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Math Help - Domain/Range of Inverse Trig Function

  1. #1
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    Domain/Range of Inverse Trig Function

    This question is difficult to put into words, but here we go...

    1) If the domain of y = sin(x) is all possible x values...
    2) And if the range of y = sin(x) is all possible y values...

    ...When looking at the inverse of this function...

    3) Wouldn't the domain of x = sin(y) be all possible y values?...
    4) And, similarly, the range of x = sin(y) be all possible x values?...

    ...Regardless of the notation being y = arcsin(x) or y = sin^-1(x)

    I ask because I'm confused as to why the y values of x = sin(y) are always referred to as the range in this situation (and x values as the domain). Maybe I'm missing something huge, but it seems to me that the common notation (y = arcsin(x)) should not take precedence over the actual meaning (x = sin(y)) when deciding upon the domain and range of a function. Wow, that was difficult to express!
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  2. #2
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    Quote Originally Posted by cusseta View Post
    This question is difficult to put into words, but here we go...

    1) If the domain of y = sin(x) is all possible x values...
    2) And if the range of y = sin(x) is all possible y values...

    ...When looking at the inverse of this function...

    3) Wouldn't the domain of x = sin(y) be all possible y values?...
    4) And, similarly, the range of x = sin(y) be all possible x values?...

    ...Regardless of the notation being y = arcsin(x) or y = sin^-1(x)

    I ask because I'm confused as to why the y values of x = sin(y) are always referred to as the range in this situation (and x values as the domain). Maybe I'm missing something huge, but it seems to me that the common notation (y = arcsin(x)) should not take precedence over the actual meaning (x = sin(y)) when deciding upon the domain and range of a function. Wow, that was difficult to express!
    If you want the inverse function \sin^{-1}(x) to exist, then the domain of sin(x) must be restricted so that it is a 1-to-1 function. The conventional restriction is \displaystyle -\frac{\pi}{2} \leq x \leq \frac{\pi}{2}.
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