How would you evaluate $\liminf\limits_{n\to\infty} \ n \,|\mathopen{}\sin n|$

Question

How would you evaluate the limit inferior of the sequence $n\,|\mathopen{}\sin n|$? That is, $$\liminf\limits_{n\to\infty} \ n \,|\mathopen{}\sin n|$$

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Edit. Let $\mu$ be the irrationality measure of $\pi$. Since $\mu$ is not known, I will split the question:

1. Assuming $\mu > 2$, what is $\liminf\limits_{n\to\infty} \ n\,|\mathopen{}\sin n|$?
2. Assuming $\mu = 2$, what is $\liminf\limits_{n\to\infty} \ n\,|\mathopen{}\sin n|$?

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I kind of like its graph...