# 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?

##### 1 Answer

#### Answer:

Please see below.

#### Explanation:

As

According to De-Moivre's theorem

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

and

i.e.

and