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

Question

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.

Answer 1

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

Answer 2

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

Answer 3

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

Answer 4

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.