How are arcsine and cosecant different mathematically if cosecant is $\frac{1}{\sin(x)}$ and arcsine is $\sin^{-1}(x)$ which is $\frac{1}{\sin(x)}$? I have tried to find an answer before but nobody explained it well enough.
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4I think the key thing to remember here is that the notation is genuinely confusing (it's very much not just you!). We're very happy to write $\sin^2(x)$ to mean $(\sin(x))^2$ and $\sin^3(x)$ to mean $(\sin(x))^3$, but when we write $\sin^{-1}(x)$, we don't mean $(\sin(x))^{-1}$! – Theo Bendit Aug 30 '18 at 02:09
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@TheoBendit: "We're very happy"? No, not at all. I never write "$\sin^2(x)$" to mean "$(\sin(x))^2$". I don't think convention is a sufficient reason for preserving inconsistent notation. After all, it is just as easy to write "$\sin(x)^2$", which is consistent with the standard function notation and arithmetical syntax. – user21820 Aug 30 '18 at 03:29
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1@user21820 Fair enough. I do see some advantages of writing $\sin^2$; for example, when you wish to refer to the function $\sin \cdot \sin$, without having to specify a variable. I think there's a slight bump in readability when you substitute some long expression into $\sin^2$. Personally, I don't like the notation $\sin^{-1}$; it's not like $\sin$ has an inverse anyway. That's why I use $\arcsin$ instead. – Theo Bendit Aug 30 '18 at 03:34
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@TheoBendit: Hmm yes so in my opinion the correct question is whether we should define $f^p$ to mean the $p$-th iterate of $f$ or the pointwise $p$-th power of $f$, where $f$ is a function of an appropriate type. I lean towards the first, because we already use it extensively for operators (like transformations), and because I have not yet seen an example where the other is really useful. (Though I agree with you about $\arcsin$ and the tiny bump in readability.) – user21820 Aug 30 '18 at 03:45
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@user21820 Also in the linear algebra/functional analysis setting, it's helpful to, say, refer to things like $|\sin^2|_\infty$ in function spaces; saying $\sin(x)^2$ introduces a variable which just confuses the matter, especially if $x$ is not a number (e.g. $C^*$ algebra continuous function calculus)! There arguably doesn't need to be one notation that trumps the other, so long as newbies are informed of the traps. – Theo Bendit Aug 30 '18 at 03:53
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1@TheoBendit: Well yea there are trade-offs if one wants to pick one. I still strongly prefer not having notation that cannot be disambiguated, and so if one wants to keep both notations, one should define/specify it before using. =) – user21820 Aug 30 '18 at 04:00
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Related: https://math.stackexchange.com/questions/30317/arcsin-written-as-sin-1x – Hans Lundmark Aug 30 '18 at 06:30
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1Possible duplicate of What's the difference between arccos(x) and sec(x) – Hans Lundmark Aug 30 '18 at 06:32
4 Answers
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They are two totally different functions with different domains and ranges and different definitions.
The notation is confusing and it takes a while for students to master the concepts.
Note that the arcsine function or $\sin ^{-1} x$ as it is common to write is the inverse function under composition not under multiplication.
That is $$\sin ( \sin ^{-1} x )=x$$ is true on the domain of $ \sin ^{-1} x$
While for $\csc x$ the story is different because it is the multiplicative inverse of $\sin x$ which is $$ ( \csc x)\times (\sin x) =1$$
Mohammad Riazi-Kermani
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$\sin^{-1}(x)$ and $\frac{1}{\sin x}$ are not same
$\sin^{-1}(x)$ is equal to some angle $\theta $ such that $\sin \theta =x $
where $-1\le x \le 1$ and Range $\in [-\pi /2,\pi /2]$
Here is the graph of $\sin^{-1}(x)$
while $\frac{1}{\sin x} $ is simply 1 divided by $\sin x $ $\hspace{20pt}$here $x \in R$ and Range $\in (-\infty,\infty)$
Here is the graph of $\frac{1}{\sin x}$
Deepesh Meena
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$\sin^{-1}(x)$ doesn't mean $\frac{1}{\sin(x)}$. It means the inverse function of the $\sin$ (when restricted to a conventional interval).
If I have some invertible function $f$ on some interval, so that I can write say $y=f(x)$ then we use $f^{-1}$ to represent the inverse function; we can write $x=f^{-1}(y)$. However, this can sometimes cause confusion with the multiplicative inverse, which can also sometimes be written using a $\cdot^{-1}$-power.
See https://en.wikipedia.org/wiki/Inverse_function ... (and note the "not to be confused with" link right under the title)
So $\arcsin$ is the functional inverse of $\sin$ while $\operatorname{cosec}(x)$ is the multiplicative inverse of $\sin(x)$.
The functional inverses of trig functions are discussed here:
https://en.wikipedia.org/wiki/Inverse_trigonometric_functions
Mathematicians are often inconsistent; when they write $\sin^2(x)$ they mean $(\sin(x))^2$ (rather than say functional composition, $\sin(\sin(x))$), and similarly for other powers - even $-2$ (!) ... but not if that power is $-1$. It's a matter of getting used to the convention, and sticking to $1/\sin(x)$ or $(\sin(x))^{-1}$ when you mean to talk about the multiplicative inverse.
Deepesh Meena
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Glen_b
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@James If you're going to change that in my answer, change it in the question (as well as the other time I used it); I deliberately used that form to be consistent with the question. – Glen_b Aug 30 '18 at 02:13
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done but if you are not happy with the edit tell me I will undo it – Deepesh Meena Aug 30 '18 at 02:15
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I am happy for you to try to improve the answer (that's how the system is supposed to work) but it's better to be consistent about it. – Glen_b Aug 30 '18 at 02:17
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@James unfortunately when you edited the question, you left one of the $\sin$ functions as $sin$ (just as you did in your answer); this is too small an edit for me to be able to fix in the question. You need to take care with edits. – Glen_b Aug 30 '18 at 02:19
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To avoid any ambiguity between the reciprocal and the inverse, one may write $$f^{-1}$$ for the reciprocal and $$\mathop f^{-1}$$ for the inverse.
Michael Hoppe
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