How do you find the coordinates of the terminal points corresponding to the following arc length on the unit circle: -25π/6?

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How do you find the coordinates of the terminal points corresponding to the following arc length on the unit circle: -25π/6?

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Answer 1 · 2015-07-20T19:45:48

Find coordinates of terminal arc $x = \frac{- 25 π}{6}$ $x = (-25pi)/6$
Answers:
$x = \frac{\sqrt{3}}{2}$ $x = (sqrt3)/2$
$y = - \frac{1}{2}$ $y = - 1/2$

Explanation:

$\frac{- 25 π}{6} \to (- \frac{π}{6} - 24 \frac{π}{6}) \to (- \frac{π}{6} - 4 π) \to (- \frac{π}{6})$ $(-25pi)/6 -> (-pi/6 - 24pi/6) -> (-pi/6 - 4pi) -> ( -pi/6)$

The terminal M of the arc $x = \frac{- 25 π}{6}$ $x = (-25pi)/6$ has as coordinates:

$x = cos (- \frac{π}{6}) = cos (\frac{π}{6}) = \frac{\sqrt{3}}{2}$ $x = cos (-pi/6) = cos (pi/6) = (sqrt3)/2$

$y = sin (- \frac{π}{6}) = - sin (\frac{π}{6}) = - \frac{1}{2}$ $y = sin (-pi/6) = -sin (pi/6) = - 1/2$

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How do you find the coordinates of the terminal points corresponding to the following arc length on the unit circle: -25π/6?

1 Answer

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