What is cos(npi/2) and sin(npi/2) in fourier series cos(npi) is (-1)^n and sin(npi)=0 ?

The problem ask to fourier series of the trigonometric function base on the data you have been given in the problem, so the answer would be cos(pi/2n) = {(-1)^n/2 - if n is even, 0 - if n is odd} and sin(pi/2n) = {0- if n is even, (-1)^(n-1)/2 if n is odd}. I hope you are satisfied with my answer