Question

What are the components of the vector between the origin and the polar coordinate `(7, (-7pi)/12)`?

Answer 1

Rectangular Coordinates:
`x=7/4*(sqrt(6-3sqrt3)-sqrt(2+sqrt3))=-1.81173`
`y=-7/4*(sqrt(6+3sqrt3)+sqrt(2-sqrt3))=-6.76148`

Polar coordinates: `(1.81173, pi)`and`(6.76148, (3pi)/2)`

Explanation:

`x=rcos theta`
`x=7*cos((-7pi)/12)`
if we add 1 revolution `=2pi` to the angle then
`2pi+(-7pi)/12=(17pi)/12`
`x=7*cos((17pi)/12)`
using double angle formulas

`x=7*cos((17pi)/12)=7*cos(pi+(5pi)/12)`

`x=7*cos(pi+(5pi)/12)`
`x=7[cos pi *cos ((5pi)/12)- sin pi *sin ((5pi)/12)]`

`x=7*(-cos((5pi)/12)-0)`

`x=7*(-1)*cos ((5pi)/12)`
Using double angle again

`x=-7*cos(pi/3+pi/12)`
`x=-7*[cos (pi/3) *cos (pi/12)-sin (pi/3) *sin (pi/12)]`

Recall the special angles `60^@=pi/3` and `30^@=pi/6` and `pi/12=1/2*pi/6` Use Half-Angle formulas for functions of `pi/12`

`cos (pi/12)=sqrt((1+cos pi/6)/2)=sqrt((1+sqrt3/2)/2)=1/2sqrt(2+sqrt3)`
`sin (pi/12)=sqrt((1-cos pi/6)/2)=sqrt((1-sqrt3/2)/2)=1/2sqrt(2-sqrt3)`

so that

`x=-7*[cos (pi/3) *cos (pi/12)-sin (pi/3) *sin (pi/12)]`

becomes

`x=-7*[1/2 *(sqrt(2+sqrt3))/2-sqrt3/2 *sqrt(2-sqrt3)/2]`

`x=7/4*(sqrt(6-3sqrt3)-sqrt(2+sqrt3))=-1.81173`

Do the same for `y` and come up with

`y=-7/4*(sqrt(6+3sqrt3)+sqrt(2-sqrt3))=-6.76148`