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Formula product of sines

  1. Jan 14, 2012 #1
    1. The problem statement, all variables and given/known data

    Prove the following equation

    [tex]2^{n-1}\prod_{k=0}^{n-1}\sin(x+\frac{k\pi}{n}) = \sin(nx)[/tex]


    2. Relevant equations

    3. The attempt at a solution

    Below you find my unsuccessfull attempt using complex numbers.

    When you convert it to complex numbers the equality can be rewritten as

    [tex]2^{n-1}\prod_{k=0}^{n-1}\frac{e^{i(x+\frac{k\pi}{n})}-e^{-i(x+\frac{k\pi}{n})}}{2i} = \frac{e^{inx}-e^{-inx}}{2i}[/tex]
    [tex]i^{-n+1}\prod_{k=0}^{n-1}e^{i(x+\frac{k\pi}{n})}-e^{-i(x+\frac{k\pi}{n})} = e^{inx}-e^{-inx}[/tex]
    [tex]e^{i\frac{(-n+1)\pi}{2}}\prod_{k=0}^{n-1}e^{i(x+\frac{k\pi}{n})}-e^{-i(x+\frac{k\pi}{n})} = e^{inx}-e^{-inx}[/tex]

    The LHS of this equation contains terms in [tex]e^{inx}[/tex], [tex]e^{i(n-1)x}[/tex], [tex]e^{i(n-2)x}[/tex], ..., [tex]e^{-inx}[/tex].
    Calculate the coefficient for each term.

    [tex]\underline{e^{inx}}[/tex]

    [tex]e^{i\frac{(-n+1)\pi}{2}}\frac{(-n+1)\pi}{2}\prod_{k=0}^{n-1}e^{i(x+\frac{k\pi}{n})}=e^{i\frac{(-n+1)\pi}{2}}e^{inx+i\sum_{k=0}^{n-1}\frac{k\pi}{n}}=e^{inx}[/tex]


    [tex]\underline{e^{-inx}}[/tex]

    [tex]e^{i\frac{(-n+1)\pi}{2}}\prod_{k=0}^{n-1}-e^{-i(x+\frac{k\pi}{n})}=(-1)^ne^{i(-n+1)\pi}e^{-inx}=(-1)^{n}e^{-inx} = -e^{-inx}[/tex]

    Now we still have to prove that for -n<k<n the coefficient of [tex]{e^{ikx}}[/tex] equals 0 to conclude the proof. But I don't know how to do this.

    All suggestions, ideas are welcome. Proofs not using complex numbers will be appreciated as well.
     
    Last edited: Jan 14, 2012
  2. jcsd
  3. Jan 15, 2012 #2
    Taylor expansion of Sin(x+kpi/n) about the point x=kpi/n ?
     
  4. Jan 15, 2012 #3

    Dick

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    Science Advisor
    Homework Helper

    k is an index of the product. Of course it's not random. It goes from 0 to n-1.
     
  5. Jan 15, 2012 #4
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