Describing a point trigonometrically 
Let us show that the transformation that reflects a point through a line through
  the origin is linear. This is the transformation that takes a point on one side
  of the line and moves it perpendicular to the line, crosses it, and continues the
  same distance away from the line.
  We will first assume that the transformation $T$ is linear, and thus given by a
  matrix whose $i$th column is $T(\vec e_i)$. Again, all we have to do is figure out what
  the $T$ does to $\vec e_1$ and $\vec e_2$. We can then apply that transformation to any point we like, by multiplying it by the matrix. There's no need to do an elaborate
  computation for each point.
  To obtain the first column of our matrix we thus consider where $\vec e_1$ is mapped
  to. Suppose that our line makes an angle $\theta$ with the $x$-axis. Then $\vec e_1$ is mapped to $\begin{bmatrix}\cos2\theta\\ \sin2\theta\end{bmatrix}.$

I am struggling with seeing how they came up with the mapping in the last line of the quote above. Going by the description of that mapping, I think they treat $\vec e_1$ as a point and mark another one on the other side of the line, the same distance away. So that other point is described as $\begin{bmatrix} \cos2\theta\\ \sin2\theta\end{bmatrix}.$ I forgot a lot of trig, so I am not sure where that description comes from. Please, explain.
 A: Your visualization of how $\vec e_1$ is transformed geometrically is correct.
So you now have the original vector, $\vec e_1$, which we treat as a point on the $x$-axis,
you have the line of the reflection, which makes an angle $\theta$ with
the $x$-axis, and we have $T(\vec e_1),$ which is the reflected image of  $\vec e_1$.
We also have the point on the line of reflection through which $\vec e_1$
was reflected. That point is the right angle of a right triangle whose other vertices
are $\vec e_1$ and the origin of the plane;
and it is also at the right angle of a congruent right triangle whose other vertices
are $T(\vec e_1)$ and the origin of the plane.
It follows that the line from the origin through $T(\vec e_1)$ makes an angle
$\theta$ with the line of reflection, so it makes an angle $2\theta$
with the $x$-axis.
The distance from the origin to $T(\vec e_1)$ is $1$, so $T(\vec e_1)$
is a point on the unit circle at an angle of $2\theta$ from the $x$-axis.
Some people actually define $\sin(2\theta)$ and
$\cos(2\theta)$ in terms of the $x$- and $y$-coordinates of that point;
in any case it should not be hard to see that those are indeed that point's coordinates.
A: $\vec e_1, \vec e_2$ are the $2$ trivial vectors, which together span $\mathbb R^2$. So basically:
$$
\vec e_1 = \begin{bmatrix}1\\0\end{bmatrix}, \vec e_2 = \begin{bmatrix}0\\1\end{bmatrix}
$$
$T(\vec e_1)$ describes what happens to $\vec e_1$ when it is transformed by $T$.
You know that $T$ takes points and reflects them with a given line with angle $\theta$. So the point $(1, 0)$ in the $xy$ plane represented by $\vec e_1$ should be transformed by $T$ to the point with angle $2\theta$ with distance $1$ from the origin, or $(cos2\theta,  sin2\theta)$. 
This gives you the first column of the matrix representing $T$. Try to think what $T$ does to the point $(0, 1)$.
