Complex numbers lead a trigonometric equality Times ago, I used to think about some trigonometric equalities. Now, I have faced a new one with different one:

Show that if $z^7+1=0$ then cos$(\frac{\pi}{7})$+cos$(\frac{3\pi}{7})$+cos$(\frac{5\pi}{7})=\frac{1}{2}$ wherein $z\in\mathbb C$.

Thanks

if $z^7+1=0$  then $$(z+1)(z^6-z^5+z^4-z^3+z^2-z+1)=0$$
Putting $z=e^{\frac{i \pi}{7}}$ we have $z + 1 \neq  0$  and  $z^7 +1 =0$, then :
$$z^6-z^5+z^4-z^3+z^2-z+1=0$$
This  gives :
$$\mathcal Re(z^6-z^5+z^4-z^3+z^2-z+1)=0$$
Since : $\cos \frac{\pi}{7} = - \cos \frac{6\pi}{7}$ and $\cos \frac{2\pi}{7} = - \cos \frac{5\pi}{7}$ abd $\cos \frac{3\pi}{7} = - \cos \frac{4\pi}{7}$  we  have  :
$$2\left(- \cos \frac{\pi}{7}- \cos \frac{3\pi}{7}- \cos \frac{5 \pi}{7} \right) +1 = 0$$ which gives the desired relation.
