Thank you very much, I understand now.
cos vπ cos θ + sin vπ sin θ
Since sin n*pi is always zero, we can omit this.
when,
v=0 : cos(0π)cosθ + sin0 = cosθ
v=1 : cos(1π)cosθ + 0 = -1 cosθ
v=2 : cos(2π)cosθ + 0 = 1 cosθ
v=3 : cos(3π)cosθ + 0 = -1 cosθ
so in general we can write it as (-1)to...