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Just to prove them wrong: Trigonometry

  1. May 4, 2015 #1
    1. The problem statement, all variables and given/known data
    There is a part of solution of a question which says that:
    ##cot|cot^{-1}x|=cot cot^{-1}x=x##

    2. Relevant equations


    3. 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?
     
  2. jcsd
  3. May 4, 2015 #2
    You should know the domain and ranges.
    What is the range for cot-1x ?
     
  4. May 4, 2015 #3
    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)
     
  5. May 5, 2015 #4

    epenguin

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    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:
     
  6. May 5, 2015 #5

    SammyS

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    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: May 5, 2015
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