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I want to prove the asymptotes for the inverse cotangent.

  1. Jan 28, 2015 #1
    1. The problem statement, all variables and given/known data
    arccot x = (π/2) - arctan x
    arccot x =/= π
    arccot x =/= 0

    2. Relevant equations
    arccot x = 1/arctan x (if x > 0)
    arccot x = 1/arctan x + π (if x < 0)
    arccot x = π/2 (if x = 0)

    3. The attempt at a solution
    π and 0 are the horizontal asymptotes, the values for which y (sine) cannot be.

    arccot x = (π/2) - arctan x
    arccot x =/= π
    π =/= (π/2) - arctan x
    (π/2) =/= -arctan x
    -(π/2) =/= arctan x

    arccot x = (π/2) - arctan x
    arccot x =/= 0
    0 =/= (π/2) - arctan x
    -(π/2) =/= -arctan x
    (π/2) =/= arctan x

    Therefore, arctan x cannot be equal to negative or positive π/2, because if it were, then adding it to the original equation would produce π or 0, respectively. This cannot work because these are the values that it cannot be. As I write this, I'm thinking that I'm only establishing the fact that 0 and π are the asymptotes. But how would I prove that they're the asymptotes? If it's beyond the scope of precalculus, please say so.
     
    Last edited: Jan 28, 2015
  2. jcsd
  3. Jan 28, 2015 #2

    Mark44

    Staff: Mentor

    What is there to prove? You can reflect the graph of y = cot(x) across the line y = x to get the graph of y = arccot(x).
     
    Last edited: Jan 28, 2015
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