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Math Help - Simulatenous intersections.

  1. #1
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    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!
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  2. #2
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    Quote Originally Posted by Rica View Post

    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

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

    now make a = \sin \theta giving

    1-a = 1-2a^2

    Can you solve this?
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  3. #3
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    Quote Originally Posted by Rica View Post
    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,

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

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

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

    hence we have

    1-sin\theta-cos2\theta=0

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

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

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

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

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

    sin\theta=\frac{1}{2}\ \Rightarrow\ \theta=\frac{{\pi}}{6}+2n{\pi},\ \frac{5{\pi}}{6}+2n{\pi}
    Last edited by Archie Meade; April 19th 2010 at 03:43 PM.
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  4. #4
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    Oh! So then we get:

    (1, 0)
    (1, π)
    (.5, π/6)
    (.5, 5π/6)

    Thank you both!
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