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Math Help - Inverse hyperbolic question

  1. #1
    Junior Member rednest's Avatar
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    Cool Inverse hyperbolic question

    Hi
    Basically I got cosh(arsinh 2)

    How do I instantly know that the answer is \sqrt{5} ?

    PS: I actually worked it out myself the long way by letting arsinh2 = ln(2+\sqrt{5}). However, I want to know a faster way of doing this.
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  2. #2
    Super Member flyingsquirrel's Avatar
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    Hi

    Simply use \cosh x =\sqrt{1+\sinh^2x}.
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  3. #3
    Junior Member rednest's Avatar
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    Thumbs up

    Quote Originally Posted by flyingsquirrel View Post
    Hi

    Simply use \cosh x =\sqrt{1+\sinh^2x}.
    THANK YOU! Seriously why didn't I think of that.... I must be so dumb.
    Merci (if you are French)

    Thanks
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