1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: 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]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


    User Avatar
    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
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook