Return to Answer

In view of a result by Khinchine (see e.g. Theorem 3.6(a)), 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, $f(1)=1$.

In view of a result by Khinchine (see e.g. Theorem 3.6(a)), 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, $f(1)=1$.

As noted by Gerald Edgar, 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.

In view of a result by Khinchine (see e.g. Theorem 3.6(a), 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, $f(1)=1$.

In view of a result by Khinchine (see e.g. Theorem 3.6(a)), 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, $f(1)=1$.

In view of a result by Scott, there exist infinitely many coprime natural numbers $p,q$ such that $q$ is odd and $|p-\tfrac\pi2\,q|<1/q$. 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, $f(1)=1$.

In view of a result by Khinchine (see e.g. Theorem 3.6(a), 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, $f(1)=1$.