Simplifying a product of sin functions

[barbiemathgurl]

can someone please simplify?

$$\sin \frac{\pi}{n} \sin \frac{2\pi}{n} ... \sin \frac{(n-1)\pi}{n}$$

 

[bob1182006]

distribute the pi on (n-1) and simplify the fraction you'll get. and use a trig identity to separate the new fraction

 

[Kummer]

can someone please simplify?

$$\sin \frac{\pi}{n} \sin \frac{2\pi}{n} ... \sin \frac{(n-1)\pi}{n}$$

Complex Numbers are the key here.

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

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

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

$$|n| = \prod_{k=1}^{n-1} \left| 1 - \cos \frac{2\pi k}{n} - i \sin \frac{2\pi k}{n} \right|$$

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

Thus,
$$|n|=n = \prod_{k=1}^{n-1} 2\sin \frac{\pi k}{n}$$

Thus,
$$n = 2^{n-1} \sin \frac{\pi }{n} \cdot \sin \frac{2\pi}{n} \cdot ... \cdot \sin \frac{\pi (n-1)}{n}$$

That means,
$$\sin \frac{\pi }{n} \cdot \sin \frac{2\pi}{n} \cdot ... \cdot \sin \frac{ (n-1)\pi}{n} = \frac{n}{2^{n-1}}$$