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Hyperbolic cosine identity help!

  1. Jan 24, 2012 #1
    1. The problem statement, all variables and given/known data

    Show that cosh^2(x) = (cosh(2x) - 1)/2

    2. Relevant equations

    cosh(x) = (e^x + e^-x)/2

    3. The attempt at a solution

    I have attempted this multiple times and get the same results every time.
    Squaring cosh(x) I get 1/4(e^2x + e^-2x +2), which is i believe 1/4(cosh(2x) +2).

    Maybe i just can't see it but how it that equivalent to the identity above??
     
    Last edited: Jan 24, 2012
  2. jcsd
  3. Jan 24, 2012 #2

    lanedance

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    wait, so what are you trying to show?

    the first is just normal double angle cos and can be shown by looking at Re and Im of (e^(itheta))^2
     
    Last edited: Jan 24, 2012
  4. Jan 24, 2012 #3

    tiny-tim

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    welcome to pf!

    hi s.perkins! welcome to pf!:smile:
    erm :redface:

    cosh(x) = (e^x plus e^-x)/2 :wink:
     
  5. Jan 24, 2012 #4

    lanedance

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    looks like the correct form has been used in the square, but also don't forget the factor of 2 as well...
    cosh(2x)=(e^(x)+e^(-2x))/2
     
  6. Jan 24, 2012 #5
    Sorry fixed the typo. I did include the 1/2, when its squared you get a 1/4 in front.
     
  7. Jan 24, 2012 #6

    lanedance

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    might as well use equals as its a bit clearer what you're trying

    here should be

    cosh(x)^2
    = (e^2x + e^-2x +2)/4
    = (2(e^2x + e^-2x)/2 +2)/4
    = (2cosh(2x) +2)/4
    = (cosh(2x) +1)/2

    which is a valid identity, as shown by adding th two below together
    cosh(x)^2- sinh(x)^2=1
    cosh(x)^2+sinh(x)^2=cosh(2x)
    which gives
    2cosh(x)^2=cosh(2x)+1
     
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