I have a problem solving the following exercise.

i have to prove that $\displaystyle \sum_{j=0}^{n-1} cos (2j \pi /{n}})^2$ = $\displaystyle \sum_{j=0}^{n-1} sin (2j \pi /{n}})^2$ = 1/2

I have to prove that each sum is equal to 1/2 !!

(n is odd and I know that both sums together are 1, but I cannot prove why each is 1/2)

Help is appreciated!