Question

GRAPHING POLAR EQUATIONS
Determine if the graph is symmetric about the x-axis, the y-axis, or the origin.

r = 2 cos 5θ

Answer 1

Here's the graph: http://www.wolframalpha.com/input/?i=r+%3D+2+cos+5+theta

Answer 2

using the Pascal's triangle, or the Tartaglia's triangle, and the formula of De Moivre, we can write:
\[\cos \left( {5\theta } \right) = {\left( {\cos \theta } \right)^5} - 10{\left( {\cos \theta } \right)^3}{\left( {\sin \theta } \right)^2} + 10\left( {\cos \theta } \right){\left( {\sin \theta } \right)^4}\]

Answer 3

so, using the formulas from polar to cartesian coordinates:
\[\Large \begin{gathered}
{r^2} = {x^2} + {y^2} \hfill \\
\cos \theta = \frac{x}{r},\quad \sin \theta = \frac{y}{r} \hfill \\
\end{gathered} \]
we can rewrite your equation, as below:
\[\Large r = 2\left\{ {\frac{{{x^5}}}{{{r^5}}} - 10\frac{{{x^3}{y^2}}}{{{r^5}}} + 5\frac{{x{y^4}}}{{{r^5}}}} \right\}\]
or:
\[\Large {\left( {{x^2} + {y^2}} \right)^3} = 2\left( {{x^5} - 10{x^3}{y^2} + 5x{y^4}} \right)\]

Answer 4

now, if I change: y---> -y, I get:
\[\Large {\left( {{x^2} + {{\left( { - y} \right)}^2}} \right)^3} = 2\left( {{x^5} - 10{x^3}{{\left( { - y} \right)}^2} + 5x{{\left( { - y} \right)}^4}} \right)\]
so, what can you conclude?

Answer 5

I don't seem to understand what you stated above. I know the De Moivre Formula, but the way I was taught about this lesson is different from what you said.

Answer 6

Based on my textbook, to test for the symmetry, I need to replace (r, \(\theta \)) by (-r, \(-\theta \)) and if the equation is the same as the first polar equation, then it is symmetric about the y-axis.

Answer 7

\[r=2\cos5\theta\]\[-r=2\cos(-\theta)\]\[-r=-2\cos5\theta\]\[r=2\cos5\theta\]

Answer 8

So I think it is symmetric about the y-axis.

Answer 9

Sorry, I wrote your equation using cartesian coordinates, so as you can see that equation is unchanged when we make this variable change:
y---> -y
then your graph is symmetric with respect to the x-axis

Answer 10

It's totally fine. I don't know if I did something wrong but based on the graph, it is symmetric about the x-axis.

Answer 11

please, note my graph:

Answer 12

\(cos (- \theta) = cos \theta\)

Answer 13

That was the first graph that appeared in wolframalpha! But then when I entered it again, I got the different one.

Answer 14

yes! nevertheless we have cos(5 \theta) @IrishBoy123

Answer 15

same difference, as they say

Answer 16

@IrishBoy123 Thanks for the correction! I forgot about that. :) But sin(-theta)=-sin(theta), right?

Answer 17

Anyway, thanks for the help! I really appreciate it.

Answer 18

I have made typo, here is the right formula:
\[\Large \cos \left( {5\theta } \right) = {\left( {\cos \theta } \right)^5} - 10{\left( {\cos \theta } \right)^3}{\left( {\sin \theta } \right)^2} + 5\left( {\cos \theta } \right){\left( {\sin \theta } \right)^4}\]

Answer 19

yes

https://gyazo.com/5f1871923b6020ebe449d395b2e01585

same as @Michele_Laino

Answer 20

\[\large \cos \left( {5\theta } \right) = {\left( {\cos \theta } \right)^5} - 10{\left( {\cos \theta } \right)^3}{\left( {\sin \theta } \right)^2} + 5\left( {\cos \theta } \right){\left( {\sin \theta } \right)^4}\]

Answer 21

for clarity, here is what is being used as an alternative to CArtesian:
https://gyazo.com/9d5ca77c20b40c9d74452f6922539076