The question in the title is asked by an undergraduate student, and a general version is $\sin n^p$ diverges for any $p>1$.
For the case of $0<p<1$, it is clear by using the fact $\lim\limits_{n\rightarrow \infty}((n+1)^p-n^p)=0$. Weyl's equidistriibution criterion can be applied here while I do not feel it is proper for an undergraduate student, see (4) of The limit of $\sin(n^\alpha)$.
For the case of $p$ rational, please see $\sin (n^2)$ diverges.
Someone said that I did not show my work. Actually I tried to use the approximation method but I failed.
For the case of $p$ irrational, we may assume that there exists a rational sequence $\{p_i\}$, such that $\{p_i\}$ is monotone increasing, and $\lim\limits_{n\rightarrow \infty} p_i=p$.
First, we can choose $\hat{\varepsilon}=\frac{\varepsilon}{3\ln k_1 \cdot k_1^p}$, then there exists $M_1$, such that $\forall i>M_1$, $$|\sin k_1^{p_i}-\sin k_1^p|<|k_1^{p_i}-k_1^p|\leq \ln k_1\cdot k_1^p |p_i-p|<\frac{\varepsilon}{3}.$$ Similarly, there exists $M_2$, such that $\forall i>M_2$, $$|\sin k_2^{p_i}-\sin k_2^p|<|k_2^{p_i}-k_2^p|\leq \ln k_2\cdot k_2^p |p_i-p|<\frac{\varepsilon}{3}.$$ Therefore, $\forall \varepsilon>0$, $\exists K>0$, such that $\forall k_1, k_2>K$, $\exists M>0$, such that $\forall i>M$, $$|\sin k_1^{p_i}-\sin k_2^{p_i}|\leq |\sin k_1^{p_i}-\sin k_1^p|+|\sin k_1^{p}-\sin k_2^{p}|+|\sin k_2^{p}-\sin k_2^{p_i}|<\frac{\varepsilon}{3}+\frac{\varepsilon}{3}+\frac{\varepsilon}{3}.$$ Then there is a rational number $\frac{a}{b}=p_i$, such that $\{\sin n^{\frac{a}{b}}\}$ is a Cauchy sequence, which is a contradiction.
The problem here is that $i$ is depend on $k_1$ and $k_2$, therefore, we may not have $i$, such that $\{\sin n^{p_i}\}$ is a Cauchy sequence.
However, I still want to find an easier way to answer a freshman in my class.
