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Trigonometric proof using derivatives

  1. Jan 20, 2012 #1
    1. The problem statement, all variables and given/known data
    Prove that (for every x smaller than -1)
    [itex]\displaystyle \frac{1}{2}arctan\frac{2x}{1-x^{2}}+arccotx=\pi[/itex]


    2. Relevant equations



    3. The attempt at a solution
    So i split the formula into two parts:
    [itex]\displaystyle \frac{1}{2}arctan\frac{2x}{1-x^{2}}[/itex] and [itex]\displaystyle \pi-arccotx[/itex]
    Calculated their derivatives separately
    And they are both equal to [itex]\displaystyle \frac{1}{x^{2}+1}[/itex]

    And I'm wondering whether it's the end of the task or should I prove some other things, especially something connected to the main assumption of x<-1 or any other.

    Thanks in advance!
     
  2. jcsd
  3. Jan 20, 2012 #2

    micromass

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    If [itex]f^\prime=g^\prime[/itex], then this doesn't mean f=g necessarily. But it does mean that f=g+C with C a real constant/

    Note that the above only holds if the domain of f and g is an interval.
     
  4. Jan 21, 2012 #3
    Thank You, now I know that it isn't sufficient.
    But I can't really come up with a complete solution.
    What should I add ? So that the task would be marked well
    How to calculate the value of that constant?
     
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