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Interesting yet challenging proof

  1. Mar 24, 2008 #1
    how can I show that ... arctan(1/v) = (π/2) - arctan(v) ???
  2. jcsd
  3. Mar 24, 2008 #2
    well,try giving and playing with the variables untill you get a result.

    More of a trial and error method.

  4. Mar 24, 2008 #3
    I know I've been doing that, moving around variables and using the unit circle and right triangles, but I cannot seem to come across a substantial reason why they are equal. Any thoughts?
  5. Mar 24, 2008 #4
    Do v and n represent anything in particular?

    Because your equation only works when you have certain values for v and n.
  6. Mar 24, 2008 #5
    i knew this would confuse someone... that "n" you see is π, or pi... sorry. Does that help?
  7. Mar 26, 2008 #6
    if you use the log representation for artan (1/x) and artan (x) so

    [tex] artan(x)= (2i)^{-1}(log(1+ix)-log(1-ix)) [/tex] and the same replacing x--> 1/x you

    get the accurate result.
  8. Mar 26, 2008 #7


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    Are you allowed to use x = arctan(v) => v = tanx ? :smile:
  9. Mar 26, 2008 #8


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    I'm going to point significantly (*points significantly*) to my signature. The clue is given by tiny-tim (and that is what they want you to use): make a diagram of a right triangle with x as one of the non-right angles and use v, written as v/1 , as the ratio of the sides that would come from finding the tangent of angle x (label the sides of the triangle appropriately).

    Now, in the same triangle, what angle has a tangent of 1/v ? What is the relationship between that angle and angle x ?

    (And, with all due respect to mhill, while that relationship is true, the math is probably way beyond what is being done in Calixto's course...)
    Last edited: Mar 26, 2008
  10. Mar 26, 2008 #9
    Thanks, I understand it now... But could you explain some more what you wrote about the logs and stuff? Just maybe explain where that comes from so I can impress my teacher :)
  11. Mar 27, 2008 #10


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    If x = tany, then y = arctanx, and so:

    log(1+ix)-log(1-ix) = log[(1+ix)/(1-ix)]

    = log[(cosy + isiny)/(cosy - isiny)]

    = log[e^{2iy}]

    = 2iy

    :smile: = 2i.arctan(x). :smile:

    (You see how, to prove anything with arctan(x), you always convert to x = tany?)
  12. Mar 28, 2008 #11

    Gib Z

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    You basically want to show that

    [tex]\arctan (1/v) + \arctan (v) = \frac{\pi}{2}[/tex].

    Draw a right angled triangle, with the smaller sides length 1 and v. What does [itex]\arctan v[/itex] represent here?
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