What are the components of the vector between the origin and the polar coordinate $(- 16, \frac{- π}{4})$ $(-16, (-pi)/4)$ ?

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What are the components of the vector between the origin and the polar coordinate $(- 16, \frac{- π}{4})$ $(-16, (-pi)/4)$ ?

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Answer 1 · 2018-02-27T05:06:53

$x = - 16 \cdot cos (- \frac{π}{4}) = - 16 \cdot (\frac{1}{\sqrt{2}}) = - 8 \sqrt{2}$ $x = -16 * cos (-pi/4 )= -16 * (1/sqrt2) = color (green)(-8sqrt2$

$y = - 16 \cdot sin (- \frac{π}{4}) = - 16 \cdot (- \frac{1}{\sqrt{2}}) = 8 \sqrt{2}$ $y = -16 * sin (-pi/4 )= -16 * (-1/sqrt2) = color (green)(8sqrt2$

Explanation:

We can use the following formulas to convert polar to Cartesian coordinates.

$x = r cos θ, y = r sin θ$ $x = r cos theta, y = r sin theta$

$G i v e n : [r, θ] = [- 16, - \frac{π}{4}]$ $Given : [r,theta] = [-16, -pi/4]$

$x = - 16 \cdot cos (- \frac{π}{4}) = - 16 \cdot (\frac{1}{\sqrt{2}}) = - 8 \sqrt{2}$ $x = -16 * cos (-pi/4 )= -16 * (1/sqrt2) = color (green)(-8sqrt2$

$y = - 16 \cdot sin (- \frac{π}{4}) = - 16 \cdot (- \frac{1}{\sqrt{2}}) = 8 \sqrt{2}$ $y = -16 * sin (-pi/4 )= -16 * (-1/sqrt2) = color (green)(8sqrt2$

$tan θ = tan (- \frac{π}{4}) = - 1$ $tan theta = tan (-pi/4) = -1$

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What are the components of the vector between the origin and the polar coordinate $(- 16, \frac{- π}{4})$ $(-16, (-pi)/4)$ ?

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