Proving that $\lim_{n\to\infty} \frac{\text{cos}(n)+\text{sin}(n)}{n^2}=0$ using the definition of the limit

Question

Let $(a_n)$ be the sequence defined by $$a_n=\frac{\text{cos}(n)+\text{sin}(n)}{n^2}.$$ Now, we can easily see that it converges to $0$ because a sequence that converges to $0$ times a bounded sequence is a sequence that converges to $0$. However, I want to prove this limit using the defintion. Let $\epsilon>0$. From the Archimedean property we know there exists an $N$ such that $$\frac{1}{N^2}<\frac{\epsilon}{2}.$$ For every $n>N$, since $$-1\leq \text{sin}(n)\leq 1$$ and $$-1\leq \text{cos}(n)\leq 1$$ we obtain $$-2\leq \text{sin}(n) + \text{cos}(n)\leq 2$$ and therefore $$|\text{sin}(n) + \text{cos}(n)|\leq 2.$$ Since $$\frac{1}{n^2}<\frac{1}{N^2}$$ we conclude that $$\left|\frac{\text{sin}(n) + \text{cos}(n)}{n^2}-0\right|=\frac{|\text{sin}(n) + \text{cos}(n)|}{n^2}<\frac{2}{N^2}<2\left(\frac{\epsilon}{2}\right)=\epsilon$$ and from the definition of the limit we get that $$\lim_{n\to\infty} a_n=0.$$ Can you please check if what I did is correct. I think it is, but I'm not sure if there is enough detail or maybe there is too much.