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Thread: Hyperbolic Trig Proof Help

  1. #1
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    Hyperbolic Trig Proof Help

    Im stuck on the following problem and would greatly appreciate any help.

    Using the exponential definitions of coshx and sinhx


    $\displaystyle sinhx = (e^x - e^-x)/2$ and $\displaystyle coshx = (e^x + e^-x)/2
    $

    show that

    $\displaystyle sinhx + sinhy = 2sinh((x+y) /2)cosh((x-y)/2)$

    I know that it obviously involves use of the double angle formula but just cant get it work. Cheers
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  2. #2
    MHF Contributor red_dog's Avatar
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    $\displaystyle 2\sinh\frac{x+y}{2}\cosh\frac{x-y}{2}=\displaystyle 2\cdot\displaystyle\frac{e^{\frac{x+y}{2}}-e^{\frac{-x-y}{2}}}{2}\cdot\frac{e^{\frac{x-y}{2}}+e^{\frac{-x+y}{2}}}{2}=$

    $\displaystyle =\frac{1}{2}\left(e^{\frac{x+y}{2}+\frac{x-y}{2}}+e^{\frac{x+y}{2}+\frac{-x+y}{2}}-e^{\frac{-x-y}{2}+\frac{x-y}{2}}-e^{\frac{-x+y}{2}+\frac{-x+y}{2}}\right)=$

    $\displaystyle =\frac{1}{2}(e^x+e^y-e^{-y}-e^{-x})=\frac{e^x-e^{-x}}{2}+\frac{e^y-e^{-y}}{2}=\sinh x+\sinh y$
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  3. #3
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    awesome thanks very much
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