Polar Coordinates Problem

**Kasper** · January 17th 2010, 02:08 PM

Hey, I have a polar coordinate problem here that feels like it should be much easier than I am making it out to be.

Find all points of intersection of the following curves, given in Polar Coordinates. $r^2 = cos(2\theta) \ and \ r = 1 - sin(\theta)$

So I started by setting them equal to each other to find the angles $\theta$ at which they intersect.

$cos(2\theta) = (1-sin\theta)^2$

$1-2sin^2\theta = 1-2sin\theta + sin^2\theta$

$sin\theta = \frac{2}{3}$

$\theta \approx 0.729 rad$

The problem is that by graphing it appears that there is 5 points of intersection, and I only have 1 floating point approximation of theta. What did I miss? Thanks for the help!

**TKHunny** · January 17th 2010, 05:44 PM

Your algebra missed a beat. What did you do with $\sin(\theta) = 0$ ?

By symmetry, you can get another one from the one you have. Think about Reference Angles.

By more symmetry, you can get two more from the one you missed. That makes five (5). Your graphing is exemplary.

Note: Check your Domain. You may not get all five (5).

**Kasper** · January 17th 2010, 07:09 PM

Originally Posted by TKHunny

Your algebra missed a beat. What did you do with $\sin(\theta) = 0$ ?

By symmetry, you can get another one from the one you have. Think about Reference Angles.

By more symmetry, you can get two more from the one you missed. That makes five (5). Your graphing is exemplary.

Note: Check your Domain. You may not get all five (5).

Wooow, what a miss that was... Gotta watch for that in the future.

So with $sin(\theta)=0 \ \ \rightarrow \ \ \theta = 0, \pi, 2\pi$ , we are at 4. Now by graphing $y=\frac{2}{3} \ and \ y = sin(\theta)$ I see that the other intersection point lies at $\theta \approx 2.41 rad$ . But I have no idea how to use symmetry to get that, as it doesn't seem to be an addition of any proper fraction of pi. Any hints on how to grab that last intersection algebraically?

Thank you for your help!

**TKHunny** · January 18th 2010, 02:58 PM

$\sin(\theta) = \sin(\pi - \theta)$

Reference Angles!!

**Kasper** · January 19th 2010, 08:25 PM

Note to self - Review high school trig.

Thanks very much

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