What are the components of the vector between the origin and the polar coordinate (-8, (-7pi)/6)(8,7π6)?

1 Answer
Feb 17, 2018

The formula is (r*cos(theta),r*sin(theta))(rcos(θ),rsin(θ)).

Explanation:

Calculate -8*cos((-7pi)/6)8cos(7π6) first. Do you understand the angle and its position? It is in Quadrant II, like -210 ^@210 or 150^@150.https://www.wyzant.com/resources/lessons/math/trigonometry/unit-circlehttps://www.wyzant.com/resources/lessons/math/trigonometry/unit-circle
The acute angle is 30^@30, and the associated cosine value is -sqrt(3)/232 .
https://www.wyzant.com/resources/lessons/math/trigonometry/unit-circlehttps://www.wyzant.com/resources/lessons/math/trigonometry/unit-circle
Take -8 times the cosine value: (-8*-sqrt(3))/2832 =(8sqrt3)/2832=4sqrt343.

Repeat the procedure for -8*sin((-7pi)/6)8sin(7π6)=-8*1/2812=-4.