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$\sin^{-1}(-\sin(x))$ = $-\sin^{-1}(\sin(x))$

Can the minus be taken out like this?

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  • $\begingroup$ It can be done only if you are using a specific domain/range in defining arcsine. See my answer below. Many of the other answers make this assumption, but it is patently false except in a very specific circumstance. When you work with inverses of functions that are not 1-1, you have to be very very careful or you can gloss over something and arrive at a wrong answer. $\endgroup$
    – MPW
    Feb 15 '14 at 14:47
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Generally, it is not true that $$f^{-1}(-f(x)) = - f^{-1}(f(x))$$ Try to cook up a counter-example!

However, in this case, since $\sin$ is odd we get that $$\sin^{-1}(-\sin(x)) = \sin^{-1}(\sin(-x)) = -x = - \sin^{-1}(\sin(x))$$

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    $\begingroup$ I think principal value should find a place in the answer $\endgroup$ Feb 15 '14 at 14:41
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    $\begingroup$ It is not true that $\sin^{-1}(\sin(x)) = x \quad \forall x$. $\endgroup$
    – Ben Voigt
    Feb 15 '14 at 18:11
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Yes, that is valid, since the $\arcsin$, like $\sin$ is an odd function.

Here, that means that $\sin^{-1}(-f(x)) = -\sin^{-1}(f(x))$, and in this case, $f(x) = \sin x$. So ultimately, we have that $$\sin^{-1}(-\sin(x)) = -x$$

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  • $\begingroup$ Is arccos even and arctan odd? $\endgroup$
    – salman
    Feb 15 '14 at 14:21
  • $\begingroup$ Think about the range of the functions: $\arccos(1/2) = \pi/3$, but $\arccos(-1/2) = 2\pi/3$. Moral: $\arccos$ is not even (although $\cos$ is). However, $\arcsin$ and $\arctan$ are both odd. $\endgroup$ Feb 15 '14 at 14:29
  • $\begingroup$ Salman: you can always graph a function to help determine whether whether even, or odd, etc. $\arccos$ is not even, but it is true that $\arctan,$ like $\arcsin$, is odd. $\endgroup$
    – amWhy
    Feb 15 '14 at 14:40
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    $\begingroup$ It is not true that $\sin^{-1}(\sin(x)) = x \quad \forall x$. arcsin is odd, and so the manipulation the question asks about is correct, but it is not a true inverse of sin, because the sine function is not invertible. $\endgroup$
    – Ben Voigt
    Feb 15 '14 at 18:10
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It is only valid if you take the domain of $\sin^{-1}$ to be $[-1,1]$ and the range $[-\pi/2,\pi/2]$. There are other definitions for which this doesn't hold. For example, you could take the range to be $[\pi/2,3\pi/2]$.

Any restriction to a symmetric subinterval (open or closed) will work as well.

Comment: From my comment above on the OP's question:

"It can be done only if you are using a specific domain/range in defining arcsine. See my answer below. Many of the other answers make this assumption, but it is patently false except in a very specific circumstance. When you work with inverses of functions that are not 1-1, you have to be very very careful or you can gloss over something and arrive at a wrong answer"

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Using the definition of principal value of inverse sine function,

we can always find an integer $n$ such that $\displaystyle x=n\pi+(-1)^n y$ with $\displaystyle -\frac\pi2\le y\le\frac\pi2$

$\displaystyle\implies-\frac\pi2\le-y\le\frac\pi2 $

$\displaystyle\implies\sin x=\sin(n\pi+(-1)^n y)=\sin y $

$\displaystyle\implies-\sin x=-\sin y=\sin(-y)$

$\displaystyle\implies\sin^{-1}(\sin(-y))=-y$ as $\displaystyle-\frac\pi2\le-y\le\frac\pi2 $

Similarly, $\displaystyle\sin^{-1}(\sin x)=\sin^{-1}(\sin y)=y$ as $\displaystyle-\frac\pi2\le y\le\frac\pi2 $

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Yes you are correct. It can be taken out since the value of $sin(x)$ is taken as an angle.

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