Question

How does one find the rectangular coordinates of a point with polar coordinates (-2, 11pi/6)?

Answer 1

`(-sqrt(3),1)`

Explanation:

In general polar coordinates are `(r,theta)`
for some radius `r` and angle `theta` counter-clockwise from the `x`-axis. This forms a right triangle with hypotenuse `r` connecting the origin and `(r,theta)`.

The Pythagorean trigonometric identity is:
`sin^2(theta)+cos^2(theta)=1`
Multiplying both sides by `r` gives us :
`rsin^2(theta)+rcos^2(theta)=r`

Going back to the triangle, this tells us that if the hypotenuse is `r` then the side going vertically up (`y`-coordinate) is `rsin(theta)`
and the side going horizontal (`x`-coordinate) is `rcos(theta)`.
Therefore `(x,y) = (rcos(theta),rsin(theta))`.
Since the angle is given in a multiple of pi, I hope you mean radians.

We have `(r,theta)` = `(-2,(11pi)/6)` so `(x,y) = (-2cos((11pi)/6),-2sin((11pi)/6))` = `(-sqrt(3),1)`