# Polar Symmetry

1. Apr 30, 2013

### whatlifeforme

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. Apr 30, 2013

### 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).

3. Apr 30, 2013

### whatlifeforme

can you please explain how to evaluate if f(θ) = f(∏-θ) ? i always get confused with this one.

4. Apr 30, 2013

### Staff: Mentor

I defined f(θ) = 8 + 7 sin(θ), so f($\pi$ - θ) = 8 + 7sin($\pi$ - θ) = ?

5. Apr 30, 2013

### whatlifeforme

it must equal 8 + 7sin(θ) but i'm not sure how. are you using a trig identity?

6. Apr 30, 2013

### 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.

7. Apr 30, 2013

### whatlifeforme

i realize sin(θ) = sin(∏- θ) for values on in the interval 0<θ<pi by looking at the unit circle.

8. Apr 30, 2013

### Staff: Mentor

And using the difference formula, you can show that it's true for all real values of θ.

9. Apr 30, 2013

### whatlifeforme

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. Apr 30, 2013

### 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?

11. Apr 30, 2013

### whatlifeforme

yes. thanks.

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