Manually calculate Arccosine and Arctangent

[Philosophaie]

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}})

 

[arildno]

Make a finite Taylor series expansion, for example.

 

[arildno]

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]

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]

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]

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.)