We have $\phi:$${R/H\to G}$, where $(R,+)$ is a group, $H = ${$2n$$\pi$$:n$$\epsilon$$Z$} is a subgroup and $G$ is a group of matrices given as $\begin{bmatrix}cos{\theta}&-sin{\theta}\\sin{\theta}&cos{\theta}\end{bmatrix},$ where ${\theta}$$\epsilon$$R$ and the group operation is matrix multiplication.
I want to give an explicit formula for the map $\phi.$ So, I started by saying let $y$$\epsilon$$R/H.$ Then, $y=x+H$ for $x$ $\epsilon R$. This means $\phi(x+H)$ $=$ $\phi(x)$$=g,$ for $g$$\epsilon$$G.$ I don't know how to express the matrix part. That is, I don't know what $g$ should be. I am thinking of taking it as a matrix whose determinant is always 1 but I still don't know what to do with that.
Any help will be appreciated.
If you want to define a map $\phi:R\to G$ with the property $\phi(x+H)=\phi(x)$, then this leads to a map $\phi:R/H\to G$ (where calling both maps $\phi$ seems like an abuse of notation). You just need $\phi(x)$ to be $2\pi$-periodic in $x$. You can write $\phi(x)$ as \begin{equation*} \phi(x)=\left(\begin{matrix}\cos f(x) & -\sin f(x) \\ \sin f(x) & \cos f(x)\end{matrix}\right) \end{equation*} where $f(x)$ is any function such that $\cos f(x)$ and $\sin f(x)$ are both $2\pi$-periodic. The necessary and sufficient conditions for the composition of two functions to be periodic do not seem to be known. For any natural number $N$, any $(2\pi/N)$-periodic function $f(x)$ will do, but there are others.
To give three examples, the function $f(x)=x$ leads to possibly the most interesting map, which is an homomorphism (i.e., it preserves the group structure, meaning $\phi(x+y)=\phi(x)\phi(y)$): \begin{equation*} \phi(x)=\left(\begin{matrix}\cos x & -\sin x \\ \sin x & \cos x\end{matrix}\right) \end{equation*} the function $f(x)=0$ leads to the less interesting map \begin{equation*} \phi(x)=\left(\begin{matrix}1 & 0 \\ 0 & 1\end{matrix}\right) \end{equation*} and the function $f(x)=\sin(2x)$ also leads to a valid map: \begin{equation*} \phi(x)=\left(\begin{matrix}\cos \left[\sin(2x)\right] & -\sin \left[\sin(2x)\right] \\ \sin \left[\sin(2x)\right] & \cos \left[\sin(2x)\right]\end{matrix}\right). \end{equation*}