How do you convert θ =(5pi)/6 to rectangular form?

1 Answer
Dec 26, 2016

y=-1/sqrt 3((x-|x|)/2). See the Socratic graph, for verification.

Explanation:

graph{-(x-|x|)/(2sqrt3) [-10, 10, -5, 5]}

theta = 5/6pi in Q_2 represents the half line in Q_2, from the

pole r = 0, in that direction. Only r = sqrt(x^2+y^2)>=0 varies..

The conversion formula is

(x. y)=r(cos theta, sin theta)=r(cos (5/6pi), sin(5/6pi))=r(-sqrt3/2, 1/2),

where r = sqrt(x^2+y^2)>=0.

In #Q_2, x <=0 and y >=0.

It is not correct to write y/x=-1/sqrt 3 ( using tan(5/6pi)=-1/sqrt3 )# that

represents the full line, in Q_2 and Q_4..

The correct cartesian equation to theta =5/6pi is

y=-1/sqrt 3((x-|x|)/2)