Why $\sin(n\pi) = 0$ and $\cos(n\pi)=(-1)^n$? I am working out a Fourier Series problem and I saw that the suggested solution used
$\sin(n\pi) = 0$ and $\cos(n\pi)=(-1)^n$ to simply the expressions while finding the Fourier Coefficients $a_0$, $a_n$, $b_n$. 

I am aware that the $\sin(x)$ has a period of $2\pi$. So I am thinking that every half of period, the graph of $\sin(x)$ has to cut through the $x$ axis thus giving us the value $0$. Am I right to think that way or is there some more important reason for that?
Also, how do they come up with $\cos(n\pi) = (-1)^n$?

Since:
$sin(x) = \frac{exp(ix) - exp(-ix)}{2i} $
for x = n$\pi$:
$ =>  sin(n\pi) = \frac{exp(in\pi) - exp(-in\pi)}{2i}$ 
You can think of the positive exponential term as rotating counterclockwise, and the negative exponential term rotating clockwise. Since both exponential terms are of equal value (i.e. 1), but only moving in opposite directions, they cancel out at all points n
$cos(x) = \frac{exp(ix) + exp(-ix)}{2} $, the physical intuition follows from above
