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Domain of sin(arcsin(x))

  1. Oct 3, 2012 #1
    1. The problem statement, all variables and given/known data
    I need to find the domain of sin(arcsin(x)). Now I "know" that this is xε(-1,1), but I don't understand why. If the range of arcsin(x) is (-∏/2,∏/2), then shouldn't this be the domain of sin(arcsin(x))?

    Help appreciated.
     
  2. jcsd
  3. Oct 3, 2012 #2
    sin (arcsin(x)) means that first arcsin is applied to x, then the result of that is passed on to sin. In what set must x be for this to make sense?
     
  4. Oct 3, 2012 #3

    HallsofIvy

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    Since sine has all real numbers as domain, the domain of this function is the domain of arcsine. Do you understand why its domain is [-1, 1]?
     
  5. Oct 3, 2012 #4
    I understand why the domain of arcsin is [1,-1]. But if sine is the inverse of arcsine, shouldn't the domain of sine be the range of arcsine? I thought that was the rule where inverse functions are concerned.
     
  6. Oct 3, 2012 #5

    SammyS

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    The sine function is not a one-to-one function, so technically, there is no function which is the inverse of sine function.

    So, where do we get the arcsine function from? If we restrict the sine function to a domain of [-π/2, π/2], then that restricted sine function is one-to-one. (By the Way: The range of this restricted sine function is [-1, 1], the same as for the "unrestricted" sine function.) The arcsine function is the inverse of this restricted sine function.

    As it turns out, sin(arcsin(x)) has the same domain and range regardless of which of these sine functions is used.
     
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