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question_about_arccos[cos(x)]

How did we get $2\pi - x$? Kindly provide a general answer because many other similar questions have the same issue.

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    $\begingroup$ For the same reason leading to $\sqrt{x^2}=|x|$, i.e. we have to pay attention to the domain and range of the functions we are manipulating, and their (piecewise) monotonicity. $\endgroup$ – Jack D'Aurizio Oct 5 '16 at 21:43
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First, $\arccos (\cos x)\ne x$ in your case because $x$ is defined to be from $(\pi, 2\pi)$ and range of $\arccos x$ is $[0,\pi]$, that is $\arccos x$ cannot return values you want - it cannot return $x$.

That is why we have to shift $x$ in range of $[0,\pi]$. We can do that by letting $y=2\pi-x$. Remember that $\cos(2\pi-x)=\cos x$, so we did not change original equation.

Now, since $y\in [0,\pi]$ we can write $\arccos (\cos y)=y$ or $\arccos(\cos (2\pi-x))=2\pi-x$ or (final answer) $\arccos (\cos x)=2\pi-x$.

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  • $\begingroup$ I got it and thank you very much. Yet, i'm not confidence to answer every question about the same issue like : arcsin(sinx) if pi / 2 < x < 3pi / 2 $\endgroup$ – MMM Oct 5 '16 at 9:30
  • $\begingroup$ arctan(tanx) when -3pi / 2 < x < -pi / 2 $\endgroup$ – MMM Oct 5 '16 at 9:31
  • $\begingroup$ What general solution i should think of whenever i see such examples? $\endgroup$ – MMM Oct 5 '16 at 9:31
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    $\begingroup$ You have to check domain and range of each arc function, then shift $x$ accordingly. $\endgroup$ – MaliMish Oct 5 '16 at 9:39

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