# Intuition around why domain of x of arcsine and arccosine is [-1;1] for “real result” & domain for arctangent is all real numbers

Context

I'm working my way through basic trig (this question has a focus on inverse trig functions, specifically arcsine, arccosine and arctangent ), using Khan Academy, wikipedia and some of "trig without tears" - http://oakroadsystems.com/twt/

What I think I understand

My understanding is that the range of the usual principal value of arcsine, arccosine and arctangent is defined by "convention". Which reads to me, to mean a consensus of mathematicians agreed upon these principle values (I'm certain with some underlying reasoning, which I don't know of yet) to solve the issue of potential multiple results from a the same input to one of these functions.

What I don't get

Domain of arcsine & arccosine, why?

When discussing the domain of "x" for these 3 functions, as shown in this table on wikipedia: http://en.wikipedia.org/wiki/Inverse_trigonometric_function#Principal_values I see the domain of arcsine and arccosine is −1 ≤ x ≤ 1.

I've watched a number of basic tutorials on the unit circle and how it can be used to help solve these functions. So of course, I can see and visualize the fact that the maximum and minimum values of X on the unit circle are 1 & -1.

But I'm struggling to understand the intuition behind the restriction of the domain? My understanding right now is that the unit circle has x value [1;-1], so is that why the range is that?

Domain of arctangent, why?

Also the domain for arctangent is "all real numbers" - in the video on Khan Academy http://www.khanacademy.org/video/inverse-trig-functions--arctan?playlist=Trigonometry Sal (main teacher on Khan Academy) talks about how tangent of something also represents that slope of a line (I guess the hypotenuse) and how that could have infinite results.

I don't really understand this. If slope is rise over run - isn't there a limit to that ratio?

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If the the sine (and thus the cosine as well) is always between $-1$ and $1$ for all real angles, then it stands to reason that the domain of the inverse is $[-1,1]$... – J. M. Dec 16 '11 at 5:29
"If slope is rise over run - isn't there a limit to that ratio?" - a vertical line is infinitely steep, no? – J. M. Dec 16 '11 at 5:30

## 2 Answers

Remember that $\arcsin$ is supposed to be the inverse function of $\sin$ (or at least, of a "restricted sine function"; that's the choice of 'principal value').

The way that $\arcsin$ works is supposed to be: you plug in the value somebody else got out of the sine function, and $\arcsin$ will tell you what number was put into the sine function to get that value. It's like a "reverse telephone directory": you look up the phone number and find out the person it belongs to, instead of the usual way of looking up the person and finding their phone number.

But that means that the only things that you can put into the $\arcsin$ functions are real numbers that actually come out of the sine function (the only numbers you can look up in a reverse telephone directory are telephone numbers, so you can't look up "000-0000").

What are the numbers that can come out of the sine function? Every number between $-1$ and $1$ (inclusively), but only those numbers: the sine function will never give a result that is greater than $1$ or smaller than $-1$. That means that the numbers you can plug into $\arcsin$ are only the numbers that come out of the sine function: the numbers between $-1$ and $1$.

The same thing is true for $\arccos$: you can only put into $\arccos$ numbers that may come out of the cosine function, and the only numbers that may come out of the cosine function are the numbers on $[-1,1]$.

However when we come to the $\arctan$ function, things are different: what are the numbers that may come out of the tangent function? Every real number! Every real numbers is the tangent of some angle, so now we can put into $\arctan$ any real number, because any real number is, potentially (and in actuality) the result of applying the tangent function.

Note. Your final paragraph seems to be confusing the "trigonometric tangent function" with the tangent line to the graph of a function. The "trigonometric tangent function" is the function defined by $$\tan(x) = \frac{\sin(x)}{\cos(x)}.$$ The "tangent line to the graph of $f(x)$ at $x=a$" is a straight line that has certain properties (it goes through $(a,f(a))$, and is the straight line that "best approximates" the graph of $y=f(x)$ near the point $(a,f(a))$). The derivative, which is a key concept in calculus, is the slope of the tangent-line-to-the-graph-of-$f(x)$ (which is defined as a limit of a certain ratio), not to the trigonometric tangent function which is what $\arctan(x)$ is related to.

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thanks. WHy will the sine function never give a result greater than 1 or smaller than -1? – drc Dec 16 '11 at 21:01

1
@drc: Because of how it is defined. If you define it in terms of right triangles, the sine of $x$ is the length of the opposite side divided by the length of the hypothenuse; since the hypothenuse is always at least as long as the opposite side, then the quotient is never greater than $1$; taking into account signs give you that is never smaller than $-1$. If you define the sine function as the $y$-coordinate of a point on the unit circle, then the coordinates on the unit circle satisfy $x^2+y^2=1$, and so they must satisfy $|y|=\sqrt{y^2}=\sqrt{1-x^2}\leq \sqrt{1} = 1$. So $|\sin t|\leq 1$. – Arturo Magidin Dec 17 '11 at 3:40
Thanks this is great, the key thing I missed her was the fact that the hypothenuse is at least as long as the opposite side. Awesome stuff! – drc Dec 20 '11 at 3:13

If $f(x)$ and $f^{-1}(x)$ are inverse functions (meaning that $(f \circ f^{-1})(x) = (f^{-1} \circ f)(x) = x$ on the respective domains), then the domain of $f$ is the range of $f^{-1}$ and the domain of $f^{-1}$ is the range of $f$.

In your case, since the range of $y = \sin x$ is $[-1,1]$, then the domain of $y = \arcsin x$ is $[-1,1]$. Since the range for $y = \tan x$ is $(-\infty, \infty)$, then the domain for $y = \arctan x$ is $(-\infty, \infty)$.

Finally, when we discuss the functions $y = \arcsin x$ and $y = \arccos x$, we restrict the range of these functions so that the functions are one-to-one. Being one-to-one is what allows us to rightfully describe these functions as inverses.

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From my reading, you are stating the convention, but I don't really see where you have showing why. – drc Dec 16 '11 at 20:51
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