc \frac{1}{x}\\ \arccos x &{}= -i\,\ln\left(x+\sqrt{x^2-1}\right) = \frac{\pi}{2}\,+i\ln\left(i\,x+\sqrt{1-x^2}\right) = \frac{\pi}{2}-\arcsin x &{}= \arcsec \frac{1}{x}\\ \arctan x &{}= \frac{i}{2}\left(\ln\left(1-i\,x\right)-\ln\left(1+i\,x\right)\right) &{}= \arccot \frac{1}{x}\\ \arccot x &{}= \frac{i}{2}\left(\ln\left(1-\frac{i}{x}\right)-\ln\left(1+\frac{i}{x}\right)\right) &{}= \arctan \frac{1}{x}\\ \arcsec x &{}= -i\,\ln\left(\sqrt{\frac{1}{x^2}-1}+\frac{1}{x}\right) = i\,\ln\left(\sqrt{1-\frac{1}{x^2}}+\frac{i}{x}\right)+\frac{\pi}{2} = \frac{\pi}{2}-\arccsc x &{}= \arccos \frac{1}{x}\\ \arccsc x &{}= -i\,\ln\left(\sqrt{1-\frac{1}{x^2}}+\frac{i}{x}\right) &{}= \arcsin \frac{1}{x} \end{align}$