How to prove $\limsup_{n \to \infty} |\sin(n)| = 1$?

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Does decimal expansion of $\pi$ contain blocks of zeroes of any integer length? I.e. $0$, $00$, $000$, $\ldots$

I discovered this question, when trying to prove $$\limsup_{n \to \infty} |\sin(n)| = 1.$$ Or is there any idea how to find subsequence $k_n$ such that $\sin(k_n) \rightarrow 1$?

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See this article, a special case is that there is two increasing sequences of odd positive integers $(p_n),(q_n)$ such that $$ \left|\frac{\pi}{2} - \frac{p_n}{q_n} \right| \leq \frac{1}{q_n^2},\quad n>1.$$

Note that $\sin |x| = |\sin x |$ for $x\in [0,\pi]$, then $$\sin \left|\frac{q_n\pi }{2} - p_n\right|= \left|\cos p_n\right| <\frac{1}{q_n} \to 0.$$ therefore $|\sin p_n|\to 1$.