How do you convert #r² = costheta# into polar form?

1 Answer
Tony B
May 31, 2016

#(sqrt(cos(theta)),theta)#

Explanation:

Standard polar form is of format #(r,theta)#

Where #y=rsin(theta)" and "x=rcos(theta)#
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Notice that in the question we have #r^2 -> r=sqrt(r^2)=sqrt(cos(theta))#

Implying that standard form of #y=rsin(theta)# translates into #y=sqrt(cos(theta))xxsin(theta)#

Implying that the standard form of #x=rcos(theta)# translates into #x=sqrt(cos(theta))xxcos(theta)#

So in polar form of type #(r,theta)# we have:

#(sqrt(cos(theta)),theta)#

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