Find the coordinates of the point on the unit circle where the angle with the positive x-axis is $frac{5pi}{6}$ radians.

Given the angle $\theta = \frac{5\pi}{6}$ radians, we need to find the coordinates of the point on the unit circle.

On the unit circle, the coordinates of a point at an angle $\theta$ are $(\cos(\theta), \sin(\theta))$.

Therefore,

$x = \cos\left(\frac{5\pi}{6}\right) = -\cos\left(\frac{\pi}{6}\right) = -\frac{\sqrt{3}}{2}$

$y = \sin\left(\frac{5\pi}{6}\right) = \sin\left(\frac{\pi}{6}\right) = \frac{1}{2}$

So, the coordinates are $\left(-\frac{\sqrt{3}}{2}, \frac{1}{2}\right)$.

To find the coordinates of the point where the angle with the positive x-axis is $frac{5pi}{6}$ radians, we use the unit circle properties.

The coordinates are given by $(cos heta, sin heta)$.

For an angle $ heta = frac{5pi}{6}$,

$x = cosleft(frac{5pi}{6}
ight)$

$y = sinleft(frac{5pi}{6}
ight)$

We know, $cosleft(frac{5pi}{6}
ight) = -frac{sqrt{3}}{2}$ and $sinleft(frac{5pi}{6}
ight) = frac{1}{2}$.

Hence, the coordinates are $left(-frac{sqrt{3}}{2}, frac{1}{2}
ight)$.

Find the coordinates on the unit circle at angle $frac{5pi}{6}$:

$x = cosleft(frac{5pi}{6}
ight) = -frac{sqrt{3}}{2}$

$y = sinleft(frac{5pi}{6}
ight) = frac{1}{2}$

Coordinates: $left(-frac{sqrt{3}}{2}, frac{1}{2}
ight)$.

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