Question

Prove that the product of a rotation and a translation is a rotation.

Answer 1

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 RR is rotation by angle θθ (in the counterclockwise direction; measured in radians) around the origin, and TT is translation vector v⃗ =(v1,v2)v→=(v1,v2). A good representation for the plane is as the complex numbers. Then RR is multiplication by a=exp(iθ)a=exp⁡(iθ) and TT is addition by v=v1+iv2v=v1+iv2. Then T∘RT∘R is the function that takes zz to v+azv+az.

For a≠1a≠1, this transformation has a fixed point pp, which can be found by solving p=v+app=v+ap, from which we see that p=v1−ap=v1−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∘RT∘R is rotation by angle θθ around pp.

When a=1a=1, so that θθ is an integer multiple of 2π2π, the composition T∘RT∘R is not a rotation, but a translation, and the problem as stated is false.

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.

Let's suppose that T(x⃗ )=x⃗ +b⃗ T(x→)=x→+b→ for some b⃗ ≠0⃗ b→≠0→ and that RR is the rotation of the plane by θθ about the origin, where θθ is not an integer multiple of 2π.2π.
Note in particular that R(x⃗ )=Mx⃗ ,R(x→)=Mx→, where
M=[cosθ−sinθsinθcosθ].
M=[cos⁡θsin⁡θ−sin⁡θcos⁡θ].
In order for T∘RT∘R to be a rotation, it must either fix the whole plane (a trivial rotation) or a single point. Since (T∘R)(0⃗ )=b⃗ ≠0⃗ ,(T∘R)(0→)=b→≠0→, then T∘RT∘R does not fix the whole plane, so we must show that it fixes a unique point. That is, there is some unique x⃗ x→ such that (T∘R)(x⃗ )=x⃗ .(T∘R)(x→)=x→. In other words, we must show that there is a unique solution to Mx⃗ +b⃗ =x⃗ .Mx→+b→=x→. This is equivalent to Ax⃗ =b⃗ ,Ax→=b→, where
A=[1−cosθ−sinθsinθ1−cosθ],
A=[1−cos⁡θsin⁡θ−sin⁡θ1−cos⁡θ],
and noting that det(A)=2−2cosθ≠0det(A)=2−2cos⁡θ≠0 (since θθ is not an integer multiple of 2π2π), a unique solution exists (and is readily found by inverting AA). Say that p⃗ p→ is the unique fixed point of T∘R.T∘R.
Now, consider any x⃗ x→ in the plane, and note that
(T∘R)(x⃗ +p⃗ )=R(x⃗ +p⃗ )+b⃗ =M(x⃗ +p⃗ )+b⃗ =Mx⃗ +Mp⃗ +b⃗ =R(x⃗ )+R(p⃗ )+b⃗ =R(x⃗ )+(T∘R)(p⃗ )=R(x⃗ )+p⃗ ,
(T∘R)(x→+p→)=R(x→+p→)+b→=M(x→+p→)+b→=Mx→+Mp→+b→=R(x→)+R(p→)+b→=R(x→)+(T∘R)(p→)=R(x→)+p→,
and so T∘RT∘R is the rotation of the plane by angle θθ about the point p⃗ p→.

As a side note, the claim still holds when b⃗ =0⃗ ,b→=0→, regardless of whether or not θθ is an integer multiple of 2π.2π. However, if the translation is non-trivial, then we cannot let θθ be an integer multiple of 2π,2π, for then T∘RT∘R is a translation, and not a rotation.