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It seems that $\sin^2(x)$ is used to denote the square of whatever value $\sin(x)$ is, instead of the expected $(\sin(x))^2$.

Based on that, I would assume that $\sin^{-1}(x) = \frac{1}{\sin(x)}$, but it turns out that $\sin^{-1}(x)$ is used to indicate the inverse function of sin, arcsin.

What are the reasons for this apparent lack (to me, at least) of notational coherence, and how do you write $\frac{1}{\sin(x)}$ in exponential notation?

I saw this: $\sin^2$ notation and uses of the alternative. , but I don't think it really answers what I'm asking (specifically, answers are along the lines of "it is what it is").

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    $\begingroup$ Notation is not always unique. Is $n\choose k$ a binomial coefficient or a two-dimensional row vector? - Here we have the problem that functions may form a multiplicative algebraic structure both under composition and under pointwise multiplication ... $\endgroup$ – Hagen von Eitzen May 4 '15 at 20:55
  • $\begingroup$ This question doesn't even deserve a separate answer. You may interpret $\sin^2(x)$ as the function $\sin$ squared, then applied to $x$. $\endgroup$ – Alex M. May 4 '15 at 20:55
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    $\begingroup$ Since we already have a function $\csc(x) \overset{\text{def}}{=} \frac{1}{\sin x}$, it's safe to assume that $\sin^{-1}$ is never referring to cosecant. But you can take solace in knowing it will always cause confusion to the uninitiated :) $\endgroup$ – pjs36 May 4 '15 at 20:56
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    $\begingroup$ It's much more common nowadays for $f^n(x)$ to mean $\underbrace{f \circ f \circ \cdots \circ f}_{n}(x)$ than to mean $f(x)^n$, and $f^{-1}(x)$ almost invariably means the inverse function. My impression is that $\sin^2 x$ is an historical exception, and is still around because it's so common to square $\sin x$. Should you ever need to repeatedly compose $\sin x$, you can say: let $f(x) = \sin x$ and consider $f^n(x)$. Or, at grave risk of being misread, I suppose you could write $\sin^{\circ n} (x)$. $\endgroup$ – aes May 4 '15 at 21:10
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    $\begingroup$ Personally, I never write $\sin^{-1}x$ as it's too ambiguous. I write the inverse function as $\arcsin x$ and the multiplicative inverse as $\csc x$ $\endgroup$ – Dylan May 5 '15 at 1:49
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There are two uses of notation here: writing $f^2(x)$ for $ f(f(x)) $, and writing $f^2(x)$ for $(f(x))^2$. The latter is basically only standard for trigonometrical functions, because it saves time over writing $(\sin{x})^2$. I would assume this is historically based, given how frequently we write powers of trigonometrical functions.

The other notation, $\sin^{-1}{x}$, is a consequence of the modern function notation, where $f^{-1}$ is the inverse of $f$, that is, a function so that $f(f^{-1}(y)) = y$, and $f^{-1}(f(x)) = x $. Why this was ever introduced is quite puzzling, to say the least, considering that we have the perfectly comprehensible notation $\arcsin{x}$ for the inverse of the sine function. What is worse, the inverses of the trigonometrical functions are not actually inverses, since these functions are periodic, and therefore not injective. Throwing out the $\arcsin$ notation would be much the same as throwing out the notation $\log{x}$ for the inverse of the exponential function.

All you can do is encourage the use of unambiguous notation, and explain that the notation for trigonometrical functions is not standard functional notation: for generic functions, most mathematicians do not leave out the brackets around the arguments, but this is entirely standard for trigonometrical functions, for example.

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I agree that this is very confusing.

Luckily there seem to be a certain consistency, at least within the class of trigonometric functions, that the notation $\sin^{-1}(x)$ always means the inverse function of $\sin(x)$.

If you want to denote $\frac{1}{\sin(x)}$ in a similar way, I suggest writing $\sin(x)^{-1}$, or maybe even $(\sin(x))^{-1}$, to avoid ambiguity.

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    $\begingroup$ That's why I like to use $\arcsin$ instead of $\sin^{-1}$ $\endgroup$ – steven gregory May 4 '15 at 21:02
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    $\begingroup$ Some people write $\frac1{\sin x}$ as $\csc x$. $\endgroup$ – MJD May 4 '15 at 21:06

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