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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- April 19th 2011, 04:37 PMhomeylova223Graphing 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? - April 19th 2011, 05:03 PMmr fantastic
- April 19th 2011, 05:19 PMSoroban
Hello, homeylova223!

Quote:

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