Showing diffeomorphism between $S^1 \subset \mathbb{R}^2$ and $\mathbb{RP}^1$ I am trying to construct a diffeomorphism between $S^1 = \{x^2 + y^2 = 1; x,y \in \mathbb{R}\}$ with subspace topology and $\mathbb{R P}^1 = \{[x,y]: x,y \in \mathbb{R}; x \vee y \not = 0  \}$ with quotient topology and I am a little stuck.
I have shown that both are smooth manifolds, and I used stereographic projection for $S^1$, but now I am runing into trouble when I give the homeomorphism between $S^1$ and $\mathbb{RP}^1$ as the map that takes a line in $\mathbb{RP}^1$ to the point in $S^1$ that you get when letting the parallel line go through the respective pole used in the stereographic projection. 
If I use the south and north poles I get a potential homeomorphism, but I cannot capture the horizontal line in my image, but when I pick say north and east then my map is not well defined as I get different lines for the same point in $S^1$. Can somebody give me a hint how to make this construction work, or is it better to move to a different representation of $S^1$ ?

The easiest explicit map I know is:
$$(\cos(\theta), \sin(\theta))\mapsto [\cos(\theta/2):\sin(\theta/2)]$$
Note that although $\cos(\theta/2)$ and $\sin(\theta/2)$ depend on $\theta$ and not just $\sin(\theta)$ and $\cos(\theta)$, the map is well-defined so long as we use the same value of $\theta$ when computing coordinates in $S^1$.That is,  $$\cos(\frac{\theta+2\pi}{2})=-\cos(\theta/2) \text{ and } \sin(\frac{\theta+2\pi}{2})=-\sin(\theta/2)$$ So the choice of $\theta$ modulo $2\pi$ does not affect $[\cos(\theta/2):\sin(\theta/2)]$, since $[x,y]=[-x,-y]$.
