# Derive a formula for a function F

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

• $t$ is fixed. So, $y = at-1$ is not a line. Or what is $a$? – amsmath Oct 18 '19 at 16:27
• $a$ is the slope – JOJO Oct 18 '19 at 16:43
• 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. – amsmath Oct 18 '19 at 17:07
• 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. – 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 $$y^2=\frac{(t-1)^2}{1+t^2}.$$

• and what this tell me about the formula of F? – JOJO Oct 18 '19 at 16:36
• Just complete the start I gave you, please. – Michael Hoppe Oct 18 '19 at 16:40
• I dont know what to do – JOJO Oct 18 '19 at 16:46
• You basically know the $y$ already by taking the appropriate square root. From $x^2+y^2=1$ calculate $x$. – Michael Hoppe Oct 18 '19 at 16:54
• 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? – JOJO Oct 18 '19 at 17:38

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:

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

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