# By considering the roots of z^5 =1, how do you show that cos(2π/5) + cos(4π/5) + cos(6π/5) + cos(8π/5) = -1?

Feb 25, 2017

#### Explanation:

As ${z}^{5} = 1 = \left(\cos 0 + i \sin 0\right)$

According to De-Moivre's theorem

$z = {\left({z}^{5}\right)}^{\frac{1}{5}} = \left(\cos \left(\frac{2 n \pi}{5}\right) + i \sin \left(\frac{2 n \pi}{5}\right)\right)$ and we get five roots of $1$ by changing value of $n$ from $0$ to $4$ and we get five roots as

$\cos 0 + i \sin 0$or $1 + i 0$, $\cos \left(\frac{2 \pi}{5}\right) + \sin \left(\frac{2 \pi}{5}\right)$, $\cos \left(\frac{4 \pi}{5}\right) + \sin \left(\frac{4 \pi}{5}\right)$, $\cos \left(\frac{6 \pi}{5}\right) + \sin \left(\frac{6 \pi}{5}\right)$ and $\cos \left(\frac{8 \pi}{5}\right) + \sin \left(\frac{8 \pi}{5}\right)$

and these five roots, when graphed on the complex plane, are equally spaced around a circle of unit radius as shown below.

and it is apparent that their sum will be zero.

As such comparing real and imaginary parts we should have

$1 + \cos \left(\frac{2 \pi}{5}\right) + \cos \left(\frac{4 \pi}{5}\right) + \cos \left(\frac{6 \pi}{5}\right) + \cos \left(\frac{8 \pi}{5}\right) = 0$

and $0 + \sin \left(\frac{2 \pi}{5}\right) + \sin \left(\frac{4 \pi}{5}\right) + \sin \left(\frac{6 \pi}{5}\right) + \sin \left(\frac{8 \pi}{5}\right) = 0$

i.e. $\cos \left(\frac{2 \pi}{5}\right) + \cos \left(\frac{4 \pi}{5}\right) + \cos \left(\frac{6 \pi}{5}\right) + \cos \left(\frac{8 \pi}{5}\right) = - 1$

and $\sin \left(\frac{2 \pi}{5}\right) + \sin \left(\frac{4 \pi}{5}\right) + \sin \left(\frac{6 \pi}{5}\right) + \sin \left(\frac{8 \pi}{5}\right) = 0$