Prove that range of $e^{2\pi i t\sin (1/t)}=S^1$. Let $\gamma(t)=e^{2\pi i t\sin (1/t)}$, where $t\in [0,2\pi]$
Since $2\pi t\sin (1/t)$ is continuous, we can apply intermediate value theorem, I can find some $t_0\in [0,2\pi]$ such that $\gamma(t_0)=e^{iy_0}$, where $y_0=2\pi t_0\sin (1/t)$.
But how to prove that the range is actually $S^1?$
 A: The idea is the function $e^{2i\pi t}$ is periodic of period 1. And it maps the interval (0,1] (bijection) into the $S^1$ so what you need to do is to show that the function $t\sin(1/t)$ for $t\in [0,2\pi]$ image is an interval of at least length 1. Let
$g(t)=t\sin (1/t)$ for $t\in [0,2\pi]$
And we have
$g(\frac{6}{\pi}) = \frac6\pi \frac{1}{2}=\frac{3}{\pi} $
$g(\frac{2}{3\pi})=-\frac{2}{3\pi}$
As you said it is continuous function so will cover any $\frac{-2}{3\pi}\leq y \leq \frac{3}{\pi}$
And that interval have length $\frac{3}{\pi} + \frac{2}{3\pi} = \frac{11}{3\pi} > \frac{11}{3(3.5)} =\frac{11}{10.5}>1$
Thus even if you look at $t\in [\frac{2}{3\pi} , \frac{6}{\pi} ] \subset [0,\pi]$ the function $\gamma (t) = e^{2i\pi t\sin (1/t)}$ will cover $S^1$
You can think about $\gamma$ as composition of two functions
$\gamma(s) = e^{2i\pi s}$ and
$s(t) = t\sin(1/t)$
$s(t)$ maps the interval $[\frac2{3\pi}, \frac6\pi]$ into the interval $[-\frac{2}{3\pi} , \frac6\pi]$ and gamma maps that to the unit circle. Hope it is clear now.
A: *

*$\{\gamma (t): t\in (0,2\pi]\}\subseteq$ $\{\exp ( i x)_:x\in \Bbb R\}=$ $\{\cos x+i\sin x:\;x\in \Bbb R\}=$ $=\{z\in \Bbb C:|z|=1\}=S^1.$


*The  function  $g(t)=t \sin 1/t$ is continuous for $t\in (0,2\pi).$
Let $t_0=1/\pi.$ Then $g(t_0)=0.$
We have $\lim_{t\to 0^+}g(t)=1.$ And we have $0<g(t)=\frac {\sin (1/t)}{(1/t)}<1$ for all $t\in (0,t_0)=(0,1/\pi).$ So for any $r\in (0,1)$ there exists $t_r\in (0,t_0)$ with $r\le g(t_r)<1.$
So for any $r\in (0,1)$ we have $\{g(t): t\in [t_r,t_0]\}\supseteq [g(t_0),g(t_r)]\supseteq [0,r]$.
So $\{g(t): t\in (0,2\pi]\}\supseteq \cup\{[g(t_0),g(t_r)]:r\in (0,1)\}\supseteq\cup \{[0,r]:r\in (0,1)\}=[0,1).$
So $\{\gamma (t):t\in (0,2\pi]\}=$ $\{\exp (\,2\pi ig(t)\,):t\in (0,2\pi]\}\supseteq $ $\{\exp (2\pi i r): r\in [0,1)\}=$ $=\{\cos x+i\sin x: x\in [0,2\pi)\}=$ $\{\cos x+i\sin x:x\in \Bbb R\}=S^1.$
