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Atan(a/b) + atan(b/a) = pi/2

  1. May 19, 2008 #1
    Hey,

    I came across this 'identity' today and was wondering if there was any algebraical explanation to this...

    Basically I had to show:
    [tex]\arctan( \frac{a}{b}) + \arctan( \frac{b}{a}) = \frac{\pi}{2}[/tex]
    (And if necessary, (a,b) > 0 )

    It is pretty easy to show when you draw a right triangle:
    Code (Text):
         /|
        /c|
       /  |
      /   |a
     /d   |
    /_____|
       b
    (a and b are the sides while c and d are the angles)

    Now, [tex]\tan d = \frac{a}{b}[/tex] and [tex]\tan c = \frac{b}{a}[/tex] and because it is a right triangle, [tex]c + d = \frac{\pi}{2} = \arctan( \frac{a}{b}) + \arctan( \frac{b}{a})[/tex].


    But I was wondering if you can also proof this algebraically?
    I typed it into Maple and got [tex]\frac{1}{2} \text{signum}(\frac{a}{b}) \pi[/tex]. If my memory serves me well, signum is always 1 if a/b > 0 and -1 if a/b < 0, so if a/b > 0 this holds...

    So yeah, just wondering... I can't see any way to do it algebraically...
     
  2. jcsd
  3. May 19, 2008 #2

    rock.freak667

    User Avatar
    Homework Helper

    Let [itex]tan\alpha =\frac{a}{b}[/itex] and [itex]tan\beta =\frac{b}{a}[/itex]

    Then consider

    [tex]tan(\alpha + \beta)=\frac{tan\alpha +tan\beta}{1-tan\alpha tan\beta}[/tex]

    and what that gives.
     
  4. May 20, 2008 #3
    Thanks, I understand now!

    [tex]\tan \alpha = \frac{a}{b} \text{ , } \tan \beta = \frac{b}{a}[/tex]
    [tex]\arctan \frac{a}{b} + \arctan \frac{b}{a} = \alpha + \beta[/tex]

    [tex]\tan(\alpha + \beta) = \frac{ \frac{a}{b} + \frac{b}{a} }{ 1 - \frac{ab}{ab}} = \frac{ \frac{a}{b} + \frac{b}{a} }{0} = \infty[/tex]*

    If [tex]\tan(\alpha + \beta) = \infty[/tex] then [tex]\alpha + \beta = \frac{\pi}{2}[/tex].

    *Can you say this so easily? Shouldn't you handle the divide by zero better? I know if a denominator tends to zero, the fraction tends to infinity, but you can't divide by 0 exactly... I also can't see any way I could take a limit here?
     
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