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Math Help - proving a hyperbolic trig identity

  1. #1
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    proving a hyperbolic trig identity

    how would you prove that:

    sinh (x+y) = sinhx coshy + coshx sinhy ??

    this is a little bit tricky (for me!) since it involves hyperbolic functions
    Last edited by nmanik90; January 23rd 2008 at 02:46 PM. Reason: wrong/misleading title
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  2. #2
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    It's not that bad. Actually, it seems easier to prove than the standard trig identity.

    Use cosh(x)+sinh(x)=e^{x}, \;\ cosh(x)-sinh(x)=e^{-x}

    sinh(x+y)=\frac{e^{x+y}-e^{-(x+y)}}{2}=\frac{e^{x}e^{y}-e^{-x}e^{-y}}{2}

    =\frac{1}{2}[(cosh(x)+sinh(x))(cosh(y)+sinh(y))-(cosh(x)-sinh(x)) (cosh(y)-sinh(y))]

    =sinh(x)cosh(y)+cosh(x)sinh(y)
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  3. #3
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    Quote Originally Posted by nmanik90 View Post
    how would you prove that:

    sinh (x+y) = sinhx coshy + coshx sinhy ??

    this is a little bit tricky (for me!) since it involves hyperbolic functions
    1. \displaystyle \sinh (x + y) = \frac{1}{2} \left( e^{x + y} - e^{-(x + y)}\right) = \frac{1}{2} \left( e^{x + y} - e^{-x - y} \right)


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


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


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


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


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


    So, putting these two results together ........

    In answer to your question "How did you know to do that?", it helps to know what the answer has to be
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  4. #4
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    Here is another way to look at the question.
    \left( {e^x  - e^{ - x} } \right)\left( {e^y  + e^{ - y} } \right) = e^{x + y}  + e^{x - y}  - e^{-x + y}  - e^{ - x - y}
    \left( {e^x  + e^{ - x} } \right)\left( {e^y  - e^{ - y} } \right) = e^{x + y}  - e^{x - y}  + e^{ - x + y}  - e^{ - x - y}
    Add these two together.
    Do you realize 2cosh(x)= {e^x  + e^{ - x} } and 2sinh(y)= \left( {e^y  - e^{ - y} } \right)?
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