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The function $f(x)=\sin(x)$ is monotone in $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$ thus it is invertible and its inverse function ${\rm arcsin}(x)$ is defined from $[-1, 1]$ to $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$.

Anyway, the function $f(x)=\sin(x)$ is monotone also in $\left[\frac{\pi}{2}, \frac{3\pi}{2}\right]$ so why don't define its inverse from $[-1, 1]$ into $\left[\frac{\pi}{2}, \frac{3\pi}{2}\right]$?

I am sorry if it is a dumb question. Thank you in advance!

N. F. Taussig
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C. Bishop
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    It is not a dumb question. You could also define an inverse there. What's nice about $[-\frac\pi2,\frac\pi2]$ is that it is symmetric about $0$, and $\arcsin\colon[-1,1]\to[-\frac\pi2,\frac\pi2]$ is odd. I guess your $\sin^{-1}\colon[-1,1]\to[\frac\pi2,\frac{3\pi}2]$ function is simply equal to $\sin^{-1}(x)=\pi-\arcsin(x)$. – nejimban Oct 27 '21 at 15:08
  • You can define a different pseudo-inverse. There really isn’t an inverse function of a continuous function which is not bijective, but we can define “local” inverses. We define $\arcsin$ to have the range $[-\pi/2,\pi/2]$ for simplicity, not because the choice has a mathematical significance. – Thomas Andrews Oct 27 '21 at 15:13
  • Of course you can define such a function! Let $f:[\pi/2,3\pi/2]$ be given by $f(x)=\sin x$. The function $f^{-1}$ is the inverse of $\sin$ on the interval $[\pi/2,3\pi/2]$. Of course, the function $f^{-1}$ is less often used by the mathematical community than $\arcsin$ (the fact that $\arcsin$ is a "named" function is already indicative of this). But that does not make it wrong to use a different inverse for your purposes. – Joe Oct 27 '21 at 15:22
  • It's the same story with division by zero. We say that "division by zero is undefined", but really what we should say is "in most cases, mathematicians find it more convenient to not define division by zero than to define it". But if you find an instance where it is useful to define the division function differently, then there is nothing wrong with that. As J.G. puts it succintly here, "definitions aren't right or wrong; claims about their consequences are". – Joe Oct 27 '21 at 15:22
  • The main thing we want is continuity, because we could also choose $\arcsin(x)$ to be in $(\pi,3\pi/2]$ when $x<0$ and in $[0,\pi/2]$ when $x\geq 0.$ We want continuity, and we want $[-1,1]$ to be the domain, and we want $\sin(\arcsin x)=x.$ We might prefer $\arcsin x$ increasing, which would reject $[\pi/2,3\pi/2],$ but would not reject $[3\pi/2,5\pi/2].$ – Thomas Andrews Oct 27 '21 at 15:24
  • One of the interesting cases is $\arctan.$ Because we often really want to get the full angle in $(-\pi,\pi]$ for polar coordinate reasons, we actually define a special function of two variables, $\operatorname{arctan2}(x,y),$ which, when $x>0,$ give us $\operatorname{arctan2}(x,y)=\arctan(y/x).$ This also let’s us deal with the case $\operatorname{arctan2}(0,1)=\frac{\pi}2,$ which would require us to write $\arctan(+\infty)=\pi/2,$ which is technically incorrect. But here, we could take any range $[\alpha,2\pi+\alpha)$ or $(\alpha,2\pi+\alpha].$ – Thomas Andrews Oct 27 '21 at 15:33
  • There really isn’t an easy way to do a similar $\operatorname{arcsin2}(x,y)$ where $x$ is the hypotenuse, since the hypotenuse is always positive. – Thomas Andrews Oct 27 '21 at 15:38
  • Note the Analytic Continuation for sine. – Тyma Gaidash Oct 27 '21 at 16:00
  • It used to be common in school mathematics to speak of branches of inverse for (e.g.) trig functions. The inverse of the sine function you mention is sometimes called the principal branch. In fact, as you note, there is a (non-principal) branch of arcsin taking values in $[\frac{\pi}{2}, \frac{3\pi}{2}]$, as well as infinitely many other branches. – Andrew D. Hwang Oct 28 '21 at 11:11

1 Answers1

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If $f:\mathbb R\to \mathbb R,$ and $f$ is not injective, we can’t find an inverse function.

We can find inverse functions to the restriction of $f$ to a subset $S\subseteq\mathbb R,$ when $f$ is injective on $S.$

Ideally, we’d find an $S$ such that $S$ is simple (say an interval) and $f(S)=f(\mathbb R).$ Thus we’d have a bijection $S$ with the entire range of $f.$

A common example of this is $f(x)=x^2.$ We choose to define a pseudo-inverse using a range $S=[0,\infty),$ and we call the function $\sqrt x:[0,\infty)\to[0,\infty).$ We could have chosen the negative inverses, instead, but there is a stronger reason for picking the positive square roots - we get $\sqrt{xy}=\sqrt{x}\sqrt y$ for $x,y\geq 0.$ If we choose the negative square root $g:[0,\infty)\to (-\infty,0]$ we’d get $g(xy)=-g(x)g(y),$ which is just slightly more complicated.

The positive square root also composes well with $\sqrt{\sqrt x}$ being a suitable definition of $\sqrt[4]x.$

There are other reasons we choose the positive square root, but I’ll skip them here.

There are less compelling reasons for specifically choosing $[-\pi/2,\pi/2]$ for the range of $\arcsin x.$ At heart, we choose this range just because we prefer numeric symmetry around zero. It is almost entirely an aesthetic decision.

One aesthetic reason is that our first definition of $\sin x$ and $\cos x$ in terms of geometry, is often only on $[0,\pi/2],$ in terms of right triangles and side lengths. So we are most comfortable thinking about those angles. The largest interval containing $[0,\pi/2]$ on which $\sin$ is $1-1$ is $[-\pi/2,\pi/2].$

The same is true for $\arccos.$ The largest interval we could use for the domain of $\arccos$ containing $[0,\pi/2]$ is $[0,\pi].$

Sometimes, $\arcsin$ is written $\sin^{-1},$ which I dislike, because it implies this is somehow defined entirely by $\sin$ and thus involves no choice. At least the name $\arcsin$ doesn’t exactly give that impression.

Thomas Andrews
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