# $\mathbb{R}^2$ rotations

Considering $\mathbb{R}^2$ for the two rotations $g_1,g_2$ with centers of rotation $x_1$ and $x_2$ by $\theta_1$ and $\theta_2$ I have to show that $g_1\circ g_2$ is a rotation iff $\theta_1+\theta_2\notin\{0,2\pi\}$, the rotation angle is given by $\theta_1+\theta_2\mod 2\pi$ and I have to calculate the center of rotation.

A rotation is defined by $v\mapsto d_\theta(v-x)+x$ with $d_\theta=\begin{pmatrix}\cos(\theta) & -\sin(\theta) \\ \sin(\theta) & \cos(\theta)\end{pmatrix}$.

Also if $\theta_1+\theta_2\in\{0,2\pi\}$, $g_1\circ g_2$ is a translation by identity. Is there a nice proof of this?

• I am not sure if I understand the question correctly. But if the case is general true - it should be true for $x_1 = \vec{0}$ and $x_2 = \vec{0}$ - but in that case you always end up with a rotation, given by $R(\theta_1) R(\theta_2) = R(\theta_1 + \theta_2)$... – johannesvalks Jun 23 '14 at 16:10
• Yep - I did not understand the question. I now understand it... – johannesvalks Jun 23 '14 at 16:14

## 3 Answers

So $$g_1(g_2(v)) = d_{\theta_1}(d_{\theta_2}(v-x_2)+x_2-x_1) + x_1 = d_\theta(v) + y ,$$ where $\theta = \theta_1 + \theta_2$, and $y = d_{\theta_1}(d_{\theta_2}(-x_2) + x_2-x_1) + x_1$. If $\theta$ is a multiple of $2\pi$, then $d_{\theta}$ is the identity, and hence $g_1 \circ g_2$ is a translation. If $\theta$ is not a multiple of $2\pi$, then $I - d_\theta$ is an invertible matrix, and hence there exists a vector $z$ such that $z - d_\theta(z) = y$, and hence $$g_1(g_2(v)) = d_\theta(v-z) + z .$$ By the way, to show that $I - d_\theta$ is invertible when $\theta$ is not a multiple of $2\pi$, show that there is only one solution to $(I - d_\theta) w = 0$.

In the context of your question, the nicest proof is to insert the coordinates of the column vectors $v$ and $x$ into the formula for the rotation $v \mapsto d_\theta(v-x)$, and calculate.

• There are other proofs available in synthetic geometry, starting from synthetic definitions of translations and rotations, but the context of your question is analytic geometry. – Lee Mosher Jun 23 '14 at 15:46
• Yes I already did that and I applied the definition of $g_1\circ g_2(v) = g_2(g_1(v)) = d_{\theta_2}((d_{\theta_1}(v-x_1)+x_1)-x_2)+x_2$ and calculated, it gets very big and I get to no nice result in the end... – sj134 Jun 23 '14 at 15:48

There is a standard way to combine rotation and translation by $(a, b)'$:

$$A_\theta=\begin{pmatrix}\cos(\theta) & -\sin(\theta) & a \\ \sin(\theta) & \cos(\theta) & b \\ 0 & 0 & 1 \end{pmatrix}$$ which is applied to the vector $\begin{pmatrix}x & y & 1 \end{pmatrix}'$

You can first apply the translation, then rotation, then translation (each of those amounts to a matrix multiplication) and see what the result looks like.