Arc Tangent

Information, Videos , News and Images about Arc Tangent

SAIT-GeoCon19 - arc tangent between 2 circles
Create an arc that is tangent between two existing circles.
ENGD206 Geometric Constructions arc tangent between circles
SAIT-GeoCon16 - arc tangent to line and circle
Create an arc of a given radius that is tangent to an existing circle and tangent to an existing line.
ENGD206 Geometric constructions arc tangent to line and circle Line geometry
SAIT-GeoCon20 - arc tangent to line passing point
Create an arc that is tangent to a line and passes through a point.
ENGD206 Geometric constructions arc tangent to line passing point
Tangent and Arcs Problem Solving
Learn about solving for tangent and arcs with these helpful problems. Learn more about online education at www.studyatapu.com
APU American Public University Math Geometry Tangent and Arcs Problem Solving
Lesson 7 ArcTangent
This lesson shows how the Arctangent trigonometric function is used to convert x, y coordinates into an angle. MST Workshop is an Interactive Graphic Programing Language. Download it from home.comcast.net
MST Workshop Interactive Math Science Technology arctangent
Proof: Derivative of Arctangent
Prerequisites: Know your trigonometry! Derivatives. I describe a simple way to derive the derivative of arctan(x) or any other trigonometric arc-function. Kinda nifty if you've forgotten and have no way to look it up.
calculus derivative arctan arcsin tangent arctangent BC AB AP calc II math mathematics proof
€199,- Powerful drafting 2d CAD software
www.tensorcad.com: Easy in use CAD program to make professional mechanical drawings. Import and export of DXF and Iges. Perfect for technical documentation.
free trial download dxf iges easy in use software pc import export CNC90 Alberi
Trigonometry: Values of Arcsine, Arccosine, Arctangent
Watch more at www.educator.com Other subjects include Algebra 1/2, Basic Math, Pre-Calculus, Geometry, Pre-Algebra, Calculus, Statistics, Biology, Chemistry, Physics, Organic Chemistry, and Computer Science. -All lectures are broken down by individual topics -No more wasted time -Just search and jump directly to the answer
trigonometry trig inverse trigonometric functions arcsine arccosine arctangent math example
Tangent-Secant and Tangent-Chord Angles
Powered by www.tenmarks.com . Learn about Tangent-Secant and Tangent-Chord Angles.
Tenmarks Education Math Circle Angle Measure Arc tangent Secant Chord Theorem Video What How
Google SketchUp Toolbar Series: Arc
Shows how to use the arc tool in Google SketchUp with tips and best practices. Video Outline: - How to draw arcs - Drawing tangent arcs - Drawing continuous smooth arcs - Defining number of arc segments
Google sketchup arc tool 3d sofware training tutorial
Andrew Gorny - Arctangent
If you HATE this video, you don't have to SUBSCRIBE. But YT supports users who get subscribers, so if you liked this, please help me by subscribing! Tuning is CADEBE Inspired by Richard Wagner's "Forest Murmurs" From Andrew Gorny - Identity Coming in 2010! Guitar: 1984 Martin D-28 Friend me on these sites I love talking to people!!! www.myspace.com/andrewgorny www.facebook.com/agorny www.twitter.com/andrewgorny
andrew gorny michael hedges andy mckee don ross erik mongrain antoine dufour fingerstyle guitar tapping harmonics acoustic gorney
10.6 More Angle-Arc Theorems (Practice)
A lesson on 1.) the relationship of inscribed and tangent-chord angles with the same or congruent arcs 2.) the measure of an inscribed angle whose intercepted arc is a semicircle and 3.) the relationship of tangent-tangent angles and their minor arcs
geometry math mathematics homestead high school tangent tangent angles minor arc major arc tangent chord angle inscribed angle intercepted arc semicircle high school math
Trigonometry
History
Usage
Functions
Generalized
Inverse functions
Further reading
Reference
Identities
Exact constants
Trigonometric tables
Laws and theorems
Law of sines
Law of cosines
Law of tangents
Law of cotangents
Pythagorean theorem
Calculus
Trigonometric substitution
Integrals of functions
Derivatives of functions
Integrals of inverse functions

In mathematics, the inverse trigonometric functions (occasionally called cyclometric functions[1]) are the inverse functions of the trigonometric functions with suitably restricted domains .

