What is $\cos(k \pi)$? I want to ask question for which I have been finding answer for.
Please could anyone explain me why $\cos(k \pi) = (-1)^k$ and also explain me same for $\sin(k \pi)$?

Let $k\in\mathbb Z$. Then
$$\cos(0)=1,~~\text{for}~~k=0,$$
$$\cos(\pi)=\cos(-\pi)=-1,~~\text{for}~~k=\pm 1,$$
$$\cos(2\pi)=\cos(-2\pi)=1,~~\text{for}~~k=\pm 2,$$
and so on, where the first equalities hold as $\cos(\cdot)$ is an even function.
Every time $k$ is even then we get $\cos(k\pi)=1$. When $k$ is odd, then $\cos(k\pi)=-1$. You can summarize these considerations into
$$\cos(k\pi)=(-1)^k. $$
Can you apply the same lines to $\sin(k\pi)$ now?
