# 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

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But there isn't any $z$ in the equation you are trying to establish, so what is the relevance of $z^7+1=0$? –  Gerry Myerson Jun 19 '12 at 9:53
a better manner ask this can be :Let $z =e^{\frac{i \pi}{7}}$. Vérifie that $z^7+1=0$ and deduce teh trigonometric relation : ... –  Mohamed Jun 19 '12 at 10:12
@Mohamed: Thanks –  Babak S. Jun 19 '12 at 10:57

Hint: write the cosines in terms of $e^{\pi i/7}$ and look for a geometric sequence.
Can we have the inverse problem as above. I mean; does it make sense to have, for example an equality like $8$sin$(\frac{\pi}{7})$sin$(\frac{2\pi}{7})$sin$(\frac{3\pi}{7})$=$\sqrt{7}$ and search for an equation? –  Babak S. Jun 19 '12 at 10:15
Sorry, I don't know what you mean. But those sines can also be written in terms of $e^{\pi i/7}$. –  Gerry Myerson Jun 19 '12 at 12:59
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.