Some Subgroup of Dihedral Group is Normal

If $D_n=\langle r,s\mid r^n=s^2=1,srs=r^{-1}\rangle$, then $D_n$ has order $2n$ and the group generated by $r$ has order $n$.

Therefore the index of $\langle r\rangle$ in $D_n$ is equal to two, and it is a general fact that if $H\leq G$ is a subgroup with $[G:H]=2$ then $H$ is a normal subgroup of $G$.

The index $2$ suggestion works, but you can also show this directly. One can check that the generators $R$ and $F$ of the dihedral group conform to the rule $RF = FR^{-1}$. From this, we see that any element in $D_n$ can be written as $R^jF^k$ where $0 \leq j \leq n-1$ and $0 \leq k \leq 1$.

A subgroup $N \leq G$ is normal whenever, given any $n \in N$ and $g \in G$, we have $gng^{-1} \in N$. In this case, any element of the rotation subgroup looks like $R^m$ for $1 \leq m \leq n-1$. Considering any arbitrary element $R^jF^k$ of $D_n$, we just need to show that $(R^jF^k)R^m(R^jF^k)^{-1} \in \langle R \rangle$. Clearly this is true if $k=0$, so assume $k=1$. Now look to the helpful rule in the first paragraph to conclude that this is indeed an element of $\langle R \rangle$.

How many rotations are there, and how does this compare to the total number of elements? You may have shown as an exercise previously that a subgroup of index $2$ is normal. That is relevant here. If not, you should prove it, because using this fact is the easiest way I see to solve your problem.