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Math Help - hyperbolic functions

  1. #1
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    Post hyperbolic functions

    hi all,

    how do you prove the following:

    Prove  sinh(x + y) = sinh x cosh y + cosh x sinh y


    thanks.
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  2. #2
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    Quote Originally Posted by dadon View Post
    hi all,

    how do you prove the following:

    Prove  sinh(x + y) = sinh x cosh y + cosh x sinh y


    thanks.
    ----sinh u = (1/2)[e^u -e^(-u)]
    ----cosh u = (1/2)[e^u +e^(-u)]
    ----a^b * a^c = a^(b+c)

    sinh(x+y) = sinh x cosh y + cosh x sinh y

    We develop the RHS.
    RHS =
    = {(1/2)[e^x -e^(-x)] *(1/2)[e^y +e^(-y)]} +{(1/2)[e^x +e^(-x)] *(1/2)[e^y -e^(-y)]}

    = (1/4)[e^x -e^(-x)][e^y +e^(-y)] +(1/4)[e^x +e^(-x)][e^y -e^(-y)]

    Doing FOILs in the expansion,

    = (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)]

    = (1/2)[e^(x+y) -e^(-x-y)]

    = (1/2)[e^(x+y) -e^[-(x+y)]

    = sinh (x+y)

    = LHS

    Therefore, proven.
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  3. #3
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    thank you

    Nicely explained!
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