Is this equation correct for all values of x?

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Homework Help Overview

The discussion revolves around the equation tan^{-1}|tan x| = |x| and its validity for permissible values of x. Participants are exploring the properties of the tangent function and its inverse.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Some participants question the meaning of "permissible" values and suggest alternative expressions for the equation. Others provide counterexamples to illustrate that the tangent function is not one-to-one, raising concerns about the validity of the original equation.

Discussion Status

The discussion is active, with participants offering different perspectives on the equation's correctness. Some have provided examples and reasoning that challenge the original poster's assumptions, while others are exploring the implications of the tangent function's properties.

Contextual Notes

There is an ongoing debate about the definition of permissible values and the behavior of the tangent function, particularly regarding its range and the effects of absolute values.

utkarshakash
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Homework Statement


Is this equation correct for permissible values of x

tan^{-1}|tan x| = |x|

Homework Equations



The Attempt at a Solution


I assume LHS to be θ.
Then tanθ=|tanx|
The original equation becomes
tan^{-1}tan \theta = |x|
 
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hi utkarshakash! :smile:

(what do you mean by "permissible"? :confused:)

wouldn't it be easier to start by saying |tanx| = tan|x| ? :wink:
 
No, that is not correct because tangent is not a "one-to-one function". For example, if \theta= 5\pi/4 then tan(\theta)= tan(5\pi/4)= 1 so that tan^{-1}(tan(\theta))= tan^{-1}(tan(5\pi/4))= tan^{-1}(1)= \pi/4, not 5\pi/4. Since everything is positive, the absolute value is irrelevant.
 
tiny-tim said:
hi utkarshakash! :smile:

(what do you mean by "permissible"? :confused:)

wouldn't it be easier to start by saying |tanx| = tan|x| ? :wink:


Even if so, it is still wrong, as tangent can be negative for positive values of x. For instance, \tan(7\pi/4)=-1. It is easy to observe from here that |\tan(7\pi/4)|\neq \tan|7\pi/4|, as 1\neq-1.
 
Thanks all for your answers.
 
Another way to look at it is that atan(|y|) always produces a result in the range [0,pi/2). So the statement must be false for any |x| outside that range.
 

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