Is it possible to prove that every point on the circle can be expressed as $(\cos \theta,\sin \theta)$ using this approach?

Question

Assume the IVP or powerseries definitions of $\sin$ and $\cos$ for this question.

It's a basic geometric fact that:

Theorem. Given a pair $(x,y)$ of real numbers, the following are equivalent:

1. $\exists \theta \in \mathbb{R}(x = \cos \theta \wedge y = \sin \theta)$
2. $x^2+y^2 = 1$

However, proving this rigorously seems to be rather non-trivial. The direction $(1) \rightarrow (2)$ is straightforward; it's equivalent to $\cos^2\theta +\sin^2\theta = 1,$ which can itself be proved by differentiating $\cos^2 \theta+\sin^2\theta$ and obtaining $0$, thereby showing that it's a constant function. However, showing $(2) \rightarrow (1)$ seems to be much harder. One idea I had was to first show that the equation $x^2+y^2=1$ defines a smooth $1$-dimensional submanifold of $\mathbb{R}^2$. We can then use:

Conjecture. Let $\gamma : \mathbb{R} \rightarrow X$ denote a path in a connected $1$-dimensional Riemannian manifold with constant non-zero speed. Then $\gamma$ is surjective.

My questions are twofold:

Questions.

Q0. Is this even true?

Q1. If so, are there generalizations to the case where the domain of $\gamma$ is replaced by $\mathbb{R}^n$?

Answer 1

The geometrical angle bisector is the line through $(x,y)+(1,0)=(x+1,y)$. This ray has $$\frac{(x+1,y)}{\sqrt{2+2x}}=\left(\sqrt{\frac{x+1}2},\frac{y}{\sqrt{2(x+1)}}\right)$$ as its point on the unit circle. The sequence of angle bisectors $(x_0,y_0)=(x,y)$, $$(x_{n+1},y_{n+1})=\frac{(x_n+1,y_n)}{\sqrt{2+2x_n}}$$ converges to $(1,0)$ and will thus reach a neighborhood of that point where you can easily prove that it is covered by the definitions of the trigonometric functions around $θ=0$.

Answer 2

Here's a proof sketch of your conjecture.

Let $A$ be the image of $\gamma$ and $B$ its complement. Suppose $B$ is nonempty.

Since $X$ is connected, there exists a point $P \in \overline{A} \cap \overline{B}$.

Let $U$ be any neighborhood of $P$ homeomorphic to $\mathbb{R}$.

$U$ must contain a point of $A$ and a point of $B$.

The closed interval from $A$ to $B$ must have finite length, and thus $\gamma$ traverses the interval from $A$ to $B$ in finite time, and consequently it passes through $B$, giving a contradiction.