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Hi,

I am concerned with the first mathematical statement at the top in the photo of wiki.

eg, sin inverse (sin(2.5pi)) = pi/2 if you put it into calculator. But 2.5pi is in the domain of sin.

Also, of equal importance, how could I predict beforehand that is would be pi/2 without the calculator? What is the general way to determine the result of some "inverse function applied to the original function of the input x, where x is not in the range of the inverse function"?

I am sorry if this is confusing.

yourhuckleberryfriend
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    Note that $\arcsin$ is not the inverse of $\sin$ in the sense described e.g. in the referenced Wikipedia article, it is only the rightsided inverse, i.e., for every $y\in[-1,1]$, we know that $\sin(\arcsin(y))=y$. We also know that $\arcsin(\sin(x))=x$ for every $x\in[-\pi/2,\pi/2]$, hence $\arcsin$ and $\sin$ are inverse of each other if we "artificially" restrict the domain of $\sin$. Without that, all we know about $x':=\arcsin(\sin(x))$ is that it is some real number in $[-\pi/2,\pi/2]$ such that $\sin(x')=\sin(x)$. If you know that $\sin(x+n\pi)=(-1)^n\sin(x)$, this may help you shortcut – Hagen von Eitzen Jul 28 '24 at 12:17
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    $\sin$ is not invertible over $\mathbb{R}$ – Soham Saha Jul 28 '24 at 12:19
  • @HagenvonEitzen so is there a thing called a "both-handed" inverse?? W – yourhuckleberryfriend Jul 28 '24 at 13:41
  • @thetrueembodimentofstupidity The right inverse exists when $f$ is surjective; the function $f: \mathbb{R} \to [0, 1]$ given by $f (x) = \sin x$ is surjective. The left inverse exists when $f$ is injective. The function $f (x) = \sin x$ is not injective because, e.g., $\sin (0) = \sin (\pi) = 0$. An invertible function (also called bijection) is one in which both the left and right inverses exist. – K. Jiang Jul 28 '24 at 17:31

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