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Math Help - Polar Equation: Draw and Fill in chart

  1. #1
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    Exclamation Polar Equation: Draw and Fill in chart

    Draw and fill in the chart that has a row for "Theta" and a row for "r" for this polar equation: r=1-3cos(theta)

    I have to fill in the 24 spots (12 for theta and 12 for r) on the chart using this polar equation: r=1-3cos(theta).

    The picture given is a unit circle counting by pi/12. Please help???
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  2. #2
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    Re: Polar Equation: Draw and Fill in chart

    Hve you been introduced to ...polar plot r=1-3cos(t) - Wolfram|Alpha ?
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    Re: Polar Equation: Draw and Fill in chart

    Thank you very much. How would I go about solving for R when cos(theta)= pi/12?
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    Re: Polar Equation: Draw and Fill in chart

    Quote Originally Posted by Binary View Post
    Thank you very much. How would I go about solving for R when cos(theta)= pi/12?
    I think you mean ... "how can I evaluate r when \theta = \frac{\pi}{12}" ?

    r = 1 - 3\cos\left(\frac{\pi}{12}\right)

    ... get out your calculator

    r \approx 0.966


    by hand ...

     \cos\left(\frac{\pi}{12}\right) = \cos\left(\frac{\pi}{3} - \frac{\pi}{4}\right)

    ... use the difference identity for cosine.
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    Re: Polar Equation: Draw and Fill in chart

    Quote Originally Posted by skeeter View Post
    I think you mean ... "how can I evaluate r when \theta = \frac{\pi}{12}" ?

    r = 1 - 3\cos\left(\frac{\pi}{12}\right)

    ... get out your calculator

    r \approx 0.966


    by hand ...

     \cos\left(\frac{\pi}{12}\right) = \cos\left(\frac{\pi}{3} - \frac{\pi}{4}\right)

    ... use the difference identity for cosine.
    Yeah I figured it out finally then just saw your post. Thank you very much. I was trying to do it by hand btw.
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    Re: Polar Equation: Draw and Fill in chart

    When I use my calculator I dont get r=0.966. I get approx r=-1.897
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  7. #7
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    Re: Polar Equation: Draw and Fill in chart

    Quote Originally Posted by Binary View Post
    When I use my calculator I dont get r=0.966. I get approx r=-1.897
    you're correct ... I only punched in \cos\left(\frac{\pi}{12}\right) , my mistake.
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