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Math Help - Hyperbolic functions

  1. #1
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    Hyperbolic functions

    If 3tanhx = 1 ,determine exact values of sinhx and coshx.

    Can anybody help with this one!

    The ans is,

    1/(2(sqrt 2)) and 3/(2(sqrt2))
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  2. #2
    Super Member Aryth's Avatar
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    Well, first you'd need to move the 3 to the other side and then take the inverse:

    x = arctanh\left(\frac{1}{3}\right)

    There is a formula for solving the arctanh, and that is:

    arctanh(u) = \frac{1}{2}ln\left(\frac{1 + u}{1 - u}\right)

    Plugging in u = 1/3, you get:

    x = arctanh\left(\frac{1}{3}\right) = \frac{1}{2}\left[ln\left(1 + \frac{1}{3}\right) - ln\left(1 - \frac{1}{3}\right)\right]

    x = \frac{1}{2}\left[ln\left(\frac{4}{3}\right) - ln\left(\frac{2}{3}\right)\right]

    This ends up being:

    x \approx 0.347

    Plug this into sinh and cosh to get:

    sinh(0.347) \approx 0.354

    \frac{1}{2\sqrt{2}} \approx 0.354

    That checks out with the answer.

    cosh(0.347) \approx 1.061

    \frac{3}{2\sqrt{2}} \approx 1.061

    This also checks out.

    Hope this helps!
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  3. #3
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    Grandad's Avatar
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    Hello heatly
    Quote Originally Posted by heatly View Post
    If 3tanhx = 1 ,determine exact values of sinhx and coshx.

    Can anybody help with this one!

    The ans is,

    1/(2(sqrt 2)) and 3/(2(sqrt2))
    A more direct way is to use the relationships:
    \tanh x = \dfrac{\sinh x}{\cosh x} ...(1)
    and
    \cosh^2x-\sinh^2x = 1 ...(2)
    So
    3\tanh x=1

    \Rightarrow\tanh x=\dfrac{\sinh x}{\cosh x} = \frac13, using (1)

    \Rightarrow \sinh x = \frac13\cosh h
    Then substitute into (2):
    \cosh^2x-\frac19\cosh^2x = 1

    Can you complete it from here?

    Grandad
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  4. #4
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    Thanks for that aryth !
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  5. #5
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    Thanks ............... "grandad''
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