1. ## Sketching polar graph

Sketch the curve $r = 2cos4\theta$

Attempt:

when $\theta$ = 0, r = 2

when $\theta$ = $\frac{\pi}{6}$, r = -1

when $\theta$ = $\frac{\pi}{4}$, r = -2

The graph I got was a circle (not ment to look like an ellipse in the drawing), but the correct answer was a flower

2. Originally Posted by SyNtHeSiS
Sketch the curve $r = 2cos4\theta$

Attempt:

when $\theta$ = 0, r = 2

when $\theta$ = $\frac{\pi}{6}$, r = -1

when $\theta$ = $\frac{\pi}{4}$, r = -2

The graph I got was a circle (not ment to look like an ellipse in the drawing), but the correct answer was a flower
You need more points.

Hint: Find the points where r=0.

3. Hello, SyNtHeSiS!

$\text{Sketch the curve: }\;r \:=\: 2\cos4\theta$

We are expected to recognize this as a "rose curve", a flower-shaped curve.
. . [But it looks more like a daisy.]

There are two basic forms: . $\begin{Bmatrix}r &=& a\sin n\theta \\ r &=& a \cos n\theta \end{Bmatrix}$

$\,a$ is the length of a "petal".

If $\,n$ is odd, there are $\,n$ petals.
If $\,n$ is even, there are $2n$ petals.
. . The petals are equally spaced about the center.

We have: . $r \:=\:2\cos4\theta$

Since $n = 4$ (even), there are 8 petals, spaced 45 degrees apart.
Each petal is 2 units long.

Now where are the petals?
. . When is $r = 2$ ?

We have: . $2\cos4\theta \:=\:2 \quad\Rightarrow\quad \cos4\theta \:=\:1 \quad\Rightarrow\quad 4\theta \:=\:\cos^{\text{-}1}(1)$

. . . . . . . . $4\theta \:=\:0^o \quad\Rightarrow\quad \theta\:=\:0^o$

Hence, the "first" petal is on the $0^o$-line (positive $\,x$-axis).

You should be able to sketch the graph now.