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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
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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triciaal
 one year ago
Best ResponseYou've already chosen the best response.0dw:1439090016277:dw

triciaal
 one year ago
Best ResponseYou've already chosen the best response.0hope something helped

triciaal
 one year ago
Best ResponseYou've already chosen the best response.0@jim_thompson5910 @UsukiDoll what am I missing?

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.0I'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.

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.0OH GAWD y'all scared me *faints*

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.0anyway \[\tan(\theta) = 1 \] is negative 45 degrees. so counter clockwise.

triciaal
 one year ago
Best ResponseYou've already chosen the best response.0@UsukiDoll thanks I have all that what am I missing (except r = 2sqrt2)

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.0polar coordinates \[(r, \theta) \] since \[r = 2 \sqrt{2} , 2 \sqrt{2} \] \[(2 \sqrt{2} , \theta) \], \[(2 \sqrt{2}, \theta) \] now to find theta....

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.0hmmmmm... if we have negative 45 degrees... maybe 360  45 = 315 degrees for the fourth quadrant and 18045 = 135 degrees for the second quadrant. I might be a bit off on this one since it has been a while.

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.0oh wait... maybe.. dw:1439093531632:dw

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.0dw:1439093615408:dw a bit sketchy on this part though.

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.0uh 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

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.0If \[r = 2\sqrt{2}\] 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

triciaal
 one year ago
Best ResponseYou've already chosen the best response.0@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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