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Proving arcsin(somevalue) = 2arctan(rootx) - value

  1. Mar 20, 2009 #1
    1. The problem statement, all variables and given/known data

    prove the identity:

    arcsin( x-1/x+1 ) = 2 * arctan( sqrt(x) ) - pi/2


    2. Relevant equations
    if f'x = g'x for all x in an interval (a,b) then f - g is constant on (a,b); that is, fx = gx + c where c is constant.

    this material above is in the same section as rolles thm and the mean value thm if that helps at all.


    3. The attempt at a solution
    Im assuming that i use arcsin(value) as fx and 2arctan(value) as gx and verify that their derivatives are equal, however i cant seem to do this. this is what i got... :

    d/dx arcsin( x-1/x+1 ) = 1/( sqrt( 1 - ( x-1/x+1 )^2 ) ) * ( x+1 - x-1 / ( x+1 )^2

    and

    d/dx 2arctan(sqrt( x) ) = 2/(1+ sqrt(x^2)) * 1/(2 * sqrt(x)) = 2/(1+x) * 1/(2 * sqrt(x) )

    my real questions here is what are the real derivatives of these two functions and can someone please right down the solution to that explicitly so i am not confused, THANK YOU PHYSICS FORUMS MEMBERS.
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Mar 21, 2009 #2

    Mark44

    Staff: Mentor

    You didn't get the chain rule part quite right in the work above.
    d/dx arcsin( x-1/x+1 ) = 1/( sqrt( 1 - ( x-1/x+1 )^2 ) ) * ( x+1 - x+1) / ( x+1 )^2
    I fixed an incorrect sign right here --------------------------------^
    There's a lot of simplification that you can do, to get it to 1/(sqrt(x) * (x + 1)), which is the same as what you get in the next part.
     
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