Question

if cos(n*pi)=(-1)^n, sin(n*pi)=0
what is cos(n*pi/2)
sin(n*pi/2)
sin(n+1)*pi/2
cos(n+1)*pi/2
sin(n-1)*pi/2
cos(n-1)*pi/2
sin(n-1)*pi
cos(n-1)*pi

Answer 1

sin(n*pi)=0
sin(n+1)*pi/2=(-1)^n for n=odd >>0
cos(n+1)*pi/2=0
sin(n-1)*pi/2=(-1)^(n-1) for n=odd >>0
cos(n-1)*pi/2 =0
sin(n-1)*pi=0
cos(n-1)*pi=(-1)^(n-1)

Answer 2

cos(n*pi/2) = (i^n)[1+(-1)^n]/2

Answer 3

sin(n*pi/2) = (i^n-1)[1-(-1)^n]/2