In a book its said that for $f:(-\pi/4,\pi/2)\to\Bbb R^2,\,t\mapsto\sin 2t(-\sin t,\cos t)$ then $f^{-1}|_{{\rm img}(f)}$ is not continuous, but it is not clear how to see it. I guess that this is related to the factor $\sin 2t$ that is not invertible in $(-\pi/4,\pi/2)$.
Then setting $x:=-\sin 2t\sin t$ and $y:=\sin 2t\cos t$ then $g(x,y):=\arctan(-x/y)$ is a candidate for the inverse of $f^{-1}|_{{\rm img}(f)}$. I dont see clearly how to continue from here. Some help will be appreciated, thank you.
Ok, I solved it. It is true that $f^{-1}|_{{\rm img}(f)}$ is not continuous because there are two paths in ${\rm img}(f)$ that converge to $(0,0)$ but there is a unique $f^{-1}(0,0)=0$. That is: when $t\to \pi/2$ then $f(t)\to 0$.
Thus, from the definition of continuity, if $h:=f^{-1}|_{{\rm img}(f)}$ is continuous at $(0,0)$ then for an arbitrarily small $\epsilon>0$ there is a $\delta>0$ such that
$$(x,y)\in \Bbb B((0,0),\delta)\cap{\rm img}(f)\implies h(x,y)\in\Bbb B(0,\epsilon)\tag1$$
However this is not true because we can find some $t_0$ arbitrarily close from the left to $2\pi$ such that $(x_0,y_0):=(f_1(t_0),f_2(t_0))\in\Bbb B((0,0),\delta)\cap{\rm img}(f)$ for any chosen $\delta>0$ and $h(x_0,y_0)$ will be arbitrarily close to $2\pi$, thus for $1>\epsilon>0$ we can see that $(1)$ cannot hold.