How do you graph #r=4cos6theta#?

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Answer 1 · 2016-12-27T02:20:53

Th Socratic graph of this 6-petal rose is inserted.

#r=4 cos 6theta>=0#. So, #cos 6theta >=0#.

It follows that #6theta in Q_1 or Q_4#.

The period for the graph is #(2pi)/6 =pi/3#.

So, there are 6 cycles for #theta in [-pi, pi]#.

For one cycle #theta in [-pi/6, pi/6]#

Yet, #r >=0#, only in the middle half #[-pi/12, pi/12]#, wherein one

petal is completed..

In the remaining half# r <0# and the graphics software designers

have kept off negative r.

For Socratic graphing, I have used the cartesian form of the

equation obtained from

#cos 6theta#

#=cos^6theta-6C_2cos^4thetasin^2theta#

#+6C_4cos^2thetasin^4theta-6C_6sin^6theta#

graph{(x^2+y^2)^3.5-4(x^6-y^6-15x^2y^2(x^2-y^2))=0 [-10, 10, -5, 5]}