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Math Help - Graphing a polar equation?

  1. #1
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    Graphing a polar equation?

    1= r cos ( theta- pi/6)

    Now I know how to graph a polar equation when r is on the other side of the equation when you plug in a value for theta.

    But how would I move r to the other side of the equation?
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  2. #2
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    Quote Originally Posted by homeylova223 View Post
    1= r cos ( theta- pi/6)

    Now I know how to graph a polar equation when r is on the other side of the equation when you plug in a value for theta.

    But how would I move r to the other side of the equation?
    r = 1/cos ( theta- pi/6).
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  3. #3
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    Hello, homeylova223!

    r cos (θ - π/6) .= .1

    We have: .r [cosθ cos(π/6) + sinθ sin(π/6)] .= .1
    . . . . . . . . . . . . . . . . . ._
    . . . . . . . . . . . . r cosθ(√3/2) + r sinθ(1/2) .= .1
    . . . . . . . . . . . . . . . . . . _
    . . . . . . . . . . . . . . x * (√3/2) + y * (1/2) . = .1
    . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ._
    . . . . . . . . . . . . . . . . . . . . . . . . . . . . . y .= .-√3 x + 2

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