$\cos(\theta_n) \to \cos(\theta)$ 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?
 A: 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 
$$f^{-1}(f(\theta_n)) = \theta_n \to f^{-1}(f(\theta)) = \theta. $$
A: Suppose on the contrary that $\theta_{n} $ does not tend to $\theta$. Then there is an $\epsilon>0$ such that for any given positive integer $n$ there is a positive integer $m>n$ such that $|\theta_{m} - \theta|>\epsilon$. Note that taking a smaller $\epsilon$ does not affect the truth of the above statement so we can assume $\epsilon <\pi+|\theta|$ without any loss of generality. 
Then we can see that $$2-2\cos(\theta_{n}-\theta) =(\cos\theta_{n} - \cos\theta) ^{2}+(\sin\theta_{n} - \sin\theta)^{2}\to 0$$ ie $\cos(|\theta_{n} - \theta|) \to 1$. Now from the last paragraph there are infinitely many values of $n$ for which $0<\epsilon<|\theta_{n} - \theta|<\pi+|\theta |<2\pi$ and thus $\cos(|\theta_{n} - \theta|) $ stays away from $1$ for infinitely many values of $n$ which is contrary to the fact that $\cos(|\theta_{n} - \theta|) \to 1$.

To expand on "stays away from $1$" in previous paragraph just consider the function $f(x)=1-\cos x$ on interval $[a, b] \subset (0,2\pi)$. The minimum value of $f$ on this interval is $\min(f(a), f(b)) >0$ (verify this).

Your approach using inverse trigonometric functions will work only if the variables $\theta_{n}, \theta$ simultaneously lie in the range of these inverse functions. In this manner we can't cover the interval $(-\pi, \pi)$ of length $2\pi$ which is given in question. 
A: Hint: For large $n$, $|e^{i\theta_n}-e^{i\theta} | = |e^{i(\theta_n-\theta)}-1|<\epsilon$, for given $\epsilon$. It's almost immediate now.  
A: Let $p(t) = (\cos t, \sin t)$. If $x \in D= S^1 \setminus \{(-1,0)\}$, then there is a unique $t$ with $|t| < \pi$ such
that $x= p(t)$ and we define $\arg x = t$. The goal is to show that $\arg$ is continuous on $D$.
Suppose $x\in D$ is such that $x_2 \neq 0$ and $\theta$ (with $|\theta| < \pi$) is such that $x_2 = \sin \theta$. Since $\cos' \theta \neq 0$, the inverse function theorem
shows that there is some neighbourhood of $x_1$ and a continuous (in fact $C^1$) function $\alpha$ defined in this neighbourhood such that $\alpha(\cos t) = t$ (this is only valid for $t$ in a suitable neighbourhood of $\theta$). In particular, we have $\arg x = \alpha(x_1)$, and hence $\arg$ is continuous at $x \in D$ when $x_2 \neq 0$.
A similar argument applies for $x \in D$ such that $x_1 \neq 0$.
A: 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 $n\rightarrow \infty$.
