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Polar Symmetry

  • #1

Homework Statement


Identify the symmetries of the curve.


Homework Equations


r = 8 + 7sinθ


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?
 

Answers and Replies

  • #2
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5,294

Homework Statement


Identify the symmetries of the curve.


Homework Equations


r = 8 + 7sinθ


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?
Writing r = 8 + 7sinθ as r = f(θ), the graph will be symmetric across the y-axis if f(##\pi## - θ) = f(θ).
 
  • #3
can you please explain how to evaluate if f(θ) = f(∏-θ) ? i always get confused with this one.
 
  • #4
33,634
5,294
I defined f(θ) = 8 + 7 sin(θ), so f(##\pi## - θ) = 8 + 7sin(##\pi## - θ) = ?
 
  • #5
it must equal 8 + 7sin(θ) but i'm not sure how. are you using a trig identity?
 
  • #6
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5,294
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.
 
  • #7
i realize sin(θ) = sin(∏- θ) for values on in the interval 0<θ<pi by looking at the unit circle.
 
  • #8
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5,294
And using the difference formula, you can show that it's true for all real values of θ.
 
  • #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
 
  • #10
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5,294
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?
 
  • #11
yes. thanks.
 

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