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Simple question about the domain of sec x

  1. Jan 24, 2008 #1
    1. The problem statement, all variables and given/known data
    A problem in my textbook gives the domain of sec x as [0,(pi/2)) U ((pi/2), pi]. This makes perfect sense of course. However, it is asking for an alternate domain. I've looked at the graphs of both sec x and arcsec x to try to figure out a different domain for sec x(I looked at the range of arcsec x obviously), and I can't seem to come up with anything. Anyone know a different domain that can apply to the sec x? It seems like it shouldn't be hard, but for some reason I cannot get it. I've searched the internet for a different domai and kept finding the same as the one in my textbook.


    2. Relevant equations
    y = sec x


    3. The attempt at a solution
    I could only restate the domain as 0<= x <= pi, where x =/ (pi/2)
     
    Last edited: Jan 24, 2008
  2. jcsd
  3. Jan 24, 2008 #2

    NateTG

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    I would think that the domain of [itex]\sec x[/itex] includes a whole lot more than that...
    http://mathworld.wolfram.com/Secant.html

    I suspect that the question is asking for an alternative range for [tex]\sec^{-1}[/itex] instead.
     
  4. Jan 24, 2008 #3
    That was something I was somewhat confused on, whether it was asking for a new domain for the sec x or a new range arcsec x. I was looking at the domain of sec x in terms of it being a one-to-one function. So the domain previously stated creates that. I also focused on a different range for arcsec, but I still can't seem to figure anything out. I actually visited that same site earlier, before posting here.
     
  5. Jan 24, 2008 #4

    NateTG

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    Let's say I know that:
    [tex]\sec x = 1[/tex]
    can you list the possible values of [itex]x[/itex]?

    How does this compare to your notion of
    [tex]\sec^{-1} 1[/tex]
     
  6. Jan 24, 2008 #5
    If sec x = 1, then the possible values for x are pi/4 and 5pi/4.
    Arcsec 1 would yield the same thing wouldn't it? Arcsec 1 is the same thing as saying what angle yields a sec of 1, which is the same thing as the sec x = 1. I know you're trying to steer me in a certain direction, but I'm not sure I'm seeing that direction. Sorry about the delayed response, I had a few classes.
     
  7. Jan 25, 2008 #6

    HallsofIvy

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    Plus all multiples of [itex]2\pi[/itex]! There are an infinite number of values of x such that sec(x)= 1.

    No, it's not. Arcsec(x) is a function and can return only one value. WHAT value that is depends on how you restrict sec(x) so that it is one-to-one (that's the whole point of this exercise). The standard choice is restrict sec(x) to x between [itex]-\pi/2[/itex] to [itex]3\pi/2[/itex].

     
  8. Jan 25, 2008 #7
    I'm confused as to why you'd restrict the domain of the sec x to -pi/2 to 3pi/2. If you used that domain as the range of the arcsec x, would that even work? The range of the arcsec x is 0 to pi. Perhaps I'm misunderstanding the question. Here it is, from the book:

    In the definition above, the inverse secant function is defined by restricting the domain of the secant function to the intervals [0, pi/2)U(pi/2, pi]. Most other texts and references agree with this, but some disagree. What other domains might make sense?
     
  9. Jan 25, 2008 #8

    HallsofIvy

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    Well, I wrote that very quickly, after looking at a rough graph! 0 to pi, skipping, of course, pi/2 looks right. The other possible values would then be pi to 2pi, skipping, 3pi/2, or -pi to 0, skipping -pi/2. On each of those sec is single valued so arcsec exists. Any of those can be used to define arcsin. The "usual rule" is to stay as close to 0 as possible.
     
  10. Jan 26, 2008 #9
    That's why I'm confused. It seems too simple to just change the domain in that manner and say it is a "new" domain, much less have some text books "disagree" with the stated domain. I'm going to e-mail my professor about this as well. Thanks for your help.
     
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