Geometry troubles Prove that a composition of a rotation and a translation
is a rotation. (This is true for any order of the rotation/translation; for
concreteness, you can assume that the rotation is performed first.)
I know that it could be considered a bunch of reflections.  That a translation is just a composition of two reflections over parallel lines.  A rotation is a composition of two reflections over lines who cross.  So reflection over ||, then |/ should be the same as just a reflection over |/
 A: A rotation is two reflections, with mirrors intersecting along the axis of rotation.  You can pick one mirror to be at any angle, and the other must go at (angle)/2 from the first.
A translation is also two reflections, with parallel mirrors perpendicular to the axis of motion.  You can put the first mirror anywhere you want along that axis, even very far away, but the second mirror must go at (distance)/2 from the first.
Can you move one mirror of the rotation and one mirror of the translation so that they coincide?  Then the one undoes the effect of the other, cancelling out, leaving two mirrors total, quite likely at some angle and therefore equivalent to a rotation.
A: Consider a rotation of $\theta$ about the point $p$
$$
R_p^\theta(x)=\begin{bmatrix}
\cos(\theta)&-\sin(\theta)\\
\sin(\theta)&\cos(\theta)
\end{bmatrix}(x-p)+p
$$
and translation by $h$
$$
T_h(x)=x+h
$$

First
$$
R_p^\theta\circ T_h(x)=\begin{bmatrix}
\cos(\theta)&-\sin(\theta)\\
\sin(\theta)&\cos(\theta)
\end{bmatrix}(x+h-p)+p
$$
Let
$$
q=\begin{bmatrix}
1-\cos(\theta)&\sin(\theta)\\
-\sin(\theta)&1-\cos(\theta)
\end{bmatrix}^{-1}
\begin{bmatrix}
\cos(\theta)&-\sin(\theta)\\
\sin(\theta)&\cos(\theta)
\end{bmatrix}h
+p
$$
Then
$$
\begin{align}
R_q^\theta(x)
&=\begin{bmatrix}
\cos(\theta)&-\sin(\theta)\\
\sin(\theta)&\cos(\theta)
\end{bmatrix}(x-q)+q\\
&=\begin{bmatrix}
\cos(\theta)&-\sin(\theta)\\
\sin(\theta)&\cos(\theta)
\end{bmatrix}x
+\begin{bmatrix}
1-\cos(\theta)&\sin(\theta)\\
-\sin(\theta)&1-\cos(\theta)
\end{bmatrix}q\\
&=\begin{bmatrix}
\cos(\theta)&-\sin(\theta)\\
\sin(\theta)&\cos(\theta)
\end{bmatrix}(x+h)
+\begin{bmatrix}
1-\cos(\theta)&\sin(\theta)\\
-\sin(\theta)&1-\cos(\theta)
\end{bmatrix}p\\
&=\begin{bmatrix}
\cos(\theta)&-\sin(\theta)\\
\sin(\theta)&\cos(\theta)
\end{bmatrix}(x+h-p)
+p\\
&=R_p^\theta\circ T_h(x)
\end{align}
$$
To assure the invertibility of $\begin{bmatrix}
1-\cos(\theta)&\sin(\theta)\\
-\sin(\theta)&1-\cos(\theta)
\end{bmatrix}$, its determinant is $2(1-\cos(\theta))$, which is $0$ iff $\theta$ is a multiple of $2\pi$.

Next
$$
T_h\circ R_p^\theta(x)=\begin{bmatrix}
\cos(\theta)&-\sin(\theta)\\
\sin(\theta)&\cos(\theta)
\end{bmatrix}(x-p)+p+h
$$
Let
$$
r=\begin{bmatrix}
1-\cos(\theta)&\sin(\theta)\\
-\sin(\theta)&1-\cos(\theta)
\end{bmatrix}^{-1}
h
+p
$$
Then
$$
\begin{align}
R_r^\theta(x)
&=\begin{bmatrix}
\cos(\theta)&-\sin(\theta)\\
\sin(\theta)&\cos(\theta)
\end{bmatrix}(x-r)+r\\
&=\begin{bmatrix}
\cos(\theta)&-\sin(\theta)\\
\sin(\theta)&\cos(\theta)
\end{bmatrix}x+h
+\begin{bmatrix}
1-\cos(\theta)&\sin(\theta)\\
-\sin(\theta)&1-\cos(\theta)
\end{bmatrix}r\\
&=\begin{bmatrix}
\cos(\theta)&-\sin(\theta)\\
\sin(\theta)&\cos(\theta)
\end{bmatrix}(x-r)+r+h\\
&=T_h\circ R_p^\theta(x)
\end{align}
$$

Thus, as long as $\theta$ is not a multiple of $2\pi$, we have
$$
R_p^\theta\circ T_h=R_q^\theta\quad\text{and}\quad T_h\circ R_p^\theta=R_r^\theta
$$
