Speaking loosely, we’d like for the arctangent to be a function which “undoes” the tangent.
Figure6.4.1.Notice that the \(x\) and \(y\) axes have been swapped in the graph at the left.
Figure 6.4.1 shows the graph of \(x=\tan(y)\) (in red) and the same graph (in blue) reflected about the graph of the line \(y=x\) (dashed, in black). Reflecting in this manner swaps the coordinates of each point. For example, the point \((\pi, 0)\) is on the graph of \(x=\tan(y)\text{,}\) because \(\tan\left(\pi\right)=0\text{.}\) Similarly, the point \(\left(0,\pi\right)\) is on the graph of \(y=\arctan(x)\text{.}\)
But is it true that \(\arctan(0)=\pi\text{?}\) If we are to believe our graph, it could be true. But couldn’t it also be true that \(\arctan(0)=2\pi\text{,}\) or even \(\arctan(0)=-\pi\text{?}\) The blue graph above is clearly the graph of the arctangent as we have defined it but we seem to have a choice for the output of \(\arctan(x)\text{.}\) Many choices, in fact.
Having more than one output for a given input clearly violates what we mean when we use the word “function.” A mathematical function returns exactly one output for a given input. There is no choice. The sketch of the arctangent (in blue) is not the graph of a function. It is the graph of the multifunction \(\arctan(x)\text{.}\) Each of the blue curves on the right side of Figure 6.4.1 is one branch of the multifunction, \(\arctan(x)\text{.}\) In fact all of the “arc” functions from trigonometry are multifunctions. Multfunctions are interesting objects and are well worth studying, but this is not the time for that study. Right now we are only interested in the properties of the inverse function of \(\tan(y)\text{.}\) Since \(\arctan(x)\) is not a function, it can’t be the inverse function of \(\tan(y)\) so we will stop thinking about it as soon as we can.
But if \(\arctan(x)\) is not the inverse of \(\tan(y)\) what is? Since we only allow one output for a given input, could it be as simple as choosing just one of the branches of the \(\arctan(x)\text{,}\) say the one that lies between \(y=-\frac{\pi}{2}\) and \(y=\frac{\pi}{2}\text{,}\) shown below, and calling it the the inverse of \(\tan(y)\text{?}\)
Actually, yes. It is just that simple. And the usual choice is the one we’ve indicated. In this text we will designate this branch of \(\arctan(x)\) with \(\inverse\tan(x)\text{.}\) Thus \(\inverse\tan(x)\) denotes the inverse tangent function. Because the multifunction arctangent is not a function we will henceforth ignore it as much as possible.
DIGRESSION: Inverse Function Notation.
The use of \(-1\) as an exponent is probably the most common notation used to indicate function inversion. Unfortunately, from the standpoint of a student it is also probably the worst notation we could have possibly invented because It is so very easily confused with reciprocation. While it is true that
\begin{equation}
\inverse{\tan}(x) \text{ is not equal to } \frac{1}{\tan(x)}.\tag{6.13}
\end{equation}
That the “\(-1\)” notation is used for both comes from the fact that both \(2^{-1}\) and \(\inverse\tan(x)\) really are inverses. But they are different kinds of inverses: \(\inverse2\) is the multiplicative inverse of \(2\text{,}\) which means simply that
\begin{equation*}
\frac12\cdot2=1 \text{ and that } 2\cdot\frac12 =1.
\end{equation*}
On the other hand \(\inverse\tan(x)\) is the functional inverse of \(x=\tan(y)\text{,}\) which means that if \(-\frac{\pi}{2}\lt y\lt \frac{\pi}{2}\) then
\begin{equation*}
\inverse\tan(\tan(y)) = y,
\end{equation*}
In equation (6.12) you can think of the \(-1\) in the exponent as an operator. It operates on \(2\) by taking its reciprocal: \(\inverse{2}=\frac12\text{.}\) In equation (6.13) the \(-1\) is not an operator. It is part of the symbol, \(\inverse\tan\text{,}\) that we use to denote the inverse tangent.
With practice this all gets easier, but in the beginning it is very troublesome. Be careful.
But which function exactly, is \(\inverse\tan(x)\) the inverse of? From the sketch above it should be clear that restricting the range of \(\inverse\tan(x)\) to values in the interval \((-\pi/2, \pi/2)\) forced us to restrict the domain of \(\tan(y)\text{.}\) So
\begin{equation*}
\inverse\tan(x) \text{ with domain all real numbers, } (\RR)
\end{equation*}
is the inverse of \(\tan(y) \text{ with domain, } (-\pi/2, \pi/2),\) and it is not the inverse of \(\tan(y) \text{ with domain, } \RR\) because that function has no inverse.
