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Math Help - hyperbolic functions

  1. #1
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    hyperbolic functions

    Hi all,

    I am stuck with this question:

    Using coshx and sinhx to verify the identity:

    (coshx + sinhx)^n = cosh(nx) + sinh(nx)

    I have tried using coshx = (e^x + e^(-x))/2 and sinhx = (e^x - e^(-x))/2... and i plugged it in and got (e^x)^n but lol how can i get cosh(nx) + sinh(nx)?

    Thanks in advance,
    ArTiCK
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by ArTiCK View Post
    Hi all,

    I am stuck with this question:

    Using coshx and sinhx to verify the identity:

    (coshx + sinhx)^n = cosh(nx) + sinh(nx)

    I have tried using coshx = (e^x + e^(-x))/2 and sinhx = (e^x - e^(-x))/2... and i plugged it in and got (e^x)^n but lol how can i get cosh(nx) + sinh(nx)?

    Thanks in advance,
    ArTiCK
    plug in the identities you have into the right side first and simplify as much as possible. then do the same for the left side, and show that they simplify to the same thing.
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  3. #3
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    Quote Originally Posted by ArTiCK View Post
    By writing \cosh x and \sinh x in exponential form prove the identity:

    (\cosh x + \sinh x)^n = \cosh nx + \sinh nx<br />
    \cosh x = \frac{1}{2}(e^{x} + e^{-x}) and \sinh x = \frac{1}{2}(e^{x} - e^{-x})

    \therefore coshx + sinhx = \frac{1}{2}(e^{x} + e^{-x}) + \frac{1}{2}(e^{x} - e^{-x})
    \Rightarrow e^x

    (coshx + sinhx)^n = e^{nx}

    \Rightarrow \frac{1}{2}e^{nx} + \frac{1}{2}e^{nx}  +  \underbrace{\frac{1}{2}e^{-nx} -\frac{1}{2}e^{-nx}}
    . . . . . . . . . . . adding "zero"

    \Rightarrow \frac{1}{2}(e^{nx} +e^{-nx}) + \frac{1}{2}(e^{nx} -e^{-nx})

    \Rightarrow \cosh nx + \sinh nx

    Bobak
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  4. #4
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    Quote Originally Posted by ArTiCK View Post
    Hi all,

    I am stuck with this question:

    Using coshx and sinhx to verify the identity:

    (coshx + sinhx)^n = cosh(nx) + sinh(nx)

    I have tried using coshx = (e^x + e^(-x))/2 and sinhx = (e^x - e^(-x))/2... and i plugged it in and got (e^x)^n but lol how can i get cosh(nx) + sinh(nx)?

    Thanks in advance,
    ArTiCK
    Note: \cosh (ax) + \sinh (ax) = \frac{e^{ax} + e^{-ax}}{2} + \frac{e^{ax} - e^{-ax}}{2} = e^{ax}.

    On the left hand side you have a = 1. On the right hand side you have a = n. The proof now falls out in a handfull of lines at most ......
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