Sketching Polar Graphs

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    Sketch the curve of r=1+2costheta
    intercepts at 1 because cos90=0 and -1 since cos is an 'even function', meaning its symmetrical with the horizontal polar axis as the mirror line.
    cos theta takes values between -1 and 1 where 1 (at theta=0) gives 3 to be biggest value for r
    r=0 is when costheta is 2*pi/3. plot the rest symmetrically and it gives a graph like so
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    the inside loop is when r<0, would you need to plot this on an exam question?
    since cos is an even function, what is sin? how would you solve this with cartesian coordinates, or is it best untouched? am i missing anything out in my working/any way to work it out faster (other than using a graphical calculator)
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    (Original post by Daiblain)
    Sketch the curve of r=1+2costheta
    intercepts at 1 because cos90=0 and -1 since cos is an 'even function', meaning its symmetrical with the horizontal polar axis as the mirror line.
    cos theta takes values between -1 and 1 where 1 (at theta=0) gives 3 to be biggest value for r
    r=0 is when costheta is 2*pi/3. plot the rest symmetrically and it gives a graph like so

    the inside loop is when r<0, would you need to plot this on an exam question?
    since cos is an even function, what is sin? how would you solve this with cartesian coordinates, or is it best untouched? am i missing anything out in my working/any way to work it out faster (other than using a graphical calculator)
    Cartesian are a mess, and yes you need to include the inside loop. Sine is an odd function, it will be symmetrical about the 'y' axis (so to speak, but you know what I mean since there is y-axis). To sketch it you just test the function are different values of \theta and mark them on, you should have an idea of what it looks like once you do enough of these and plot enough points. The symmetry helps.
 
 
 
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