Understanding trig equality

Question

I was working on a math question and I was wondering if this is true

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

Thanks.

Answer 1

This is false, but I can see where you got it from. In short:

$$\sin^{-1}(x)=y\to x=\sin(y)$$

$$\csc(x)=y\to \frac{1}{\sin(x)}=y\to x=\sin^{-1}(\frac1y)$$

While in most cases, raising a function to the power $-1$ means flipping it onto the denominator, here $\sin^{-1}(x)$ represents the inverse of $\sin(x)$, as opposed to the reciprocal. In a lot of basic cases, they mean the same thing, but here they don't.

Answer 2

Pretend notation wasn't invented.

$\frac 1{\sin x}$ (no matter what notation is given) a different concept than "for a given $x$ find the angle $\theta$ so that $\sin \theta = x$" (no matter what we call) and these values are basically unrelated. And if we had any doubt we'd just have to try a few examples to get that this is ... absurd.

($\sin 60 = \frac {\sqrt 2}3$ and so $\frac 1{\sin 60} = \frac 3{\sqrt 2}$. So... if $\sin \theta = 60$ [which is absurd and impossible; $\sin \theta < 1$] then $\theta = \frac 3{\sqrt 2}$???? That's just nutty. It makes zero sense. Zero sense at all.)

Then it becomes an unhappy ill thought out occurence that we frequently take $f^k(x)$ to mean "raise $f(x)$ to the $k$-th power" (if we aren't using to mean, instead "take the $k$-th derivative of $f(x)$"). And we take $f^{-1}(x)$ to mean "the inverse of $f(x)$ " i.e. "For $x$ find the value of $w$ so that $f(w) = x$".

This leads to $f^{-1}(x)$ being ambiguous. You will almost never see $f^{-1}(x)$ to mean $\frac 1{f(x)}$ but you might see in $f^{-3}(x)$ to mean $(\frac 1{f(x)})^3$. Is this consistent and unambiguous? Not in the least. It's ambiguous and quite frankly a mess.

However, it's worth noting that if you ever see $\sin^{-1} x$, it will almost certainly mean $\arcsin x$.... unless it doesn't.... I can't say you'll never see it to mean $\frac 1{\sin x}$. You might. I hope not but it could happen.

What can I say but.... sorry?