Find the angle in radians and degrees for the point $ left(-frac{1}{2}, -frac{sqrt{3}}{2}
ight) $ on the unit circle

We need to find the angle corresponding to the point $ \left(-\frac{1}{2}, -\frac{\sqrt{3}}{2}\right) $ on the unit circle. This point lies in the third quadrant where both sine and cosine are negative. The reference angle is given by:

$ \text{Reference angle} = \arccos\left( \frac{1}{2} \right) = \frac{\pi}{3} $

Since the point is in the third quadrant, the angle in radians is:

$ \theta = \pi + \frac{\pi}{3} = \frac{4\pi}{3} $

To convert this to degrees:

$ \theta = \frac{4\pi}{3} \times \frac{180}{\pi} = 240^{\circ} $

Hence, the angle is $ \frac{4\pi}{3} $ radians or $ 240^{\circ} $.

To find the angle for the point $ left(-frac{1}{2}, -frac{sqrt{3}}{2}
ight) $, observe that it lies in the third quadrant with a reference angle:

$ ext{Reference angle} = arccosleft( frac{1}{2}
ight) = frac{pi}{3} $

The actual angle in radians is:

$ heta = pi + frac{pi}{3} = frac{4pi}{3} $

In degrees:

$ heta = 240^{circ} $

For the point $ left(-frac{1}{2}, -frac{sqrt{3}}{2}
ight) $, the angle is:

$ heta = frac{4pi}{3} ext{ radians or } 240^{circ} $

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