Thread: 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?

Originally Posted by homeylova223

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).

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