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The Domain of $f(x)$ is $(0,1)$.

When finding the domain of $f(\ln(x))$ We can say: $0<\ln x<1$ and apply the inverse $\ln$ function $\exp$.

We get $e^0 < e^{\ln(x)} < e^1$ which leads to $1<x<e$.

When trying to apply the same method on $f(\cos(x))$ we get : $\cos^{-1}0 < x < \cos^{-1}1$ which leads to $\frac{\pi}{2} < x < 0$ .

I know that this is flawed because the domain of $f(\cos(x))$ while : $0 < \cos(x) < 1$ should be: $-\frac{\pi}{2} < x < \frac{\pi}{2}$ (neglecting periodicity).

Where is the flaw in this method for the use on $\cos$ function?

Mostafa Ayaz
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queency3 jones
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  • $0$ can't be in the domain as $\cos(0)=1$. The answer you expect is probably like $(-\frac\pi2, 0) \cup (0, \frac\pi2)$ if you restrict the study to $[-\pi,\pi]$. – Lelouch May 07 '23 at 16:24
  • You could ask a similar question about $f(1-x)$, incorrectly saying $1<x<0$ based on the false assumption $1-x$ is increasing. – coiso May 07 '23 at 16:29

1 Answers1

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$\cos$ is not bijective. When you use the inverse cosine function, you implicitly restrict your study on a half-period, for example on $[0, \pi]$ (if you want to use the classic definition of $\arccos$ or $\cos^{-1}$), where $\cos$ is strictly decreasing. Note that $\arccos$ is also a decreasing function therefore you have to reverse the order of the inequality.

$ 0 < \cos(x) < 1 \Leftrightarrow \arccos(0) > x > \arccos(1) $

And $\arccos(0)=\frac\pi2$ and $\arccos(1)=0$

Which gives you the final inequality $\frac\pi2 > x > 0.$

And this is expected, since these are truly the only values of $x$ that satisfty $0< \cos(x) < 1$ when $x$ belongs to $[0,\pi]$.

If you want to extend the result to all real values, then you have to use the symmetry ($\cos$ is even) to see that the answer is $(-\frac\pi2,0)\cup(0,\frac\pi2)$ on $[-\pi,\pi]$ and then use the periodicity to see that every interval $(2k\pi-\frac\pi2, 2k\pi)\cup(2k\pi, 2k\pi+\frac\pi2)$ works for $k$ integer.

Lelouch
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