How do you convert #r = 1-2 cosθ# into rectangular forms?

Question

Benjamin Achtenberg · Accepted Answer

#(x^2+y^2- bx)^2 = a^2(x^2+y^2)#
a = 1; b = -2
#(x^2+y^2+ 2x)^2 = (x^2+y^2)# use the Polar to Cartesian transformation: #r^2 = x^2 + y^2# #x = rcostheta # # r = a + bcostheta# #r^2 = ar + rbcostheta# #r^2- rbcostheta =ar# #x^2+y^2- bx = ar# square the entire thing and write #(x^2+y^2- bx)^2 = (ar)^2# #(x^2+y^2- bx)^2 = a^2(x^2+y^2)# a = 1; b = -2 #(x^2+y^2+ 2x)^2 = (x^2+y^2)#

Christian Antunez · Answer

undefined To convert the polar equation $ r = 1 - 2\cos(\theta) $ into rectangular form, we can use the relationships between polar and rectangular coordinates:

$ x = r \cos(\theta) $
$ y = r \sin(\theta) $

Substituting $ r = 1 - 2\cos(\theta) $ into these equations:

$ x = (1 - 2\cos(\theta)) \cos(\theta) $
$ y = (1 - 2\cos(\theta)) \sin(\theta) $

We then simplify these expressions:

$ x = \cos(\theta) - 2\cos^2(\theta) $
$ y = \sin(\theta) - 2\sin(\theta)\cos(\theta) $

Using the trigonometric identity $ \cos^2(\theta) = 1 - \sin^2(\theta) $, we can rewrite $ \cos^2(\theta) $ as $ 1 - \sin^2(\theta) $:

$ x = \cos(\theta) - 2(1 - \sin^2(\theta)) $
$ y = \sin(\theta) - 2\sin(\theta)\cos(\theta) $

Further simplifying:

$ x = \cos(\theta) - 2 + 2\sin^2(\theta) $
$ y = \sin(\theta) - 2\sin(\theta)\cos(\theta) $

Using the trigonometric identity $ 2\sin^2(\theta) = 1 - \cos(2\theta) $:

$ x = \cos(\theta) - 2 + 1 - \cos(2\theta) $
$ y = \sin(\theta) - 2\sin(\theta)\cos(\theta) $

Simplifying further:

$ x = -1 + \cos(\theta) - \cos(2\theta) $
$ y = \sin(\theta) - 2\sin(\theta)\cos(\theta) $

These are the rectangular forms of the given polar equation.