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Thread: ArcTan(x,y) formula

  1. #1
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    Post ArcTan(x,y) formula

    Hi,

    I need an idiot's guide on how to calculate ArcTan(x,y), can somebody please help?

    I have a working example (below) but have no idea how it works:

    α = arctan(sin(12.0321°) * cos(23.45°), cos(12.0321°)) = 11.0639°

    Incidentally, in order to calculate sin and cos (in VB Code) I have to convert the degrees to radians first - will this mess up the output or can I just convert it back into degrees again?

    Thanks for looking,

    Chris.
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  2. #2
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    Opalg's Avatar
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    Quote Originally Posted by Chris147 View Post
    I need an idiot's guide on how to calculate ArcTan(x,y), can somebody please help?

    I have a working example (below) but have no idea how it works:

    α = ArcTan(sin(12.0321°) * cos(23.45°), cos(12.0321°)) = 11.0639°

    Incidentally, in order to calculate sin and cos (in VB Code) I have to convert the degrees to radians first - will this mess up the output or can I just convert it back into degrees again?
    This is a weird definition of ArcTan, certainly not a standard one. Normally, arctan is a function of one variable, not two, and arctan(x) basically means the angle whose tangent is x. I'll refer to this function of one variable as arctan (with lower case a and t) to distinguish it from that function ArcTan of two variables.

    If you work out sin(12.0321°) * cos(23.45°), it comes to 0.1912425; and cos(12.0321°) is 0.9780309. So we are told that ArcTan(0.1912425,0.9780309) = 11.0639°. What I found by trial and error is that if I divide 0.1912425 by 0.9780309 it comes to 0.1955383. Then when I press the \tan^{-1} button on my calculator (remember that \tan^{-1} is just another name for arctan), it gives the answer as 11.0639°, which is what we want.

    So it seems to me as though your ArcTan is related to the conventional arctan by the relation ArcTan(x,y) = arctan(x/y).
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  3. #3
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    ArcTan(x/y) not (x,y)

    Hi Opalg

    First of all, thank you.

    I don't think I would have figured that out in a month of Sundays.

    All I need to do now is create a ArcTan function (of the standard ArcTan(x) variety) - unfortunately Access VBA doesn't support or list it in the derived Math Functions). This shouldn't prove too difficult though - I've already done the same ArcSin and ArcCos.

    If anyone has an outline of how ArcTan(x) is calculated I'd appreciate it, meanwhile - thanks again, your help is greatly appreciated.

    Chris.
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  4. #4
    Super Member ebaines's Avatar
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    If you already have a function that determines arcsin(x), then you can determine arctan(x) as follows:

    arctan(x) = arcsin (x/sqrt(x^2+1))

    For example: arctan(3/4) = arcsin[(3/4)/(5/4)] = arcsin(3/5)

    Another approach would be to use the sequence:

    arctan(x) = x - (x^3)/3 + (x^5)/5 - (x^7)/7 + ...

    which converges for x<=1. If x>1, just remember that arctan(x) = pi/2-arctan(1/x).
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  5. #5
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    ArcTan Formula

    Thanks ebaines,

    that's pretty much fixed the whole thing :-).

    Thanks also to the Math Help Forum for a great service.

    Chris.
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