How do you graph $r = 4 cos 3 θ$ $r = 4 cos 3 theta$ ?

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Answer 1 · 2018-07-16T06:53:01

See graph and explanation.

Using

$(x, y) = r (cos θ, sin θ), r = \sqrt{x^{2} + y^{2}}$ $( x, y ) = r ( cos theta, sin theta ), r = sqrt ( x^2 + y^2 )$ and

$cos n θ$ $cos ntheta$

$= \sum {(- 1)}^{r} n C_{2 r} {cos}^{n - 2 r} θ {sin}^{2 r} θ$ $= sum (-1)^r nC_(2r) cos^(n-2r)thetasin^(2r) theta$ ,

r from 0 to $[\frac{n}{2}]$ $[n/2]$ (integer part of n/2).

the equation in Cartesian form for $r = a cos n θ$ $r = a cos ntheta$ can be obtained.

Here, n = 3 and the Cartesian equation is

#( x^2 +y^2 )^2 = 4 (x^3 - 3 x y^2),

The Socratic graph is immediate.

graph{( x^2 +y^2 )^2 - 4 (x^3 - 3 xy^2)=0[-8 8 -4 4]}

How do you graph r=4cos3θr = 4 cos 3 theta?