I am having trouble understanding the following problem:

Assuming k is a whole number, evaluate the following:

a.) sin (π/2 + 2kπ)
b.) cos (π/2 + 2kπ)
c.) sin kπ
d.) cos kπ

The only instruction given in my book is:

"Since θ and θ + 2π correspond to the same point on the unit circle, we have sinθ = sin (θ + 2π). This means that the graphs repeat over and over ad infinitum. In fact, sinθ is periodic with period 2π. The same is true of cosθ. Remember that 2πk radians, where k is a whole number, is exactly k revolutions."

The answers given in the book:

a.) 1
b.) 0
c.) 0
d.) -1 if k is odd, and 1 if k is even.

Thanks!

a) sin (π/2 + 2kπ) = sin (π/2) = 1
b) cos (π/2 + 2kπ) = cos (π/2) = 0

If you're still confused about the above problems, simply draw the unit circle. You'll see how the values keep repeating over and over again. Because 2kπ basically means you're going in another round around circle.

c) sin kπ =0 (try different values for k and you'll see that the answer is always 0)

d) cos kπ= -1 if odd, 1 if even(again try different values for k and you'll see that the answer is -1 or 1)

explanation:
Go back to the unit circle. In the c) and d) you're basically going around the unit circle beginning with 0 and leaving π between every 2 angles (you're going in half circles). so sin(0) = 0 and sin(π)=0.....etc.
Cos(0)= 1, cos(π)=-1,...etc.