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anonymous

  • one year ago

Precalculus question: Determine two pairs of polar coordinates for the point (2, -2) with 0° ≤ θ < 360°. WILL MEDAL

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  1. triciaal
    • one year ago
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    |dw:1439090016277:dw|

  2. triciaal
    • one year ago
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    hope something helped

  3. triciaal
    • one year ago
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    @jim_thompson5910 @UsukiDoll what am I missing?

  4. UsukiDoll
    • one year ago
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    I've seen similar questions like these online (2,-2) means that we are at the fourth quadrant and only cosine is positive. (x,y) ->(2,-2) is the rectangular coordinate so we need to switch to polar coordinates \[x^2+y^2=r^2 \] \[(2)^2+(-2)^2=r^2 \] \[4+4=r^2 \] \[8=r^2 \] \[2 \sqrt{2}, -2 \sqrt{2}=r \] the value of tangent is indeed negative in the second and fourth quadrants.

  5. UsukiDoll
    • one year ago
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    OH GAWD y'all scared me *faints*

  6. UsukiDoll
    • one year ago
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    anyway \[\tan(\theta) = -1 \] is negative 45 degrees. so counter clockwise.

  7. triciaal
    • one year ago
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    @UsukiDoll thanks I have all that what am I missing (except r = -2sqrt2)

  8. UsukiDoll
    • one year ago
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    polar coordinates \[(r, \theta) \] since \[r = 2 \sqrt{2} , -2 \sqrt{2} \] \[(2 \sqrt{2} , \theta) \], \[(-2 \sqrt{2}, \theta) \] now to find theta....

  9. UsukiDoll
    • one year ago
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    hmmmmm... if we have negative 45 degrees... maybe 360 - 45 = 315 degrees for the fourth quadrant and 180-45 = 135 degrees for the second quadrant. I might be a bit off on this one since it has been a while.

  10. UsukiDoll
    • one year ago
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    oh wait... maybe.. |dw:1439093531632:dw|

  11. UsukiDoll
    • one year ago
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    |dw:1439093615408:dw| a bit sketchy on this part though.

  12. UsukiDoll
    • one year ago
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    uh oh... I missed something no wonder XD \[x=rcos(\theta) \] if \[r = 2\sqrt{2}\] \[2=2\sqrt{2}cos(\theta) \] \[\frac{1}{\sqrt{2}}=cos(\theta) \] that's 45 degrees \[y=rsin(\theta) \] if \[r = 2\sqrt{2}\] \[-2=2\sqrt{2}sin(\theta) \] \[-\frac{1}{\sqrt{2}}=sin(\theta) \] that's negative 45 degrees

  13. UsukiDoll
    • one year ago
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    If \[r = -2\sqrt{2}\] and (2,-2) \[x = r \cos(\theta)\] \[2 = -2\sqrt{2} \cos(\theta)\] \[-\frac{1}{\sqrt{2}} = \cos(\theta)\] that's negative 45 degrees \[-2 = -2\sqrt{2} \sin(\theta)\] \[\frac{1}{\sqrt{2}} = \sin(\theta)\] that's 45 degrees

  14. triciaal
    • one year ago
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    @UsukiDoll thanks for coming and agreeing @Jim_thompson5910 thanks for coming @jdoherty solution 2rt2, 135 2rt2, 315 the angle was -45 degrees

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