How do you graph $4cos(3theta)$ ?

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Answer 1 · 2018-08-14T13:01:09

See explanation and graph.

$0 <= 4 sqrt (x^2 + y^2 ) = 4 cos 3theta in [ 0, 4 ]$

The period for this cosine function $= ( 2 pi )/3$ .

$r >= 0$ for $theta in [ 0, pi/6 ], [ pi/2, 2/3pi ], [ 7/6pi, 3/2pi ] and$

$[ 11/6pi, 2pi ]$ , in one revolution $theta in [ 0, 2pi ]$ . The first

and the last are halves, from the same loop. For subsequent

revolutions, these three three loops are redrawn.

For Socratic graphic utility that keeps off r-negative pixels, the

Cartesian form.

$sqrt ( x^2 + y^2 )$

$= 4 = cos 3theta =4 ( cos^3theta - 3 cos theta sin^2theta )$

$= 4 ( x^3/r^3- (3xy^2)/r^3)$ , giving

$( x^2 + y^2 )^2 = 4 ( x^3 - 3 xy^2 )$

is used. See graph.
graph{( x^2 + y^2 )^2 - 4 ( x^3 - 3 xy^2 )=0}

How do you graph 4cos(3theta) 4cos(3theta)?