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Homework Help: Complex Numbers

  1. Feb 16, 2013 #1
    1. The problem statement, all variables and given/known data


    2. Relevant equations

    [tex]sin\frac{\pi }{7}.sin\frac{2\pi }{7}.sin\frac{3\pi }{7}[/tex]

    3. The attempt at a solution

    [tex]Using[/tex] [tex]z^{7}-1[/tex] [tex]got:[/tex]
    [tex]cos\frac{\pi }{7}.cos\frac{2\pi }{7}.cos\frac{3\pi }{7}=\frac{1}{8}[/tex]
    Last edited: Feb 16, 2013
  2. jcsd
  3. Feb 16, 2013 #2
    Seems correct!
  4. Feb 16, 2013 #3
    But my solution is for the cosine product. How can I find the solution to the question from what I got?
  5. Feb 16, 2013 #4
    Can't you repeat much of the same proof? Where does it fail?
  6. Feb 16, 2013 #5
    The way I got: [tex]cos\frac{\pi }{7}.cos\frac{2\pi }{7}.cos\frac{3\pi }{7}=\frac{1}{8}[/tex] is by solving the equation [tex]z^{7}-1=0[/tex]
    [tex]z=1,cis\pm\frac{2\pi }{7},cis\pm\frac{4\pi }{7},cis\pm\frac{6\pi }{7}[/tex]
    [tex]w+w^{-1}=cis\frac{2\pi }{7}+cis\frac{-2\pi }{7}=2cos\frac{2\pi }{7}[/tex]
    [tex]w^{2}+w^{-2}=-2cos\frac{3\pi }{7}[/tex]
    [tex]w^{3}+w^{-3}=-2cos\frac{\pi }{7}[/tex]
    , since [tex]w^{6}+w^{5}+w^{4}+w^{3}+w^{2}+w+1=0 \therefore cos\frac{\pi }{7}.cos\frac{2\pi }{7}.cos\frac{3\pi }{7}=\frac{1}{8}[/tex]
    So I don't think I can do the same to get the produce for sine.
  7. Feb 16, 2013 #6
    It might be easier to determine

    [tex]\prod_{k=0}^6 \sin(k\pi/7)[/tex]

    Do this by expressing the sine as complex exponentials. Also try to make use of the following formula

    [tex]\frac{z^7-1}{z-1}=\prod_{k=1}^6 (1- \zeta^k)[/tex]

    with [itex]\zeta = \cos(\pi/7) + i\sin(\pi/7)[/itex]. Let z go to 1.
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