Manually calculate Arccosine and Arctangent

  1. 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]
     
  2. jcsd
  3. arildno

    arildno 12,015
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    Make a finite Taylor series expansion, for example.
     
  4. arildno

    arildno 12,015
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    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.
     
  5. UltrafastPED

    UltrafastPED 1,919
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    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.
     
  6. arildno

    arildno 12,015
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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".
     
    Last edited: Oct 1, 2013
  7. 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.)
     
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