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I've come across different uses and meanings for $\sin^{-1}(x)$ by various authors:

  • $\sin^{-1}(x) = \arcsin(x)$
  • $\sin^{-1}(x) = \frac{1}{\sin(x)} = \csc(x)$

The Wikipedia article "Sin-1" says:

$\sin^{−1}y = \sin^{−1}(y)$, sometimes interpreted as $\arcsin(y)$ or arcsine of y, the compositional inverse of the trigonometric function sine (see below for ambiguity)

$\sin^{-1}x = \sin^{−1}(x)$, sometimes interpreted as $(sin(x))^{−1} = \frac{1}{\sin(x)} = \csc(x)$ or cosecant of x, the multiplicative inverse (or reciprocal) of the trigonometric function sine (see above for ambiguity)

When I look on Khan Academy, it tells me that $\sin^{-1}(x)$ does not stand for $\csc(x)$, instead:

If a number or variable is raised to the $-1$ power, then this refers to the multiplicative inverse, or the reciprocal. For example, $3^{-1} = \frac{1}{3}$. In general, if $a$ is a nonzero real number, then $a^{-1} = \frac{1}{a}$.

However, this is not the case for $\sin^{-1}(x)$. This is because the sine is a function, not a quantity!

In general, whenever you see a raised $-1$ after a function name, it refers to the inverse function.

Is it therefore considered a notational mistake to use a $-1$ exponent to denote the multiplicative inverse of a function?

Community
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Mandelmus100
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    No, it's abuse of Notation, that's correct. It should be clear from context, which of the two is meant. If not, it should be stated somewhere. Some Authors also make the difference (which is better imo), that $\sin^{-1}(x)$ is the inverse sine function and $\sin(x)^{-1}$ is the reciprocal of the sine function. Note that there is another ambiguity, that is, $f^{-1}(x)$ could also mean the preimage of x, if f is not invertible. That should also be clear from context though. – Luke Feb 11 '20 at 11:06
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    I think it's origin is just plain lazyness. But I have no sources to back it up. – mathreadler Feb 11 '20 at 11:13
  • Some authors adopt the notation $\dfrac{1}{f}(x)$ for $\dfrac{1}{f(x)}$. We usually use $f^{-1}$ as an inverse operator. See my discussion here. – Ng Chung Tak Feb 11 '20 at 11:16
  • Yes, it is considered a mistake to use a $-1$ exponent to denote the multiplicative inverse of a function. – Servaes Feb 11 '20 at 11:39
  • See the Linked and Related questions for previous discussions of this issue. – Gerry Myerson Feb 11 '20 at 12:11
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    Really, it's using $\sin^2(x)$ as shorthand for $\sin(x)\cdot \sin(x)$ that is the real abuse of notation, assuming exponents on functions means composition. Which generally speaking, is more common I think. At least in algebra. – Arthur Feb 11 '20 at 12:25
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    "authoritative consensus" seems to be a bit of a contradiction. Either you look to authorities as authoritative, or you look for a consensus. The Wikipedia article shows that there's no consensus about this notation. – joriki Feb 11 '20 at 12:59
  • It is worth pointing out that while in the United States it is far more common to see $\sin^{-1}(x)$ refer to $\arcsin(x)$... my experience with foreign students from Asia in particular was that $\sin^{-1}(x)$ more often refers to $\csc(x)$ in many of the regions they come from. For that reason, when working with a group of people of mixed backgrounds, I always used $(\sin(x))^2$ or $\arcsin$ or similar to make it clear what was meant. – JMoravitz Feb 11 '20 at 13:32

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Yes, I think it's fair to say it's wrong to write $\sin^{-1}(x)=1/\sin(x)$, unless you specify at the start that you're using the notation this way.

But the reasons you cite from Wikipedia and Khan are bogus! The reason the above is wrong is simply because it's a standard convention that $\sin^{-1}(x)=\arcsin(x)$. If the reasons given were valid they would show just as well that $\sin^2(x)=\sin(\sin(x))$, while in fact the standard convention there is $\sin^2(x)=(\sin(x))^2$.

David C. Ullrich
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