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Math Help - Hyperbolic Function proof

  1. #1
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    Hyperbolic Function proof

    I'm required to prove that:

    \sinh ^{-1}\left( x \right)\; =\; \ln \left( x\; +\; \sqrt{x^{2}\; +\; 1} \right)

    The furthest I can get is this:

    \sinh ^{-1}\left( x \right)\; =\; \frac{1}{\frac{1}{2}\left( e^{x}-e^{-x} \right)}

    I have no idea what to do Any kind of nudge in the right direction would be greatly appreciated.
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  2. #2
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by DJ Hobo View Post
    I'm required to prove that:

    \sinh ^{-1}\left( x \right)\; =\; \ln \left( x\; +\; \sqrt{x^{2}\; +\; 1} \right)

    The furthest I can get is this:

    \sinh ^{-1}\left( x \right)\; =\; \frac{1}{\frac{1}{2}\left( e^{x}-e^{-x} \right)}

    I have no idea what to do Any kind of nudge in the right direction would be greatly appreciated.
    \sinh(x)=\frac{e^x-e^{-x}}{2}. So, use the normal method. x=\frac{e^{y}-e^{-y}}{2}\implies xe^y=\frac{e^{2y}-1}{2}. Call z=e^{y} and see if anything pops out at you.
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  3. #3
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    Quote Originally Posted by DJ Hobo View Post
    I'm required to prove that:

    \sinh ^{-1}\left( x \right)\; =\; \ln \left( x\; +\; \sqrt{x^{2}\; +\; 1} \right)

    The furthest I can get is this:

    \sinh ^{-1}\left( x \right)\; =\; \frac{1}{\frac{1}{2}\left( e^{x}-e^{-x} \right)}

    I have no idea what to do Any kind of nudge in the right direction would be greatly appreciated.
    Incorrect.

    The notation \sinh^{-1}{x} does NOT mean \frac{1}{\sinh{x}}. It stands for the inverse hyperbolic sine function.

    It's the same sort of notation used in trigonometric functions to show the inverse trigonometric functions. E.g. \sin^{-1}x = y implies x = \sin{y}.


    Suppose you had

    y = \sinh{x}

     = \frac{1}{2}(e^x - e^{-x}).


    Its inverse function is

    x = \sinh{y} (in other words, y = \sinh^{-1}{x})

    x = \frac{1}{2}(e^y - e^{-y})

    2x = e^y - e^{-y}

    2x = e^y - \frac{1}{e^y}

    2x = \frac{e^{2y} - 1}{e^y}

    2x\,e^y = e^{2y} - 1

    0 = e^{2y} - 2x\,e^y - 1

    0 = e^{2y} - 2x\,e^y + (-x)^2 - (-x)^2 - 1

    0 = \left(e^y - x\right)^2 - x^2 - 1

    x^2 + 1 = \left(e^y - x\right)^2

    \sqrt{x^2 + 1} = e^y - x

    x + \sqrt{x^2 + 1} = e^y

    y = \ln{\left(x + \sqrt{x^2 + 1}\right)}.


    Therefore \sinh^{-1}{x} = \ln{\left(x + \sqrt{x^2 + 1}\right)}.
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  4. #4
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    ah yes, I made a very stupid mistake to start with. Thanks for the help, guys
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