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

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Answer 1 · 2016-12-26T05:20:38

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

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)#