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 $\cos(n\pi) = (-1)^n$?

If you go through the unit circle process, then you see how they are equivalent. But if you want a proof then:
You may observe the fact that $\sin(0)=0$.
Then you may prove that $\sin(x+\pi)=\sin(x)$ and $\sin(x-\pi)=\sin(x)$ using angle addition formulas. Having done this you prove:
$$\sin(0 \pm \pi n)=0,  n \in  \mathbb{N}$$
Now for cosine you use similar logic. observe  $\cos(0)=1$ and $\cos(x \pm \pi)=-\cos x$...
