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.

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.
