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.

How to derive a formula for $F$? If I get the line passing through $p$ and $(t,0)$, I get $(D): y = at-1$, right?

And I know that to show that $\phi$ : $\mathbb R$$S$ \ {$p$}, I have to show that $\phi$ is injective and surjective.

Please I need help solving this

  • $\begingroup$ $t$ is fixed. So, $y = at-1$ is not a line. Or what is $a$? $\endgroup$ – amsmath Oct 18 '19 at 16:27
  • $\begingroup$ $a$ is the slope $\endgroup$ – JOJO Oct 18 '19 at 16:43
  • $\begingroup$ Slope of what? Your task is to find it. And -- as I said -- $t$ is fixed. So, $y = at-1$ is a constant, not a line. $\endgroup$ – amsmath Oct 18 '19 at 17:07
  • $\begingroup$ You have a line that passes through $(t,0)$ and $(0,-1)$. I hope you are able to find the formula for the line. It has the form $z=as+b$, where $a$ and $b$ are to be found. $\endgroup$ – amsmath Oct 18 '19 at 17:10

For a start: From the intercept theorem we have $$\frac y1=\frac{\sqrt{(x-t)^2+y^2}}{\sqrt{t^2+1}}.$$ Squaring and using $x^2+y^2=1$ we arrive in $$y^2=\frac{(t-1)^2}{1+t^2}.$$

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  • $\begingroup$ and what this tell me about the formula of F? $\endgroup$ – JOJO Oct 18 '19 at 16:36
  • $\begingroup$ Just complete the start I gave you, please. $\endgroup$ – Michael Hoppe Oct 18 '19 at 16:40
  • $\begingroup$ I dont know what to do $\endgroup$ – JOJO Oct 18 '19 at 16:46
  • $\begingroup$ You basically know the $y$ already by taking the appropriate square root. From $x^2+y^2=1$ calculate $x$. $\endgroup$ – Michael Hoppe Oct 18 '19 at 16:54
  • $\begingroup$ ok so then $y = \frac{|t-1|}{1+t^2}$? and $x = \sqrt(1- \frac{(t-1)^2}{1+t^2})$, and then how do I get the formula of F? $\endgroup$ – JOJO Oct 18 '19 at 17:38

If we express $S$ in polar coordinates:


then, from simple trigonometric relations, the inverse function $\Phi$ is:


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:

$$F(t)=(\frac{2 t}{t^2 + 1},\frac{1 - t^2}{t^2 + 1})$$

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  • $\begingroup$ and so this is the formula of F? $\endgroup$ – JOJO Oct 19 '19 at 8:06
  • $\begingroup$ Remark: $F$ is not the inverse. $\endgroup$ – JOJO Oct 19 '19 at 8:11
  • $\begingroup$ @JOJO $F$ is the function, or the inverse of the inverse $\endgroup$ – mik Oct 21 '19 at 11:59

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