# Difference between $\sin^{-1}(x)$ and $\frac1{\sin(x)}$? [closed]

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.

## closed as off-topic by user21820, user91500, Hans Lundmark, Gibbs, José Carlos SantosAug 30 '18 at 13:02

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• I 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 2:09
• @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 3:29
• @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 3:34
• @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 4:00
• Possible duplicate of What's the difference between arccos(x) and sec(x) – Hans Lundmark Aug 30 '18 at 6:32

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 $$( \csc x)\times (\sin x) =1$$

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

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

• @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 2:13
• done but if you are not happy with the edit tell me I will undo it – Deepesh Meena Aug 30 '18 at 2:15
• 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 2:17
• Ok so what do you want – Deepesh Meena Aug 30 '18 at 2:18
• @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 2:19

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.