Parametrization of a circle $\gamma(t)= (\cos(t), \sin(t))$

Question

In my multivariable calculus class, we are learning about parametrization. In my class, my teacher said that the circle $$\gamma(t)= \left(\cos(t), \sin(t)\right)$$

with $t \in [0, \frac{\pi}{2}]$ is a parametrization of $$\sigma(u)= \left(\frac{1-u^2}{1+u^2}, \frac{2u}{1+u^2}\right)$$

with $u \in [0, 1]$, but he never really explained why.

Could someone explain to me why?

Answer 1

This important parametrization is known as tangent half-angle formula and we can explain it at least that in two ways.

Algebraically we have

$$\sigma(u)_x^2+\sigma(u)_y^2=\left(\frac{1-u^2}{1+u^2}\right)^2+\left(\frac{2u}{1+u^2}\right)^2=1$$

with $\sigma(0)=(1,0)$ and $\sigma(1)=(0,1)$.

The other way is geometrically, according to the following sketch

we have

$$u=\tan \frac t 2 =\frac{\sin t}{1+\cos t} \implies u^2=\frac{\sin^2 t}{(1+\cos t)^2}=\frac{1-\cos t}{1+\cos t}$$

from which we obtain

$$\cos t=\frac{1-u^2}{1+u^2} \implies \sin t=\frac{2u}{1+u^2} $$