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When graphing polar plots, one way to make our job easier is to determine if the graphic is symmetric somehow. Regarding polar plots, we have

Symmetry wrt (with respect to) the polar axis: When the expression remains unchanged if $r(-\theta) = r(\theta)$

Symmetry wrt the pole: When the expression remains unchanged if $-r= r$

Symmetry wrtto the line $\theta = \frac{\pi}{2}$: When the expression remains unchanged if $r(\pi-\theta) = r(\theta)$

That worked fine until I encountered the following example:

$$r=\sin(2\theta)$$

When plotting the graph, we can clearly see that its symmetry "works" for all three conditions listed above.

enter image description here

But, knowing that

$$r = \sin(2\theta) = 2\sin(\theta)\cos(\theta)$$

the three conditions do not look verified, since

$$r(-\theta) = 2 \sin(-\theta) \cos(-\theta) = - 2 \sin(\theta)\cos(\theta)\\ -r = - 2 \sin(\theta) \cos(\theta)\\ r(\pi-\theta) = 2 \sin(\pi-\theta) \cos(\pi-\theta) = - 2 \sin(\theta)\cos(\theta)$$

So my question is: is this graph really symmetrical given the three "conditions"? If so, how can it be explained that algebra isn't in agreement with that fact?

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    $r(\pi-\theta)=-r(\theta)$ is clearly another translation of this symmetry. – 2017-02-11
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    I'm not sure what you mean – 2017-02-11
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    You established $r(\pi-\theta)$=-r(\theta)=r(-\theta)$ w.r.t. the vertical axis (actually, there are 4 axes of symmetry and a central symmetry). – 2017-02-11

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