anonymous
  • anonymous
Find all polar coordinates of point P where P = (3, -pi/3)
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
@le0n any idea how to solve this one? :D
anonymous
  • anonymous
this is in cartesian right?
anonymous
  • anonymous
and you want in polar or other way?

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anonymous
  • anonymous
I found this on google, hopefully it explains a little bit: You have to add 2nπ where n is an integer, because you're looking for all polar coordinates and not just some. often adding 2nπ means the original angle is included, as well as any angles that are at a distance of 2π away, which will have the same sin and cos and tan values and therefore will be included in the solution. For example, π/4 and π/4 + 2π (note this is equal to 9π/4) both have a cos value of sqrt(2) / 2 and a sin value of sqrt(2) / 2.
anonymous
  • anonymous
I'm still a little confused :O
anonymous
  • anonymous
aaa now i understand what you were usking :D
anonymous
  • anonymous
is this multiple choice btw?
anonymous
  • anonymous
|dw:1377899913875:dw| in polar coordinates you denote point by its distance from the origin and angle from the x-axis. but if you will add 2pi radians to an angle this means that you went in a circle and came back to the same point
anonymous
  • anonymous
yeah it is multiple choice!! I understand that it will circle around the circle but I'm still confused on how exactly to read it. I'll post the answers
anonymous
  • anonymous
A (3, negative pi divided by 3 + 2nπ) or (-3, negative pi divided by 3 + (2n + 1)π) B(3, negative pi divided by 3 + 2nπ) or (-3, negative pi divided by 3 + 2nπ) C(3, negative pi divided by 3 + 2nπ) or (3, negative pi divided by 3 + (2n + 1)π) D (3, negative pi divided by 3 + (2n + 1)π) or (-3, negative pi divided by 3 + 2nπ)
anonymous
  • anonymous
|dw:1377900588362:dw| here is a hint
anonymous
  • anonymous
by the way sorry for my awful drawings :)
anonymous
  • anonymous
@le0n HEY! i'm back!! sorry i atedinner :) and okay so what;s up with that??? lol!

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