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anonymous

  • one year ago

Trig/ Pre Cal How do I determine what quadrant this is in? 2 owlbucks! \[ \tan^-1(\tan(\frac{5\pi}{6})) \] Thank you.

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  1. anonymous
    • one year ago
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    It was suppose to be \[ \tan^1(\tan\frac{5\pi}{6}) \] I don't know how to make the -1 power so 1 = -1. It is an inverse

  2. liamschumm
    • one year ago
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    I'm not sure what you mean... You mean where it is on the unit circle?

  3. liamschumm
    • one year ago
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    Just a number does not have a "quadrant".

  4. anonymous
    • one year ago
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    Btw to make an exponent negative, just do tan^{-1} in latex :)

  5. anonymous
    • one year ago
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    Yes. For instance, we have a domain of \( -\frac{\pi}{2}< x< \frac{\pi}{2} \) and range of \( -infinity< x< infinity \)

  6. anonymous
    • one year ago
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    How do I find the quadrant that \( \frac{5\pi}{6} lies in?\)

  7. anonymous
    • one year ago
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    Easy way is to just look at the unit circle http://upload.wikimedia.org/wikipedia/commons/thumb/4/4c/Unit_circle_angles_color.svg/2000px-Unit_circle_angles_color.svg.png

  8. anonymous
    • one year ago
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    I know by looking at the unit circle but lets say we had a strange angle like \( \frac{4\pi}{5} \)

  9. anonymous
    • one year ago
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    \[\large \frac{\pi}{5} = \frac{180}{5} = 36~degrees\]So \[\large \frac{4\pi}{5} = 4 \times \frac{\pi}{5} = 4 \times 36 = 144\] Since 144 is above 90 but below 180, it would be in quadrant 2 :) Also if the numerator is lower than the denominator, you know it will be in it either the first or second quadrants

  10. anonymous
    • one year ago
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    Thank you!!

  11. anonymous
    • one year ago
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    I sent your owlbucks :-)

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