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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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