The notations sin−1, cos−1, etc. are often used for arcsin, arccos, etc., but this convention logically conflicts with the common semantics for expressions like sin2(x), which refer to numeric power rather than function composition, and therefore may result in confusion between multiplicative inverse and compositional inverse.

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].

Contents

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 ranges of the inverse functions are subsets of the domains of the original functions

For example, just as the square root function y = \sqrt{x} is defined such that y2 = x, the function y = arcsin(x) is defined so that sin(y) = x. There are multiple numbers y such that sin(y) = x; for example, sin(0) = 0, but also sin(π) = 0, sin(2π) = 0, etc. It follows that the arcsine function is multivalued: arcsin(0) = 0, but also arcsin(0) = π, arcsin(0) = 2π, 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.

Name Usual notation Definition Domain of x for real result Range of usual principal value
(radians)		 Range of usual principal value
(degrees)
arcsine y = arcsin x x = sin y −1 ≤ x ≤ 1 −π/2 ≤ y ≤ π/2 −90° ≤ y ≤ 90°
arccosine y = arccos x x = cos y −1 ≤ x ≤ 1 0 ≤ y ≤ π 0° ≤ y ≤ 180°
arctangent y = arctan x x = tan y all real numbers −π/2 < y < π/2 −90° < y < 90°
arccotangent y = arccot x x = cot y all real numbers 0 < y < π 0° < y < 180°
arcsecant y = arcsec x x = sec y x ≤ −1 or 1 ≤ x 0 ≤ y < π/2 or π/2 < y ≤ π 0° ≤ y < 90° or 90° < y ≤ 180°
arccosecant y = arccsc x x = csc y x ≤ −1 or 1 ≤ x −π/2 ≤ y < 0 or 0 < y ≤ π/2 -90° ≤ y < 0° or 0° < y ≤ 90°

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

Relationships among the inverse trigonometric functions

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:

\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 (1/x) \,= \arcsec x \,
\arcsin (1/x) \,= \arccsc x \,
\arctan (1/x) = \tfrac{1}{2}\pi - \arctan x =\arccot x,\text{ if }x > 0 \,
\arctan (1/x) = -\tfrac{1}{2}\pi - \arctan x = -\pi + \arccot x,\text{ if }x < 0 \,
\arccot (1/x) = \tfrac{1}{2}\pi - \arccot x =\arctan x,\text{ if }x > 0 \,
\arccot (1/x) = \tfrac{3}{2}\pi - \arccot x = \pi + \arctan x,\text{ if }x < 0 \,
\arcsec (1/x) = \arccos x \,
\arccsc (1/x) = \arcsin x \,

If you only have a fragment of a sine table:

\arccos x = \arcsin \sqrt{1-x^2},\text{ if }0 \leq x \leq 1
\arctan x = \arcsin \frac{x}{\sqrt{x^2+1}}

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

Relationships between trigonometric functions and inverse trigonometric functions

\sin (\arccos x) = \cos(\arcsin x) = \sqrt{1-x^2}
\sin (\arctan x) = \frac{x}{\sqrt{1+x^2}}
\cos (\arctan x) = \frac{1}{\sqrt{1+x^2}}
\tan (\arcsin x) = \frac{x}{\sqrt{1-x^2}}
\tan (\arccos x) = \frac{\sqrt{1-x^2}}{x}

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 where k is some integer:

\sin(y) = x \ \Leftrightarrow\ y = \arcsin(x) + 2k\pi \text{ or } y = \pi - \arcsin(x) + 2k\pi
\cos(y) = x \ \Leftrightarrow\ y = \arccos(x) + 2k\pi \text{ or } y = 2\pi - \arccos(x) + 2k\pi
\tan(y) = x \ \Leftrightarrow\ y = \arctan(x) + k\pi
\cot(y) = x \ \Leftrightarrow\ y = \arccot(x) + k\pi
\sec(y) = x \ \Leftrightarrow\ y = \arcsec(x) + 2k\pi \text{ or } y = 2\pi - \arcsec (x) + 2k\pi
\csc(y) = x \ \Leftrightarrow\ y = \arccsc(x) + 2k\pi \text{ or } y = \pi - \arccsc(x) + 2k\pi

