How does one determine the rectangular coordinates of polar coordinates (-2, 11pi/6)?

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Answer 1 · 2018-04-24T00:43:48

#(x,y)=(-sqrt{3}, 1)#

Those aren't really valid polar coordinates because #r < 0#. Nonetheless we can convert:

#(r,theta)=(-2, {11pi}/6)#

#x = r cos theta = -2 cos({11pi}/6) = -2 cos(pi/6) = -sqrt{3} #

# y = r sin theta = -2 sin({11 pi}/ 6)= 2 sin(pi/6) = 1 #

#(x,y)=(-sqrt{3}, 1)#

Not sure if you're there yet, but we can also write this

# -2 e^{i ( {11 pi}/6) } = -sqrt{3} + i#