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How can a $\sin$ have inverse function?

Let's say we have $f:A\rightarrow B$, which means the domain of function $f$ is set A, and the range is set $B$. If we wanted to find $f^{-1}(x)$, that function would be $f^{-1}\colon B\rightarrow A$.

Now, if we apply same logic to $\sin$ we get that $\sin\colon(-\infty, +\infty)\rightarrow [-1,1]$. So how can that have an inverse function?

We would have situations where for the same input we get different outputs, and by definition, a function for every input can have one and only one output.

Sebastiano
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cooooool
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    You are right: $\sin$ does not have an inverse for the domain $(-\infty,\infty)$. However, if you restrict $\sin$ to say, $[-\pi/2,\pi/2]$, then it has an inverse (which is usually denoted by $\arcsin\colon [-1,1]\to [-\pi/2,\pi/2]$). See the section on "Principal values" in this Wiki article. – projectilemotion Jan 19 '23 at 20:42
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    Ask yourself how you feel about square root being considered an "inverse" of square. – JonathanZ Jan 19 '23 at 20:51
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    See $\sin x=\sin y$ mis-understanding for similar remarks. – peek-a-boo Jan 19 '23 at 20:56
  • It has a partial inverse function. – Dan Jan 19 '23 at 20:58
  • @JonathanZsupportsMonicaC OP probably feels uneasy about it too, and it would be totally understandable since it takes time to get used to such abuses of language – peek-a-boo Jan 19 '23 at 20:58
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    In summary, $\arcsin$ is by convention the inverse of the function $\sin|_{[-\pi/2,\pi/2]}\colon [-\pi/2,\pi/2]\to [-1,1]$. Similarly, the square root function is by convention the inverse of the function $f(x)=x^2$ with $f\colon [0,\infty)\to \mathbb{R}$ (not $f\colon \mathbb{R}\to \mathbb{R}$). – projectilemotion Jan 19 '23 at 21:02
  • @peek-a-boo - that's quite possibly the case. And i do think it's useful for people to see examples of something that they might be familiar with, and understand how to work with, that is similar to their current situation. – JonathanZ Jan 19 '23 at 21:03
  • In addition to the topic linked by peek-a-boo, there are several related and similar questions on the site, e.g. https://math.stackexchange.com/q/4221047/ , https://math.stackexchange.com/q/3989971/ , and https://math.stackexchange.com/q/4289059/ . – Xander Henderson Jan 19 '23 at 21:19
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    It's a right-inverse, not a left-inverse. $\sin(\arcsin(x))=x$, but $\arcsin(\sin(x))\neq x$. – mr_e_man Jan 19 '23 at 21:30
  • @projectilemotion yes, that is also interesting, as a beginner I do not quite understand, but I guess if that is "convention" than it is true. But then what is the correct answer to the question does sin have inverse function? It depends. – cooooool Jan 19 '23 at 21:37
  • @JonathanZsupportsMonicaC Interesting, so is it still called inverse function, since if that is inverse, $f(x)=|x|$ can also have "inverse" that only works in specific conditions, and what about that rule $(f \circ f^{-1})(x)=(f^{-1} \circ f)(x)=I$? – cooooool Jan 19 '23 at 21:49
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    @FikiCar: Well, despite $\sin$ not being injective, there's often a need to find an angle such that angle $x$ such that $\sin(x) = y$. By convention, we pick the one between $\frac{-\pi}{2}$ and $\frac{\pi}{2}$. Of course, if $x$ is a solution to $\sin(x) = y$, so is the supplementary angle $\pi - x$, and so is either one shifted by $\pm 2\pi k$ for any integer $k$. But it would be tedious to point that out all the time. – Dan Jan 19 '23 at 21:49
  • @Dan I understand now that there is convention in place for convenience, but if we can pick solution that we want, we can say that inverse of $f(x)=|x|$ is $k*x$ where k is -1 or 1, and we pick the one we need. – cooooool Jan 19 '23 at 21:53
  • @FikiCar: Well, a more practical example is the inverse of $f(x) = x^2$, aka "square root". We say that $\sqrt{16} = 4$ even though $(-4)^2 = 16$ as well. – Dan Jan 19 '23 at 22:02
  • @Dan I agree, I just find it all confusing, so I came here to see if someone can explain it to me. So far I understood that arcsin is there for convenience and that it is not "true" inverse of sin. And some of you gave me some more inconsistencies, so I guess I will have something to think about. – cooooool Jan 19 '23 at 22:06
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    I think a good take-away is to realize that a function might not be one-to-one (which is required if an inverse is going to exist), but you can make it one-to-one of you make the domain smaller. (Technically, the reduced domain version is a different function, as the domain is an essential part of a function.) – JonathanZ Jan 19 '23 at 22:30
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    You should go read the linked article on "partial inverses", and also keep in mind that mathematicians can be a bit lazy, and might call something an inverse, and depend on you, the reader, to use the context to understand that strictly it's only a partial inverse. – JonathanZ Jan 19 '23 at 22:33

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You're technically right: $\sin$ is not a one-to-one (“injective”) function, and thus is not invertible.

However, $\sin$ does have a partial inverse. By convention, we restrict $\sin$'s domain to $\left[- \frac{\pi}{2}, \frac{\pi}{2}\right]$, in order to define the partial inverse $\arcsin: \left[-1, 1\right] \to \left[- \frac{\pi}{2}, \frac{\pi}{2}\right]$.

It just makes things easier for computer programmers and publishers of trigonometric tables if we pick just one “standard” solution to $\sin(y) = x$. If you ever need the full solution set, it's:

$$y \in \{ \theta + 2\pi k : \theta \in \{ \arcsin(x), \pi - \arcsin(x)\}, k \in \mathbb{Z} \}$$

It's similar to how we say that the square root of $16$ is $4$, even though $-4$ is also a square root of $16$.

Dan
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