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Math Help - show cosh and cos

  1. #1
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    show cosh and cos

    Hi, can someone please help me prove this:

     cos(iy) = cosh (y)

    I used eulers formula for complex numbers but i ended up with
     cos (y) = cosh (iy)



    please help or provide any guidance

    thanks\ you!!
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  2. #2
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    Quote Originally Posted by Dgphru View Post
    Hi, can someone please help me prove this:

     cos(iy) = cosh (y)

    I used eulers formula for complex numbers but i ended up with
     cos (y) = cosh (iy)



    please help or provide any guidance

    thanks\ you!!

    Ok, so Euler it is: for x\in\mathbb{R}\,,\,\,e^{ix}=\cos x+i\sin x\,,\,\,e^{-ix}=\cos x-i\sin x . Now sum up both these equations:

    2\cos x=e^{ix}+e^{-ix}\Longrightarrow \cos x=\frac{e^{ix}+e^{-ix}}{2} . OTOH, the definition of the hyperbolic cosine is:

    \cosh x=\frac{e^x+e^{-x}}{2} . Well, now just put ix intead of x and compare with the above...

    Tonio
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  3. #3
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    Let y=ix

    Then:

    \cos(y)=\cosh(iy) ~\implies~ \cos(ix)=\cosh(i^2x)=\cosh(-x)=\cosh(x)

    because cosh is an even function.

    So basically, since both cos and cosh are even functions, we have that:

    \cos(x) = \cos(-x) = \cosh(ix) = \cosh(-ix)

    and

    \cos(ix) = \cos(-ix) = \cosh(x) = \cosh(-x)
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