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Deriving trig identity

  1. Sep 17, 2011 #1

    zcd

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    How would I go about showing [tex]\prod_{k=1}^{N-1} \sin{\frac{k\pi}{N}} = \frac{N}{2^{N-1}}[/tex]

    I've tried using Euler's equation to substitute sin, but it just gets messy.
     
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  3. Sep 17, 2011 #2

    micromass

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    Do start with

    [tex]\sin(x)=\frac{e^{ix}-e^{-ix}}{2i}[/tex]

    and try to get things into the following shape

    [tex]\prod_{k=1}^{n-1}{1-\xi^k}[/tex]

    where [itex]\xi[/itex] is the n-th root of unity. Then you can use the identity

    [tex]z^n-1=\prod_{k=1}^n{z-\xi^k}[/tex]

    somehow.
     
  4. Sep 17, 2011 #3

    epenguin

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    "Be a country boy!"

    What is sinΠ, what is sin2Π, what is sin3Π,... ?

    Are you sure this is the question?

    :surprised
     
    Last edited: Sep 17, 2011
  5. Sep 17, 2011 #4

    HallsofIvy

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    Well, the question you ask is not the question posed in the original post because you have ignored the "N" in the denominator.
     
  6. Sep 17, 2011 #5

    epenguin

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    Ah you're right. :redface:
     
  7. Sep 17, 2011 #6

    zcd

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    I substituted it in to get
    [tex]\prod_{k=1}^{N-1} \frac{e^{i\frac{k\pi}{N}}-e^{-i\frac{k\pi}{N}}}{2i}[/tex]
    after a bit of manipulation I arrived at
    [tex]\prod_{k=1}^{N-1} \frac{-1}{2i e^{i\frac{k\pi}{N}}}\prod_{k=1}^{N-1} 1-(e^{i\frac{\pi}{N}})^k[/tex]
    which remotely resembles the form. From here, how would I apply the identity and how do I get rid of that other product?
     
  8. Sep 17, 2011 #7

    micromass

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    To get rid of the other product is easy. Just apply your exponential laws: [itex]e^xe^y=e^{x+y}[/itex].

    To apply the other identity, you have that

    [tex]\prod_{k=1}^{N-1}{z-(e^{i\pi /N})^k}=\frac{z^k-1}{z-1}=...[/tex]

    Our ultimate goal is to put z=1 in the above equality.
     
  9. Sep 17, 2011 #8

    zcd

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    I managed to change
    [tex]\prod_{k=1}^{N-1} \frac{-1}{2i e^{i\frac{k\pi}{N}}}=\frac{1}{2^{N-1}}\prod_{k=1}^{N-1}\frac{i}{\exp(i\frac{k\pi}{N})}=\frac{1}{2^{N-1}}(\frac{i}{i})^{N-1}=\frac{1}{2^{N-1}}[/tex]
    but I'm still trying to see how the roots of unity identity can be applied. Also, it seems that if
    [tex]z^N-1=\prod_{k=1}^N{z-(e^{i\pi /N})^k}[/tex] then
    [tex]z^N-1=({z-(e^{i\pi /N})^N})\prod_{k=1}^{N-1}{z-(e^{i\pi /N})^k}[/tex] so
    [tex]\prod_{k=1}^{N-1}{z-(e^{i\pi /N})^k}=\frac{z^N-1}{z-(-1)}[/tex]
     
  10. Sep 17, 2011 #9

    micromass

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    I missed a 2, there. It has to be

    [tex]z^N-1=\prod_{k=1}^N{z-(e^{2i\pi /N})^k}[/tex]


    I think you're missing a 2 as well in your form. So it would have to be

    [tex]\prod_{k=1}^{N-1}{z-(e^{2i\pi /N})^k}=\frac{z^N-1}{z-1}[/tex]

    Now, try to use the identity

    [tex]z^N-1=(z-1)(z^{N-1}+z^{N-2}+...+z+1)[/tex]
     
  11. Sep 17, 2011 #10

    zcd

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    I took [tex]z=1+\epsilon[/tex] and then took [tex]\lim_{\epsilon\to 0} \frac{(1+\epsilon)^N-1}{(1+\epsilon)-1}=\frac{d}{dx}(x^N)=N[/tex]for x=1
     
  12. Sep 17, 2011 #11

    zcd

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    or I could do
    [tex]\prod_{k=1}^{N-1}{z-(e^{2i\pi /N})^k}=\frac{z^N-1}{z-1}=\frac{(z-1)(z^{N-1}+z^{N-2}+...+z+1)}{z-1}=z^{N-1}+z^{N-2}+...+z+1[/tex] which sums up to N
     
  13. Sep 17, 2011 #12

    micromass

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    Indeed, it seems you've got it!!
     
  14. Sep 17, 2011 #13

    zcd

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    Thanks for the help :)
     
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