Calculate the coordinates of the point on the unit circle corresponding to an angle of $frac{5pi}{6}$ radians.

To find the coordinates of the point on the unit circle corresponding to the given angle, we use the cosine and sine functions:

$x = \cos\left(\frac{5\pi}{6}\right)$

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

Since $\frac{5\pi}{6}$ is in the second quadrant, where cosine is negative and sine is positive, we have:

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

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

Therefore, the coordinates are:

$\left( -\frac{\sqrt{3}}{2}, \frac{1}{2} \right)$

To determine the coordinates for the angle $frac{5pi}{6}$, we need to calculate the x and y coordinates using the unit circle’s properties. The coordinates are given by:

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

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

In the second quadrant, cosine is negative, and sine is positive. So for the angle $frac{5pi}{6}$:

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

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

The coordinates of the point are:

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

For the angle $frac{5pi}{6}$ on the unit circle, the coordinates are:

$left(cosleft(frac{5pi}{6}
ight), sinleft(frac{5pi}{6}
ight)
ight)$

Since it is in the second quadrant:

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

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

Thus, the point is:

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

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