Fan + Medal Determine two pairs of polar coordinates for the point (2, -2) with 0° ≤ θ < 360°.

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Fan + Medal Determine two pairs of polar coordinates for the point (2, -2) with 0° ≤ θ < 360°.

Mathematics
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\[x = rcos \theta\] \[y = rsin \theta\] For polar to cartesian coordinates. For cartesian to polar: \[r^2 = x^2+y^2 ~~~~ \theta = \tan ^{-1} \left( \frac{ y }{ x } \right)\]
Note you want r so \[r = \sqrt{x^2+y^2}\]
r = sqrt(8) or 2.828

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Sure, now find theta
theta = 45
now how do we find which quadrant it is in?
Yeah but notice |dw:1435521133327:dw| theta gives us the value for the first and we need the fourth quadrant.
Nice, you figured that out
oh that makes sense
so 315
Yeah that seems reasonable so 7pi/4
which is \[\large(\frac{ \sqrt{2} }{ 2 }, \frac{ \sqrt{-2} }{ 2 })\]
on the unit circel
circle*
When you're looking for theta, you need unit circle :)
Seems you got a good grasp on it though, nice work.
\[\sqrt{8} \implies 2\sqrt{2}\]
so what is the answer exactly? I need 2 points in Polar Coordinates of the original thing
these are my options: (2 square root of 2, 225°), (-2 square root of 2, 45°) (2 square root of 2, 135°), (-2 square root of 2, 315°) (2 square root of 2, 315°), (-2 square root of 2, 135°) (2 square root of 2, 45°), (-2 square root of 2, 225°)
Oh it wants two pairs
So now what do you think the other will be?
Im blanking @Astrophysics
I think it's the third option. Am I right @zepdrix ?
Oh I see a mistake
What does \[\theta = \tan^{-1}\left( \frac{ -2 }{ 2 } \right)\] give?
|dw:1435522015498:dw|So you found your first polar pair by rotating 315, and then extending outward 2sqrt2? Looks good.
|dw:1435522105721:dw|For your second option, you're rotating 180 degrees less around, and moving radially backwards.
ya third option looks good bro!
for the second coordinate pair i meant*
Thanks both of you!
I have another question but ill post a new form.
If we could only give more than one medal :P
haha true

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