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Thread: Hyperbolic trig identities?

  1. #1
    Senior Member Stroodle's Avatar
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    Hyperbolic trig identities?

    Hi,

    I'm not sure how or which identities to use to solve the following question:

    Express $\displaystyle sinh^5x$ in terms of hyperbolic sines of multiples of $\displaystyle x$.

    The answer is: $\displaystyle \frac{1}{16}(sinh\ x-5sinh\ 3x+10sinh\ x)$

    I have a fair few similar problems, so if you could please show me the steps for this one I should hopefully be able to solve the others.

    Thanks heaps for your help.
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  2. #2
    MHF Contributor Also sprach Zarathustra's Avatar
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    Re: Hyperbolic trig identities?

    Quote Originally Posted by Stroodle View Post
    Hi,

    I'm not sure how or which identities to use to solve the following question:

    Express $\displaystyle sinh^5x$ in terms of hyperbolic sines of multiples of $\displaystyle x$.

    The answer is: $\displaystyle \frac{1}{16}(sinh\ x-5sinh\ 3x+10sinh\ x)$

    I have a fair few similar problems, so if you could please show me the steps for this one I should hopefully be able to solve the others.

    Thanks heaps for your help.




    $\displaystyle \sinh{x}=\frac{e^x-e^{-x}}{2}$


    $\displaystyle \sinh^5{x}=(\frac{e^x-e^{-x}}{2})^5$

    $\displaystyle =\frac{1}{32}(e^{5x}-5e^{3x}+10e^{x}-10e^{-x}+5e^{-3x}-e^{-5x})$

    $\displaystyle =\frac{1}{32}((e^{5x}-e^{-5x})-5(e^{3x}-e^{-3x})-10(e^{x}-e^{-x}))$

    So, what can you say?
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  3. #3
    Senior Member Stroodle's Avatar
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    Re: Hyperbolic trig identities?

    Awesome. Thanks for that.
    I think I'm meant to use the hyperbolic trig identities though (I might be wrong, but it's in that section of my text).
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  4. #4
    MHF Contributor Also sprach Zarathustra's Avatar
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    Re: Hyperbolic trig identities?

    Quote Originally Posted by Stroodle View Post
    Awesome. Thanks for that.
    I think I'm meant to use the hyperbolic trig identities though (I might be wrong, but it's in that section of my text).


    By definition:

    $\displaystyle \sinh{x}=\frac{e^x-e^{-x}}{2}$

    $\displaystyle \cosh{x}=\frac{e^x+e^{-x}}{2}$

    Now, with the above you can prove:

    1) $\displaystyle \cosh^2{x}-\sinh^2{x}=1$

    2) $\displaystyle \sinh{2x}=2\sinh{x}\cosh{x}$

    3) $\displaystyle 1-\tanh^2{x}=\frac{1}{\cosh^2{x}}$

    4) $\displaystyle \cosh{x+y}=\cosh{x}\cosh{x}+\sinh{x}\sinh{y}$

    5) $\displaystyle \sinh{x+y}=\sinh{x+y=\sinh{x}\cosh{y}+\cosh{x}\sin h{y}$

    6) $\displaystyle \sinh{x}+\sin{y}=2\sinh{(\frac{x+y}{2})}\cosh (\frac{x-y}{2})$



    AND MORE...

    Try to prove 1-6, try to find more identities...
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