We’re all familiar with the usual trigonometric parameterization of the unit circle: Each point on $x^2 + y^2 = 1$ is given by $(\cos \theta, \sin \theta)$ for some real $\theta$. Less well-known is the parameterization of the unit circle by rational functions.

The line through the point $(-1, 0)$ with slope $m$ is given by $y = m(x+1)$. This line intersects the unit circle in one other point and as we vary $m$ we strike every point on the unit circle. Here’s an illustration for a few values of $m$:

What are the coordinates of this point? Well, the point $(x,y)$ satisfies both $x^2 + y^2 = 1$ and $y = m(x+1)$ so we have

$x^2 + (m(x+1))^2 = 1$

$\implies x^2 + m^2(x^2+2x+1) = 1$

$\implies (1+m^2)x^2 + 2m^2x + (m^2-1) = 0$

$\displaystyle{ \implies x = \frac{-2m^2 \pm \sqrt{4m^4 - 4(1+m^2)(m^2-1)}}{2(1+m^2)} }$

$\displaystyle{ \implies x = \frac{-2m^2 \pm \sqrt{4}}{2(1+m^2)} }$

$\displaystyle{ \implies x = -1,\ \ \frac{1-m^2}{1+m^2} }$

$x = -1$ corresponds to the point $(-1, 0)$. When $x = \frac{1-m^2}{1+m^2}$ we have

$\displaystyle{ y^2 = 1 - x^2 }$

$\displaystyle{ y^2 = 1 - \left( \frac{1-m^2}{1+m^2} \right)^2 }$

$\displaystyle{ y^2 = \frac{(1+m^2)^2 - (1-m^2)^2}{(1+m^2)^2} }$

$\displaystyle{ y^2 = \frac{(1+2m^2 + m^4) - (1-2m^2 + m^4)}{(1+m^2)^2} }$

$\displaystyle{ y^2 = \frac{4m^2}{(1+m^2)^2} }$

$\displaystyle{ \implies y = \pm\frac{2m}{1+m^2} }$

and we want the $\pm$ to be $+$ so that positive slopes correspond to the upper half of the circle as we illustrated above.

Therefore every point on the unit circle (other than $(-1, 0)$) is of the form

$\displaystyle{ \left(\frac{1-m^2}{1+m^2},\ \frac{2m}{1+m^2}\right) }$ for some $m$. As $m \rightarrow \pm \infty$ we have $\left(\frac{1-m^2}{1+m^2},\ \frac{2m}{1+m^2}\right) \rightarrow (-1, 0)$.

We now have two parameterizations of the unit circle. How are they connected? For every $m$ there is a $\theta$ such that

$\displaystyle{ \cos \theta = \frac{1-m^2}{1+m^2} \text{ and } \sin \theta = \frac{2m}{1+m^2} }$

which gives

$\displaystyle{ \tan \theta = \frac{2m}{1-m^2} }$.

This calls to mind the tangent addition formula $\tan(A+B) = \frac{\tan A + \tan B}{1-\tan A\tan B}$. This suggests $m = \tan(\theta/2)$ and $\theta = 2\tan^{-1}(m)$. The usual calculations show that this is correct: If we start with $m = \tan^{-1}(\theta/2)$ we get the desired $\cos \theta$ and $\sin \theta$.

There is no non-constant real polynomial parameterization. You can see this by considering the degree of $x^2 + y^2$.