How do you find all polar coordinates of point #P# where #P =(9, (2pi)/3)#?

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Answer 1 · 2018-01-26T11:40:47

Polar coordinates of #color(green)(P (9.2405, 13.1^0))#

#P (9, (2pi)/3)#

#x = r cos theta, y = r sin theta#

Hence #r = sqrt(x^2 + y^2), = sqrt(9^2 + ((2pi)/3)^2) = 9.2405#

#tan theta = y / x = ((2pi)/3) / 9 = (2pi) / 27#

#theta = tan ^(-1) ((2pi) / 27) = 13.1^0#

Polar coordinates of #color(green)(P (9.2405, 13.1^0))#

Answer 2 · 2018-01-26T13:51:24

#(9,(2pi)/3+2pin)# or #(-9, (5pi)/3+2pin)#, #n in ZZ#.

Or in one shot: #((-1)^n*9, (2pi)/3 + pi*n)#, #n in ZZ#

Given #(9,(2pi)/3)#, any angle coterminal to #(2pi)/3# will put us in the same place, so:

#(9,(2pi)/3+2pin)#, #n in ZZ# works.

But because of how polar works, we could also have #r=-9# and #theta=(5pi)/3# and it's coterminal angles, so:

#(-9, (5pi)/3+2pin)#, #n in ZZ# works as well.

We can actually combine these if we want:

#((-1)^n*9, (2pi)/3 + pi*n)#, #n in ZZ#