Results 1 to 4 of 4

Math Help - Hyperbolic function..

  1. #1
    Member
    Joined
    Sep 2007
    Posts
    76

    Hyperbolic function..

    Hi, I'm struggling with proof for this hyperbolic function, I just need a bit of explination if someone can help??

    Let y=arccosh(x), then x=coshy
    x = (e^y+e^-y)/2
    2x = e^y+e^-y
    e^(2y)-2xe^y+1 = 0
    e^y = [2x +/- sqrt(4x^2-4)]/2

    For the next step I understand it to be..
    e^y = x +/- sqrt(2x^2-2)

    but apparentley the correct answer is
    e^y = x +/- sqrt(x^2-1)

    How is this? surely your only dividing by 2... can someone please explain where I'm going wrong

    Thanks..

    Heres the rest, but i get that...

    y = ln[x +/- sqrt(x^2-1)]
    arccosh(x)=ln[x +/- sqrt(x^2-1)]
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Joined
    Mar 2007
    Posts
    1,240

    Question

    What were the instructions? What are you supposed to be doing with the "Let y =" bit?

    Thank you!
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Sep 2007
    Posts
    76
    I'm trying to prove that arccosh(x)=ln[x +/- sqrt(x^2-1)]...
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Member
    Joined
    Sep 2007
    Posts
    76
    Its the formulae for the inverse hyperbolic function arccos(x)
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 2
    Last Post: March 20th 2011, 02:29 AM
  2. hyperbolic function
    Posted in the Calculus Forum
    Replies: 1
    Last Post: September 14th 2009, 02:37 PM
  3. Hyperbolic Function & e
    Posted in the Calculus Forum
    Replies: 5
    Last Post: May 3rd 2008, 02:38 PM
  4. hyperbolic function
    Posted in the Calculus Forum
    Replies: 1
    Last Post: January 23rd 2008, 03:03 PM
  5. hyperbolic function
    Posted in the Calculus Forum
    Replies: 1
    Last Post: October 11th 2007, 03:36 AM

Search Tags


/mathhelpforum @mathhelpforum