3

On my calculator $\tan^{-1}$ is used to calculate the $\arctan$, but $\tan^{-1}$ actually is $\cot$. $\cot$ and $\arctan$ are not the same thing though. Am I missing something or is the labeling of my Casio fx-991ES really wrong?

To make the question more clear: Is $\arctan = \tan^{-1}$ correct?

  • 2
    All calculators I know, among them the Casio ones, use $;\tan^{-1};$ to denote $;\arctan;$ , not $;\cot;$ .. – Timbuc Feb 07 '15 at 17:48
  • What do you get when you punch in $\tan^{-1}(1)$ or $\tan^{-1}(\sqrt{3})$? – Mike Pierce Feb 07 '15 at 17:49
  • I have that one besides me and it is working fine, except the fact that you keep record whether the measurement of angles is set in degrees or radians. – RE60K Feb 07 '15 at 17:53
  • I have a 991-MS, same story. It's not wrong. It's just inverse notation of a function. – AvZ Feb 07 '15 at 18:14
  • Related: http://math.stackexchange.com/questions/30317/arcsin-written-as-sin-1x – Hans Lundmark Feb 08 '15 at 10:59

3 Answers3

7

Calculators have to save space on the labels, therefore $\tan^{-1}$ is more convenient than $\arctan$.

Moreover, the notation $f^{-1}$ conventionally denotes the functional inverse of $f$. It's rare to write $f^{-1}$ meaning $1/f$.

On my calculator, for example, $\sin^{-1}$ and $\cos^{-1}$ are used in place of the more correct (in my opinion, since less confusing) $\arcsin$ and $\arccos$.

rubik
  • 9,344
  • So you're saying $\tan^{-1}$ is not wrong but not exactly correct either? Mathematically speaking... – Stacksatty Feb 07 '15 at 18:28
  • @Stacksatty: Exactly. It's an established convention, but unfortunately it overlaps with other notations, thereby generating confusion. It's the reason I much prefer the arc-functions, as they have an unique and explicit meaning. – rubik Feb 07 '15 at 19:15
4

The notation $\tan^{-1}$ is often reserved for the canonical functional inverse of the $\tan$ version rather than its reciprocal. You're right that it's confusing, but that's the conventional mathematical notation.

Yuval Filmus
  • 57,157
1

Indeed this is ambiguous, but the reason for that is that $f^{-1}$ is the inverse element of $f$ by composition law ($(f \circ f)(x) = f(f(x))$ : $$f \circ f^{-1} = \text{id}$$ Where id is the identity function, the neutral of $(F(\mathbb{R},\mathbb{R}), \circ)$.

servabat
  • 1,509