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anonymous
 4 years 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?
anonymous
 4 years 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?

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lgbasallote
 4 years ago
Best ResponseYou've already chosen the best response.1but the positive and negative reference angles are in Quadrant I and IV yes?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Yes? Not entirely sure if I understand the concept of this.

lgbasallote
 4 years ago
Best ResponseYou've already chosen the best response.1what don't you understand?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Well, 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?

lgbasallote
 4 years ago
Best ResponseYou've already chosen the best response.1now....im not entirely sure what you're asking

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Well, 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?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I mean \(\frac π2 ≤ \theta ≤ \frac π2\)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1348274826087:dw

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0NEver mind. I have to go. Thanks.
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