Just to prove them wrong: Trigonometry

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    Trigonometry
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Homework Help Overview

The discussion revolves around the properties of the cotangent and arccotangent functions, specifically examining the equation cot|cot-1x| = cot cot-1x = x. Participants are exploring the implications of substituting specific values into this equation, particularly focusing on the case when x = -1.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants are questioning the validity of the equation by substituting values and analyzing the results. There is a focus on understanding the domain and range of the arccotangent function, with some participants suggesting that the range is (0, π) and discussing its implications on the equation.

Discussion Status

The discussion is active, with participants providing insights into the properties of the cotangent and arccotangent functions. Some participants suggest that the original equation may not hold true for all values of x, while others are reconsidering their understanding of the range of the arccotangent function. There is no explicit consensus yet, but various interpretations are being explored.

Contextual Notes

There is mention of potential confusion regarding the range of the arccotangent function, with some participants noting that it is not universally established. This highlights the need for clarity on definitions and assumptions in the problem context.

mooncrater
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Homework Statement


There is a part of solution of a question which says that:
##cot|cot^{-1}x|=cot cot^{-1}x=x##

Homework Equations

The Attempt at a Solution


If we put ##x=-1## in this equation then ##cot^{-1}(-1)=-\pi /4## which will have a modulus =##\pi /4##. And ##cot \pi /4=1## which is not equal to the ##x## we took. So this equation seems wrong to me. Am I correct here?
 
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You should know the domain and ranges.
What is the range for cot-1x ?
 
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Raghav Gupta said:
You should know the domain and ranges.
What is the range for cot-1x ?
Hmmmm... I think I have lost it... it's domain is ## R## and has a range ## (0, \pi)##. So ##cot^{-1}(-1)=3\pi /4## not ##-\pi /4##.
I forgot its range...
So now this equation is true for all values of ##x##.
Thanks (##R\rightarrow R## help)
 
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mooncrater said:

Homework Statement


There is a part of solution of a question which says that:
##cot|cot^{-1}x|=cot cot^{-1}x=x##

Homework Equations

The Attempt at a Solution


If we put ##x=-1## in this equation then ##cot^{-1}(-1)=-\pi /4## which will have a modulus =##\pi /4##. And ##cot \pi /4=1## which is not equal to the ##x## we took. So this equation seems wrong to me. Am I correct here?

If it is not equal to the x you took, it is equal to
| (the x you took) |
and there is a | | in the problem.

I think the answer you have been given is wrong and the answer is | x | .

You seem to have been given a few wrong answers and even wrong questions, or at least at the limit thereof. :oldwink:
 
epenguin said:
If it is not equal to the x you took, it is equal to
| (the x you took) |
and there is a | | in the problem.

I think the answer you have been given is wrong and the answer is | x | .

You seem to have been given a few wrong answers and even wrong questions, or at least at the limit thereof. :oldwink:
The standard range used for the arccotangent is (0, π). Thus the arccotangent function always returns a positive value.

So we have | cot-1(x) | = cot-1(x)

Added in Edit:
However, the range used for the arccotangent function is not universally established.
 
Last edited:
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