Using a Table of Values to Sketch this Curve

What if you can't use a computer to draw the graph?

You'll need to set up a table of values, as follows. I've put degrees and the radian equivalents.

θ (degrees) 30° 60° 90° 120° 150° 180°
θ (radians) `0` `π/6`  `π/3` `π/2` `(2π)/3` `(5π)/6` `π`

`r = 3\ cos\ 2θ`

`3` `1.5` `-1.5` `-3` `-1.5` `1.5` `3`

θ (degrees) 210° 240° 270° 300° 330° 360°
θ (radians) `(7π)/6` `(4π)/3` `(3π)/2` `(5π)/3` `(11π)/6` `2π`

`r = 3\ cos\ 2θ`

`1.5` `-1.5` `-3` `-1.5` `1.5` `3`

The first 7 points from this table are (3, 0°), (1.5, 30°), (-1.5, 60°), (-3, 90°), (-1.5, 120°), (1.5, 150°), and (3, 180°).

Placing those first 7 points on a polar coordinate grid gives us the following:

Graph ata points r = 3 cos (2theta) on polar axes

We start at Point 1, (3, 0°), and move around the graph by increasing the angle and changing the distance from the origin (determined by substituting the angle into r = 3 cos 2θ. I have drawn arrows to indicate the basic direction we have to head in to get to the next point.

Recall: A negative "r" means we need to be on the opposite side of the origin.

I have only plotted the first 7 points above to keep the graph simple. Clearly, we would need to calculate more than this number of points to get a good sketch. (You would need at least twice as many points as I have in the table above - every 15° would be adequate.)

Here's the complete graph.


Graph of r = 3 cos (2θ).

[In the above graph, angles are in radians, where π radians = 180°. To learn more, see: Radians.]

Notice the curve is fully drawn once θ takes all values between 0 and 2π.

Conversion from Polar to Rectangular Coordinates

Next, here's the answer for the conversion to rectangular coordinates.

Why? We convert the function given in this question to rectangular coordinates to see how much simpler it is when written in polar coordinates.

To convert `r = 3\ cos\ 2θ` into rectangular coordinates, we use the fact that

cos 2θ = cos2 θ - sin2 θ.

So r = 3 cos 2θ = 3(cos2 θ - sin2 θ).

Now since `cos\ theta=x/r`, `sin\ theta=y/r` and `r^2=x^2+y^2`, we have




Taking the positive square root of `r^2=x^2+y^2` gives us:


So `r = 3 cos\ 2θ` in polar coordinates is equivalent to





in rectangular coordinates.

We see that our equation in polar coordinates, r = 3 cos 2θ, is much simpler than the rectangular equivalent.

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