anonymous
  • anonymous
GRAPHING POLAR EQUATIONS Determine if the graph is symmetric about the x-axis, the y-axis, or the origin. r = 2 cos 5θ
Mathematics
  • Stacey Warren - Expert brainly.com
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
Here's the graph: http://www.wolframalpha.com/input/?i=r+%3D+2+cos+5+theta
Michele_Laino
  • Michele_Laino
using the Pascal's triangle, or the Tartaglia's triangle, and the formula of De Moivre, we can write: \[\cos \left( {5\theta } \right) = {\left( {\cos \theta } \right)^5} - 10{\left( {\cos \theta } \right)^3}{\left( {\sin \theta } \right)^2} + 10\left( {\cos \theta } \right){\left( {\sin \theta } \right)^4}\]
Michele_Laino
  • 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: \[\Large {\left( {{x^2} + {y^2}} \right)^3} = 2\left( {{x^5} - 10{x^3}{y^2} + 5x{y^4}} \right)\]

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Michele_Laino
  • 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?
anonymous
  • 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.
anonymous
  • 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.
anonymous
  • anonymous
\[r=2\cos5\theta\]\[-r=2\cos(-\theta)\]\[-r=-2\cos5\theta\]\[r=2\cos5\theta\]
anonymous
  • anonymous
So I think it is symmetric about the y-axis.
Michele_Laino
  • 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
anonymous
  • 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.
Michele_Laino
  • Michele_Laino
please, note my graph:
1 Attachment
IrishBoy123
  • IrishBoy123
\(cos (- \theta) = cos \theta\)
anonymous
  • anonymous
That was the first graph that appeared in wolframalpha! But then when I entered it again, I got the different one.
Michele_Laino
  • Michele_Laino
yes! nevertheless we have cos(5 \theta) @IrishBoy123
IrishBoy123
  • IrishBoy123
same difference, as they say
anonymous
  • anonymous
@IrishBoy123 Thanks for the correction! I forgot about that. :) But sin(-theta)=-sin(theta), right?
anonymous
  • anonymous
Anyway, thanks for the help! I really appreciate it.
Michele_Laino
  • Michele_Laino
I have made typo, here is the right formula: \[\Large \cos \left( {5\theta } \right) = {\left( {\cos \theta } \right)^5} - 10{\left( {\cos \theta } \right)^3}{\left( {\sin \theta } \right)^2} + 5\left( {\cos \theta } \right){\left( {\sin \theta } \right)^4}\]
IrishBoy123
  • IrishBoy123
yes https://gyazo.com/5f1871923b6020ebe449d395b2e01585 same as @Michele_Laino
Michele_Laino
  • Michele_Laino
\[\large \cos \left( {5\theta } \right) = {\left( {\cos \theta } \right)^5} - 10{\left( {\cos \theta } \right)^3}{\left( {\sin \theta } \right)^2} + 5\left( {\cos \theta } \right){\left( {\sin \theta } \right)^4}\]
IrishBoy123
  • IrishBoy123
for clarity, here is what is being used as an alternative to CArtesian: https://gyazo.com/9d5ca77c20b40c9d74452f6922539076

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