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Cot^(-1)(0) = pi/2, why?

  1. Aug 4, 2011 #1
    1. The problem statement, all variables and given/known data

    I don't understand why cot^(-1)(0) = pi/2 and was hoping someone could explain this to me. cot(theta)=1/tan(theta)
    because tan^(-1)(0) is undefined
    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Aug 4, 2011 #2

    Pengwuino

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    I think you're confusing [itex]cot^{-1}(\theta) = {{1}\over{cot(\theta)}}[/itex] with the Arc-cotangent or "Inverse cotangent", which is the inverse function of [itex]cotan(\theta)[/itex]. The inverse cotangent is the function that tells you what values of [itex]\theta[/itex] give you the value 'x' in [itex]cotan(\theta) = x[/itex].

    So in other words, [itex]cot^{-1}(0)[/itex] is asking what values of [itex]\theta[/itex] give you [itex]cot(\theta) = 0[/itex].

    EDIT: Ok I think I got that squared away correctly.
     
    Last edited: Aug 4, 2011
  4. Aug 4, 2011 #3
    oh so cot(0) is undefined because
    cot(0) = 1/tan(0) = 1/0 = undefined, makes sense
    cot(pi/2) = 1/tan(pi/2) = 1/undefined =/= 0
    I don't see how the two are equal, and ya I think I may be getting some things mixed up as I haven't dealt with basic trig in several years lolz
     
  5. Aug 4, 2011 #4

    Pengwuino

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    No, cot(pi/2) = 0.

    Why would they be equal in the first place?
     
  6. Aug 4, 2011 #5
    cot(theta)=1/tan(theta)
    cot(pi/2) should then be equal to 1/tan(pi/2)
    cot(pi/2) = 0 = 1/tan(pi/2)

    I just don't understand why
    1/tan(pi/2) is equal to zero
    because tan(pi/2) = undefined
    so 1/tan(pi/2) = 1/undefined
    how is this equal to zero?[/quote]
    Rather than think of cot(x) as 1/tan(x), a more fundamental definition is: tan(x)= sin(x)/cos(x) and cot(x)= cos(x)/sin(x). [itex]sin(\pi/2)= 1[/itex] and [itex]cos(\pi/2)= 0[/itex] so [itex]tan(\pi/2)[/itex] is undefined (the denominator is 0) while [itex]cot(\pi/2)= 0[/itex] (the numerator is 0).
     
    Last edited by a moderator: Aug 5, 2011
  7. Aug 4, 2011 #6

    SammyS

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    cot-1(x) is another way to write the arccot(x) function.

    cot-1(x) ≠ 1/cot(x) .
     
  8. Aug 5, 2011 #7

    Mentallic

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    It's undefined in the sense that division by zero is not allowed. Think about it this way, if [tex]\frac{a}{\left(\frac{b}{c}\right)}=\frac{ac}{b}[/tex] then [tex]\frac{1}{\left(\frac{1}{0}\right)}=0[/tex]

    Or you can even think of division by zero as being [itex]\pm\infty[/itex] so when we divide a finite value by this amount, we get 0.
     
  9. Aug 5, 2011 #8

    uart

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    Hi Greenprint. Mentallic has the correct answer here. [itex]1/0[/itex] is undefined however [itex]0/1[/itex] is perfectly well defined and is equal to zero.
     
  10. Aug 6, 2011 #9
    [itex]\cot^{-1}(0) = \pi/2 \Rightarrow \cot(\pi/2) = 0[/itex]

    Trying to rearrange it so you can use the more familiar tan doesn't really help since it brings in division by 0. Take a look at the graph of cot-1x and see what the value is when x = 0
     

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  11. Aug 6, 2011 #10

    HallsofIvy

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    I think it would help to point out that the basic definition of [itex]cot(x)[/itex] is [itex]cos(x)/sin(x)[/itex], not [itex]1/tan(x)[/itex]. At [itex]x= \pi/2[/itex], [itex]tan(x)[/itex] is not defined but [itex]cos(\pi/2)= cos(\pi/2)/sin(\pi/2)= 0/1= 0[/itex]
     
  12. Aug 6, 2011 #11

    SammyS

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    O.P. acknowledged that [itex]\cot(\pi/2)= 0[/itex] in Post #5.
     
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