Finding the Value of a Triginometric Function Without a Calculator

DCircuit
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How can you find, say, sin(41.3°) without using a calculator? Or maybe a better question is: How does a calculator find that value when you punch it in? Also, what about arcsin, arccos, etc... How does a calculator find those values?
 
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They use Taylor series (http://en.wikipedia.org/wiki/Taylor_series) expansions to get an answer that is correct up to a limited number of digits. That way only repeated additions and multiplications are needed.
 
They for sure don't use Taylor because that is much too slow in convergence.
For trigonometric functions in particular some simple systems might use
http://en.wikipedia.org/wiki/CORDIC
In general there are iterative algorithms for each elementary function which use only the basic operation (multiplication,...). I once heard a lecture but can't find the notes anymore :(
Maybe you find some information on
http://www.math.niu.edu/~rusin/known-math/index/65-XX.html
http://www.math.niu.edu/~rusin/known-math/98/special.func.comp
 
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Actually, Euler's Formula gives the exact values of the Trig functions and is very easy to use.

e^ix = cos(x) + isin(x)
 
:-D

How exactly would you calculate exp(i*41.3°) on paper? (with multiplication and addition only) ;-)
 
And furthermore, I believe that this formula applies only to real numbers (i.e., radian measure), not angles in degree measure, so you would need to convert 41.3° to radians.
 
41.3° = 41.3pi/180 That is easy.

<br /> sinx\ =\ \frac{1}{2i}\left(e^{ix}\ -\ e^{-ix}\right)<br />
 
So go ahead and do the complex exponential :D
How would you think calculators do it?
(Hint: They use the trig functions!)
 
They can't use the trig functions to get the trig functions. I guess the only way to evaluate complex exponents is with the Taylor's power series, which will only give you a decimal approximation depending on how far you want to go.
 
  • #10
Good point. And the conclusion is?
(Hint: Calculators don't use the complex exponential to find trig functions)

If you go back to Taylor series, then you start over at Post #2. And the answer is post #3 which says Taylor series converge too slowly to be useful.
 
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