What are the components of the vector between the origin and the polar coordinate $(- 8, \frac{- 7 π}{6})$ $(-8, (-7pi)/6)$ ?

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What are the components of the vector between the origin and the polar coordinate $(- 8, \frac{- 7 π}{6})$ $(-8, (-7pi)/6)$ ?

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Answer 1 · 2018-02-17T13:52:23

The formula is $(r \cdot cos (θ), r \cdot sin (θ))$ $(r*cos(theta),r*sin(theta))$ .

Explanation:

Calculate $- 8 \cdot cos (\frac{- 7 π}{6})$ $-8*cos((-7pi)/6)$ first. Do you understand the angle and its position? It is in Quadrant II, like $- 210^{\circ}$ $-210 ^@$ or $150^{\circ}$ $150^@$ .![https://www.wyzant.com/resources/lessons/math/trigonometry/unit-circle]( useruploads.socratic.org )
The acute angle is $30^{\circ}$ $30^@$ , and the associated cosine value is $- \frac{\sqrt{3}}{2}$ $-sqrt(3)/2$ .
![https://www.wyzant.com/resources/lessons/math/trigonometry/unit-circle]()
Take -8 times the cosine value: $\frac{- 8 \cdot - \sqrt{3}}{2}$ $(-8*-sqrt(3))/2$ = $\frac{8 \sqrt{3}}{2}$ $(8sqrt3)/2$ = $4 \sqrt{3}$ $4sqrt3$ .

Repeat the procedure for $- 8 \cdot sin (\frac{- 7 π}{6})$ $-8*sin((-7pi)/6)$ = $- 8 \cdot \frac{1}{2}$ $-8*1/2$ =-4.

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What are the components of the vector between the origin and the polar coordinate $(- 8, \frac{- 7 π}{6})$ $(-8, (-7pi)/6)$ ?

1 Answer

Explanation:

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What are the components of the vector between the origin and the polar coordinate $(- 8, \frac{- 7 π}{6})$ $(-8, (-7pi)/6)$ ?