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Is this equation correct

by utkarshakash
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utkarshakash
#1
Jan8-13, 06:38 AM
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1. The problem statement, all variables and given/known data
Is this equation correct for permissible values of x

[itex]tan^{-1}|tan x| = |x|[/itex]

2. Relevant equations

3. The attempt at a solution
I assume LHS to be θ.
Then tanθ=|tanx|
The original equation becomes
[itex]tan^{-1}tan \theta = |x|[/itex]
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tiny-tim
#2
Jan8-13, 06:50 AM
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hi utkarshakash!

(what do you mean by "permissible"? )

wouldn't it be easier to start by saying |tanx| = tan|x| ?
HallsofIvy
#3
Jan8-13, 06:53 AM
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No, that is not correct because tangent is not a "one-to-one function". For example, if [itex]\theta= 5\pi/4[/itex] then [itex]tan(\theta)= tan(5\pi/4)= 1[/itex] so that [itex]tan^{-1}(tan(\theta))= tan^{-1}(tan(5\pi/4))= tan^{-1}(1)= \pi/4[/itex], not [itex]5\pi/4[/itex]. Since everything is positive, the absolute value is irrelevant.

Millennial
#4
Jan8-13, 07:41 AM
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Is this equation correct

Quote Quote by tiny-tim View Post
hi utkarshakash!

(what do you mean by "permissible"? )

wouldn't it be easier to start by saying |tanx| = tan|x| ?
Even if so, it is still wrong, as tangent can be negative for positive values of x. For instance, [itex]\tan(7\pi/4)=-1[/itex]. It is easy to observe from here that [itex]|\tan(7\pi/4)|\neq \tan|7\pi/4|[/itex], as [itex]1\neq-1[/itex].
utkarshakash
#5
Jan9-13, 06:24 AM
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Thanks all for your answers.
haruspex
#6
Jan9-13, 02:46 PM
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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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