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 [tex]\sin(2\pi-2\theta)[/tex] so it's just in terms of [tex]\sin(2\theta)[/tex].

 

[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 [itex]\theta=0[/itex] 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 [itex]\theta[/itex].
Now take a look at these two cases:

[tex]r^2=25\sin(-\theta)[/tex] and [tex]r^2=-25\sin(\theta)[/tex]

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

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

 

[PuddySporty]

Okay, I got it! Thanks for your help!