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Why is the range of \(y = \sin^{1}\theta\) in Quadrants I and IV? I thought that the principal values were in Quadrants I and III?
 one year ago
 one year ago
Why is the range of \(y = \sin^{1}\theta\) in Quadrants I and IV? I thought that the principal values were in Quadrants I and III?
 one year ago
 one year ago

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lgbasalloteBest ResponseYou've already chosen the best response.1
but the positive and negative reference angles are in Quadrant I and IV yes?
 one year ago

CalcmathleteBest ResponseYou've already chosen the best response.0
Yes? Not entirely sure if I understand the concept of this.
 one year ago

lgbasalloteBest ResponseYou've already chosen the best response.1
what don't you understand?
 one year ago

CalcmathleteBest ResponseYou've already chosen the best response.0
Well, I don't quite understand what the range comes from? I understand that the principal values are the values that make \(y = \sin^{1}\theta\) a function, but I don't quite understand what the change of domain and range is? Since I don't seem to understand the refernece angles you mentioned above for this, could you explain that?
 one year ago

lgbasalloteBest ResponseYou've already chosen the best response.1
now....im not entirely sure what you're asking
 one year ago

CalcmathleteBest ResponseYou've already chosen the best response.0
Well, going back to the original question, I was confused on why the range would be in Quadrants I and IV. Would I be getting the range of which quadrants it's in based on the principal values? For instance, the principal values are \(\frac π2 ≤ \theta ≤  \frac π2\) Would it mean that the range is in quadrants I and IV because if \(\theta\) is within that range, it would lie in Quadrants I and IV?
 one year ago

CalcmathleteBest ResponseYou've already chosen the best response.0
I mean \(\frac π2 ≤ \theta ≤ \frac π2\)
 one year ago

CalcmathleteBest ResponseYou've already chosen the best response.0
dw:1348274826087:dw
 one year ago

CalcmathleteBest ResponseYou've already chosen the best response.0
NEver mind. I have to go. Thanks.
 one year ago
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