Inverse trigonometric function

From Wikipedia, the free encyclopedia

In mathematics, the inverse trigonometric functions or cyclometric functions are the inverse functions of the trigonometric functions. The principal inverses are listed in the following table.

Name	Usual notation	Definition	Domain 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 ∞	0 ≤ y < π/2 or π/2 < y ≤ π
arccosecant	y = arccsc(x)	x = csc(y)	−∞ to −1 or 1 to ∞	−π/2 ≤ y < 0 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.

The notations sin⁻¹, cos⁻¹, etc are often used for arcsin, arccos, etc

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.

Principal values of the f(x) = arcsec(x) and f(x) = arccsc(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 General solutions
3 Derivatives of inverse trigonometric functions
4 Expression as definite integrals
5 Infinite series
6 Continued fraction for arctangent
7 Indefinite integrals of inverse trigonometric functions
- 7.1 Example proof
8 Recommended method of calculation
9 Two argument variant of arctangent
10 Logarithmic forms
- 10.1 Example proof
11 Arctangent Addition Formula
- 11.1 Proof
12 Practical usage
13 See also
14 External links

[edit] Relationships among the inverse trigonometric functions

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:

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

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

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

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

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

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

Reciprocal arguments:

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

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

$\arctan \frac{1}{x} = \frac{\pi}{2} - \arctan x =\arccot x, \$ if $\ x > 0$

$\arctan \frac{1}{x} = -\frac{\pi}{2} - \arctan x = -\pi + \arccot x, \$ if $\ x < 0$

$\arccot \frac{1}{x} = \frac{\pi}{2} - \arccot x =\arctan x, \$ if $\ x > 0$

$\arccot \frac{1}{x} = \frac{3\pi}{2} - \arccot x = \pi + \arctan x,\$ if $\ x < 0$

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

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

If you only have a fragment of a sine table:

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

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

Notice that 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).

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] General solutions

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:

sin y = x if and only if y = arcsin x + 2kπ or y = π − arcsin x + 2kπ for some integer k.

cos y = x if and only if y = arccos x + 2kπ or y = 2π − arccos x + 2kπ for some integer k.

tan y = x if and only if y = arctan x + kπ for some integer k.

cot y = x if and only if y = arccot x + kπ for some integer k.

sec y = x if and only if y = arcsec x + 2kπ or y = 2π − arcsec x + 2kπ for some integer k.

csc y = x if and only if y = arccsc x + 2kπ or y = π − arccsc x + 2kπ for some integer k.

[edit] Derivatives of inverse trigonometric functions

Main article: Differentiation of trigonometric functions

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

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

Only for real values of $x$ :

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

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:

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

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

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

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

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

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

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

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

(Notice that the term in the sum for n= 0 is the empty product which is 1.)

Alternatively, this can be expressed:

$\arctan x = \sum_{n=0}^\infty \frac{2^{\,2n}\,(n!)^2}{\left(2n+1\right)!} \; \frac{x^{\,2n+1}}{\left(1+x^2\right)^{n+1}}$

[edit] Continued fraction for arctangent

An 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

For real and complex values of $x$ :

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

For real x≥1:

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

All of these can be derived using integration by parts and the simple derivative forms shown above.

[edit] Example proof

Using $\int u\,\mathrm{d}v = u v - \int v\,\mathrm{d}u$ , set

$\begin{align} u &{}=&\arcsin x &\quad\quad\mathrm{d}v = \mathrm{d}x\\ \mathrm{d}u &{}=&\frac{\mathrm{d}x}{\sqrt{1-x^2}}&\quad\quad{}v = x \end{align}$

Then

$\int \arcsin x\,\mathrm{d}x = x \arcsin x - \int \frac{x}{\sqrt{1-x^2}}\,\mathrm{d}x$

Substitute $k = 1 - x^2\,$ . Then $\mathrm{d}k = -2x\,\mathrm{d}x$ and

$\int \frac{x}{\sqrt{1-x^2}}\,\mathrm{d}x = -\frac{1}{2}\int \frac{\mathrm{d}k}{\sqrt{k}} = -\sqrt{k}$

Back-substitute for x to yield

$\int \arcsin x\,\mathrm{d}x = x \arcsin x + \sqrt{1-x^2}$

[edit] Recommended method of calculation

To calculate arcsine, use:

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

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

The two-argument atan2 function computes the arctangent of $y / x$ given y and x, but with a range of $( - π,π]$ . It was introduced first in many computer programming languages but is now common in all fields of science and engineering too.

