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[tex]\theta = atan(\frac{y}{x})[/tex]

[tex]\theta2 = acos(\frac{z}{\sqrt{x^2+y^2+z^2}})[/tex]

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- Thread starter Philosophaie
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[tex]\theta = atan(\frac{y}{x})[/tex]

[tex]\theta2 = acos(\frac{z}{\sqrt{x^2+y^2+z^2}})[/tex]

- #2

arildno

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Make a finite Taylor series expansion, for example.

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arildno

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UltrafastPED

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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.

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arildno

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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".

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".

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- #6

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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.)

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