How do you convert # r= cos^2(theta/2)# into cartesian form?

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Answer 1 · 2016-06-30T23:31:45

#2(x^2 + y^2) = x + sqrt{x^2 + y^2}#

With #r= cos^2(theta/2)#, we can get that #theta/2# back into a #theta#

as we have the double angle formula

#cos 2Q = 2 cos^2 Q -1#

so

# cos^2 Q = (cos 2Q +1)/2 #

here that means that

#r = (cos theta +1)/2 implies 2 r = cos theta +1#

now we have #x = r cos theta# so #cos theta = x/r#

meaning that

#2 r = x/r +1#

#2 r^2 = x + r#

in cartesian, that is very very ugly

#2(x^2 + y^2) = x + sqrt{x^2 + y^2}#