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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

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

  2. Michele_Laino
    • one year ago
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    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
    • one year ago
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    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
    • one year ago
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    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
    • one year ago
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    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
    • one year ago
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    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
    • one year ago
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    \[r=2\cos5\theta\]\[-r=2\cos(-\theta)\]\[-r=-2\cos5\theta\]\[r=2\cos5\theta\]

  8. anonymous
    • one year ago
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    So I think it is symmetric about the y-axis.

  9. Michele_Laino
    • one year ago
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    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
    • one year ago
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    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
    • one year ago
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    please, note my graph:

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  12. IrishBoy123
    • one year ago
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    \(cos (- \theta) = cos \theta\)

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

  14. Michele_Laino
    • one year ago
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    yes! nevertheless we have cos(5 \theta) @IrishBoy123

  15. IrishBoy123
    • one year ago
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    same difference, as they say

  16. anonymous
    • one year ago
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    @IrishBoy123 Thanks for the correction! I forgot about that. :) But sin(-theta)=-sin(theta), right?

  17. anonymous
    • one year ago
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    Anyway, thanks for the help! I really appreciate it.

  18. Michele_Laino
    • one year ago
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    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
    • one year ago
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    yes https://gyazo.com/5f1871923b6020ebe449d395b2e01585 same as @Michele_Laino

  20. Michele_Laino
    • one year ago
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    \[\large \cos \left( {5\theta } \right) = {\left( {\cos \theta } \right)^5} - 10{\left( {\cos \theta } \right)^3}{\left( {\sin \theta } \right)^2} + 5\left( {\cos \theta } \right){\left( {\sin \theta } \right)^4}\]

  21. IrishBoy123
    • one year ago
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    for clarity, here is what is being used as an alternative to CArtesian: https://gyazo.com/9d5ca77c20b40c9d74452f6922539076

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