Question

How do you graph `r=1-2costheta`?

Answer 1

`" "`
Please read the explanation.

Explanation:

`" "`
We have the Polar equation: `color(red)(r=1-2 cos(theta)`

Polar Equations are graphed on the two-dimensional Polar Coordinate System.

The point `color(red)(Z=(r, theta)`, where `color(red)(r` refers to the distance from the origin.

`color(red)(theta` refers to the angle from the positive x-axis, measured counter-clockwise.

In the next step, we will create a Data Table of values for the Polar equation: `color(red)(r=1-2 cos(theta)`:

To calculate `color(blue)(Cos(theta))`, we use the following values for `color(blue)(theta`:

`color(red)(theta = 0^@, 30^@, 45^@, 60^@, 90^@, 135^@ " and " 180^@`

The Data Table is below:

Use the table of values to generate the following graph:

Two graphs are drawn:

One for the parent function

`color(red)(r=Cos(theta)`

and the other for

`color(red)(r=1-2 cos(theta)`

We can understand the behavior of the graph for `color(red)(r=1-2 cos(theta)`
`" "`
when we compare the two graphs:
`" "`

Hope it helps.

Answer 2

I have used Socratic graphic facility that discards r-negative pixels.

Explanation:

`r = 1 - 2 cos theta >= 0`, for `theta in [ pi/3, 5pi/3]`.

The pole r = 0 is a node, with two distinctive tangents, in the

directions `theta = pi/3` and, upon completing the curve ( in the

anticlockwise sense ) and returning to the pole, theta = `(2pi)/3`.

In exactitude, the Cartesian equation is

x^2+y^2 = sqrt(x^2+y^2)+2x=0.

The Socratic graph is immediate, for `r >= 0` only. .
graph{x^2+y^2-sqrt(x^2+y^2)+2x=0[-5 5 -2.5 2.5]}.

See r-positive graph of r = - ( 1 + 2 cos theta )
graph{x^2+y^2+sqrt(x^2+y^2)+2x=0[-5 5 -2.5 2.5]}

See the r-positive combined graph for `r = +-1 - 2 cos theta`. .
graph{(x^2+y^2+2x)^2-(x^2+y^2)=0[-5 5 -2.5 2.5]}.

I use Mathematical graphic plotting, for `r >= 0.`