Polar Coordinate Symmetry with Double Angles

[PuddySporty]

Homework Statement

Given the equation r²=25sin2Θ Asked to find symmetry with respect to line Θ = pi/2

Homework Equations

w.r.t. Θ = pi/2: (r,Θ) - (r, pi-Θ) and (r, Θ) - (-r,-Θ)

The Attempt at a Solution

For the first case, I plugged in (pi-Θ) for Θ, but I'm confused about what to do with the double angle. Does It become: r² = 25 sin (2pi - 2Θ) and therefore does not match the original equation?

 

[Mentallic]

Simplify \sin(2\pi-2\theta) so it's just in terms of \sin(2\theta).

 

[PuddySporty]

So, since it's (2pi - 2Θ) it becomes r² = 25 sin (-2Θ) which is r² = -25 sin 2Θ because is negative is quadrants III and IV where 2Θ would have to be?

 

[Mentallic]

Think about it like this, the graph is created by starting at \theta=0 and the necessary distance from the origin, in this case 0. Then you trace out the pattern of the graph as you increase the value of \theta.
Now take a look at these two cases:

r^2=25\sin(-\theta) and r^2=-25\sin(\theta)

The first one tells us that whatever pattern you were tracing when increasing the values of \theta (thus, turning counter-clockwise starting at \theta=0) now you will be tracing that same pattern, except you'll be turning clockwise (because of the -\theta). What symmetry would this create?

Now look at the second one, the negative value means that whatever pattern you're drawing as \theta increases is now flipped onto the other side of the origin.

 

[PuddySporty]

Okay, I got it! Thanks for your help!