$x^2+y^2=1\Rightarrow \exists\theta\in (-\pi,\pi]: x=\sin\theta, y=\cos\theta$

Question

I'm working on a proof of the following fact:

"Let $x,y$ be real numbers such that $x^2+y^2=1$.

Show that there is exactly one real number $\theta \in (-\pi,\pi]$ such that $x=\sin\theta$ and $y=\cos\theta$."

Now, according to the book from which I've taken this problem a way to prove this statement would be to divide the proof into cases depending on whether $x,y$ are positive, negative, or zero.

My question is: Is there a way to prove this in a more straightforward way? I haven't seen another way yet, so I'd appreciate any hint.

Answer 1

Since $x^2+y^2=1$, $x\in[-1,1]$ and therefore, by the intermediate value theorem, there is a $\theta\in\left[0,\pi\right]$ such that $\cos\theta=x$. Since$$\sin^2\theta=1-\cos^2\theta=1-x^2=y^2,$$$\sin\theta=\pm y$. If $\sin\theta=y$, we're done. Otherwise, just replace $\theta$ by $-\theta$.

Answer 2

It is enough to show that $\psi:(-\pi,\pi]\to S^1$ given by $\psi(\theta)=(\cos\theta,\sin\theta)$ is bijective.
It is clearly surjective since it is a parametrization with constant speed, in particular it is a parametrization with respect to the arc length. Let us assume $\theta,\varphi\in (-\pi,\pi]$ and $\psi(\theta)=\psi(\varphi)$.
We have: $$\begin{eqnarray*} 0&=&\left(\cos\theta-\cos\varphi\right)^2+(\sin\theta-\sin\varphi)^2\\&=&2-2\cos(\theta-\varphi) \end{eqnarray*}$$ from which $\cos(\theta-\varphi)=1$, implying $(\theta-\varphi)\in2\pi\mathbb{Z}$ and $\theta=\varphi$ (i.e. injectivity) since there are no elements of $(-\pi,\pi]$ having a distance $\geq 2\pi$.