Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
The inverse trigonometric functions have restricted domains (…), while the standard trigonometric functions take arguments from the whole of $\mathbb{R}$. – in_wolframAlpha_we_trust May 23 '12 at 10:54
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. – Dejan Govc May 23 '12 at 10:56
Actually, this comment is a bit imprecise, so I expanded it to an answer, see below. – Dejan Govc May 23 '12 at 11:22
$\arcsin$ 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 $[-\pi/2,\pi/2]$. – Arturo Magidin May 23 '12 at 18:14
up vote 6 down vote accepted

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 $y = x$.

share|cite|improve this answer
Perhaps a closer analogy would consider $y=\left(\sqrt x\right)^2$ and $y=\sqrt{x^2}$. – MJD May 23 '12 at 15:52
@MarkDominus I agree. It fits better with the question. – Inquest May 23 '12 at 15:53

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 $\arcsin\sin x$.

share|cite|improve this answer


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

share|cite|improve this answer

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.

share|cite|improve this answer

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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.