Arc Tangent in Small Basic

Question

I have an equation that is relatively simply using the ArcTan function (angle between two points) and nearly impossible without it. Without accessing an external library, can someone give me a SmallBasic equivalent for ArcTan?

Answer 1

Using radians

For |x| < 1 : arctan(x) ~ x/(1+0.28*x^2)

For |x| > 1 : arctan(x) ~ pi/2 - x/(x^2 + 0.28)

Answer 2

Using a convergent series, this gives a "close-enough" answer

  Sub ArcTan
    If x > 1 then
      x = 1 / x
      ArcTanInternal()
      y = Math.Pi / 2 - y
    Else
      ArcTanInternal()
    EndIf
  EndSub

  Sub ArcTanInternal
    y = x - Math.Power(x, 3) / 3 + Math.Power(x, 5) / 5 - Math.Power(x, 7) / 7
  EndSub

Sample program published as: QHL110

Answer 3

The resulting angle is in Radians, you can use Math.GetDegrees to convert the angle to degrees.

Answer 4

Using a convergent series, this gives a "close-enough" answer

I don't find that a "close-enough" answer :)

It's only a rough approximation... This is the best thing we can do for the moment but it's not really a pretty way to do the thing.

---

Fremy - Developer in VB.NET, C# and JScript ... - Feel free to try my extension

Answer 5

Thank you.

And add the Arc-Trig functions to the wish list.

Answer 6

That convergent series can be run out there a long ways to get an even closer "close-enough" value.

Answer 7

Here is a simple way if getting that arctan value.  I just ran that Taylor series out to 99.  It is the same as my TI-89 to at least 6 decimal places.  We got men on the moon with less accuracy.

x = 1.5
ArcTan()
TextWindow.WriteLine(y)

x = 0.6 
ArcTan()
TextWindow.WriteLine(y)

Sub ArcTan
  If x > 1 then
    x = 1 / x
    ArcTanInternal()
    y = Math.Pi / 2 - y
  Else
    ArcTanInternal()
  EndIf
EndSub

Sub ArcTanInternal
  'y = x - Math.Power(x, 3) / 3 + Math.Power(x, 5) / 5 - Math.Power(x, 7) / 7
  sign = -1
  y = 0
  For p = 1 To 99 Step 2
    sign = - sign
    y = y + sign * Math.Power(x, p)/p
  EndFor
EndSub