Why is #r=3cos2theta# not symmetric over #theta=pi/2#?

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1 Answer
Jul 19, 2017

The graph IS symmetric about that line.

Explanation:

You already see the graph, so you were able to observe its symmetry.
One test to determine symmetry about #theta = pi/2# is to substitute
#theta - pi# for #theta#.

#3cos(2(theta -pi)) = 3cos(2theta -2pi)#
#=3cos2thetacos2pi+sin2thetasin2pi#
#=3cos2theta#.

Therefore, the function is symmetric about #theta = pi/2#.