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anonymous

  • 4 years ago

Help with finding principal value. I've memorized the unit circle , but my teacher was very vague over principal values.

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  1. anonymous
    • 4 years ago
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    i dont get how \[\sin^{-1} (-\sqrt{2}/2) is -\pi/4\]

  2. anonymous
    • 4 years ago
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    because \[\sin(-\frac{\pi}{4})=-\frac{\sqrt{2}}{2}\]

  3. anonymous
    • 4 years ago
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    you are looking for the angle (number) in the interval \[[-\frac{\pi}{2},\frac{\pi}{2}]\] where the sine is your number

  4. anonymous
    • 4 years ago
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    there are an infinite number of numbers whose sine is anything between -1 and 1 because it is periodic, but only one in the above mentioned interval

  5. anonymous
    • 4 years ago
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    So it sin inverse would only be in the first quadrant of the unit circle?

  6. Mertsj
    • 4 years ago
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    No. It could also be in the 4th quadrant. All values for sin^-1 are, by definition, in quadrants I or IV

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