# Parameterization of Unit Circle is Simple Loop

## Theorem

Let $\mathbb S^1$ denote the unit circle whose center is at the origin of the Euclidean space $\R^2$.

Let $p: \closedint 0 1 \to \R^2$ be defined by:

$\forall t \in \closedint 0 1 : \map p t = \tuple {\map \cos {2 \pi t}, \map \sin {2 \pi t} }$

Then $p$ is a simple loop with image equal to $\mathbb S^1$.

## Proof

Parametric Equation of Circle shows that for all $r \in \R$, the point $\tuple {\map \cos r, \map \sin r}$ lies on the unit circle $\mathbb S^1$.

Parametric Equation of Circle also shows that for all points $\tuple {x, y}$ on the unit circle $\mathbb S^1$, the point can be expressed as $\tuple {x, y} = \tuple {\map \cos r, \map \sin r}$ for some $r \in \R$.

Sine and Cosine are Periodic on Reals shows that sine and cosine functions are periodic functions with period equal to $2 \pi$.

It follows that the image of $p$ is equal to $\mathbb S^1$.

Let $t_1, t_2 \in \closedint 0 1$ such that $\tuple {\map \cos {2 \pi t_1}, \map \sin {2 \pi t_1} } = \tuple {\map \cos {2 \pi t_2}, \map \sin {2 \pi t_2} }$ with $t_1 \ne t_2$.

Shape of Cosine Function shows that the cosine function is strictly decreasing on the interval $\closedint 0 \pi$, and strictly increasing on the interval $\closedint \pi {2 \pi}$.

Strictly Monotone Real Function is Bijective shows that with $t_1 \ne t_2$, we cannot have:

both $t_1, t_2 \in \closedint 0 {\dfrac 1 2}$, as this would imply $\map \cos {2 \pi t_1} \ne \map \cos {2 \pi t_2}$
or both $t_1, t_2 \in \closedint {\dfrac 1 2} 1$, as this would imply $\map \cos {2 \pi t_1} \ne \map \cos {2 \pi t_2}$

Without loss of generality, we have $t_1 \in \closedint 0 {\dfrac 1 2}$, and $t_2 \in \closedint {\dfrac 1 2} 1$.

Sine and Cosine are Periodic on Reals/Corollary shows that the sine function is strictly positive on the interval $\openint 0 \pi$, and strictly negative on the interval $\openint \pi {2 \pi}$.

It follows that we have $\map \sin {2 \pi t_1} \ne \map \sin {2 \pi t_2}$, unless we have $t_1 = 0, t_2 = 1$, in which case:

$\tuple {\map \cos 0, \map \sin 0} = \tuple {\map \cos {2 \pi}, \map \sin {2 \pi} } = \tuple {1, 0}$

We have that Real Sine Function is Continuous and Cosine Function is Continuous.

$\map \cos {2 \pi t}, \map \sin {2 \pi t}$

are continuous.

Hence from Continuity of Composite with Inclusion: Inclusion on Mapping and Continuous Mapping to Product Space, it follows that $p$ is continuous.

By definition of simple loop, it follows that $p$ is a simple loop.

$\blacksquare$