1. ## Simulatenous intersections.

Hey, everyone! We're on our Polar Unit in Pre-Calc and I'm completely lost when it comes to finding simultaneous points of intersections. I understand the method, but I just can't seem to get past step 1 much to my chagrin.

Take for example,

r = 1 - sinθ
r = cos2θ

Now, I set the two equal to each other.

1 - sinθ = cos2θ

And that's about as far as I get despite trying to substitute cos2θ with 1-2sin^2θ.

Any help would be greatly appreciated!

2. Originally Posted by Rica

1 - sinθ = cos2θ

And that's about as far as I get despite trying to substitute cos2θ with 1-2sin^2θ.
That's a good idea

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

now make $\displaystyle a = \sin \theta$ giving

$\displaystyle 1-a = 1-2a^2$

Can you solve this?

3. Originally Posted by Rica
Hey, everyone! We're on our Polar Unit in Pre-Calc and I'm completely lost when it comes to finding simultaneous points of intersections. I understand the method, but I just can't seem to get past step 1 much to my chagrin.

Take for example,

r = 1 - sinθ
r = cos2θ

Now, I set the two equal to each other.

1 - sinθ = cos2θ

And that's about as far as I get despite trying to substitute cos2θ with 1-2sin^2θ.

Any help would be greatly appreciated!
Hi Rica,

$\displaystyle \frac{1}{2}(1-cos2\theta)=sin^2\theta$

$\displaystyle 1-cos2\theta=2sin^2\theta$

$\displaystyle cos2\theta=1-2sin^2\theta$

hence we have

$\displaystyle 1-sin\theta-cos2\theta=0$

$\displaystyle 1-sin\theta-(1-2sin^2\theta)=0$

$\displaystyle 2sin^2\theta-sin\theta=0$

$\displaystyle sin\theta(2sin\theta-1)=0$

$\displaystyle sin\theta=0,\ sin\theta=\frac{1}{2}$

$\displaystyle sin\theta=0\ \Rightarrow\ \theta=n{\pi}$

$\displaystyle sin\theta=\frac{1}{2}\ \Rightarrow\ \theta=\frac{{\pi}}{6}+2n{\pi},\ \frac{5{\pi}}{6}+2n{\pi}$

4. Oh! So then we get:

(1, 0)
(1, π)
(.5, π/6)
(.5, 5π/6)

Thank you both!