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Homework Help: Inverse problem

  1. Dec 17, 2007 #1
    1. The problem statement, all variables and given/known data

    cot x= 0

    How do I figure this out? I can't remember how to do it.
    It makes sense the answer would be zero. But I got my test back and the answer was 3pi/2 +2pik. This answer doesn't make sense.

    2. Relevant equations

    This is what I remember. cot^-1=x I don't know how to do it on my calculator.

    3. The attempt at a solution
  2. jcsd
  3. Dec 17, 2007 #2


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    Emmm what does k stand for?

    (Sorry I don't know why it isn't zero either, but I'd like to know why?) Not much help though sorry . .

    Maybe if we try to expand it into a series or something? ??
    Last edited: Dec 17, 2007
  4. Dec 17, 2007 #3
    K represents every time you go around the unit circle.
  5. Dec 17, 2007 #4


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    How are you asking about an inverse? The cotangent function is not necessarily an inverse.
    [tex] \[
    \cot (x) = 0 = \frac{{\cos (x)}}{{\sin (x)}}

    or did you want the angle which has cotangent of 0?
  6. Dec 17, 2007 #5


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    Most modern calculators have either a "2nd" function or "inverse function" key and that accesses the function just above or to the side of the original function. You should find "TAN-1" function just above or beside the button labeled "TAN". As I said, you access that by pressing the "2nd" key and then the "TAN" key.

    However, you shouldn't need that. You should know that [itex]cot(\pi/2)= 0[/itex] and that cotangent and tangent have period [itex]\pi[/itex]. I don't know why you were told that solutions to cot(x)= 0 are "3pi/2 +2pik". They are, in fact, [itex]\pi/2+ k\pi[/itex] where k is any integer. (Which would be the same as [itex]\pi/2+ k\pi[/itex].)
  7. Dec 17, 2007 #6
    What about to find cot^-1 on a calculator? The problem is cot x=0. I am trying to take the inverse cotangent of both sides to solve for x.,
    Last edited: Dec 17, 2007
  8. Dec 17, 2007 #7
    That button usually isn't there, but you can quickly get around it

    if you have

    [tex]\cot x = y[/tex]

    instead of going [tex]x = \cot^{-1} y[/tex]

    you can do

    [tex]\frac{1}{\tan x} = y[/tex]

    [tex]\tan x = \frac{1}{y}[/tex]

    [tex]x = \tan^{-1}\frac{1}{y}[/tex]
  9. Dec 17, 2007 #8
    I don't understand. At that point, you would have zero in the denominator.
  10. Dec 17, 2007 #9


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    Yes, and for what [itex]\theta[/itex] is [itex]tan(\theta)[/itex] undefined?
  11. Dec 17, 2007 #10
  12. Dec 17, 2007 #11
    anywhere else?

    or what other condition could you apply to pi/2? such as what you were confused about earlier, with the "k"
  13. Dec 17, 2007 #12
  14. Dec 18, 2007 #13
    So lim x-> 0 1/x would give you infinity and the calc. will give you an error. Try putting a very large number like 999999999 (fill up all the digits on the clac.) That should also give you your answer.

    You can also graphically get the answer by plotting the tan function and doing a 1/tan to get the cot function. Graphically is the best way as even if you forget, you can always derive it. This way you won't need to memorize anything.
  15. Dec 19, 2007 #14
    This question has more than 2 answers

    Ok, first of all: cot (data)= 0. But, cot= cos/sin (data)
    so, on the Unit Circle there woul be Pi/2 (0,1), and 3Pi/2 (0,-1). And that would make the fuction to be equal 0. However, k is any integer that -Pi/2<data<Pi/2, so data has to lie in QD IV and I. For a cot fucn, the period goes to 2Pi, therefore:
    data= Pi/2 +2KPi and data= 3Pi/2 +2KPi. Let k=1,2,and 3, then solve for data. If data is in degree, covert 2KPi to 360Pi. Good luck
  16. Dec 19, 2007 #15
    [tex]cot x= 0 = \frac {cos x}{sin x}[/tex]

    So where ever cos x = 0 so will cot x = 0. If you're not given any domain then you'll need to find ALL the values where cos x = 0.

    Read ThienAn's post^^
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