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enter image description here

(source for above graph)

enter image description here

(source for above graph)

Both functions simplify to x, but why aren't the graphs the same?

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    $\begingroup$ The inverse trigonometric functions have restricted domains (en.wikipedia.org/wiki/…), while the standard trigonometric functions take arguments from the whole of $\mathbb{R}$. $\endgroup$ May 23, 2012 at 10:54
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    $\begingroup$ The two functions are inverse to each other only on the interval $[-1,1]$, so they simplify to $x\mapsto x$ only on that interval. Outside of that interval you get the behaviour you see on the pictures. $\endgroup$
    – Dejan Govc
    May 23, 2012 at 10:56
  • $\begingroup$ Actually, this comment is a bit imprecise, so I expanded it to an answer, see below. $\endgroup$
    – Dejan Govc
    May 23, 2012 at 11:22
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    $\begingroup$ $\arcsin$ is the inverse of the restricted sine, not of the usual sine. (The restricted sine is like the sine function, but its domain is only $[-\pi/2,\pi/2]$. $\endgroup$ May 23, 2012 at 18:14

5 Answers 5

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Well, it depends on what you define as "same".

For $\sin(\arcsin(x))$, the domain of the function is domain of $\arcsin$ which is [-1,1]. So the graph is strictly defined between -1 and 1. It would be mathematically incorrect to substitute x as 3 since there is no such thing as $\arcsin(3)$.

The domain of $\arcsin(\sin(x))$ is entire of $\mathbb{R}$, so you can substitute any (real) number you want and you'd get an answer.

So, are they same? If you zoom into your graph such that you look only between -1 and 1, there is no difference. Beyond these limits, the term "same" is meaningless since one of them ($\sin(\arcsin(x))$ doesn't even exist!

You can argue a similar case for $y = (\sqrt{x})^2 \text{and } y = x$. In the latter, you have a straight line passing through the origin from $-\infty$ to $+\infty$. In the former, the function is not defined for negative x although the simplification leads to $y = x$.

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    $\begingroup$ Perhaps a closer analogy would consider $y=\left(\sqrt x\right)^2$ and $y=\sqrt{x^2}$. $\endgroup$
    – MJD
    May 23, 2012 at 15:52
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    $\begingroup$ @MarkDominus I agree. It fits better with the question. $\endgroup$
    – Inquest
    May 23, 2012 at 15:53
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The domain of the function $\arcsin x$ is the interval $[-1,1]$; it isn’t defined for any value of $x$ outside that interval. You can then take the sine of $\arcsin x$, but the domain of the composite function $\sin\circ\arcsin$ is still just $[-1,1]$; no other ‘inputs’ are meaningful. That’s why the graph of $y=\sin\arcsin x$ is chopped off at $x=-1$ and $x=1$.

The function $\sin x$, on the other hand, is defined for all real numbers $x$. Moreover, it’s always between $-1$ and $1$, so it makes sense to take its arcsine. However, the function $\arcsin x$ always returns the angle between $-\pi/2$ and $\pi/2$ whose sine is $x$, so the composite function $\arcsin\circ\sin$ always ‘outputs’ a value between $-\pi/2$ and $\pi/2$. Because of the way the sine function works, these values oscillate between $-\pi/2$ and $\pi/2$ as in your other graph.

Both functions give you simply $x$ as long as you stay within appropriate limits, $[-1,1]$ for $\sin\arcsin x$ and $[-\pi,2,\pi/2]$ for $\arcsin\sin x$.

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$$[-1,1]\stackrel{\textrm{arcsin}}{\longrightarrow}[-\frac{\pi}{2},\frac{\pi}{2}]\stackrel{\textrm{sin}}{\longrightarrow}[-1,1]$$

$$ \mathbb{R} \stackrel{\textrm{sin}}{\longrightarrow} [-1,1]\stackrel{\textrm{arcsin}}{\longrightarrow}[-\frac{\pi}{2},\frac{\pi}{2}] $$

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Actually, my comment above is a bit imprecise. I shall make it more precise here.

The motivation is clear. We want to invert the function $\sin:\Bbb R\to \Bbb R$. But this function isn't injective, nor is it surjective, and that is a problem, since a function is invertible if and only if it is bijective.

So if we want to invert $\sin$, we first have to restrict its domain and codomain appropriately to make it bijective. The restriction $\sin:[-\frac{\pi}{2},\frac{\pi}{2}]\to[-1,1]$ is a bijective function, so it has an inverse. (Stricly speaking, this is a new function, but we usually use the same name for it, which might add to the confusion.) And it is the inverse of this function that is called $\arcsin:[-1,1]\to [-\frac{\pi}{2},\frac{\pi}{2}]$. That's why $\arcsin$ is defined only on the interval $[-1,1]$ and the functions $\sin:[-\frac{\pi}{2},\frac{\pi}{2}]\to[-1,1]$ and $\arcsin:[-1,1]\to [-\frac{\pi}{2},\frac{\pi}{2}]$ are inverse to each other.

Now, we can still look at what happens if we take the original function $\sin:\Bbb R\to\Bbb R$ and compose it with $\arcsin:[-1,1]\to [-\frac{\pi}{2},\frac{\pi}{2}]$, but then the functions are not inverse anymore and we get the behaviour you describe above.

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the domain of arcsin(x) only goes from -1 to 1.

Although the functions reduce to the same, they're different in non-reduced form.

This is similar to functions with removable discontinuities: f(x) = (x - 1)^2 / (x - 1) for instance.

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