How do you graph the polar equation $2 = r cos (θ + 60^{\circ})$ $2=rcos(theta+60^circ)$ ?

Question

2211 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-07-27T13:52:15

$x - \sqrt{3} y = 4$ $x - sqrt 3 y = 4$ ..

Use $r = \sqrt{x^{2} + y^{2}} \geq 0$ $r = sqrt ( x^2 + y^2 ) >= 0$

$and r (cos θ, sin θ) = (x, y)$ $and r( cos theta, sin theta ) = ( x, y )$

$2 = r (cos θ cos (\frac{π}{3}) - sin θ sin (\frac{π}{3}))$ $2 = r (cos theta cos (pi/3) - sin theta sin (pi/3 ))$

$= \frac{r}{\sqrt{2}} (cos θ - \sqrt{3} sin θ$ $= r/sqrt2 ( cos theta - sqrt3 sin theta$ ) converts to

$x - \sqrt{3} y = 4$ $x - sqrt3 y = 4$ .

Note that the perpendicular form of the polar equation of a straight

line is

$r cos (θ - α) = p$ $r cos (theta - alpha ) = p$ , where

$(p, α)$ $( p, alpha )$ is th foot of the perpendicular to the line, from the

pole $r = 0$ $r = 0$ . See graph of the given equation, with the foot of the

perpendicular, with $p = 2 and α = - \frac{π}{3}$ $p = 2 and alpha = - pi/3$ .

graph{(x-sqrt 3 y-4)((x-1)^2+(y+sqrt3)^2-0.001)(y+sqrt3 x)=0[0 4 -2 0]}

How do you graph the polar equation 2=rcos(θ+60∘)2=rcos(theta+60^circ)?