How do you convert #Theta=-pi/6# into cartesian form?

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Answer 1 · 2016-06-03T05:01:02

I adhere to the convention that r is modulus (length), and so, r is non-negative. #y=-x/sqrt 3, y<0#.

Use theta = tan^(-1)(y/x)#

Here, #tan^(-1)(y/x)=theta=-pi/6#.

So,# y/x=tan(-pi/6)=-tan(pi/6)=-1/sqrt 3#.

This line is restricted to the 4th quadrant.

The other half is given by the polar equation #theta=(5/6)pi#.

Note that #tan ((5/6)pi)=-1/sqrt 3#.

You can include r = 0 ((x, y)=(0, 0)) as the limiting end-point, for each

half.