Equivalence classes of set of rotations Let $\varphi_\theta:\mathbb R^2\to\mathbb R^2$ be the counterclockwise rotation around the origin through an angle $\theta\in\mathbb R$. Let $\Phi=\{\varphi_\theta\mid\theta\in\mathbb R\}$ be the set of all such rotations. We have an equivalence relation on $\Phi$ such that $kRl$ iff $k(x) = l(x)$ for every $x\in\mathbb R^2$. We have to describe the set of equivalence classes on $\Phi\setminus R$. 
My answer was that the set consisted of all functions which rotated the angle by $\theta + 360^\circ n$, for $n\geq0$. Is this correct? I'm not sure what the $\setminus$ means in $\Phi\setminus R$.
 A: For $S$ a set and $R$ an equivalence relation, $S\setminus R$ is just the set of equivalence classes, that is, the unique partition of $S$ such that $a$, $b$ belong to the same element of $S\setminus R$ iff $aRb$. I've seen it more often written up as $S/R$.
Your answer is correct: let's attempt to write it up cleanly.
Two rotations $\varphi_\alpha$ and $\varphi_\beta$ will be related by $R$ iff $\alpha-\beta=2\pi k$ for some integer $k$. To see this, notice that if the latter condition holds, for every $x$, $$\varphi_\alpha(x)=(\varphi_\beta\circ\varphi_{2\pi k})(x)=(\varphi_\beta\circ\text{id})(x)=\varphi_\beta(x),$$ implying the former condition. Furthermore, if the former condition holds, $\varphi_\alpha$ will send the point $(\cos(-\alpha),\sin(-\alpha))$ to $(1,0)$, while $\varphi_\beta$ will send it to $(\cos(\beta-\alpha),\sin(\beta-\alpha))$; the distance between these two points will be $$0=\left(\cos(\beta-\alpha)-1\right)^2+\sin(\beta-\alpha)^2=2-2\cos(\beta-\alpha),$$ so that $\cos(\beta-\alpha)=1$, implying the latter condition.
This is what we wanted to prove. $\blacksquare$
