Question

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

Answer 1

See explanation and graph.

Explanation:

`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}