$\cos(\theta_n) \to \cos(\theta)$ and $\sin(\theta_n) \to \sin(\theta)$. How to show that $\theta_n \to \theta$?

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 $$f^{-1}(f(\theta_n)) = \theta_n \to f^{-1}(f(\theta)) = \theta. $$

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.