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Domain/Range for Inv. Functions

  1. Jul 15, 2010 #1
    1. The problem statement, all variables and given/known data
    Determine if each of the following is true or false. If false, give a counterexample.

    Arccot (cot x) = x for all x


    2. Relevant equations
    [tex]cot= 1/tan x[/tex]
    Principal values of tan are -90 degrees <x <90 degrees
    Capitalization of trig function indicates a function with restricted domain


    3. The attempt at a solution
    I figured since the cot = 1/tan, it would still have an infinite domain and range.
    By that same thinking, the Cot would have the same domain and range as the Tan, -90 degrees < x <90 degrees.

    But, we dont have Cot, we have the inverse of the Cot. So, do I just switch the x and y values?
    I.e., the x values become infinity, and the y values become -90 degrees < x <90 degrees?

    If so, does this make the equation true?
     
  2. jcsd
  3. Jul 15, 2010 #2

    Mark44

    Staff: Mentor

    But tan(x) is not defined for all values of x, and cot(x) is undefined at values of x for which tan(x) = 0.
    No, Cot and Tan don't have the same domains.
    Here's a similar example: Arcsin(sin(pi)) = 0. This shows a value x for which Arcsin(sin(x)) [itex]\neq[/itex] x.
     
  4. Jul 15, 2010 #3
    Ok, I must really be confused.
    Can you walk me through your thinking when determining whether the equation was true or false?
     
  5. Jul 15, 2010 #4
    I dont understand what your saying here. Your example does not look like the equation of the problem. And, by my understanding, the arcsin and sin cancel, leaving you with pi, not zero.
     
  6. Jul 15, 2010 #5

    Mark44

    Staff: Mentor

    Arcsin and sin "cancel" only under certain circumstances. You shouldn't even be thinking about cancelling, since that term refers to eliminating common factors in the numerator and denominator of a fraction or rational expression.

    sin(pi) = 0, and arcsin(0) = 0, so for this value, pi, clearly arcsin(sin(pi)) [itex]\neq[/itex] pi.

    The example I gave is similar to the one you're working. The similarity is that it is working with trig and inverse trig functions. My example shows that there are some subtleties at play here, involving domain and range.

    Speaking of which, you have some misconceptions about the domain and range of the functions you're working with, as noted in my previous post.
     
  7. Jul 15, 2010 #6
    Ok, I understand your example now.

    However, I still dont know how to determine whether the equation is true or false. The cot x is defined for 0< x <180, and y values being infinite.

    Now, to find the inverse, you just exchange the x and y values with each other, right? So, x becomes infinite and y becomes 0<y<180.

    But how does this equate to Arccot (cot x) = x for all values x?
     
  8. Jul 15, 2010 #7

    vela

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    This isn't true. The cotangent function is defined for all real numbers except multiples of pi.
     
  9. Jul 15, 2010 #8
    Ok, I guess I can see that. But, I still dont see how to go from knowing that to knowing whether or not the equation is true or false.
     
  10. Jul 15, 2010 #9

    eumyang

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    No, not for trig functions. For instance:
    f(x) = sin x
    The domain is all reals, and the range is [-1, 1]

    The inverse is the arcsin:
    g(x) = arcsin x
    BUT while the domain is [-1, 1], the range is NOT all reals. The reason is that you wouldn't have a function at all -- you would have multiple output values for a single input value. So we usually say that the range for g(x) = arcsin x is restricted to [-pi/2, pi/2].

    Given vela's correction, you'll have to do something similar for f(x) = cot x and g(x) = arccot x.


    69
     
  11. Jul 15, 2010 #10

    vela

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    As Mark mentioned in post #5, this problem has to do with the domain and range of the two functions. At this point, you just need to think about it for a while and figure it out on your own.

    Try writing down the domain and range of both functions. Perhaps seeing them on paper in front of you will make everything click into place.
     
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