If the standard unit circle is flipped over the x-axis, describe the transformation of the angle $ heta$ and calculate the new coordinates for $ heta = frac{2pi}{3}$.

When the unit circle is flipped over the x-axis, the y-coordinates of all points on the circle are inverted. Therefore, the angle $\theta$ remains the same in magnitude but the y-value of the coordinate changes sign.

For $\theta = \frac{2\pi}{3}$, the original coordinates on the unit circle are:

$\left(\cos\left(\frac{2\pi}{3}\right), \sin\left(\frac{2\pi}{3}\right)\right)$

We compute:

$\cos\left(\frac{2\pi}{3}\right) = -\frac{1}{2}$

$\sin\left(\frac{2\pi}{3}\right) = \frac{\sqrt{3}}{2}$

Therefore, the original coordinates are:
$\left(-\frac{1}{2}, \frac{\sqrt{3}}{2}\right)$

After flipping over the x-axis, the new coordinates become:
$\left(-\frac{1}{2}, -\frac{\sqrt{3}}{2}\right)$

When a unit circle is flipped over the x-axis, each point $(x, y)$ is transformed to $(x, -y)$. The angle $ heta$ does not change in measure but the sign of the sine (y-coordinate) changes.

For $ heta = frac{2pi}{3}$, the original coordinates are:

$left(cosleft(frac{2pi}{3}
ight), sinleft(frac{2pi}{3}
ight)
ight)$

Since:
$cosleft(frac{2pi}{3}
ight) = -frac{1}{2}$

$sinleft(frac{2pi}{3}
ight) = frac{sqrt{3}}{2}$

The original coordinates are:
$left(-frac{1}{2}, frac{sqrt{3}}{2}
ight)$

Flipping the circle over the x-axis, the new coordinates are:
$left(-frac{1}{2}, -frac{sqrt{3}}{2}
ight)$

For $ heta = frac{2pi}{3}$, initially:
$left(cosleft(frac{2pi}{3}
ight), sinleft(frac{2pi}{3}
ight)
ight) = left(-frac{1}{2}, frac{sqrt{3}}{2}
ight)$

After flipping over the x-axis:
$left(-frac{1}{2}, -frac{sqrt{3}}{2}
ight)$

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