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    Hi guys,

    Can someone explain why r=(1/1+2cosϴ) for -2/3Pi < ϴ , 2/3 Pidoes not have a pole? I was able to differentiate it so I'm quite confused. I got ϴ = 0

    Thanks
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    (Original post by ChemBoy1)
    Hi guys,

    Can someone explain why r=(1/1+2cosϴ) for -2/3Pi < ϴ , 2/3 Pidoes not have a pole? I was able to differentiate it so I'm quite confused. I got ϴ = 0

    Thanks
    The pole is where r=0 and clearly the function \dfrac{1}{1+2\cos \theta} is never 0
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    Thank you very much, I seem to having difficulties with this part of the topic. It said in the book that if F(Thetha) = 0 then the theta is tangent to the graph
    (Original post by RDKGames)
    The pole is where r=0 and clearly the function \dfrac{1}{1+2\cos \theta} is never 0
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    Also if there was a situation with more than 1 trig function e..g. 2+ cos theta - cos^2 theta. How are you meant to find out the tangents to the pole?
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    (Original post by ChemBoy1)
    Also if there was a situation with more than 1 trig function e..g. 2+ cos theta - cos^2 theta. How are you meant to find out the tangents to the pole?
    Try to use the correct terminology. The pole is a point, so it wouldn't make sense to have something 'tangent' to it.

    Here you are interested in straight lines going through the pole, which are tangent to your polar eq. r=f(\theta). These lines have equation \theta = \alpha for -\pi &lt; \alpha \leq \pi (or between 0 and 2 pi, if you want to use that).
    The first step is to obvious set f(\theta) = 0 and solve it for the \theta values. But make sure that f'(\theta) \neq 0.
    For the example above, we would simply need to solve 2+\cos \theta - \cos^2 \theta = 0 but note that this is just a quadratic in \cos \theta which also factorises.
 
 
 
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