Product of rotation and translation is a rotation I have a homework question that I'm not sure how to answer.
Given rotation R and translation T (neither of which are the identity), show that T(R) must be a rotation.
My guess is that we can draw a triangle, then rotate and translate it, and then find some sort of intersection by extending lines from the triangles that is the center of the overall rotation, but I don't think that works.
 A: I assume you are working on the Euclidean plane.  We can then choose coordinates so that $R$ is rotation by angle $\theta$ (in the counterclockwise direction; measured in radians) around the origin, and $T$ is translation vector $\vec v = (v_1,v_2)$.  A good representation for the plane is as the complex numbers.  Then $R$ is multiplication by $a = \exp(i\theta)$ and $T$ is addition by $v = v_1 + i v_2$.  Then $T\circ R$ is the function that takes $z$ to $v + az$.
For $a \neq 1$, this transformation has a fixed point $p$, which can be found by solving $p = v + ap$, from which we see that $p = \frac v{1-a}$.  It is well known that a rigid transformation which fixes a point is a rotation; in this case, it is not difficult to check that $T\circ R$ is rotation by angle $\theta$ around $p$.
When $a = 1$, so that $\theta$ is an integer multiple of $2\pi$, the composition $T\circ R$ is not a rotation, but a translation, and the problem as stated is false.
A: I encountered this question myself while reading about the wallpaper groups on this site in figure 2.  I have an answer but it seems I am a few years too late.  Regardless here it is.
Let $T_{\mathbf{v}}$ denote a translation by vector $\textbf{v}$ and $R(\phi)$ denote a rotation about the origin by angle $\phi$.
The idea is that a rotation about a point specified by vector $\mathbf{Q}$ can be accomplished by the sequence of transformations $T_{\mathbf{Q}}R(\theta)T_{-\mathbf{Q}}$.
Suppose we are given the rotation 
$R(\theta) = 
\begin{bmatrix}
\cos(\theta)&-\sin(\theta)\\
\sin(\theta)&\cos(\theta)
\end{bmatrix}$
along with the translation,
$ \vec{t}=
\begin{bmatrix}
t_x\\
t_y
\end{bmatrix}.$
We now wish to find $\alpha$ and $\mathbf{Q}$ such that,
$T_{\mathbf{t}}R(\theta)=T_{\mathbf{Q}}R(\alpha)T_{-\mathbf{Q}}.$
To make matters simpler lets make the clever guess that $\alpha=\theta$ if this works we are good.
Now let each side act on an arbitrary vector $\mathbf{r}$.  Writing this out in matrix form canceling terms which appear on both sides gives us the system of equations,
\begin{eqnarray}
t_x &=& Q_x (\cos(\theta)-1)-Q_y \sin(\theta)\\
t_y &=& Q_x \sin(\theta) + Q_y(\cos(\theta)-1).
\end{eqnarray}
The solutions to these equations are,
\begin{eqnarray}
Q_x &=& \frac{1}{2}t_x - \frac{1}{2}\cot\left(\frac{\theta}{2}\right)t_y \\
Q_y &=& \frac{1}{2}\cot\left(\frac{\theta}{2}\right) t_x + \frac{1}{2} t_y.
\end{eqnarray}
So yes, a rotation about some point $\mathbf{P}$ followed by a translation can be rewritten as a rotation about a different point $\mathbf{Q}$ by the same angle.
A: This only works in a plane. A rotation can be achieved by reflection in two lines intersecting in the point of rotation. The lines of reflection are otherwise arbitrary, as long as the angle between them is half the desired angle of reflection. The translation may be achieved by reflection in two parallel lines, each perpendicular to the translation, and separated by half the final translation. All of this can be arranged so two of these lines are coincdent, and the corresponding reflections cancel. The remaining two lines of reflection intersect, resulting in a rotation.
A: Let's suppose that $T(\vec x)=\vec x+\vec b$ for some $\vec b\ne\vec 0$ and that $R$ is the rotation of the plane by $\theta$ about the origin, where $\theta$ is not an integer multiple of $2\pi.$
Note in particular that $R(\vec x)=M\vec x,$ where $$M=\begin{bmatrix}\cos\theta & \sin\theta\\-\sin\theta & \cos\theta\end{bmatrix}.$$ In order for $T\circ R$ to be a rotation, it must either fix the whole plane (a trivial rotation) or a single point. Since $(T\circ R)(\vec 0)=\vec b\ne\vec 0,$ then $T\circ R$ does not fix the whole plane, so we must show that it fixes a unique point. That is, there is some unique $\vec x$ such that $(T\circ R)(\vec x)=\vec x.$ In other words, we must show that there is a unique solution to $M\vec x+\vec b=\vec x.$ This is equivalent to $A\vec x=\vec b,$ where $$A=\begin{bmatrix}1-\cos\theta & \sin\theta\\-\sin\theta & 1-\cos\theta\end{bmatrix},$$ and noting that $\det(A)=2-2\cos\theta\ne 0$ (since $\theta$ is not an integer multiple of $2\pi$), a unique solution exists (and is readily found by inverting $A$). Say that $\vec p$ is the unique fixed point of $T\circ R.$
Now, consider any $\vec x$ in the plane, and note that $$\begin{align}(T\circ R)(\vec x+\vec p) &= R(\vec x+\vec p)+\vec b\\ &= M(\vec x+\vec p)+\vec b\\ &= M\vec x+M\vec p+\vec b\\ &= R(\vec x)+R(\vec p)+\vec b\\ &= R(\vec x)+(T\circ R)(\vec p)\\ &= R(\vec x)+\vec p,\end{align}$$ and so $T\circ R$ is the rotation of the plane by angle $\theta$ about the point $\vec p$.
As a side note, the claim still holds when $\vec b=\vec 0,$ regardless of whether or not $\theta$ is an integer multiple of $2\pi.$ However, if the translation is non-trivial, then we cannot let $\theta$ be an integer multiple of $2\pi,$ for then $T\circ R$ is a translation, and not a rotation.
