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Math Help - i identity

  1. #1
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    i identity

    Demonstrate that \arccos^i(\cosh1) = e^{-\frac{\pi}{2}}

    I am not a mathematician so sorry if this is not very challenging. Basically I just found this identity to be interesting and wonder how obvious/obscure it is to real mathematicians.
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  2. #2
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    Note that cosh(1)=cos(i)

    cos^{-1}(cos(i))=i

    Therefore, we have i^{i}

    Now, show this equals e^{\frac{-\pi}{2}}
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  3. #3
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    Quote Originally Posted by galactus View Post
    Note that cosh(1)=cos(i)

    cos^{-1}(cos(i))=i

    Therefore, we have i^{i}

    Now, show this equals e^{\frac{-\pi}{2}}

    Yep.

    Just to tidy up a bit:

    \cos^{i}(\cosh1)^{-1}=i^i=e^{\frac{-\pi}{2}}
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