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