## anonymous one year ago GRAPHING POLAR EQUATIONS Determine if the graph is symmetric about the x-axis, the y-axis, or the origin. r = 2 cos 5θ

1. anonymous

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

2. Michele_Laino

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

3. Michele_Laino

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: $\Large {\left( {{x^2} + {y^2}} \right)^3} = 2\left( {{x^5} - 10{x^3}{y^2} + 5x{y^4}} \right)$

4. Michele_Laino

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?

5. anonymous

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.

6. anonymous

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.

7. anonymous

$r=2\cos5\theta$$-r=2\cos(-\theta)$$-r=-2\cos5\theta$$r=2\cos5\theta$

8. anonymous

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

9. Michele_Laino

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

10. anonymous

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.

11. Michele_Laino

12. IrishBoy123

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

13. anonymous

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

14. Michele_Laino

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

15. IrishBoy123

same difference, as they say

16. anonymous

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

17. anonymous

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

18. Michele_Laino

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

19. IrishBoy123

yes https://gyazo.com/5f1871923b6020ebe449d395b2e01585 same as @Michele_Laino

20. Michele_Laino

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

21. IrishBoy123

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