If  is a rotation and  is a reflection, then  is a reflection. 
Consider a dihedral group $D_n$. If  is a rotation and  is a
  reflection, then  is a reflection.

I'm new to algebra, so I tried to check for a specific $n$ to get an intuitive way to understand why this works, but how do you prove this for an arbitrary $n$?
 A: Reflections have order $2$ while rotations have order divisors of $n$, with the only rotation having order $2$ being $r^{\frac{n}{2}}$ if $n$ is even. The order of $rs$ is two because $(rs)^2=r(sr)s$ and you can use $rs=sr^{-1}$. If $n$ is odd, $rs$ must be a reflection. If $n=2k$, then if $rs$ is not a reflection then $rs=r^k$, thus $s=r^{k-1}$, thus $e=s^2=r^{n-2}=r^{-2}$, i.e $r$ has order $2$, which is a contradiction.
Edit: From your comment, we have to go back to the "nature" of $r$ and $s$ similarly to what Melody did in his answer.
Let us consider an $n$-gone in the plane, number his vertices from $1$ to $n$ in a counterclockwise manner. $r$ consists of the rotation that sends vertex $1$ to $2$, vertex $2$ to $3$,..., vertex $n-1$ to $n$, and vertex $n$ to $1$. $s$ consists of the reflection that fixes $1$ and $(n/2)+1$ if $n$ is even, and permutes $2$ and $n$, $3$ and $n-1$, $4$ and $n-2$, and so on... Now, if $i\in\{1,\dots,n\}$, you'll find by calculation that $(rs)^2(i)=r(s(r(s(i))))=i$. One concludes from this that $(rs)^2=e$ using the fundamental theorem of affine geometry.
A: Well, $D_n$ consists of precisely $n$ rotations and $n$ reflections, with identity considered a rotation by $0$ radians. All rotations in $D_n$ are order preserving, while all reflections are order reversing. The product of an order reversing map and an order preserving map is order reversing, hence $rs$ is order reversing, and therefore a reflection.
To state this in terms of matrices, any rotation can be represented by a matrix of the form
$$\begin{bmatrix}\cos\theta&-\sin\theta\\\sin\theta&\cos\theta\end{bmatrix}.$$
Which has a determinate of $1$. By choosing a proper basis WLOG we may assume that $s$ is represented by 
$$\begin{bmatrix}1&0\\0&-1\end{bmatrix},$$
which has determinate $-1.$ Then $\det(rs)=(1)\cdot(-1)=-1,$ so $rs$ is not a rotation, and hence is a reflection.
