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Sinh in terms of cosh

  1. May 3, 2014 #1
    Is possible express the sine in terms of cosine and vice-versa:

    ##\sin(x) = \cos(x-\frac{\pi}{2})##

    ##\cos(x) = \sin(x+\frac{\pi}{2})##

    So, of some way, is possible express the sinh in terms of cosh too and vice-versa?
     
  2. jcsd
  3. May 3, 2014 #2

    lurflurf

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    sure

    ##
    \sinh(x)=-\imath \, \cosh(x+\pi\,\imath/2)
    \\
    \cosh(x)=-\imath \, \sinh(x+\pi\,\imath/2)
    ##
     
  4. May 4, 2014 #3
    No exist an identity that no involve imaginary unit?
     
  5. May 4, 2014 #4

    HallsofIvy

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    No, because that is the whole point of the hyperbolic functions: sinh(x)= i sin(ix) and cosh(x)= cos(ix).
     
  6. May 5, 2014 #5

    Curious3141

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    Of course, there's the 'obvious' one which you can find by rearranging ##\cosh^2 x - \sinh^2 x = 1##, but I don't think that's what you're going for, is it?
     
  7. May 6, 2014 #6
    no, but thanks
     
  8. May 25, 2014 #7

    lurflurf

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    Last edited by a moderator: May 6, 2017
  9. Jun 10, 2014 #8
    Considering the graphs of sinh(x) and cosh(x), they don't have the shift property that sin(x) and cos(x) have for x in set of real numbers.
     
  10. Jun 11, 2014 #9

    mathman

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    cosh2(x) - sinh2(x) = 1.
     
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