Example similar to the take home final

By request, I present an example like the take home final.  This is the graph of r =2cos(2theta).  I calculated the values at all 24 points that have lines radiating from the origin and marked them, then I played connect the dots, smoothing the curve instead of drawing straight lines.  Here are the values at all 24 points.

r = 2cos(2*0) = 2

r = 2cos(2*pi/12) = sqrt(3)

r = 2cos(2*pi/6) = 1

r = 2cos(2*pi/4) = 0

r = 2cos(2*pi/3) =-1

r = 2cos(2*5pi/12) =-sqrt(3)

r = 2cos(2*pi/2) =-2

r = 2cos(2*7pi/12) =-sqrt(3)

r = 2cos(2*2pi/3) =-1

r = 2cos(2*3pi/4) = 0

r = 2cos(2*5pi/6) = 1

r = 2cos(2*11pi/12) = sqrt(3)

r = 2cos(pi) = 2

r = 2cos(2*13pi/12) = sqrt(3)

r = 2cos(2*7pi/6) = 1

r = 2cos(2*5pi/4) = 0

r = 2cos(2*4pi/3) =-1

r = 2cos(2*17pi/12) =-sqrt(3)

r = 2cos(2*3pi/2) =-2

r = 2cos(2*19pi/12) =-sqrt(3)

r = 2cos(2*5pi/3) =-1

r = 2cos(2*7pi/4) = 0

r = 2cos(2*11pi/6) = 1

r = 2cos(2*23pi/12) = sqrt(3)

The result is called a four leaf rose.With the problem from the take home, because we have the formula r = 2 + cos(3*theta), the radius will never get to be less than zero, so it will not cross the origin mulitple times like this one does.