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Simplifying a product of sin functions

  1. Jul 15, 2007 #1
    can someone please simplify?

    [tex]\sin \frac{\pi}{n} \sin \frac{2\pi}{n} ... \sin \frac{(n-1)\pi}{n}[/tex]
     
  2. jcsd
  3. Jul 15, 2007 #2
    distribute the pi on (n-1) and simplify the fraction you'll get. and use a trig identity to separate the new fraction
     
  4. Jul 15, 2007 #3
    Complex Numbers are the key here.

    Consider the polynomial,
    [tex]\Phi(z) = 1+z+z^2+...+z^{n-1} = (z - \zeta)(z-\zeta^2)...(z-\zeta^{n-1})[/tex] where [tex]\zeta = \cos \frac{2\pi }{n} + i\sin \frac{2\pi }{n}[/tex].

    Then,
    [tex]\Phi(1) = \overbrace{1+1+...+1}^n = \prod_{k=1}^{n-1} \left( 1 - \zeta^k \right)[/tex]

    [tex]n = \prod_{k=1}^{n-1} \left( 1 - \cos \frac{2\pi k}{n} - i \sin \frac{2\pi k}{n} \right) [/tex].

    [tex]|n| = \prod_{k=1}^{n-1} \left| 1 - \cos \frac{2\pi k}{n} - i \sin \frac{2\pi k}{n} \right| [/tex]

    Now,
    [tex]\left| 1 - \cos \frac{2\pi k}{n} - i \sin \frac{2\pi k}{n} \right| =\sqrt{ \left(1 - \cos \frac{2\pi k}{n} \right)^2 + \sin^2 \frac{2\pi k}{n} }= \sqrt{1 -2\cos \frac{2\pi k}{n}+ \cos^2\frac{2\pi k}{n} + \sin^2 \frac{2\pi k}{n}}[/tex]
    [tex] = \sqrt{2\left( 1 - \cos \frac{2\pi k}{n} \right)} = \sqrt{4\sin^2 \frac{\pi k}{n}} = 2\sin \frac{\pi k}{n}[/tex]

    Thus,
    [tex]|n|=n = \prod_{k=1}^{n-1} 2\sin \frac{\pi k}{n}
    [/tex]

    Thus,
    [tex] n = 2^{n-1} \sin \frac{\pi }{n} \cdot \sin \frac{2\pi}{n} \cdot ... \cdot \sin \frac{\pi (n-1)}{n}[/tex]

    That means,
    [tex]\sin \frac{\pi }{n} \cdot \sin \frac{2\pi}{n} \cdot ... \cdot \sin \frac{ (n-1)\pi}{n} = \frac{n}{2^{n-1}}[/tex]
     
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