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anonymous

  • one year ago

GIVING MEDAL N FAN determine if the graph is symmetric about the x-axis, the y-axis or the orgiin r= 5 cos 5theta y-axis only x-axis, y-axis x-axis, y-axis, origin x-axis only

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  1. anonymous
    • one year ago
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    @Michele_Laino

  2. welshfella
    • one year ago
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    do you know what the graph of cos x looks like?

  3. anonymous
    • one year ago
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    NO

  4. welshfella
    • one year ago
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    if f(x) = f(-x) then its symetrical about the y-axis

  5. Michele_Laino
    • one year ago
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    using the De Moivre formula (complex numbers), we get this identity: \[\large \cos \left( {5\theta } \right) = {\left( {\cos \theta } \right)^5} - 10{\left( {\cos \theta } \right)^3}{\left( {\sin \theta } \right)^2} + 5\left( {\cos \theta } \right){\left( {\sin \theta } \right)^4}\]

  6. anonymous
    • one year ago
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    y axis only @Michele_Laino ?

  7. Michele_Laino
    • one year ago
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    so, we can rewrite your equation as follows: \[\large r = \left\{ {5{{\left( {\cos \theta } \right)}^5} - 10{{\left( {\cos \theta } \right)}^3}{{\left( {\sin \theta } \right)}^2} + 5\left( {\cos \theta } \right){{\left( {\sin \theta } \right)}^4}} \right\}\]

  8. Michele_Laino
    • one year ago
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    and going to the cartesian coordinates, we can write: \[\Large r = \left\{ {5\frac{{{x^5}}}{{{r^5}}} - 10\frac{{{x^3}{y^2}}}{{{r^5}}} + 5\frac{{x{y^4}}}{{{r^5}}}} \right\}\] so what can you conclude?

  9. anonymous
    • one year ago
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    is it all 3? x-axis, y-axis, origin

  10. anonymous
    • one year ago
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  11. Michele_Laino
    • one year ago
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    oops... I have made a typo, here is the right equation: \[\Large r = 5\left\{ {\frac{{{x^5}}}{{{r^5}}} - 10\frac{{{x^3}{y^2}}}{{{r^5}}} + 5\frac{{x{y^4}}}{{{r^5}}}} \right\}\] we have a symmetry for a variable whose exponent is an even number

  12. anonymous
    • one year ago
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    omg im lost is it x-axis or y-axis, could u just tell me ? @Michele_Laino

  13. welshfella
    • one year ago
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    check if f(-60) = f(60) if so then its symmetrical over the y-axis

  14. Michele_Laino
    • one year ago
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    I think x-axis, since we have this: \[\Large f\left( {x, - y} \right) = f\left( {x,y} \right)\]

  15. anonymous
    • one year ago
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    so x-axis everyone @Michele_Laino @welshfella @Astrophysics ??

  16. welshfella
    • one year ago
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    no i think its y-axis

  17. anonymous
    • one year ago
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    both??

  18. anonymous
    • one year ago
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    or just y axis?

  19. welshfella
    • one year ago
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    just y-axis

  20. anonymous
    • one year ago
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    ok

  21. welshfella
    • one year ago
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    because its an even function and graph of cos x is |dw:1438441043404:dw|

  22. Michele_Laino
    • one year ago
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    if it is y-axis, then we shoul have this: \[\Large f\left( { - x,y} \right) = f\left( {x,y} \right)\]

  23. Michele_Laino
    • one year ago
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    should*

  24. welshfella
    • one year ago
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    yes

  25. Michele_Laino
    • one year ago
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    nevertheless x compares with odd exponent

  26. welshfella
    • one year ago
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    and that is the case , i think

  27. Michele_Laino
    • one year ago
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    \[f\left( { - x,y} \right) = 5\left\{ {\frac{{ - {x^5}}}{{{r^5}}} - 10\frac{{ - {x^3}{y^2}}}{{{r^5}}} + 5\frac{{ - x{y^4}}}{{{r^5}}}} \right\} = - f\left( {x,y} \right)\] am I right?

  28. welshfella
    • one year ago
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    would like to discuss firther but got to go

  29. Michele_Laino
    • one year ago
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    I think taht we have both symmetries with respect to x-axis and the origin

  30. Michele_Laino
    • one year ago
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    that*

  31. anonymous
    • one year ago
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    Guys can we solve this by using this|dw:1438442042055:dw| now for eq r=5cos5theta .the equation will be symmetric about Y axis if r is unchanged for theta =180-theta.. and symmetric about x axis if r is unchanged for theta =360-theta.Since the first one is not true from from basic trigonometric relation of cos(180-theta)=-costheta.Hence above equation is symmetric only about X axis for which equality still holds...Just a thought..Dont know what the direction change (negative or positive) actually means though..Oen for thoughts..

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