Cohomology of $S^1 \rtimes \mathbb{Z}/2$ I am trying to compute the mod2 cohomology of the semi direct product $G = S^1 \rtimes \mathbb{Z}/2$ using the extension
$$ 1\rightarrow  S^1   \rightarrow G \rightarrow\mathbb{Z}/2\rightarrow 1$$
and the HLS spectral sequence associated to it. 
We get then that 
$$E_2 = H^*(S^1) \otimes H^*(\mathbb{Z}/2) = \mathbb{F}_2[c,w]$$ where $|c|=2$ and $|w|=1$ are the respective generators.
the second differential depends on the value $d_2(w) = \alpha c$ where $\alpha \in \{0,1\}$.
I am stuck here and I do not know how to compute the value of $\alpha$. I know that if $\alpha = 0$ then  $H^*(G) = E_\infty = E_2$ (which is kind of strange cause this is the cohomology of the direct product though).
I really appreciate any help/comment.
 A: Thanks to the comments and edits from the people involved in the OP I got the following; if there is still any mistake please don't hesitate commenting.
To compute the cohomology of $G$, I will use the Lyndon-Hochschild-Serre  Spectral Sequence associated to the extension
$$1 \rightarrow S^1 \rightarrow G \rightarrow \mathbb{Z}/2  \rightarrow 1$$
where the $E_2$ page of such spectral sequence is given by 
$$E_2^{p,q} \cong H^p(\mathbb{Z}/2; H^q(S^1))$$
where $\mathbb{Z}/2$ acts on $H^q(S^1)$ trivially. (In general the action of $\mathbb{Z}/2 = \{\pm 1\}$ over $H^q(S^1;\mathbb{Z})$ is given by $\pm 1 \cdot c = \pm c$)**.
This implies that the $E_2$ page decomposes into the first quadrant spectral sequence
$$E_2 \cong H^*(\mathbb{Z}/2) \otimes H^*(S^1) \cong \mathbb{F}_2[w,c],$$
and the differential $d_2$ depends only on the values over $w$ and $c$ respectively since it is a derivation; indeed, $d_2(w) = 0$ since $w$ lies on the $x$-axis of the spectral sequence, and $d_2(c) = 0$ since 
$d_2(c) \in E^{2,1}_2 = H^2(\mathbb{Z}/2)\otimes H^1(S^1) = 0$.
Therefore, $d_2 = 0$ implying that the spectral sequence degenerates at page 2 and thus $E_2 \cong E_\infty \cong H^*(G)$.
**Thanks to @Hari Rau-Murthy for the clarification
