Is $\arctan ( \tan ( π + (π/4) ) )$ equal to $π/4$ or undefined

functionsinverseinverse functiontrigonometry

We know than $\arctan ( \tan ( π + (π/4) ) )$ equals $x$ when $x$ belongs to the domain of $\tan x$ (and hence range of $\arctan x$).

Where $\arctan x$ and $\tan x$ are inverses of each other, and hence the domain of $\tan x$ equals the range of arctanx and the range of $\tan x$ equals the domain of $\arctan x$ (by the definition of inverse function).

Consider the expression: $\arctan ( \tan ( π + (π/4) ) )$.

$π+π/4$ does not belong to the range of $\arctan x$ and hence does not belong to the domain of $\tan x$ (which is the inverse of $\arctan x$).

And hence this expression is undefined as, if $F$ is a function then $F$ (number not in domain) is undefined.

Is this right?

Best Answer

It is defined, $\arctan(\tan(\pi+\pi/4))=\arctan(\tan(5\pi/4))=\arctan1=\pi/4$.

Let's take a look at what is wrong with your reasoning:

$\pi+\pi/4$ does not belong to the range of $\arctan$ and hence does not belong to the domain of $\tan$ (which is the inverse of $\arctan$).

This is incorrect in two ways:

  1. The inverse of $\arctan x$ is not $\tan x$, instead, it is the part of $\tan x$ with $x$ restrited to $(-\pi/2,\pi/2)$.

  2. $x$ does not belong to the range of $\arctan$ merely means that $\arctan(\tan(x))=x$ does not hold; it does not mean $\arctan(\tan (x))$ is not defined.

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