In view of a result by Khinchine (see e.g. [Theorem 3.6(a)][1]), there exist infinitely many pairs $(p,q)$ of natural numbers such that 
$|2\pi q+\tfrac\pi2-p|<1/q$. For such $p$ and $q$, letting $q\to\infty$, we have $\sin p=1-O(1/q^2)=1-O(1/p^2)$ and hence $\sin^p p\to1$. So, $f(1)=1$. 

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As noted by [Gerald Edgar][2], the same argument shows that $f(x)=1$ for all $x\in(0,2)$. For $x=2$, it seems we have to deal with Diophantine approximation to $\pi$, rather than that to a arbitrary irrational number. This seems to boil down to an open problem; cf. [this answer](https://mathoverflow.net/a/424166/36721).


  [1]: https://www-ams-org.services.lib.mtu.edu/journals/tran/1961-099-01/S0002-9947-1961-0121357-5/
  [2]: https://mathoverflow.net/questions/450769/limsup-n-rightarrow-infty-n-in-mathbbn-sinnnx-for-various-x/450775#comment1165116_450775