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anonymous
 one year ago
GRAPHING POLAR EQUATIONS
Determine if the graph is symmetric about the xaxis, the yaxis, or the origin.
r = 2 cos 5θ
anonymous
 one year ago
GRAPHING POLAR EQUATIONS Determine if the graph is symmetric about the xaxis, the yaxis, or the origin. r = 2 cos 5θ

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Here's the graph: http://www.wolframalpha.com/input/?i=r+%3D+2+cos+5+theta

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2using the Pascal's triangle, or the Tartaglia's triangle, and the formula of De Moivre, we can write: \[\cos \left( {5\theta } \right) = {\left( {\cos \theta } \right)^5}  10{\left( {\cos \theta } \right)^3}{\left( {\sin \theta } \right)^2} + 10\left( {\cos \theta } \right){\left( {\sin \theta } \right)^4}\]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2so, 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: \[\Large {\left( {{x^2} + {y^2}} \right)^3} = 2\left( {{x^5}  10{x^3}{y^2} + 5x{y^4}} \right)\]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2now, 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?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I 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.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Based 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 yaxis.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[r=2\cos5\theta\]\[r=2\cos(\theta)\]\[r=2\cos5\theta\]\[r=2\cos5\theta\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0So I think it is symmetric about the yaxis.

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2Sorry, 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 xaxis

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0It's totally fine. I don't know if I did something wrong but based on the graph, it is symmetric about the xaxis.

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2please, note my graph:

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.0\(cos ( \theta) = cos \theta\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0That was the first graph that appeared in wolframalpha! But then when I entered it again, I got the different one.

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2yes! nevertheless we have cos(5 \theta) @IrishBoy123

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.0same difference, as they say

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@IrishBoy123 Thanks for the correction! I forgot about that. :) But sin(theta)=sin(theta), right?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Anyway, thanks for the help! I really appreciate it.

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2I have made typo, here is the right formula: \[\Large \cos \left( {5\theta } \right) = {\left( {\cos \theta } \right)^5}  10{\left( {\cos \theta } \right)^3}{\left( {\sin \theta } \right)^2} + 5\left( {\cos \theta } \right){\left( {\sin \theta } \right)^4}\]

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.0yes https://gyazo.com/5f1871923b6020ebe449d395b2e01585 same as @Michele_Laino

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2\[\large \cos \left( {5\theta } \right) = {\left( {\cos \theta } \right)^5}  10{\left( {\cos \theta } \right)^3}{\left( {\sin \theta } \right)^2} + 5\left( {\cos \theta } \right){\left( {\sin \theta } \right)^4}\]

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.0for clarity, here is what is being used as an alternative to CArtesian: https://gyazo.com/9d5ca77c20b40c9d74452f6922539076
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