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Since $\sin(\frac{\pi}{4})$ and $\sin(\frac{3\pi}{4})$ are both $\frac{\sqrt{2}}{2}$, shouldn't $\arcsin(\frac{\sqrt{2}}{2})$ map to both $\frac{\pi}{4}$ and $\frac{3\pi}{4}$ and therefore not be a function?

when I input $\arcsin(\frac{\sqrt{2}}{2})$ into Wolfram, I get back $\frac{\pi}{4}$ only, but isn't the point of an inverse function that $f^{-1}(f(x)) = x$?

In this case however, $\arcsin(\sin(\frac{3\pi}{4})) = \frac{\pi}{4}$

What am I missing?

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    $\begingroup$ I think this is just a convenient definition. No one says $\arcsin x$ is the inverse function of $\sin x$. Please point out if I made mistake. $\endgroup$ – MonkeyKing May 1 '15 at 1:41
  • $\begingroup$ I'm pretty sure I've always seen arcsin defined as the inverse function of the sin function. $\endgroup$ – jeremy radcliff May 1 '15 at 1:42
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    $\begingroup$ OK, I mean not the inverse function of $\sin x$ over $\Bbb R$. I should practice language skill more. :( $\endgroup$ – MonkeyKing May 1 '15 at 1:46
  • $\begingroup$ No problem, what you said is clear now in the context of user17762's answer, I just didn't have the knowledge to understand your reply when I first read it :) $\endgroup$ – jeremy radcliff May 1 '15 at 1:49
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The important thing to realise here is that a function is not just a rule. To specify a function properly you must also define the domain and codomain. For example, the sine function is $$\sin:{\Bbb R}\to{\Bbb R}\quad\hbox{where}\quad \sin x= \langle\hbox{insert your favourite definition}\rangle\ .$$ This function is not one-to-one (injective) and therefore has no inverse.

Here is a different function: $$f:\Bigl[-\frac\pi2,\frac\pi2\Bigr]\to[-1,1]\quad\hbox{where}\quad f(x)=\sin x=\langle\hbox{same definition as above}\rangle\ .$$ This function is one-to-one and onto and therefore has an inverse. Its inverse is commonly denoted $\sin^{-1}$ or $\arcsin$. Notice however that this terminology is not strictly accurate: $\arcsin$ is not the inverse of $\sin$, it is the inverse of $f$, which is a different function.

Hope this clears things up.

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  • $\begingroup$ Thanks David, it does clear things up. In the line $\sin:{\Bbb R}\to{\Bbb R}$, you don't limit the range to [-1, 1], but even with all the real numbers as potential inputs, in the original sin function the outputs can never be out of the range [-1, 1]. Do you not have to specify the codomain because it's implicitly limited? And in that case, do you have to specify the codomain for $f$? (not trying to nitpick, I just want to make sure I grasp the exact notational logic) $\endgroup$ – jeremy radcliff May 1 '15 at 2:02
  • $\begingroup$ There is another common imprecision involved here. (Mathematicians speaking imprecisely... shock, horror ;-) Strictly speaking a function $f:A\to B$ is invertible if and only if it is one-to-one and onto. For example,$$f:{\Bbb R}\to{\Bbb R}\ ,\quad f(x)=e^x$$is not onto and therefore not invertible. But in calculus, one commonly treats $f$ as the same function as$$g:{\Bbb R}\to{\Bbb R}^+\ ,\quad g(x)=e^x$$(which it really isn't), and $g$ is invertible. (continued...) $\endgroup$ – David May 1 '15 at 2:08
  • $\begingroup$ If my $f$ above had codomain $\Bbb R$ it would be invertible under this (mis)interpretation. However $\sin$ with domain $\Bbb R$ is not one-to-one and therefore is not invertible, no matter what you choose for the codomain. $\endgroup$ – David May 1 '15 at 2:08
  • $\begingroup$ This is great, it all makes sense now. Thanks for taking the time to clarify, I really appreciate. $\endgroup$ – jeremy radcliff May 1 '15 at 2:17
  • $\begingroup$ You're welcome, glad I could help. $\endgroup$ – David May 1 '15 at 2:20
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Recall that $\arcsin: [-1,1] \mapsto [-\pi/2,\pi2/]$, i.e., the range of $\arcsin$ is $[-\pi/2,\pi/2]$.

The function $\sin$ on the other hand is a non-injective function that maps the real line to $[-1,1]$. Defining an inverse for a non-injective functions is not possible. Hence, one route that is often taken is to restrict the domain of the original function such that the function on the new restricted domain is injective.

Hence, the function $\arcsin$ is the inverse of $\sin$ only when the domain of the function $\sin$ is restricted to $[-\pi/2,\pi/2]$.


EDIT

What mathematica does is the following. It first evaluate $\sin(3\pi/4)$, which is $1/\sqrt{2}$. Now it goes and asks the function $\arcsin$, what $\arcsin(1/\sqrt2)$ is, which is now $\pi/4$.

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  • $\begingroup$ That makes sense, but in this case isn't $\frac{3\pi}{4}$ an invalid input and shouldn't Wolfram Alpha give me an error along the lines of "invalid domain" or something? Are they just allowing it because it's convenient? $\endgroup$ – jeremy radcliff May 1 '15 at 1:45
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    $\begingroup$ @jeremyradcliff No. What mathematica does is the following. It first evaluate $\sin(3\pi/4)$, which is $1/\sqrt{2}$. Now it goes and asks the function $\arcsin$, what $\arcsin(1/\sqrt2)$ is, which is now $\pi/4$. $\endgroup$ – Leg May 1 '15 at 1:46
  • $\begingroup$ Ah ok, now all of it makes sense; thank you, it was starting to drive me insane. $\endgroup$ – jeremy radcliff May 1 '15 at 1:48
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As others have pointed out, $\arcsin x$ has the restricted domain $[-1,1]$ and range $[\frac{-\pi}{2},\frac{\pi}{2}]$. This restriction is a choice, a definition.

What's really happening is that $\arcsin x$ is multi-valued. We're only selecting what is called the principal branch, and we're doing this for convenience.

Since the trigonometric functions are periodic, a single-valued function that acts as an inverse doesn't exist. We elect to "chop up" the inversion relation into sections - the branches. In the case of $\arcsin x$ we focus on returning $\sin x$ input values $[\frac{-\pi}{2},\frac{\pi}{2}]$. This branch is the principal branch - the one we've elected to use most often. Why? Well, consider that $\sin x$ cycles through all possible outputs when we give it $[\frac{-\pi}{2},\frac{\pi}{2}]$. Also, it's "nicely packaged" - there's a distinct symmetry and accessibility in using $[\frac{-\pi}{2},\frac{\pi}{2}]$.

When evaluating a query such as yours, we compensate by finding an equivalent result in the principal branch.

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  • $\begingroup$ Thanks zahbaz, that's very helpful because it makes it clear that it's a choice of convenience and why, which is something I've been struggling with (the idea that a lot of math is based on convenience and not some absolute ideal) $\endgroup$ – jeremy radcliff May 1 '15 at 2:11

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