# Is it possible to elementarily parametrize a circle without using trigonometric functions?

Just out of curiosity: Is it possible to parametrize a full circle or part of one with elementary functions but without using trigonometric functions? If so, what are advantages/disadvantages compared to the standard parametrizations using $\cos(t)$ and $\sin(t)$?

• for half a circle we can use $x=t$ and Commented Oct 7, 2016 at 21:44
• There is a rstional parametrización, asssociated with the name of Weierstrass. See math.stackexchange.com/questions/1246832/… Commented Oct 7, 2016 at 21:45
• $y=\sqrt{R^2-t^2}$ with $-R\leq t \leq R$ Commented Oct 7, 2016 at 21:46
• You'll find some more parametrisations here: en.wikipedia.org/wiki/Circle#Equations Commented Oct 7, 2016 at 22:07
• The main advantage of $\cos(t),\sin(t)$ is that it's an arc length parametrization. Commented Oct 8, 2016 at 7:41

You may use the fact that $(1+it)/(1-it)$ has abs value 1. So: $$x(t) = \frac{1-t^2}{1+t^2}\ \ , \ \ \ y(t) = \frac{2t}{1+t^2}$$ gives a parametrization for $S^1$ minus one point $(-1,0)$ (the limit of $t\rightarrow \infty$).

There is a non-trivial application in taking a skew-symmetric (or more generally anti self adjoint matrix, or operator) $S$ on a Hilbert space $H$ and produce the following orthogonal/unitary matrix: $$U = (1+S) (1-S)^{-1}$$ It is also used in the so-called Cayley transform (see wiki) to analyse e.g. unbounded selfadjoint operators, with a factor of $i$ as well: $V=(1+iA)(1-iA)^{-1}$.

It is also used in numerical analysis, when you use a finite difference method for the wave equation and want to preserve e.g. the $L^2$ (discrete) norm.

Later edit: If you want the whole circle to be covered you may take a square before splitting into real/imag parts: $t \in {\Bbb R} \mapsto \frac{(1+it)^2}{(1-it)^2}\in {\Bbb C}$ covers the circle twice (although (1,0) only once). This gives: $$x(t) = \frac{1-6t^2+t^4}{1+2t^2+t^4} , \ \ \ y(t) = \frac{4t -4t^3}{1+2t^2+t^4}$$

• This one is also given in Mariano's link in the comments...very nice Commented Oct 7, 2016 at 21:50
• Thank you. Is there some situation where this parametrization would be more useful than the one with trigonometric functions? Commented Oct 7, 2016 at 21:51
• Yes, Victor, it leads to a formula for Pythagorean triples. Commented Oct 7, 2016 at 21:55
• @Algorithms, see this or search the site for "pythagoran triple rational" for more. Commented Oct 8, 2016 at 4:15
• I think the parametrization you give misses $(-1,0)$, not $(1,0)$. Commented Oct 8, 2016 at 9:55

What about $f(x,\pm)=\pm\sqrt{1-x^2}$, where $f(\cdot,\cdot)$ has a discrete and continous parameter defined in $[-1,1]$...

You may also use $e^{it}=\cos(t)+i\sin(t)$ to represent a circle in the complex plane. With this calculating Fourier transforms becomes handy...

Just a comment to H.H. Rugh answer that needs graphical support:

His parametrisation is the stereograhic projection which has an application in Photography:

$\hskip1.5in$

• Yes. But is there any reason to use any parametrization over another? Commented Oct 7, 2016 at 21:47
• Depends on your problem. Choose the one that suits you best! Commented Oct 7, 2016 at 21:48
• Thank you draks ... . It's too bad I can't accept two answers, as your answer is correct and useful too. Commented Oct 7, 2016 at 22:01
• @ViktorKaspervich no problem. Cheers... Commented Oct 7, 2016 at 22:04

This is an old question but I think the following is perhaps somewhat interesting:

A simple parametrization of the circle is $$q\in[-1,1]:\quad x(q)=2|q|-1\quad y(q)=-\mathrm{sign}(q)\sqrt{1-x(q)^2}=-2\,\mathrm{sign}(q)\sqrt{|q|(1-|q|)}$$ This is pretty ugly in my opinion because of the $$||$$ and $$\mathrm{sign}$$ functions. However, we can bijectively map $$[-1,1]$$ to itself using the map $$q=\mathrm{sign}(t)t^2$$ so that $$|q|=t^2$$ and $$\mathrm{sign}(q)=\mathrm{sign}(t)$$. Then $$t\in[-1,1]:\quad x(t)=2t^2-1\quad y(t)=-2\,\mathrm{sign}(t)\sqrt{t^2(1-t^2)}=-2t\sqrt{1-t^2}$$ is a parametrisation of the circle as $$\mathrm{sign}(t)\sqrt{t^2}=\mathrm{sign}(t)|t|=t$$.
However I still think there is a flaw with this $$t$$ parameter because $$y(t)$$ is not differentiable (or at least its one-sided derivatives do not exist) at the endpoints $$t=\pm 1$$. To remedy this, I tried to find another bijection from $$[-1,1]$$ to itself which has zero derivative at $$\pm 1$$ as this may cancel out the infinity from the square root (think $$\sqrt[3]{x}$$ and $$x^3$$). The simplest one I could find is: $$t=\frac{3}{2}u-\frac{1}{2}u^3$$ Let's analyse what happens just to $$\sqrt{1-t^2}$$: $$\sqrt{1-t^2}=\sqrt{1-\left(\frac{3}{2}u-\frac{1}{2}u^3\right)^2}=\frac{1}{2}\sqrt{(4-u^2)(1-u^2)^2}=\frac{1-u^2}{2}\sqrt{4-u^2}$$ Luckily $$1-u^2\geq 0$$ so that $$\sqrt{(1-u^2)^2}=1-u^2$$. Anyway, the last expression above is clearly smooth on $$[-1,1]$$. Plugging $$t(u)$$ into the last parametrisation we had, this pops out: $$u\in[-1,1]:\quad x(u)=\frac{1}{2}(u^2-2)(u^4-4u^2+1)\quad y(u)=-\frac{1}{2}u(1-u^2)(3-u^2)\sqrt{4-u^2}$$ So $$(x(u),y(u))$$ is a smooth parametrisation of the circle which is also bijective (with the exception of the point $$(1,0)$$ as the circle is a closed curve).
However, I don't really see any reason why you would want to use this parametrisation, it's more of a curiosity.
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