How do you convert the polar equations in Cartesian form $Theta=-pi/6$ ?

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Answer 1 · 2016-07-19T14:36:49

4th quadrant half-line of $y = -x/sqrt 3$ .

The line $theta=-pi/6$ is the radial line in the 4th quadrant, wherein

the polar coordinate $r >=0$ is arbitrary.

Use $(x, y)=r(cos theta, sin theta)$ .

Here, $x =r cos(-pi/6)=r sqrt 3/2 and y=r sin (-pi/6)=-r/2, r>=0$

In brief, $(x, y)=r(sqrt 3/2, -1/2)$ .

and this represents the 4th quadrant half-line of the whole line

$y = -x/sqrt 3$ .

The other half line in the 2nd quadrant is given by $theta=(5pi)/6$

and the separate cartesian (parametric) equations are

$(x, y)= r(sqrt 3/2, -1/2)$ and

. $(x, y)= r(-1/2, sqrt 3/2)$ , r>=0..

How do you convert the polar equations in Cartesian form Theta=-pi/6Theta=-pi/6?