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Arctan of a fraction of two tangents?

  1. Jan 10, 2012 #1



    f(x)/tan(g(x)) ?

    Is this correct or does the tan function at the denominator also vanish?

    Please advise
  2. jcsd
  3. Jan 10, 2012 #2
    No, both are not correct.
    Where is no simple formula of this kind.
  4. Jan 10, 2012 #3


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    Nothing as simple as the OP wants, clearly, but there is an interesting elementary way to "split" the arctangent of a ratio into a sum of arctangents.

    Start with this:

    [itex]\frac{\tan x + \tan y}{1 - {\tan x}{\tan y}} = \tan{(x+y)}[/itex]

    Put [itex]A = \tan{x}, B = \tan{y}[/itex]

    [itex]\frac{A + B}{1 - AB} = \tan{(\arctan{A}+\arctan{B})}[/itex]

    Finally take arctan of both sides:

    [itex]\arctan{(\frac{A + B}{1 - AB})} = \arctan{A}+\arctan{B}[/itex]

    This is a pretty nifty formula that works both ways: to convert the arctan of a quotient to a sum of arctans, and vice versa.

    To see how to do it the first way, let's say we want to convert the expression for [itex]\arctan({\frac{f}{g}})[/itex] into a sum of arctangents.

    We start by setting up the simult. eqn. pair:

    [itex]A + B = f[/itex] ---eqn 1

    [itex]1 - AB = g[/itex] ---eqn 2

    Solving those gives:

    [itex]A = \frac{f \pm \sqrt{f^2 + 4(g-1)}}{2}[/itex]

    and [itex]B = \frac{f \mp \sqrt{f^2 + 4(g-1)}}{2}[/itex]

    giving the result as:

    [itex]\arctan({\frac{f}{g}}) = \arctan{(\frac{f + \sqrt{f^2 + 4(g-1)}}{2})} + \arctan{(\frac{f - \sqrt{f^2 + 4(g-1)}}{2})}[/itex]

    As I said, this is probably not what the OP was looking for (or thinking of), but it's an interesting result I thought bore mentioning.
  5. Jan 10, 2012 #4
    Very nice Curious, I like the trick. +1
  6. Jan 10, 2012 #5
    If "smart" is the good word, I would say : Smart !
  7. Jan 10, 2012 #6


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    Thanks to the above 2 posters for the nice compliments. :smile:

    In fact, this is also a nice way to see this beautiful relationship:

    [tex]\arctan \phi - \arctan \frac{1}{\phi} = \arctan \frac{1}{2}[/tex]

    where [itex]\phi[/itex] is the golden ratio 1.618...

    EDIT: Sorry, made a sign error in my original post. Micromass and others - please take note.
    Last edited: Jan 10, 2012
  8. Jan 10, 2012 #7
    Very nice!! :approve:

    Yes, math can be beautiful indeed...
  9. Jan 10, 2012 #8


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    This is the correct relationship (reposted as I don't want anyone to get it wrong on account of my original typo):

    [tex]\arctan \phi - \arctan \frac{1}{\phi} = \arctan \frac{1}{2}[/tex]
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