Q:

Function in Polar Coordinates

Let $f,g:I\to\mathbb{R}$ be two function in $C^{k}(I)$, with the property that $f^2(t)+g^2(t)=1, \ \forall\ t\in I$. Is there a function $\theta: I\to\mathbb{R}$, $\theta\in C^{k}(I)$, such that:
$\left\{\begin{array}{cc} \cos(\theta(t))=f(t) \\ \sin(\theta(t))=g(t)\end{array}\right.$
?
Note that, $C^{k}(I)$ is the class of all continuous function, having derivative of order $k$ that are continuous on $I$.
Thanks a lot!

A:

Yes, by a very slight generalization of the path lifting theorem (PLT). Treat your functions $f$ and $g$ together as a path
$$
\gamma: I \to S^1 : t \mapsto (f(t), g(t))
$$
This lifts, by the PLT, to a map to the universal cover, $\mathbb R$, of $S^1$, 
$$
\tilde{\gamma}: I \to \mathbb R 
$$
with the property that 
$$
p \circ \tilde{\gamma}= \gamma
$$
where $p$ is the universal covering map, 
$$
p: \mathbb R \to S^1 : t \mapsto (\cos(t), \sin(t))
$$
You can then define $\theta(t) = \tilde{\gamma}(t)$. 
Now all that's left is to prove differentiability, which is straightforward. 

