# Polar Symmetry

## Homework Statement

Identify the symmetries of the curve.

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?

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Mark44
Mentor

## Homework Statement

Identify the symmetries of the curve.

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).

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

Mark44
Mentor
I defined f(θ) = 8 + 7 sin(θ), so f(##\pi## - θ) = 8 + 7sin(##\pi## - θ) = ?

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

Mark44
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.

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

Mark44
Mentor
And using the difference formula, you can show that it's true for all real values of θ.

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

Mark44
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?

yes. thanks.