Question on the property of algebraic number

Question

Given $x$ an algebraic number of degree 2. Anyone help me to prove

$$ |\sin(n\pi x)| \geq \frac{\mathrm{const.}}{n} $$ for all $n$ ?

Answer 1

We denote $d(y, A) = \inf \{\mid a - y \mid,\ a \in A \}$. If you consider $x$ an algebraic number of degree $2$, then there exists $C>0$ s.t. $\forall (p,q) \in \mathbb{Z}\times \mathbb{N}^*, \mid x - \frac{p}{q} \mid \ge \frac{C}{q^2}$

Hence $$\forall n \in \mathbb{N}^*,\ d(nx,\ \mathbb{Z}) \ge \frac{C}{n}$$

Thus $d(n\pi x, \pi \mathbb{Z}) \ge \frac{\pi C}{n}$ $(*)$.

Finally, you can use that for $h \in [-\frac{\pi}{2},\frac{\pi}{2}],\ \mid \sin (h) \mid \ge \frac{1}{2} \mid h\mid$ and $\mid \sin (\pi + h) \mid \ge \frac{1}{2} \mid h \mid$.

Therefore, $$\mid \sin (n \pi x) \mid \ge \frac{1}{2} \frac{\pi C}{n}$$ (simply write $x = k\pi + h$ with $k \in \mathbb{Z}$ and $h \in [-\frac{\pi}{2},\frac{\pi}{2}]$, and $(*)$ gives $\mid h \mid \ge \frac{\pi C}{n}$).

We take $A = \frac{\pi C}{2}$, and we have, for all $n$, $\mid \sin (n\pi x) \mid \ge \frac{A}{n}$.