Mathematics InsatiableSuffering:

So I have some polar symmetry questions. InsatiableSuffering:

$3. r=4\cos(2\theta)+1$ $Pole symmetry: -r=4\cos(2\theta)+1$ $\pi/2 symmetry: -r=4\cos(-2\theta)+1 \to -r=4\cos(2\theta)+1$ How is this polar equation symmetrical over pi/2 and the pole InsatiableSuffering:

$4.r ^{2}=9\sin(2\theta)$ $r=\sqrt{9\sin2\theta}$ What is the correct way to prove symmetry for this equation? I found that there's no symmetry over pi/2 and the polar axis, but I found there to be symmetry over the pole. That happens to be wrong as well, so can somebody show what steps should be taken? InsatiableSuffering:

$6.r=1+2\sin(\theta/2)$ $Polar symmetry:r=1+2\sin(-\theta/2)$ Can somebody prove how this equation is symmetrical over the polar axis? imqwerty:

3. has polar axis symmetry it doesn't have symmetry about the line $$\theta = \pi/2$$ or about the pole InsatiableSuffering:

It doesn't? I plugged in those exact equations on desmos, but they happen to overlap each other perfectly. imqwerty:

true I think I'm missing something I just replaced r with -r and theta with -theta to check for symmetry about pi/2 InsatiableSuffering:

I'm not sure what's going on either, but when I divided -1 on both sides of the equations I plugged into desmos, I ended up getting a different graph imqwerty:

oh, I forgot to check for the case when you replace $$(r, \theta)$$ with $$(r, \pi-\theta)$$ to check for symmetry about the line $$\pi/2$$ imqwerty:

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