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

##### 2 Answers
Jul 21, 2018

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Please read the explanation.

#### Explanation:

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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 $\textcolor{b l u e}{C o s \left(\theta\right)}$, 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)
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when we compare the two graphs:
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Hope it helps.

Jul 21, 2018

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

#### Explanation:

$r = 1 - 2 \cos \theta \ge 0$, for $\theta \in \left[\frac{\pi}{3} , 5 \frac{\pi}{3}\right]$.

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

directions $\theta = \frac{\pi}{3}$ and, upon completing the curve ( in the

anticlockwise sense ) and returning to the pole, theta = $\frac{2 \pi}{3}$.

In exactitude, the Cartesian equation is

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

The Socratic graph is immediate, for $r \ge 0$ 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 $r = \pm 1 - 2 \cos \theta$. .
graph{(x^2+y^2+2x)^2-(x^2+y^2)=0[-5 5 -2.5 2.5]}.

I use Mathematical graphic plotting, for $r \ge 0.$
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