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Thread: cosh

  1. #1
    Junior Member
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    Sep 2009
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    41

    cosh

    can someone please explain to me how they got from

    $\displaystyle \frac{-1}{4i}(e^{-1}+e^1) $

    to

    $\displaystyle \frac{i}{2}cosh(1) $

    coz if i did it i got

    $\displaystyle -\frac{1}{2i} cosh (1) $

    soo sorry im so lost
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  2. #2
    Senior Member
    Joined
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    Quote Originally Posted by Dgphru View Post
    can someone please explain to me how they got from

    $\displaystyle \frac{-1}{4i}(e^{-1}+e^1) $

    to

    $\displaystyle \frac{i}{2}cosh(1) $

    coz if i did it i got

    $\displaystyle -\frac{1}{2i} cosh (1) $

    soo sorry im so lost
    It's the same thing!

    $\displaystyle -\frac{1}{2i} = -\frac{i}{2i^2} = - \frac{i}{2\cdot(-1)} = \frac{i}{2}$
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  3. #3
    Junior Member
    Joined
    Sep 2009
    Posts
    41
    hahah i can be so stupid ><
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