Post History

Reflection in the plane with polar coordinates

My question comes from the academic paper: [Symmetry of solutions to semilinear elliptic equations via Morse Index.](https://www.ams.org/journals/proc/2007-135-06/S0002-9939-07-08652-2/S0002-9939-07-08652-2.pdf) The author is Filomena Pacella. 

Let $S$ be the vector of the $n$-dimensional unit sphere in $\mathbb{R}^N$, $S=\{x \in \mathbb{R}^N : |x|=1\}$. For a unit vector $e \in S$, we consider the hyperplane $H(e)=\{x\in\mathbb{R}^N : x\cdot e=0 \}$ and the open half-domain of the $n$-sphere $B(e)=\{ x\in B\ : x\cdot e > 0 \}$. We define $\sigma_{e} : B \rightarrow B$ as the reflection with respect to $H(e)$, that is, $\sigma_{e}(x)=x-2(x\cdot e)e$  for each $x \in B$.
Since $\sigma_{e}(x)$ is a reflection, the following properties hold
\begin{equation*}
				H(-e)=H(e) \quad \text{y} \quad B(-e)=\sigma_{e}(B(e))=-B(e) \quad \text{para cada} \; e\in S
		\end{equation*}
A new vector $e^{\prime}=(\cos(\theta_{0}), \sin(\theta_{0}), 0, \ldots, 0)$ is defined for some $\theta_{0} \in \left(- \frac{\pi}{2}, \frac{\pi}{2}\right)$. Since only the first two coordinates are modified, the remaining coordinates are grouped into $\tilde{x}=(x_3, \ldots, x_N)$. We convert to cartesian coordinates to polar coordinates for the first two, $x_1= r \cos(\theta)$, $x_2= r \sin(\theta)$.
We use the reflection with vector $e^{\prime}$ over the hyperplane $H(e^{\prime})$, so we name it $\sigma_{e^{\prime}}(x)$. The coordinates of the image after applying the reflection are
		\begin{equation*}
			\sigma_{e^{\prime}}(rcos(\theta),rsen(\theta),\tilde{x})=(rcos(2\theta_{0}-\theta + \pi),rsen(2\theta_0-\theta + \pi),\tilde{x})
		\end{equation*}
Do you know how it arrived at the expression $2\theta_{0} - \theta + \pi$? Is any trigonometric property being used here?
One more thing, could someone share a link with a diagram explaining this application? 
Linear transformations that are reflections are usually defined with matrices and those that are not defined this way are not similar to this one.