# Polar equation symmetry test?

Hi! How do we determine which replacement to be used for each test symmetry? Sometimes the first substitution works and the second doesn't and vice versa. (e.g. r = 1+2 cos theta, where (r,-theta) works but (-r, pi-theta) doesn't, in terms of obtaining the original equation)

### 2 Answers

- Steve4PhysicsLv 75 months agoFavourite answer
Using the (r, -θ)= replacement gives:

r = 1+2cos(-θ)

. = 1+2cos(θ) (because cosθ = cos(-θ))

This is the original equation. So it is symmetric about x-axis.

Using the (-r, π - θ) replacement gives:

-r = 1+2cos(π - θ)

. .= 1- 2cos(θ) (because cosθ = -cos(π – θ))

The source of confusion is the -r. The point (-r, θ), it is the same as the point (r, θ + π). That’s because the ‘ray’ with a negative value for r has length |r| but points in the *opposite direction* to the ray with the positive value for r. This effectively adds π to θ (or subtracts π from θ).

-r = 1- 2cos(θ) then becomes

r = 1 - 2cos(θ + π)

. = 1 + 2cosθ (because cos(θ + π) = -cosθ)

This is the original equation. So it is symmetric about x-axis.

The (r, - θ) replacement is a lot simpler and I can’t see a reason for using the (-r, π – θ) replacement.

- MelvynLv 75 months ago
if one works but the other dosnt it would suggest that you made an error....or that the simplification isnt complete