Parametrising the unit circle without sine and cosine Is there a nice way to make a smooth and periodic parametrisation $\gamma\colon\mathbb R\to S^1$ of the unit circle $S^1$ in $\mathbb R^2$ that does not somehow involve sine/cosine or (what I find to be equivalent) the complex exponential function --- and nor functions related to or derived from them. Preferably, if $\tau > 0$ is the period of $\gamma$, the restriction $\gamma\colon[0,\tau)\to S^1$ should be a bijection.
 A: I'll combine three parts.


*

*Let's start with the fractional part function, and adapt that so it maps $\mathbb R$ to $[-1,1)$: $$f_1:t\mapsto 2(t-\lfloor t\rfloor)-1$$ This ensures a period of $\tau=1$.

*The next part is a rational parametrization of the semicircle, which maps $[-1,1)$ to the right half of a circle with radius $2$ and center $(-1,0)$:
$$f_2:t\mapsto\frac1{1+t^2}\begin{pmatrix}1-3t^2\\4t\end{pmatrix}$$

*Then comes a stereographic projection to the unit circle:
$$\begin{pmatrix}x\\y\end{pmatrix}\mapsto\frac1{x^2+y^2+2x+1}\begin{pmatrix}x^2-y^2+2x+1\\2y(x+1)\end{pmatrix}$$


Combine all of this, use $t':=t-\lfloor t\rfloor$ as a shorthand, and you get
$$
\gamma:t\mapsto\frac1
{4t'^4 - 8t'^3 + 8t'^2 - 4t' + 1}
\begin{pmatrix}
4t'^4 - 8t'^3 + 4t' - 1 \\
-8t'^3 + 12t'^2 - 4t'
\end{pmatrix}
$$
For $t'=0$ and $t'=1$ the function values agree on $(-1,0)$, so it is continuous. The first derivatives agree as well on $(0,-4)$ so it is $C^1$ as well. Unfortunately, the second derivative $(16,\pm8)$ is different for both values, therefore the function is not $C^2$ and hence not smooth. At least not in the strict sense of the word. But perhaps $C^1$ is enough for you?
The circle formed by this function is perfectly smooth, and you even get the bijection you asked for. The rational representation in combination with the floor function is certainly far from any trigonometric or complex exponential functions.
I would guess that if you require the function to be not only smooth but analytic, then any possible function boils down to $\sin(g(t))$ and $\cos(g(t))$ where $g$ is some analytic function. I don't have a proof for this, only some rough ideas. So in that case the question “are there any fundamentally different parametrizations” would be answered “no”. This still leaves room for non-analytic but smooth functions.
