Question

How do you convert `(-1/2, -sqrt3/2)` into polar form?

Answer 1

`(1,(4pi)/3)`

Explanation:

To convert from `color(blue)"cartesian to polar form"`

That is `(x,y)to(r,theta)`

`color(orange)"Reminder " color(red)(bar(ul(|color(white)(2/2)color(black)(r=sqrt(x^2+y^2))color(white)(2/2)|)))`

and `color(red)(bar(ul(|color(white)(2/2)color(black)(theta=tan^-1(y/x))color(white)(2/2)|)))`

Here `x=-1/2" and " y=-sqrt3/2`

`rArrr=sqrt((-1/2)^2+(-sqrt3/2)^2)=sqrt(1/4+3/4)=1`

now `(-1/2,-sqrt3/2)` is in the 3rd quadrant so we must ensure that `theta` is in the 3rd quadrant.

`rArrtheta=tan^-1((-sqrt3/2)/(-1/2))`

`=tan^-1(sqrt3)=pi/3larr" reference angle"`

`rArrtheta=(pi+pi/3)=(4pi)/3larr" 3rd quadrant"`

`rArr(-1/2,-sqrt3/2)to(1,(4pi)/3)`

Answer 2

Please see the explanation.

Explanation:

The polar coordinate system is an ordered pair, `(r, theta)`.

To compute r from from Cartesian coordinates, `(x,y)`, use the equation:

`r = sqrt(x^2 + y^2)`

`r = sqrt((-1/2)^2 + (-sqrt(3)/2)^2)`

`r = sqrt((1/4 + 3/4)`

`r = 1`

To compute `theta` from Cartesian coordinates, `(x,y)`, use the appropriate one of the following equations:

1. If `x > 0 and y>=0`m then use: `theta = tan^-1(y/x)`
2. If `x = 0 and y > 0`, then use: `theta = pi/2`
3. If `x = 0 and y < 0`, then use: `theta = (3pi)/2`
4. If `x < 0`, then use: `theta = pi + tan^-1(y/x)`
5. If `x > 0 and y < 0`, then use: `theta = 2pi + tan^-1(y/x)`

The appropriate one is equation 4:

`theta = pi + tan^-1((-sqrt(3)/2)/(-1/2))`

`theta = pi + tan^-1(sqrt(3))`

`theta = pi + pi/3`

`theta = (4pi)/3`

The polar point is `(1, (4pi)/3)`