# Manually calculate Arccosine and Arctangent

by Philosophaie
Tags: arccosine, arctangent, manually
 P: 365 How do you calculate Arccosine and Arctangent if you do not have a scientific calculator. $$\theta = atan(\frac{y}{x})$$ $$\theta2 = acos(\frac{z}{\sqrt{x^2+y^2+z^2}})$$
 Sci Advisor HW Helper PF Gold P: 12,016 Make a finite Taylor series expansion, for example.
 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.
Thanks
P: 1,364

## 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.
 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".
 P: 10 Though it is nearly 8 weeks late, if you need only about 4 digits of accuracy, you could use $tan^{-1}\approx \frac{x(240+115x^{2})}{240+195x^{2}+17x^{4}}, x\in [-1,1]$ and for x outside that range, use the identity $tan^{-1}(x)=\frac{\pi}{2}-tan^{-1}(\frac{1}{x})$. 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 $tan^{-1}(.1)\approx \frac{.1(240+115*.01)}{240+195*.01+17*.0001}$ $= \frac{.1(241.15)}{241.9517}$ $= \frac{24.115}{241.9517}$ $\approx .0996686529$ Compare to my calculator which returns .0996686525 ;) See this post for the derivation of that formula. For arccos, you can use the identity: $cos^{-1}(x)=2tan^{-1}\left(\sqrt{\frac{1-x}{1+x}}\right), x\in (-1,1]$ 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.)

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