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Polar Symmetry

  1. Apr 30, 2013 #1
    1. The problem statement, all variables and given/known data
    Identify the symmetries of the curve.


    2. Relevant equations
    r = 8 + 7sinθ


    3. The attempt at a solution
    x-axis: (r,-θ) ---> 8 + 7sin(-θ) = 8-sinθ (not equal to ) 8+7sinθ; not symmetric about x-axis.
    y-axis: (-r,-θ) ---> -8+7sinθ (not equal to) 8+7sinθ; not symmetric about y-axis.
    origin: (-r,θ) ---> r = -8-7sinθ (not equal to) 8+7sinθ; not symmetric about origin.

    looking at the graph i know it is symmetric about y-axis, though, so what am i doing wrong?
     
  2. jcsd
  3. Apr 30, 2013 #2

    Mark44

    Staff: Mentor

    Writing r = 8 + 7sinθ as r = f(θ), the graph will be symmetric across the y-axis if f(##\pi## - θ) = f(θ).
     
  4. Apr 30, 2013 #3
    can you please explain how to evaluate if f(θ) = f(∏-θ) ? i always get confused with this one.
     
  5. Apr 30, 2013 #4

    Mark44

    Staff: Mentor

    I defined f(θ) = 8 + 7 sin(θ), so f(##\pi## - θ) = 8 + 7sin(##\pi## - θ) = ?
     
  6. Apr 30, 2013 #5
    it must equal 8 + 7sin(θ) but i'm not sure how. are you using a trig identity?
     
  7. Apr 30, 2013 #6

    Mark44

    Staff: Mentor

    You can use an identity, or you can use the unit circle. The identity would be sin(A - B) = sin(A)cos(B) - cos(A)sin(B). Hopefully, you have this one memorized.
     
  8. Apr 30, 2013 #7
    i realize sin(θ) = sin(∏- θ) for values on in the interval 0<θ<pi by looking at the unit circle.
     
  9. Apr 30, 2013 #8

    Mark44

    Staff: Mentor

    And using the difference formula, you can show that it's true for all real values of θ.
     
  10. Apr 30, 2013 #9
    so the proof using sin(A-B) would be:

    sin(∏-θ) = sin(∏) cos(θ) - cos(∏)sin(θ) = sin(θ)

    0 - (-1)(sinθ) = sin(θ) --> sinθ=sinθ
     
  11. Apr 30, 2013 #10

    Mark44

    Staff: Mentor

    I wouldn't do it that way. Here's what I would do:
    sin(##\pi## - θ) = sin(##\pi##) cos(θ) - cos(##\pi##)sin(θ) = 0 * cos(θ) - (-1)sin(θ) = sin(θ)
    This shows that sin(##\pi## - θ) = sin(θ).

    Do you see how what I did is different from what you did?
     
  12. Apr 30, 2013 #11
    yes. thanks.
     
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