It's defined using the standard arctan function (that is with range of (−π/2, π/2)) as follows:

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

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

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

provided that either x > 0 or y ≠ 0. However, in practical implementations it is cheaper and more robust to use the signs of x and y to choose the correct range. Assuming arctan(z) returns a value between −^π⁄₂ and ^π⁄₂ for all real z, we have

$\operatorname{atan2}(y, x) = \begin{cases} -\operatorname{atan2}(-y, x) & \qquad y < 0 \\ \pi - \arctan(-\frac y x) & \qquad y \ge 0 , x < 0 \\ \arctan(\frac y x) & \qquad y \ge 0 , x > 0 \\ \frac{\pi}{2} & \qquad y > 0 , x = 0 \\ \text{undefined} & \qquad y = 0, x = 0 \\ \end{cases}$

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. Also, IEEE floating point implementations must handle exceptional (non-numeric) argument values; FDLIBM (available through netlib) shows how this may be done reliably.

The function atan2 can be implemented in a numerically reliable manner by the CORDIC method. Thus implementations of atan(y) will probably choose to compute actually atan2(y,1).

[edit] Logarithmic forms

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

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

Elementary proofs of these relations proceed via expansion to exponential forms of the trigonometric functions.

[edit] Example proof

$\arcsin x\,=\,\theta$

$\frac{e^{i\,\theta}-e^{-i\,\theta}}{2i}\,=\,x$ (exponential definition of sine)

Let

$k=e^{i\,\theta}.$

Then

$\frac{k-\frac{1}{k}}{2i}\,=\,x$

$k^2-2\,i\,k\,x-1\,=\,0$ (solve for

k

)

$k\,=\,i\,x\pm\sqrt{1-x^2}\,=\,e^{i\,\theta}$ (the positive branch is chosen)

$\theta\,=\,\arcsin\,x\,=\,-i\log\left(i\,x+\sqrt{1-x^2}\right)$ Q.E.D.

**Inverse trigonometric functions in the complex plane**

$arcsin(z)$	$arccos(z)$	$arctan(z)$	$arccot(z)$	$arcsec(z)$	$arccsc(z)$

[edit] Arctangent Addition Formula

$\arctan u + \arctan v = \arctan \left( \frac{u+v}{1-uv} \right)$

[edit] Proof

start from

$\tan (\alpha + \beta) = \frac{\tan \alpha + \tan \beta}{1 - \tan \alpha \tan \beta}\,$

and let

$u\,=\,\tan\,\alpha\,,\,v =\,\tan\,\beta.$

[edit] Practical usage

Inverse trigonometric functions are useful when trying to determine the remaining two angles of a right triangle when you already know the length of the sides of the triangle. Remember the acronym SOHCAHTOA. Using inverse trigonometric functions

$\theta = \arcsin \left( \frac{\text{opposite}}{\text{hypotenuse}} \right)$

Often, the hypotenuse is unknown and would need to be calculated before using arcsin or arccos. Arctan comes in handy in this situation. You can compute the angle of the triangles without knowing the length of the hypotenuse.

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

For example, you can calculate the slope of your roof line if you know the rise and run of the roof. If your roof drops 8 feet as it runs out 20 feet then your roof is angled θ degrees up from horizontal, where θ may be computed as follows.

$\begin{align} \theta &{}= \arctan \left( \frac{\text{opposite}}{\text{adjacent}} \right) &{}= \arctan \left( \frac{\text{rise}}{\text{run}} \right) &{}= \arctan \left( \frac{8}{20} \right) &{}= 21.8^{\circ} \end{align}$

[edit] See also

[edit] External links

Eric W. Weisstein, Inverse Trigonometric Functions at MathWorld.
Eric W. Weisstein, Inverse Tangent at MathWorld.

Categories: Trigonometry | Elementary special functions

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Inverse trigonometric function

From Wikipedia, the free encyclopedia

Contents

[edit] Relationships among the inverse trigonometric functions

[edit] General solutions

[edit] Derivatives of inverse trigonometric functions

[edit] Expression as definite integrals

[edit] Infinite series

[edit] Continued fraction for arctangent

[edit] Indefinite integrals of inverse trigonometric functions

[edit] Example proof

[edit] Recommended method of calculation

[edit] Two argument variant of arctangent

[edit] Logarithmic forms

[edit] Example proof

[edit] Arctangent Addition Formula

[edit] Proof

[edit] Practical usage

[edit] See also

[edit] External links

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