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\,\sqrt{x^2-1}}\\ \frac{d}{dx} \arccsc x & {}= \frac{-1}{x\,\sqrt{x^2-1}} \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 a sample derivation: if \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}}

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\\ \arcsec x &{}= \pi + \int_x^{-1} \frac 1 {z \sqrt{z^2 - 1}}\,dz, \qquad x \leq -1\\ \arccsc x &{}= \int_x^\infty \frac {1} {z \sqrt{z^2 - 1}}\,dz, \qquad x \geq 1\\ \arccsc x &{}= \int_{-\infty}^x \frac {1} {z \sqrt{z^2 - 1}}\,dz, \qquad x \leq -1 \end{align}

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

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 {(1/z)} \\ & {}= \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 {(1/z)} \\ & {}= 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 z = \frac{z}{1+z^2} \sum_{n=0}^\infty \prod_{k=1}^n \frac{2k z^2}{(2k+1)(1+z^2)}.

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


Alternatively, this can be expressed:

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

Continued fractions for arctangent

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

\arctan z=\cfrac{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}}}}} =\cfrac{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.

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\left(1+\sqrt{{x^2-1}\over x^2}\right)\right) + C\\ \int \arccsc x\,dx &{}= x\,\arccsc x + \ln\left(x\left(1+\sqrt{{x^2-1}\over x^2}\right)\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.

Example

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}+C

Two-argument variant of arctangent

Main article: atan2

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:

\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}

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:

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

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.

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\,\ln\left(i\,x+\sqrt{1-x^2}\right) &{}= \arccsc \frac{1}{x}\\[10pt] \arccos x &{}= -i\,\ln\left(x+i\,\sqrt{1-x^2}\right) = \frac{\pi}{2}\,+i\ln\left(i\,x+\sqrt{1-x^2}\right) = \frac{\pi}{2}-\arcsin x &{}= \arcsec \frac{1}{x}\\[10pt] \arctan x &{}= \tfrac{1}{2}i\left(\ln\left(1-i\,x\right)-\ln\left(1+i\,x\right)\right) &{}= \arccot \frac{1}{x}\\[10pt] \arccot x &{}= \tfrac{1}{2}i\left(\ln\left(1-\frac{i}{x}\right)-\ln\left(1+\frac{i}{x}\right)\right) &{}= \arctan \frac{1}{x}\\[10pt] \arcsec x &{}= -i\,\ln\left(i\,\sqrt{1-\frac{1}{x^2}}+\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}\\[10pt] \arccsc x &{}= -i\,\ln\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.

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
k = ix \pm \sqrt{1-x^2} = e^{i\theta} \,

(the positive branch is chosen)

\theta = \arcsin x = -i \ln \left(ix + \sqrt{1-x^2}\right) \,
Inverse trigonometric functions in the complex plane
Complex arcsin.jpg
Complex arccos.jpg
Complex arctan.jpg
Complex ArcCot.jpg
Complex ArcSec.jpg
Complex ArcCsc.jpg
\arcsin(z) \arccos(z) \arctan(z) \arccot(z) \arcsec(z) \arccsc(z)

Arctangent addition formula

\arctan u + \arctan v = \arctan \left( \frac{u+v}{1-uv} \right) \pmod \pi, \qquad u v \ne 1 \,.

This is derived from the tangent addition formula

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

by letting

\alpha = \arctan u \,, \quad \beta = \arctan v \,.

Application: finding the angle of a right triangle

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, for example, it follows that

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

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

\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:

\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}.

Practical considerations

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). Similarly, arcsine is inaccurate for angles near −π/2 and π/2. To achieve full accuracy for all angles, arctangent or atan2 should be used for the implementation.

See also

References

  1. ^ For example Dörrie, Heinrich (1965). Triumph der Mathematik. Trans. David Antin. Dover. p. 69. ISBN 0-486-61348-8. 

External links

Twitter
News
Documents
Weblinks