Question

What is the Cartesian form of `( -2 , (13pi)/6 )`?

Answer 1

I got:

`(sqrt3,1)`

assuming `r > 0`. If `r < 0`, then we would necessarily have `r = -sqrt(x^2 + y^2)` to keep the equations below self-consistent, and then we would have `x = -sqrt3` and `y = -1`.

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Recall:

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

The coordinates you are given are in polar, and are the `r` and `theta` coordinates. Thus, you need to use the expression for `r` to calculate `x` or `y`, and the expression for `theta` to get the other coordinate.

However... `r` is only positive or `0`. You cannot have a negative radius, as it is physically impossible. So, there's a typo in the problem. It should be `(2,(13pi)/6)`.

`2 = sqrt(x^2 + y^2) => 4 = x^2 + y^2`

`y = rsin(theta) = 2sin((13pi)/6)`

`=> y^2 = 4sin^2((13pi)/6) = 4sin^2(390^@) = 4sin^2(30^@) = 1`

`=> color(green)(y = 1)`

(to keep consistent with the equations above, `y > 0`, since `costheta > 0` and `r > 0`.)

Therefore, we can solve for `x`:

`=> x^2 = 4 - y^2 = 4 - 1 = 3`

`=> color(green)(x = sqrt3)`

(to keep consistent with the equations above, `xy > 0`, since `costheta > 0` and `r > 0`.)

So, the cartesian coordinates are:

`color(blue)((x,y) = (sqrt3,1))`

To make sure it worked...

`r = sqrt(3 + 1) = 2` `color(blue)(sqrt"")`

`theta = arccos(x/r) = arcsin(y/r)`

`= arccos(sqrt3/2)`

As `costheta = sqrt3/2` when `theta = 30^@`, `arccos(sqrt3/2) = theta = 30^@`, which is coterminal with `390^@`. `color(blue)(sqrt"")`