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Math Help - [SOLVED] Proof of Double Angle

  1. #1
    Overwhelmed
    Guest

    [SOLVED] Proof of Double Angle

    Hi, I know its probably something really simple but does anyone know how I would go about proving that cosh(x+y)=coshxcoshy+sinhxsinhy ?
    I would really appreciate a fast answer!
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  2. #2
    MHF Contributor
    Joined
    Apr 2005
    Posts
    1,631
    Prove cosh(x+y) = cosh(x)cosh(y) +sinh(x)sinh(y).

    Righthand Side =
    = [(e^x +e^-x)/2]*[e^y +e^-y)/2] +[(e^x -e^-x)/2]*[e^y -e^-y)/2]
    = (1/4)[e^(x+y) +e^(x-y) +e^(-x+y) +e^(-x-y)] +(1/4)[e^(x+y) -e^(x-y) -e^(-x+y) +e^(-x-y)]
    = (1/4)[e^(x+y) +e^(x-y) +e^(-x+y) +e^(-x-y) +e^(x+y) -e^(x-y) -e^(-x+y) +e^(-x-y)]
    = (1/4)[2e^(x+y) +2e^(-x-y)]
    = (1/4)[2e^(x+y) +2e^-(x+y)]
    = (1/2)[e^(x+y) +e^-(x+y)]
    = [e^(x+y) +e^-(x+y)]/2
    = cosh(x+y)
    = Lefthand Side.

    Proven.

    ---------
    Remember,
    cosh(u) = (e^u +e^-u)/2
    sinh(u) = (e^u -e^-u)/2
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