Question

What are the components of the vector between the origin and the polar coordinate `(-8, (-7pi)/6)`?

Answer 1

The formula is `(r*cos(theta),r*sin(theta))`.

Explanation:

Calculate `-8*cos((-7pi)/6)` first. Do you understand the angle and its position? It is in Quadrant II, like `-210 ^@` or `150^@`.
The acute angle is `30^@`, and the associated cosine value is `-sqrt(3)/2` .

Take -8 times the cosine value: `(-8*-sqrt(3))/2` =`(8sqrt3)/2`=`4sqrt3`.

Repeat the procedure for `-8*sin((-7pi)/6)`=`-8*1/2`=-4.