Using a graphing utility to see that the equations r = $\displaystyle f_1(\theta) = 2cos(\theta) - 1$ and r = $\displaystyle f_1(\theta) = 2cos(\theta) + 1$ have the same graph. Explain why. Hint show that the points $\displaystyle (f_1(\theta + \pi), \theta + \pi)$ and $\displaystyle (f_2(\theta),\theta)$ coincide

f1 starts a 1 and rotates up then down. f2 starts on 3 and rotates up to the left. here is a table showing the values of f1 and f2 at four points of the graph.

th 0, pi/2, pi, 3pi/2

f1 1, -1, -3, -1

f2 3, 1, -1, 1

points of the functions for 0, pi/2, pi, 3pi/2

$\displaystyle (f_1(\theta + \pi), \theta + \pi)$ (-3,pi) (-1, 3pi/2) (1,0) (-1, pi/2)

$\displaystyle (f_2(\theta),\theta)$ (3,0) (1, pi/2) (-1, pi) (1, 3pi/2)

these are all the same points. a negative r means flip it over to the other direction(+pi degrees). Looking at the first table f1 (1,0) and f2 (-1, pi). these are the same points. Shifting by pi results in the same points. The functions share the same points but at different input values of theta.