Intuition behind applying the implicit function theorem to parameterization of unit circle This post is a modification of a previous post I have since deleted, which was much too messy and basically had two separate questions. (It received no answers.)

A common motivating example for the implicit function theorem (IFT) is the equation $x^2+y^2=1$ where $(x,y)\in \mathbb R^2$, as we can locally write $y$ as an equation of $x$ (and vice versa) in the familiar way.
Consider the following exercise from Wade's An Introduction to Analysis.

Suppose that $f:=(u,v):\mathbb R\to \mathbb R^2$ is $C^2$ and set
  $(x_0,y_0)=f(t_0)$. Also assume that $f'(t_0)\ne 0$. Show that either
  there is a $C^1$ function $g$ with $g(x_0)=t_0$ and $u(g(x))=x$ for
  any $x$ near $x_0$ or there is a $C^1$ function $h$ with $h(y_0)=t_0$
  and $v(h(x))=y$ for any $y$ near $y_0$. (Note that I think the
  textbook has a typo since we just need to assume that $f$ is $C^1$.
  Correct me if I'm wrong.)

I can do this by applying the IFT to the function $F(x,t)=x-u(t)$ and to the function $G(y,t)=y-v(t)$.
Now, the unit circle can be parameterized (infinitely many times) by $(\sin(t), \cos(t))$ for $t\in \mathbb R$. And the map $(\sin(t), \cos(t))$ is everywhere $C^1$.
Therefore, I am wondering if the exercise can be seen as describing the motivating example of the unit circle I described initially. That is, is there any way to relate this problem to the simple idea of writing the unit circle $x^2+y^2=1$ locally as a function of $x$ or $y$?
I think that the answer is "yes" in some sense but am unsure how. I just want to gain a bit of intuition.
 A: Let $f(t) = (\cos(t), \sin(t))$.  Notice that $f$ is $C^1$ and that $f'(t) \neq (0,0)$ for all $t$.  Fix $t_0 \in \mathbb{R}$, and let $(x_0, y_0) = f(t_0)$.  Thus, $(x_0, y_0)$ is a point on the unit circle.
The conclusion of the exercise is that:


*

*(a) There is a $C^1$ function $g$ having $g(x_0) = t_0$ and $\cos(g(x)) = x$ for all $x$ near $x_0$;

*Or: (b) There is a $C^1$ function $h$ having $h(y_0) = t_0$ and $\sin(h(y)) = y$ for all $y$ near $y_0$.


The intuition is this: Since $f'(t) \neq (0,0)$ for all $t$, it follows that the tangent line at any point $(x_0, y_0)$ of the unit circle is not vertical or not horizontal.  (Of course, at most points, the tangent line is both not vertical and not horizontal.)


*

*If the tangent line at $(x_0, y_0)$ is not vertical, then conclusion (a) holds, meaning that a function we might call "$\cos^{-1}(x)$" is well-defined for $x$ near $x_0$.  (Imagine projecting the unit circle up/down to the $x$-axis.  Locally, away from vertical tangent lines, this is a bijection.)

*If the tangent line at $(x_0, y_0)$ is not horizontal, then conclusion (b) holds, meaning that a function we might call "$\sin^{-1}(y)$" is well-defined for $y$ near $y_0$.  (Imagine projecting the unit circle left/right to the $y$-axis.  Locally, away from horizontal tangent lines, this is a bijection.)
