Maybe this is an easy question, but I can't figure it out!
What is the inverse mapping of:
$f(t) = (\cos(t), \sin(t))$,
where $f: [0, 2\pi) \to S^1$
Does one just invert the components separately?
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$\begingroup$Formally, you have $f^{-1}(\cos t, \sin t)=t$. Maybe more practical is the formula $$f^{-1}(x,y)= \begin{cases} \arccos x \quad \text{if }\, y\geq 0\\ 2\pi-\arccos x \quad \text{if} \, y<0, \end{cases} $$ assuming $x^2+y^2=1$ (for instance).
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