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    Another useless topic from me, but in light of next year I was advised to look into some topics, and polar coordinates just have me stumped. This is the question I'm not getting.

    Sketch the lines (or half-lines) with polar equations:

    e) rcos(theta - pi/6) = 2

    This is what I did....

    Rcos(theta - pi/6) = 2
    cos(theta - pi/6) = cos(theta)cos(-pi/6) + sin(theta)sin(-pi.6)

    = Rcos(theta)*(root3/2) - 0.5Rsin(theta)

    Now: Rcos(theta) = x, Rsin(theta) = y

    (root3/2)x - 0.5y = 2 [*2]
    root3*x - y = 4
    y = root3*x - 4

    I get a line, which I think is the answer.

    However, take a point (x,y) such as (4, 2.92)

    theta = arctan(2.92/4) = 0.631
    R = (4^2 + 2.92^2)^0.5 = 4.952

    So if we put it back into the original equation:

    4.952 * Cos ( 0.631 - pi/6) = 2

    However, the left hand side = 4.92, which is totally wrong.

    If you take theta as 1.67, this works. Can anybody explain why?

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    Maths in the middle of the summer holidays is music to my ears
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    (Original post by byb3)
    Another useless topic from me, but in light of next year I was advised to look into some topics, and polar coordinates just have me stumped. This is the question I'm not getting.

    Sketch the lines (or half-lines) with polar equations:

    e) rcos(theta - pi/6) = 2

    This is what I did....

    Rcos(theta - pi/6) = 2
    cos(theta - pi/6) = cos(theta)cos(-pi/6) + sin(theta)sin(-pi.6)

    = Rcos(theta)*(root3/2) - 0.5Rsin(theta)
    Actually,

    cos(theta - pi/6)
    = cos(theta)cos(-pi/6) - sin(theta)sin(-pi/6) (minus, not plus!)
    = cos(theta)cos(pi/6) + sin(theta)sin(pi/6)
    = (sqrt(3)/2) cos(theta) + (1/2) sin(theta)

    Substituting into r cos(theta - pi/6) = 2 we get

    (sqrt(3)/2) x + (1/2) y = 2
    ie, y = 4 - sqrt(3) x

    Jonny W
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    cheers mate, i see where I went wrong..

    Cos(A-B) = CosACosB + SinASinB

    But because I used (-B) on the right, it messed things up!

    Thanks for clearing that one up.
 
 
 
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