Why is the symbol for $\arcsin(\theta)$ also $\sin^{-1}(\theta)$ if $\sin^2(\theta)=(\sin(\theta))^2$? Why is the symbol for $\arcsin(\theta)$ also $\sin^{-1}(\theta)$ if $\sin^2(\theta)=(\sin(\theta))^2$? I understand what they represent but I am curious why are the symbols like this?
 A: Here's the answer I give to (fairly strong) students in my calculus tutorials when they ask about this. I usually offer a handout with this explanation to anyone who is frustrated by this clash in notation.
The operation of multiplying nonzero real numbers has a couple of important properties that are relevant here:


*

*for any real numbers $a,b$, the number $a \cdot b$ is still a real number (this doesn't require that they be nonzero)

*$a \cdot 1 = a$ for all real $a$ (doesn't require nonzero either)

*for all nonzero $a$, there is a number $b = 1/a$ such that $ab = 1$
(The algebraic term for a collection of numbers or operations with properties like this is a group.)  Since $a^{m+n} = a^m a^n$ for positive integers $m,n$, the third property gives us the idea to define $a^0 = 1$, which makes it natural to define $a^{-1} = 1/a$.
When it comes to invertible functions on the real line, there are two different operations that satisfy these properties: pointwise multiplication, i.e $f \cdot g$ defined by $(f \cdot g)(x) = f(x) \cdot g(x)$; and composition, i.e. $f \circ g$ defined by $(f\circ g)(x) = f(g(x))$. For composition, the "$1$" is actually the identity function, $\mathrm{id}(x) = x$. We have to be careful about dividing by zero and making sure that domains and ranges line up nicely and everything, but morally this is what is happening. 
Since both operations (pointwise multiplication and composition) mostly satisfy the group properties that motivate our use of exponent notation for real numbers, it's only natural to put exponents on functions, too. But there's an ambiguity: $f^3$, for instance, could mean
$$
f^3(x) = f(x) \cdot f(x) \cdot f(x)
$$
or 
$$
f^3(x) = f(f(f(x)))
$$
This ambiguity extends to negative exponents, too. So $g = f^{-1}$ could refer to either of two functions:


*

*the reciprocal or multiplicative inverse of $f$, i.e. the function $g$ such that $f(x) \cdot g(x) = 1$ for all $1$ (whenever $f(x) \neq 0$)

*the (compositional) inverse of $f$, i.e. the function $g$ such that $f \circ g = g \circ f = \mathrm{id}$, i.e. $f(g(x)) = g(f(x)) = x$ (whenever $x$ is in both the domain and the range of $f$)


In most situations, it's clear from context which one of these is intended, but when you're learning, it can be tough to interpret the context. The situation is particularly unfortunate for the basic trig functions, because they come in before students have had a lot of experience reasoning about functions more systematically.
A: Too many parenthesis are annoying, and this explains why $(\sin(x))^2$ is usually written $\sin^2(x)$. In the same contexts, the multiplicative inverse will often be written as a fraction $\dfrac1{\sin(x)}$, or using the corresponding function $\csc(x)$. And the negative powers give $\dfrac1{\sin^2(x)}$ or $\csc^2(x)$.
Unfortunately, for the functional inverse we only have the exponent notation and write $f^{-1}(x)$ for general functions. This comes from the notation for function iteration $f^2(x):=f(f(x))$, which enforces an additive composition law $f^n(f^m(x))=f^{n+m}(x)$. So it is coherent that $f(f^{-1}(x))=f^1(f^{-1}(x))=f^{1-1}(x)=f^0(x)=x$. 
In practice, $\sin^n(x)$ will "by default" denote a power of the sine for any $n\ne-1$, because iteration of the sine is exceptional (and iteration of the inverse virtually never used). On the opposite, $\sin^{-1}(x)$ is ambiguous and one should prefer $\arcsin(x)$ or $\csc(x)$. If you ever use $^{-1}$, make sure that the meaning is obvious from context, or announce it.
