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Manually calculate Arccosine and Arctangent 
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#1
Oct113, 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] 


#2
Oct113, 11:50 AM

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PF Gold
P: 12,016

Make a finite Taylor series expansion, for example.



#3
Oct113, 11:52 AM

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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.



#4
Oct113, 11:53 AM

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Thanks
PF Gold
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. 


#5
Oct113, 11:58 AM

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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". 


#6
Nov2313, 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{1x}{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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