Let $S$ := {$(x, y) ∈ \mathbb R^2: x^2 + y^2 = 1$} be the unit circle. Let $p = (0, -1)$ ∈ $S$.
Define a map $F$ from $\mathbb R$ × {$0$} to $S$ as follows: Given $(t, 0)$, let $F((t, 0)) = (x, y)$ ∈ $S$ be the intersection of $S$ with the line through $p$ and $(t, 0)$.
Derive a formula for $F$.
Show that $t$ → $F((t, 0))$ is a bijective map from $\mathbb R$ to $S$ \ {$p$}, and also derive a formula for its inverse.
Please I need help solving this
If we express $S$ in polar coordinates:
$$S:=\{(r,\varphi):r=1,\varphi∈(-\frac\pi{2},\frac{3\pi}{2})\}$$
then, from simple trigonometric relations, the inverse function $\Phi$ is:
$$\Phi(\varphi)=\tan\frac{\frac\pi2-\varphi}2$$
and its inverse $F$ is:
$$F(t)=\frac\pi{2}-2\arctan t$$
After changing coordinates back to cartesian, $F$ will look like:
$$F(t)=(\cos(\frac\pi{2}-2\arctan t),\sin(\frac\pi{2}-2\arctan t))$$
which can be further simplified to:
$$F(t)=(\frac{2 t}{t^2 + 1},\frac{1 - t^2}{t^2 + 1})$$
Is this the formula of $F$?
And I know that to show that $\phi$ : $\mathbb R$ → $S$ \ {$p$} is a bijective map, I have to show that $\phi$ is injective and surjective. But how do I do that? Can someone please help me in showing that $\phi$ is bijective? and how can I derive a formula for its inverse?