I was showing my neighbour's daughter how to do simple sine, cosine and tangent questions and their inverses. She could do the questions fine but she asked me why the calculator button for inverse tan says $\tan^{-1}$.

I was not sure.
I said it is not quite the same as $$ \frac{1}{x} = x^{-1}$$ but that situation is almost like an opposite.

Inverse tangent is exactly the opposite of tangent as instead of getting the ratio of the sides when you know the angle, you get the angle when you know the ratio of the sides.

How could I have explained this better?

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    $\begingroup$ most people prefer to call $tan^{-1}$ arctan instead $\endgroup$
    – Lenny
    Commented Apr 11 at 1:06
  • $\begingroup$ @Lenny, I would disagree with that, myself, though I suppose neither of us really knows what most people prefer. I prefer inverse tangent because the inverse function seems easier for students to understand than what arctangent has to do with arcs. $\endgroup$
    – Sue VanHattum
    Commented Apr 11 at 14:53

3 Answers 3


It's the functional inverse rather than the multiplicative inverse. Somewhere along the way, notation arrived at function composition being written as $f(f(x))=f^2(x)$, so the functional inverse became $f^{-1}(x)$. That's so that you still get the exponent rule $f^n\circ f^m=f^{n+m}$ for all integers. (Note the $\circ$; I mean function composition, not multiplication.) i.e. $(f^{1}\circ f^{-1})(x)=(f\circ f^{-1})(x)=f^{0}(x) =x$; the identity function.

You can consider this to be an analogy to what goes on with multiplication/exponents, where you have a very similar looking set of rules.

Whenever there is actual context, the ambiguity in notation is rarely an issue, but it can be a bit confusing to learners. I usually write $\arctan$ when lecturing, myself.

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    $\begingroup$ I'd remark that this is actually more reasonable than the unfortunately more common $\cos^2x = (\cos x)^2$ notation, which is incompatible with $\tan^{-1}$. $\endgroup$ Commented Apr 8 at 14:26
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    $\begingroup$ @leftaroundabout That's the other example where the presence of both kinds of notation for trig functions makes this particularly confusing for learners. Experienced mathematicians know that $\cos(\cos(x))$ is unlikely and that $\frac{1}{\tan x}$ has another name, so we don't worry too much about it. $\endgroup$
    – Adam
    Commented Apr 8 at 21:48

I will start by noting that I have written an answer to a similar question on Mathematics SE, and would direct you towards that answer for some additional discussion.

In my precalculus class, I tend to introduce students to just a little bit of algebra—I don't necessarily test them on it, but I weave it into my discussions of various topics.

  • Early in the semester, I talk about the natural numbers, integers, rational numbers, and so on. I try to introduce each of these as a set with some additional structure. The rationals, in particular, are a set with an addition and multiplication operation. The rationals are particularly nice, as they form a field: if $q$ is a rational number, then

    1. There is another rational number, denoted by $-q$, such that $q + (-q) = 0$ (where $0$ is the additive identity). This is called the additive inverse of $q$.

    2. There is another rational number, denoted by $q^{-1}$, such that $q \cdot q^{-1} = 1$ (where $1$ is the multiplicative identity). This is called the multiplicative inverse.

    Because students are somewhat used to the idea of fractions, I then note that the multiplicative inverse is often written as a fraction, i.e. $$ q^{-1} = \frac{1}{q}. $$ So, in the context of rational numbers, the inverse is the reciprocal.

  • Later in the semester, we start to talk about functions. In the world of functions, there is a new operation, called composition, which allows us to combine functions (in a way similar to how addition and multiplication can be used to combine numbers). Composition doesn't follow all of the same rules as either addition or multiplication over the rationals. For example, composition is not commutative. However, there is a compositional identity: the function $\operatorname{id}$, defined by the formula $$ \operatorname{id}(x) = x, $$ has the property that, when composed with any other function, it leaves that function alone. That is, $$ \operatorname{id}\circ f = f = f \circ \operatorname{id} $$ for any function $f$.

    It then turns out that there are some functions $f$ for which there exists a second function $g$ such that $$ f\circ g = g\circ f = \operatorname{id}. $$ Such a function $f$ is invertible, and the function $g$ is the inverse of $f$. Since composition of functions is a little like multiplication of numbers, it is not unreasonable to adopt the notation of a multiplicative inverse for this. Hence we write $f^{-1} = g$.

The thing to point out is that $x^{-1}$ nearly always represents the inverse of $x$, but one has to be careful about what kind of object $x$ is. If $x$ is a number (a rational or real number, for exmple), then $x^{-1}$ is the inverse of $x$ with respect to multiplication; if $x$ is a function, then $x^{-1}$ is the inverse of $x$ with respect to composition.

I think that it is useful to talk about things in these terms, as the idea of "inverting" an operation is all over the place in mathematics. It is possible to talk about the multiplicative inverses of numbers, the compositional inverses of functions, the inverse of a matrix (which is both kinds of inversion, as matrices can be multiplied; but matrices represent linear transformations, and the inverse matrix represents the compositional inverse of that linear transformation), the inverse of the Fourier or Laplace transforms, the operational inverse of a differentiation operator, and so on. I think that it is helpful to build a foundation early which will permit extension as students take more math classes in the future.

  • 1
    $\begingroup$ That has finally made sense of something which has puzzled me for years. $\endgroup$
    – mdewey
    Commented Apr 10 at 12:21
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    $\begingroup$ it makes sense if you hold that tan is strictly a function but I still think that one can treat $tan^{-1}(x)$ to be what people think as cotangent mostly because $tan^{2} (x) = tan(x)tan(x)$ $\endgroup$
    – Lenny
    Commented Apr 11 at 1:05
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    $\begingroup$ On the other hand, $\tan^{-1}(x)$, practically always means $\operatorname{arctan}$. Claiming that it could be the cotangent---particularly in an educational setting---is wrong, and nearly malpractice. $\endgroup$
    – Xander Henderson
    Commented Apr 11 at 2:52
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    $\begingroup$ @SueVanHattum Other people have made the same comment, but I think that you shouldn't read it that way---in terms of order of operations, functions bind "more tightly" than other algebraic operations. $f(x)^2$ means $[f(x)]^2$, and $\sin(x)^2$ means $[\sin(x)]^2$. The function binds more tightly to $(x)$ than the exponent. $\endgroup$
    – Xander Henderson
    Commented Apr 13 at 23:19
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    $\begingroup$ Personally (and I know that this is quite idiosyncratic), I really dislike the notation $\tan x$ (with no parentheses). I know that this is common and accepted notation, but I think it hides the fact that $\tan$ is a function acting on some object. But these are all comments I made in the answer to which I linked. :D $\endgroup$
    – Xander Henderson
    Commented Apr 13 at 23:22

Once we understand the reason $\tan^{-1}$ can mean the inverse tangent and that most people prefer to write this function as $\arctan$, your neighbor's daughter's question brings up an issue that nobody has mentioned yet: since $\arctan$ is more preferred (to avoid confusing notation), why do we see this function written as $\tan^{-1}$ on calculators, even on smartphone calculator screens? The answer is surely one of space: $\tan^{-1}$ fits in a smaller space than the expression $\arctan$.


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