Prove that the Dihedral group $D_n$ is isomorphic to $Z_n \rtimes_{\psi} Z_2$ I consider the following map $\psi : Z_2 \rightarrow Aut(Z_n)$ where we map 
the identity element 0 to the identity map and $1 \mapsto \theta : Z_n \rightarrow Z_n$ where $\theta(x) = -x$. I am not sure how to proceed further. Any help would be nice.
$D_n = ⟨ r, s : r^n = e = s^2, r^js = sr^{-j} ⟩ $

Perfectly good answer was already given by Galena Rupp, but I will try to give intuition why the given isomorphism really is a natural choice.
First of all, let us start with general semidirect product $N\rtimes_\psi H$. Instead of ordered pair $(n,h)$, I will simply write $nh$ (where concatenation is multiplication in $N\rtimes_\psi H$). So, how do we multiply $(n_1h_1)(n_2h_2)$? Well, $(n_1h_1)(n_2h_2) = n_1(h_1n_2h_1^{-1})h_1h_2 = (n_1\psi_{h_1}(n_2))(h_1h_2)$ which is precisely how we define multiplication in semidirect product. This tells us that $\psi$ is encoding how $H$ acts by conjugation on $N$ in the semidirect product - with this information we can completely determine the structure of semidirect product.
Let us get back at $\Bbb Z_n\rtimes_\psi \Bbb Z_2$. First I will switch to multiplicative notation $C_n\rtimes_\psi C_2$, denote the generator of $C_n$ with $r$ and the generator of $C_2$ with $s$, and appropriately change $\psi_s$ to $r^j\mapsto r^{-j}$. Now, if I want to give presentation of $C_n\rtimes_\psi C_2$, I note that it is generated by $r,s$ subject to conditions $r^n = e$, $s^2 = e$ and $\psi_s(r^j) = r^{-j}$, i.e. $$C_n\rtimes_\psi C_2 = \langle r,s\,|\, r^n = s^2 = e, \psi_s(r^j) = r^{-j}\rangle$$
Now, remember that $\psi_s$ really is a conjugation, so rewrite the above as $$C_n\rtimes_\psi C_2 = \langle r,s\,|\, r^n = s^2 = e, sr^js^{-1} = r^{-j}\rangle = \langle r,s\,|\, r^n = s^2 = e, r^js = sr^{-j}\rangle$$ which is precisely the presentation of $D_n$.
So, what is the isomorphism $\Bbb Z_n\rtimes_\psi \Bbb Z_2\cong D_n$? Well, from the above, we just have to rename generators to $r$ and $s$ and switch to multiplicative notation, i.e. the isomorphism is given by $(i,j)\mapsto r^is^j$ as noted in the answer by Galena Rupp.
