How might you convert r(cos (theta)-(pi)/6)=2 from polar to rectangular?

Question

rcos(θ-π/6)=2 (with steps, not just the answer)

Answer 1

Hi GL,

(r, θ) -----> (x, y)

In the 2D, the radial distance from the origin to the point (x,y) makes an angle θ - π/6

x² + y² = r²

r = √(x² + y²), and rcos θ = x  and  rsin θ = y

 

r cos( θ - π/6) = 2, Expand the left-hand side 

rcos( θ - π/6) = r cosθ cos( π/6) - r sinθ sin(π/6) ,

Substitute rsin(θ and rcosθ from above and sin(π/6) = 1/2 ,cos(π/6) =√3 /2

r cos( θ - π/6)= x√3 /2 + y(1/2), set this equal to the right hand side

x√3 /2 + y(1/2) = 2

x√3 + y = 4

y = - x√3 + 4