An article about SOME TRIGONOMETRIC IDENTITIES // hg

 See the full article in GMB 7/2005, pages 289-293.

First identity :

(is formula (5) , page 290, in the cited article)

$$\prod_{k=0}^{n-1}\cos \frac{x+2k\pi}{n}=\frac{1}{2^{n-1}}\left [\cos\frac{n\pi}{2}+(-1)^{n+1}\cdot \cos x\right ] \tag{1}$$

           For  $x=\pi$  we obtain :

$\cos \frac{\pi}{n}\cdot \cos \frac{3\pi}{n} \dots  \cos\frac{(2n-1)\pi}{n}=\frac{1}{2^{n-1}}\left [\cos \frac{n\pi}{2}+(-1)^{n+1}\cdot \cos \pi \right]=\frac{1}{2^{n-1}}\left [\cos \frac{n\pi}{2}+(-1)^n\right ]$

i.e. Exercise 9a).

          For  $x=0$  we obtain :

$\cos \frac{2\pi}{n}\cdot \cos \frac{4\pi}{n}\dots \cos \frac{2(n-1)\pi}{n}=\frac{1}{2^{n-1}} \left [\cos \frac{n\pi}{2}+(-1)^{n+1}\cdot \cos 0\right ]=\frac{1}{2^{n-1}}\left [ \cos \frac{n\pi}{2}-(-1)^n\right ]$

i.e. Exercise 9b).

In particular we have the equalities :

$\cos \frac{\pi}{5}\cdot \cos \frac{3\pi}{5}\cdot \cos \frac{5 \pi}{5}\cdot \cos \frac{7\pi}{5}\cdot \cos \frac{9\pi}{5}=\frac{1}{2^4}\left [ \cos\frac{5\pi}{2}-1\right ]=-\frac{1}{16}\;;$

$\cos \frac{2\pi}{5}\cdot \cos \frac{4\pi}{5}\cdot \cos \frac{6\pi}{5}\cdot \cos\frac{8\pi}{5}=\frac{1}{2^4}\left [\cos \frac{5\pi}{2}+1\right ]=\frac{1}{16}.$

(in construction)