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Hyperbolic functions

  1. Mar 15, 2009 #1
    1. The problem statement, all variables and given/known data
    If sinhx=tany show x=ln(tany±secy)

    2. Relevant equations
    sinhx=0.5(e^x-e^(-x))
    secy=1/cosy
    cosy=0.5(e^y+e^(-y))
    tany=(e^(jx)-e^(-jx))/(e^(jx)+e^(-jx))
    tany=siny/cosy

    3. The attempt at a solution
    0.5e^x -0.5e^-x=tany
    0.5e^(2x) -0.5=tany e^x
    e^(2x) -2tany e^x -1 = 0
    e^(2x) = (2tany±square[(-2tany)^2 + 4] / 2

    Is this good so far, and what to do with 4tan^2y?
     
  2. jcsd
  3. Mar 15, 2009 #2

    gabbagabbahey

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    This equation is quadratic in the variable e^x, not e^(2x), so this:

    should be e^(x) = (2tany±square[(-2tany)^2 + 4] / 2.

    Now, factor out a 4 from inside the square root....[itex]1+\tan^2 y=[/itex]___?
     
  4. Mar 15, 2009 #3

    tiny-tim

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    Hi Pietair! :smile:
    oooh, so complicated! :cry:

    Hint: if sinhx=tany, coshx = … ? :smile:
     
  5. Mar 15, 2009 #4
    I have no idea...

    Should be something like 0.5cos^2y or something I think.
     
  6. Mar 15, 2009 #5

    gabbagabbahey

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    Look it up in your table of trig identities. Or better yet, derive it yourself by writing tan^2y in terms of sines and cosines and placing everything over the common denominator...
     
  7. Mar 16, 2009 #6
    Thanks a lot, I succeeded by replacing:
    [itex]
    1+\tan^2 y
    [/itex]

    for: [itex]
    sec^2 y
    [/itex]
     
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