A question regarding the power of a sinus function. My calculus book says the following:
$\sin^{n}{x}$ is short for $(\sin{x})^n$, $n\neq -1$.
This is obviously trivial, however I don't get it why it says  $n\neq -1$. The above condition should for $n=-1$, right?
Thank you very much for your help!
Regards,
Charlie
 A: I would go as far as saying the notation
$$
\sin^n x = (\sin x)^n
$$
is only standard when $n$ is a positive integer. $\sin^{-1}$ means $\arcsin$, and I have never seen $\sin^{-2}$ or the like. It's for this very reason that people consider the $\sin^{-1}$ notation to be confusing, since it does not conform with how we interpret $\sin^n$ generally. There are other reasons this notation is illogical: strictly speaking, $\sin$ does not have an inverse, and so writing $\sin^{-1}$ is somewhat suspect to begin with. I go into more detail about this here.
A: It's a tricky notation problem. Generally in mathematics, $f^{-1}(x)$ means the functional inverse--not literally $\frac{1}{f(x)}$. That's why there's the distinction for $n \neq -1$. Furthermore, there is a function for $\frac{1}{\sin(x)} = \csc(x)$ (the cosecant), same for cosine (secant), and same for tangent (cotangent).
Personally, I'm not a fan of the $\sin^{-1}(x)$ notation for the inverse sine function (for the exact reason you're asking about). When I taught calculus, I tried to convince my students that this was, indeed, a confusing notation and urged them to, instead, use $\arcsin(x)$, $\arctan(x)$, and $\arccos(x)$ for the inverses.
I will add that there is the pesky notation of $\frac{df}{dx} = f'(x)$, $\frac{d^2f}{dx^2} = f''(x)$, and so on...it's gets cumbersome at a point, i.e. $\frac{d^6f}{dx^6} = f''''''(x)$??? So the general notation is a parenthesis: $\frac{d^6f}{dx^6} = f^{(6)}(x)$.
