1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Arctan of a fraction of two tangents?

  1. Jan 10, 2012 #1
    if:

    arctan[tan(f(x))/tan(g(x))]

    then,

    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

    Curious3141

    User Avatar
    Homework Helper

    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

    Curious3141

    User Avatar
    Homework Helper

    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

    micromass

    User Avatar
    Staff Emeritus
    Science Advisor
    Education Advisor
    2016 Award

    Very nice!! :approve:

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

    Curious3141

    User Avatar
    Homework Helper

    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]
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Arctan of a fraction of two tangents?
  1. A fraction (Replies: 1)

  2. Fraction of a fraction (Replies: 16)

  3. Arctan identity? (Replies: 2)

Loading...