Question

How do you graph the polar equation `1=rcos(theta-pi/6)`?

Answer 1

To graph this easily, we can convert it to rectangular form.

In order to convert, we need to the cosine angle difference formula:

`cos(color(red)A-color(blue)B)=coscolor(red)Acoscolor(blue)B+sincolor(red)Asincolor(blue)B`

Knowing that `rsintheta=y` and `rcostheta=x`, we can convert:

`1=rcos(theta-pi/6)`

`1=r(costhetacoscolor(black)(pi/6)+sinthetasincolor(black)(pi/6))`

`1=r(costheta*sqrt3/2+sintheta*1/2)`

`1=rcostheta*sqrt3/2+rsintheta*1/2`

`1=x*sqrt3/2+y*1/2`

`1-x*sqrt3/2=y*1/2`

`2-x*sqrt3=y`

`y=2-x*sqrt3`

`y=-sqrt3 x+2`

Now we can graph this linear equation like any other line.

An easy strategy would be to solve for the `x`- and `y`-intercepts, then connect the dots.

The `x`-intecept occurs when `y=0`, so:

`color(white)=>y=-sqrt3 x+2`

`=>0=-sqrt3 x+2`

`color(white)=>sqrt3 x=2`

`color(white)=>x=2/sqrt3`

`color(white)=>x=(2sqrt3)/3`

This means that the `x`-intercept is at `((2sqrt3)/3,0)`. Call this point `A`. The `y`-intercept occurs when `x=0`, so:

`color(white)=>y=-sqrt3 x+2`

`=>y=-sqrt3 *0+2`

`color(white)=>y=2`

This means that the `y`-intercept occurs at `(0,2)`. Call this point `B`. Now that we have our two points, we can graph the line:

![https://www.desmos.com/calculator]()

That's it. Hope this helped!