## anonymous 5 years ago how do i test for symmetry? r=16cos3(theta)

1. anonymous

symmetry around what?

2. anonymous
3. anonymous

symmetry with respect to theta=pie/2, the polar axis, and the pole

4. anonymous

thanks but how do i actually work it to find the answer?

5. anonymous

This question will, unfortunately, take more time to answer than I have available. I can give you the following transformations and an example of what to do for one of them, and hopefully you can see what you have to do for the others. A polar function is symmetric about the polar axis if, when you replace (r, theta) with EITHER (r,-theta) or (-r, pi-theta), you get the original function back. Here, if (r, theta) is on your graph, then$(r, \theta) \rightarrow r=16 \cos 3 \theta$$\rightarrow r=16 \cos 3 (- \theta))$$\rightarrow (r, -\theta)$is on your graph (because cosine is even (i.e. cos(theta) = cos(-theta)).

6. anonymous

The function's symmetric about the pole if when you replace$(r, \theta)$ with (-r, theta) OR (r, pi+ theta), you get the original function back. So here,$(r, \theta) \rightarrow r=16\cos 3 \theta$$\rightarrow r=16 \cos 3 (\pi+\theta)$$\rightarrow r=-16\cos 3 \theta\ne 16\cos3\theta$So neither (-r theta) or (r, pi + theta) lie on the graph (note, when I'm giving you two points to choose from in your test, these points on the graph are identical (you just take a different route getting there (draw it if you can't see it))).

7. anonymous

If the function's symmetric about pi/2, then replacing (r, theta) with either (r,pi-theta) or (-r, -theta) should get you back to (r, theta). Here,$(r, \pi-\theta) \rightarrow 16 \cos 3(\pi-\theta))=-16\cos 3 \theta=-r \ne r$i.e. if you replace with pi-theta, you don't get back to (r, theta), you get to (-r, theta). So this isn't symmetric about pi/2. There is a mistake in the reasoning above. I should have said,$(r,\pi+\theta) \rightarrow 16\cos 3 (\pi + \theta)= -16\cos 3 \theta = -r \ne r$

8. anonymous

So it's only symmetric about the polar axis.