Question

What is the Cartesian form of `( 6 , ( - 16pi)/3 )`?

Answer 1

`(x,y)=(-3,3sqrt(3))`

Explanation:

`(r, theta)=(6,-(16pi)/3)`

`x=r cos theta=6cos(-(16pi)/3)=6cdot(-1/2)=-3`

`y=r sin theta=6sin(-(16pi)/3)=6 cdot sqrt(3)/2=3sqrt(3)`

Hence, `(x,y)=(-3,3sqrt(3))`

I hope that this was clear.