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arcSin and arcCos without using library functions

Posted on 2009-05-03
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Last Modified: 2012-05-06
Hi guys i want to find arcSin and arcCos without using library functions,

Can you please provide me the algorithm or equation which gives accurate reading,

Please help,

its urgent,

Thanks,
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Question by:montu_12
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4 Comments
 
LVL 84

Accepted Solution

by:
ozo earned 1000 total points
ID: 24292794
float arctan(float x){
    if( x < 0 ){ return -arctan(-x); }
    if( x > 1 ){ return pi/2 - arctan(1/x); }
    if( x <= 1/8.0 ){ return  0.9974133042 * x; }
    if( x <= 2/8.0 ){ return  0.004072621 + 0.964989344 * x; }
    if( x <= 3/8.0 ){ return .017899968 + 0.910336056 * x; }
    if( x <= 4/8.0 ){ return .044740546 + 0.839015512 * x; }
    if( x <= 5/8.0 ){ return .084473784 + 0.759613648 * x; }
    if( x <= 6/8.0 ){ return .134708924 + 0.679214352 * x; }
    if( x <= 7/8.0 ){ return .192103293 + 0.602631128 * x; }
    if( x <= 8/8.0 ){ return .253371504 + 0.532545304 * x; }
}

float arcsin(float x){
  return arctan(x/sqrt(1-x*x));
}
float arccos(float x){
    return pi/2 - arcsin(x);
}
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LVL 27

Expert Comment

by:CaptainCyril
ID: 24293128
Arcsin(x) Calculator
http://www.analyzemath.com/Calculators_2/arcsin_calculator.html

Inverse trigonometric functions
http://en.wikipedia.org/wiki/Inverse_trigonometric_functions
You can write a loop to calculate the definite integrals. Divide the area let's say by 1000 and calculate the areas of the smaller 1000s rectangles.

Calculating ArcSin and ArcCos
http://cuinl.tripod.com/Tips/math9.htm
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LVL 27

Expert Comment

by:CaptainCyril
ID: 24297091
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LVL 22

Assisted Solution

by:NovaDenizen
NovaDenizen earned 1000 total points
ID: 24306215
I like ozo's code, except for the piecewise linear approximation (all the "if (x <=" lines).  I would instead recommend the Maclaurin series, see (17) at http://mathworld.wolfram.com/InverseSine.html.

Alternatively, you will get quicker convergence and better numerical accuracy if you also use the Maclaurin series for arccos(1 - y), for situations when y is close to 1 and convergence will be slow.
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