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Trigonometry: arctan equation?

  1. Nov 7, 2011 #1
    Can someone please let me know how to find x in

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


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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??
  4. Nov 7, 2011 #3
    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?
  5. Nov 7, 2011 #4


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

    tan(π/4) = x + arctan(x+arctan(x+...)) ???
  7. Nov 7, 2011 #6
    what is tan(tanx)??? Is above equation some kind of sum of geometric progress???
  8. Nov 7, 2011 #7
    is it so easy that no one is going even to comment?
  9. Nov 7, 2011 #8


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    tan(π/4)=tan(arctan(x+arctan(x+...)) )
  10. Nov 7, 2011 #9
    lurflurf: thank you, this is obvious and not the solution


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

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


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    arctan(x+arctan(x+...)) ?
  12. Nov 8, 2011 #11
    if I knew what "arctan(x+arctan(x+...))" equals to, I wouldn't be posting this thread...
  13. Nov 8, 2011 #12
    What? You said what this equals in your first post. It's x that is the unknown.
  14. Nov 8, 2011 #13
    spamiam: right, sorry, I'm really dumb

    x = 1 - π/4

    Thank you all for bearing with me...
  15. Nov 8, 2011 #14
    This is dumbest question I have ever asked. Hope this is never going to happen again...
  16. Nov 8, 2011 #15
    You're not dumb--this question is difficult until you see the trick to use.
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