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Trigonometry: arctan equation?

  1. Nov 7, 2011 #1

    XYZ313

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    Can someone please let me know how to find x in

    π/4=arctan(x+arctan(x+...)) ?
     
    XYZ313, Nov 7, 2011
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  3. Nov 7, 2011 #2

    micromass

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    Let me do a similar one:

    Let [itex]x=\sqrt{2+\sqrt{2+\sqrt{2+...}}}[/itex]. Then [itex]x^2-2=x[/itex].
    Solving the quadratic equation gives us the answer.

    Can you do your problem now along similar lines??
     
    micromass, Nov 7, 2011
  4. Nov 7, 2011 #3

    XYZ313

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    micromass: isn't it the use of inverse operator (tan in this case)?
    like z=arctan(x+arctan(x+...)) and make the use of tanz?

    and then use tan addition rule??? I am really confused...

    Can you explain further please?
     
    XYZ313, Nov 7, 2011
  5. Nov 7, 2011 #4

    lurflurf

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    π/4=arctan(x+arctan(x+...))
    take tan of both sides
    tan(π/4)=tan(arctan(x+arctan(x+...)) )
     
    lurflurf, Nov 7, 2011
  6. Nov 7, 2011 #5

    XYZ313

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    lurflurf: and then what?

    tan(π/4) = x + arctan(x+arctan(x+...)) ???
     
    XYZ313, Nov 7, 2011
  7. Nov 7, 2011 #6

    XYZ313

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    what is tan(tanx)??? Is above equation some kind of sum of geometric progress???
     
    XYZ313, Nov 7, 2011
  8. Nov 7, 2011 #7

    XYZ313

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    is it so easy that no one is going even to comment?
     
    XYZ313, Nov 7, 2011
  9. Nov 7, 2011 #8

    lurflurf

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    tan(π/4)=tan(arctan(x+arctan(x+...)) )
    1=x+arctan(x+arctan(x+...))
     
    lurflurf, Nov 7, 2011
  10. Nov 7, 2011 #9

    XYZ313

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    lurflurf: thank you, this is obvious and not the solution

    then

    tan(1-x) = x + arctan(x+arctan(x+...))

    and where it is leading???
     
    XYZ313, Nov 7, 2011
  11. Nov 8, 2011 #10

    lurflurf

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    arctan(x+arctan(x+...)) ?
     
    lurflurf, Nov 8, 2011
  12. Nov 8, 2011 #11

    XYZ313

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    if I knew what "arctan(x+arctan(x+...))" equals to, I wouldn't be posting this thread...
     
    XYZ313, Nov 8, 2011
  13. Nov 8, 2011 #12

    spamiam

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    What? You said what this equals in your first post. It's x that is the unknown.
     
    spamiam, Nov 8, 2011
  14. Nov 8, 2011 #13

    XYZ313

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    spamiam: right, sorry, I'm really dumb

    x = 1 - π/4

    Thank you all for bearing with me...
     
    XYZ313, Nov 8, 2011
  15. Nov 8, 2011 #14

    XYZ313

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    This is dumbest question I have ever asked. Hope this is never going to happen again...
     
    XYZ313, Nov 8, 2011
  16. Nov 8, 2011 #15

    spamiam

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    You're not dumb--this question is difficult until you see the trick to use.
     
    spamiam, Nov 8, 2011
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