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

On a unit circle
$x$ coordinate of any point on the circle is given by $\cos\theta$ and $y$ coordinate is given by $\sin\theta$
Now, $\sin(n\pi)$, where $n=0,1,2,3...$ is always the X-axis and on X-axis we have $y=0$
and $\cos(n\pi)$ assumes $x=1$ or $x=-1$ as
$\cos(0.\pi)=1=(-1)^0,\cos(1.\pi)=-1=(-1)^1,\cos(2.\pi)=1=(-1)^2...$
