$| \sin x| > \frac{\sqrt{2-\sqrt{2}}}{2}$ iff $1/8< \{ \frac{x}{\pi}\} < 7/8$ where $\{x\}$ is the fractional part of $x$

Question

This is a problem that arose while reading the book "Putnam and Beyond":

Why is $| \sin x| > \frac{\sqrt{2-\sqrt{2}}}{2}$ iff $\frac18 < \{ \frac{x}{\pi}\} < \frac78$ where $\{x\}$ is the fractional part of $x?$

This was used in proving the divergence of $\sum \frac{ |\sin n|}{n}$, although the divergence is not in question here.

Answer 1

Because we need to prove that: $$|\sin x|>\frac{\sqrt{2-\sqrt2}}{2},$$ which is $$\frac{\pi}{8}+2\pi k<x<\frac{7\pi}{8}+2\pi k$$ or $$\frac{9\pi}{8}+2\pi k<x<\frac{15\pi}{8}+2\pi k,$$ which is $$\frac{\pi}{8}+\pi k<x<\frac{7\pi}{8}+\pi k,$$ where $k$ is an integer numbers, which is true because $$\frac{1}{8}<\left\{\frac{x}{\pi}\right\}<\frac{7}{8}$$ it's $$\frac{1}{8}<\frac{x}{\pi}-\left[\frac{x}{\pi}\right]<\frac{7}{8}$$ or $$\frac{\pi}{8}+\pi\left[\frac{x}{\pi}\right]<x<\frac{7\pi}{8}+\pi\left[\frac{x}{\pi}\right].$$ Now, take $k=\left[\frac{x}{\pi}\right].$

Actually, $$\sin\frac{\pi}{8}=\sin\frac{7\pi}{8}=\sqrt{\frac{1-\cos45^{\circ}}{2}}=\sqrt{\frac{1-\frac{1}{\sqrt2}}{2}}=\frac{\sqrt{2-\sqrt2}}{2}.$$