An easy way to remember what \(\inverse{\tan}(x)\) means is to read the symbol \(\inverse{\tan}(x)\) as “the angle whose tangent is \(x\text{.}\)”
The advantage of this phrasing is that it emphasizes that \(x\) is the tangent of some angle, and \(\inverse{\tan}(x)\) is that angle. Similarly the inverse functions of the other trigonometric functions should be read as “the angle whose sine, cosine, secant, (or whatever) is \(x\text{.}\)” Each of these will also come with suitable restrictions on its range.
We should point out that reserving the “arc” notation for the trigonometric multifunctions is actually a conceit of the authors of this text. Most of the world uses \(\arctan(x)\) and \(\inverse\tan(x)\) interchangeably, and eventually you will too. But for now it will be helpful for you to keep in mind that the difference between them is that restricting the range of \(\inverse\tan(x)\) to \(-\frac{\pi}{2}\lt x\lt\frac{\pi}{2}\) guarantees that it is a function. But since the range of \(\arctan(x)\) is not similarly restricted it is not a function.
Drill6.4.3.
Suppose we had chosen \(\frac{\pi}{2}\lt \inverse\tan(x)\lt \frac{3\pi}{2}\text{.}\) What function (with domain) is that the inverse of?
All of this fussiness is really just about making our abstract definitions useful and consistent. It would be nice if these details only impinged on us in an abstract setting but unfortunately some practical difficulties do come up, as the following problem shows.
Problem6.4.4.
Find all solutions of \(\tan(y)=1\) and \(y=\inverse\tan\left(1\right)\text{.}\) Do they have the same set of solutions?
Hint.
Obviously, they do not. Otherwise we wouldn’t have asked the question. What is the difference between the two sets of solutions?
DIGRESSION: The Tangent Function Has No Inverse.
The function \(\tan(y)\) is not invertible. This is because, by custom, its domain is all real numbers (except \(\frac\pi2, \pm\pi,
\pm2\pi, \pm3\pi, \cdots\)). Thus it is not the same function as
\begin{equation*}
\tan(y) \text{ with domain } -\frac{\pi}{2}\lt
y\lt \frac{\pi}{2}.
\end{equation*}
A function has two parts: (1) The set of inputs (domain) and (2) the rule associating input with output. Properly speaking, we should always specify the domain of our functions, but typically we don’t. Most of the time the domain is the set of all real numbers (\(\RR\)), so we don’t bother to explicitly state that the domain of, say \(f(x)=x^2\) is \(\RR\text{.}\) Unless otherwise specified we simply assume that it is. Therefore the function, \(\tan(y)\) cannot be inverted because on its domain \((\RR)\) it has more than one branch.
Because we both want and need an inverse tangent function we restrict the range of \(\inverse\tan(x)\text{.}\) But this is the inverse of \(\tan(y)\text{,}\) with domain \(-\frac{\pi}{2}\lt x\lt \frac{\pi}{2}\text{,}\) which has only one branch. It is not the inverse of \(\tan(y)\text{,}\) because it has too many branches.
To use our notation as it was intended to be used, we should say that \(\inverse\tan(x)\) is the inverse of \(\tan(y)\text{,}\) with domain \(-\frac{\pi}{2}\lt y\lt \frac{\pi}{2}\text{,}\) which is true. But instead, we typically just say that it is the inverse of \(\tan(y)\text{,}\) which strictly speaking, is not true.
When we (mathematicians) talk amongst ourselves this is not a problem. We all understand what we mean. But it can be very confusing for students who usually have a more tenuous grasp of these distinctions. It is a bit unfair of us (mathematicians) to speak to students in what amount to incomplete sentences, but we tend to do it anyway, mostly out of habit.
We (the authors) apologize.
END OF DIGRESSION
The definition of the arccotangent is similar to the definition of the arctangent: If \(x\) is the cotangent of \(y\text{,}\)\(x=\cot(y)\text{,}\) then we say that \(y\) is the arccotangent of \(x\text{,}\) and write
\begin{equation*}
y=\arccot(x).
\end{equation*}
The arccotangent is also a multifunction with multiple branches so, just as before, we will have to decide which branch to use to define the inverse cotangent: \(\inverse\cot(x)\text{.}\)
We obtained our branch for the single-valued function \(y=\inverse\tan(x)\) from the multifunction \(y=\arctan(x)\) by restricting its range to \(-\frac{\pi}{2}\lt y\lt\frac{\pi}{2}\text{.}\) A similar restriction would allow us to obtain the single-valued function \(\inverse\cot(x)\) from the multifunction \(\arccot(x)\text{,}\) but it is a bit simpler to once again use a trigonometric identity.
