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Domain of identity

  1. Dec 21, 2013 #1
    1. The problem statement, all variables and given/known data

    Prove that there is a constant C such that
    [itex]arctan\sqrt{\frac{1-x}{1+x}}[/itex] = C - [itex]\frac{1}{2}arcsinx[/itex] for all x in a certain domain. What is the largest domain on which this identity is true? What is the value of the constant C?

    3. The attempt at a solution

    Now I know how to prove the initial statement (showing the derivatives are equal which implies they differ by only a constant), but I wanted to verify the largest domain and the value of C.

    For the largest domain on which this identity is true I obtained (-1, 1] (since arcsinx is defined on
    [-1, 1], and since -1 is not allowed).

    And I believe C can be any real number.

    So I'd just like to verify whether or not my interpretation is correct?
     
  2. jcsd
  3. Dec 21, 2013 #2
    As far as I can tell, your domain is correct.

    Am I right in interpreting that you are saying this equation is true for any value of C? If so, that's not true.

    To solve for C, pick any value for x in the domain, plug it into the equation, and solve for C.
     
  4. Dec 21, 2013 #3
    Yeah your right. I made a mistake there. And thanks scurty.
     
  5. Dec 21, 2013 #4
    Edit: Nevermind.

    To make this post worthwhile, I suggest x values of 0 or 1 to solve for C.
     
  6. Dec 21, 2013 #5
    Yeah I used x=0 and obtained C=[itex]\pi[/itex]/4.
     
  7. Dec 21, 2013 #6

    vanhees71

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    It's all correct, but [itex]C[/itex] is uniquely defined by the equation. Just choose a value for [itex]x[/itex], for which you know the values on both sides of the equation!
     
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