For #=3 cos theta - 3, r <=0#. and r is never #> 0#, for #theta# in one
period #[0. 2pi]#.,
I want length #r .>=0#, So, I use # r = 3 - 3 cos theta#, instead.
As r is a function of #cos(theta)=cos(-theta)#, the graph is
symmetrical about the axis #theta = 0#.
The Table for graphing this Cardioid, for one period #2pi#, is
#(r, theta): (0, 0) (3(1-sqrt3/2), pi/6) (3(1-1/sqrt2), pi/4)#
# (3/2, pi/3) (3, pi/2) (9/2, 2pi/3) (3(1+1/sqrt2), 3pi/4)#
#(3(1+sqrt3/2), 5pi/6) (6, pi)#
For the second half of the period #[pi, 2pi],# use symmetry( about
the axis #theta = 0# to draw this half of the cardioid.
The bulge of the cardioid is far away, in the direction #theta =pi#.
