$\cos(\theta_n) \to \cos(\theta)$ and $\sin(\theta_n) \to \sin(\theta)$. How to show that $\theta_n \to \theta$? $\cos(\theta_n) \to \cos(\theta)$ and $\sin(\theta_n) \to \sin(\theta)$ where $\theta_n$ , $\theta \in (-\pi, \pi)$.

How to show that $\theta_n \to \theta$ ?

Can I use the continuity of $\cos^{-1}$ or $\sin^{-1}$ (only one of them) to get the result?

If not, then how to proceed?

Edit:

As we know, range of $\arcsin $ is $[-\pi/2, \pi/2]$, we can use continuity of $\arcsin $ to get $\theta_n \to \theta$ when they belongs to  $[-\pi/2, \pi/2]$, in other cases use continuity of $\arccos$.

Is this argument correct?
Consider the map $f \colon (-\pi,\pi) \rightarrow \mathbb{R}^2$ given by $f(\theta) = (\cos(\theta), \sin(\theta))$. It maps the interval $(-\pi,\pi)$ bijectively onto the the unit circle minus the point $(-1,0)$ and you can explicitly write the inverse as

$$ f^{-1}(x,y)= 2 \arctan \left(  \frac{y}{1 + x} \right) $$

(see the entry on atan2 in wikipedia). 

From this expression, it is clear that $f^{-1}$ is continuous and so if $f(\theta_n) \to (\cos \theta, \sin \theta)$ then 
$$f^{-1}(f(\theta_n)) = \theta_n \to f^{-1}(f(\theta)) = \theta. $$
Observe that any convergent subsequence $\{\theta_{n_{k}}\}_{k=1}^{\infty}$ satisfies $\theta_{n_{k}}\rightarrow \theta$ as $k\rightarrow \infty$.

Thus, as the limit superior and the limit inferior are the same, $\theta_{n}\rightarrow \theta$ as $n\rightarrow \infty$.