Inverse Trigonometric Functions

Date:    Wed, 7 Aug 1996 14:18:30 -0700
From:    "Glenn M. Picher" <gpicher@maine.com>
Subject: Re: ** MATHS Question **

>How do you do inverse sine, tan and cosine functions using Lingo?

I think this is all correct, but there might be a faster solution. Inverse tan:


  atan()

Already built into Lingo.

Inverse sine:


  on asin ratio
    return atan(ratio/sqrt(1.0 - (ratio * ratio)))
  end

If ratio is less than or equal to -1.0, or greater than or equial to 1.0, then you get a divide by zero error. But that wouldn't be a valid triangle anyway, so an angle measurement would be meaningless.

Inverse cosine:


  on acos ratio
    return atan(sqrt(1.0 - (ratio * ratio))/ratio)
  end

If ratio is 0.0, you get a divide by zero error. But that wouldn't be a valid triangle, either.

The above functions return their answers in radians. There are (2.0 * pi()) radians in a full circle. Thus, one radian is 57.2957795786 degrees. To convert back and forth with degrees:


on degrees theRadians
  return theRadians * 57.2957795786
end

on radians theDegrees
  return theDegrees / 57.2957795786
end

Thus, as you would expect...


  put degrees(asin(sin(radians(30.0))))
  -- 30.0

  put acos(cos(1.0))
  -- 1.0

>Given a right-angled triangle and the lengths of each side, how do I
>find out the size of the other two angles?


on topAngle base, height
  return atan(base/height)
end

on sideAngle base, height
  return atan(height/base)
end

Just be sure you don't pass in 0 length base or height values, or you'll get a divide by zero error.

The answer is returned in radians.

It you need to calculate both angles, it would be faster to calculate one angle and then derive the other angle by subtracting from 90 degrees. Transcendental math functions are rather slow.