What are the components of the vector between the origin and the polar coordinate $(2, \frac{- 11 π}{6})$ $(2, (-11pi)/6)$ ?

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Answer 1 · 2016-09-26T04:46:44

$(\sqrt{3}, 1)$ $(sqrt 3, 1)$

The components of the position vector to $r, θ) a r e (r cos θ, r sin θ)$ $r, theta) are (r cos theta, r sin theta)$

Here, they are

$(2 cos (- \frac{11}{6} π), 2 sin (- \frac{11}{6} π))$ $(2 cos(-11/6pi), 2 sin (-11/6pi))$

$= (2 cos (\frac{11}{6} p) i, - 2 sin (\frac{11}{6} π))$ $=(2 cos (11/6p)i, -2 sin (11/6pi))$ ,

using $cos (- a) = cos a and sin (- a) = - sin a$ $cos(-a)=cos a and sin (-a)=-sin a$

$= (2 cos (2 π - \frac{π}{6}), - 2 sin (2 π - \frac{π}{6})$ $=(2 cos (2pi-pi/6), -2 sin (2pi-pi/6)$ .

$= (2 cos (\frac{π}{6}), 2 sin (\frac{π}{6}))$ $=(2 cos (pi/6), 2 sin (pi/6))$ ,

using sine in $Q_{4}$ $Q_4$ is negative and cosine in $Q_{4}$ $Q_4$ is positive.

$= (\sqrt{3}, 1)$ $=(sqrt 3, 1)$

You ought to note that, in effect, the directions

$θ = \frac{π}{6} and θ = - \frac{11}{6} π = - (2 π - \frac{π}{6})$ $theta = pi/6 and theta =-11/6pi=-(2pi-pi/6)$

are the same.

For positive $θ$ $theta$ , the sense of rotation is anticlockwise.

For negative $θ$ $theta$ , the sense of rotation is clockwise.

What are the components of the vector between the origin and the polar coordinate (2, (-11pi)/6)(2,−11π6)(2, (-11pi)/6)?