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Is the product of two sine functions always real-valued?

  1. Sep 14, 2009 #1
    Hi, I was playing around with Euler's Identity, and I found something (or at least I think I found something) interesting:

    It is a well known identity
    sin(z) = [exp(iz) - exp(-iz)]/(2*i), where z is any complex number, exp is the complex exponential function, and i is the imaginary constant.

    So if we bring the 2*i upstairs, we get

    2*i*sin(z) = exp(iz) - exp(-iz) (2)

    From (2), we can see that

    -4*sin(z)*sin(w) = [exp(iz) - exp(-iz)]*[exp(iw) - exp(-iw) ]

    = exp[i(z+w)] +exp[-i(z+w)] - (exp[i(z-w)] + exp[-i(z-w)] )
    Since exp[-i(z+w)] is the complex conjugate of exp[i(z+w)], and exp[-i(z-w)] is the complex conjugate of exp[i(z-w)], we have

    exp[i(z+w)] +exp[-i(z+w)] - (exp[i(z-w)] + exp[-i(z-w)] )

    =2RealPart{exp[i(z+w)]} - 2RealPart{exp[i(z-w)]}, which is real for every complex w and z.

    Is this really true? Am I making some mistake here?


    Edwin G. Schasteen
  2. jcsd
  3. Sep 14, 2009 #2


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    eix and e-ix are complex conjugates when x is real, but not often otherwise.
  4. Sep 15, 2009 #3
    Hi Hurkyl,

    Thank you! That makes a lot more sense now.

    Best Regards,

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