Register to reply

Manually calculate Arccosine and Arctangent

by Philosophaie
Tags: arccosine, arctangent, manually
Share this thread:
Philosophaie
#1
Oct1-13, 11:35 AM
P: 368
How do you calculate Arccosine and Arctangent if you do not have a scientific calculator.
[tex]\theta = atan(\frac{y}{x})[/tex]
[tex]\theta2 = acos(\frac{z}{\sqrt{x^2+y^2+z^2}})[/tex]
Phys.Org News Partner Mathematics news on Phys.org
Heat distributions help researchers to understand curved space
Professor quantifies how 'one thing leads to another'
Team announces construction of a formal computer-verified proof of the Kepler conjecture
arildno
#2
Oct1-13, 11:50 AM
Sci Advisor
HW Helper
PF Gold
P: 12,016
Make a finite Taylor series expansion, for example.
arildno
#3
Oct1-13, 11:52 AM
Sci Advisor
HW Helper
PF Gold
P: 12,016
Or, if you recognize the value inside the acos, for example, as the cosine value to some specific angle, then that specific angle is your desired answer.

UltrafastPED
#4
Oct1-13, 11:53 AM
Sci Advisor
Thanks
PF Gold
UltrafastPED's Avatar
P: 1,908
Manually calculate Arccosine and Arctangent

The series expansions (Taylor series) can be used though most computational methods first transform this into a Chebyshev expansion. You will study these if you take a course in numerical analysis.

For an introduction see: http://en.wikipedia.org/wiki/Chebyshev_polynomials It takes a while to get to the application of the polynomials to approximation theory.
arildno
#5
Oct1-13, 11:58 AM
Sci Advisor
HW Helper
PF Gold
P: 12,016
Essentially, what you want is a finite algorithm of a) minimal length and retaining b) optimal accuracy for your answer.
These two competing issues generate in general, a tricky balancing act, although much research has uncovered a lot of techniques that justifiably have become "favoured".
Zeda
#6
Nov23-13, 08:19 PM
P: 10
Though it is nearly 8 weeks late, if you need only about 4 digits of accuracy, you could use [itex]tan^{-1}\approx \frac{x(240+115x^{2})}{240+195x^{2}+17x^{4}}, x\in [-1,1][/itex] and for x outside that range, use the identity [itex]tan^{-1}(x)=\frac{\pi}{2}-tan^{-1}(\frac{1}{x})[/itex]. It might be a pain by hand, but it is doable.

Using Chebyshev polynomials or Taylor Series will take you a looong time :P

So, as an example, take x=.1
[itex]tan^{-1}(.1)\approx \frac{.1(240+115*.01)}{240+195*.01+17*.0001}[/itex]
[itex]= \frac{.1(241.15)}{241.9517}[/itex]
[itex]= \frac{24.115}{241.9517}[/itex]
[itex]\approx .0996686529[/itex]

Compare to my calculator which returns .0996686525 ;)

See this post for the derivation of that formula.

For arccos, you can use the identity:
[itex]cos^{-1}(x)=2tan^{-1}\left(\sqrt{\frac{1-x}{1+x}}\right), x\in (-1,1][/itex]

To compute square roots by hand, there are a number of methods, but the method I am most familiar with is in binary, so I don't know how useful that will be :/ (Note that the square root will be squared in two places plugging it into the arctangent approximation, so you don't technically need to compute the square root, there.)


Register to reply

Related Discussions
Calculate cross correlation manually Electrical Engineering 1
Convergence of the Arctangent Integral Calculus & Beyond Homework 11
How to calculate matrixA^100...manually Set Theory, Logic, Probability, Statistics 5
Disambiguating arccosine/arcsine functions Precalculus Mathematics Homework 7
Arctangent Series Calculus & Beyond Homework 1