Let $x,y:[a,b]\to\mathbb{R},\ a<b, a,b\in\mathbb{R}$ be two smooth functions ($x,y\in C^{\infty}([a,b])$). How can I prove that there is a unique function $\theta:[a,b]\to\mathbb{R},\ \theta\in C^{\infty}([a,b])$ such that:


$$\begin{cases} x(t)\sin\theta(t)=y(t)\cos\theta(t),\ \forall\ t\in [a,b].\\ (x(t\_0),y(t\_0))=(x\_0, y\_0)\neq (0,0)\ \text{is given}\\ \theta(t\_0)=\theta\_0 \ \text{is given too such that}\ x(t\_0)\sin\theta(t\_0)=y(t\_0)\cos\theta(t\_0) . \end{cases}$$



Uniqueness of a smooth function
You need $(x(t),y(t))$ to be non-zero for all $t$. If the curve $t\mapsto (x(t),y(t))$ spends some time at the origin, you can choose $\theta$ as you wish there. 


Assuming that $(x,y)$ is non-zero, your question boils down to representing the curve in polar coordinates. Writing $x=r\cos(\eta)$, $y=r\cos(\eta)$, we get 
$$ 
r(\cos \eta \sin \theta - \sin\eta \cos \theta) = r \sin(\theta -\eta) =0
$$
Given some initial condition, it is classical that we can choose $\eta(t)$ uniquely if $r$ is non-zero everywhere, by the fact that the standard projection ${\bf R}\mapsto {\bf S}^1$ given by $\eta\mapsto (\cos(\eta), \sin(\eta))$ is a covering map. Then $\eta(t)-\theta(t)$ must belong to $\pi\over 2$ mod $\pi$ and thus by continuity, this difference is constant, the constant being given by the initial condition.


EDIT: here is a counterexample to the existence of such $\theta$ when $(x,y)$ is allowed to vanish. Choose a $C^\infty$ function $\phi$ on ${\bf R}$ such that this function together with all its derivatives vanish at the origin. We take 
$$x(t) = \phi(t) \cos(1/t), \quad y(t)=\phi(t) \sin(1/t)$$
It is not difficult to show that $x$ and $y$ are smooth but the angle has no limit at $t = 0$. What is happening is that the vector $(x(t),y(t))$ is spinning very fast near the origin.


