To find the sine of an angle \( \theta \) given that \( \cos(\theta) = -\frac{\sqrt{3}}{2} \) and that \( \theta \) is in Quadrant III, we can use the Pythagorean identity:
\[ \sin^2(\theta) + \cos^2(\theta) = 1 \]
First, we substitute the value of \( \cos(\theta) \):
\[ \sin^2(\theta) + \left(-\frac{\sqrt{3}}{2}\right)^2 = 1 \]
Calculating \( \left(-\frac{\sqrt{3}}{2}\right)^2 \):
\[ \left(-\frac{\sqrt{3}}{2}\right)^2 = \frac{3}{4} \]
Now substitute this back into the equation:
\[ \sin^2(\theta) + \frac{3}{4} = 1 \]
Next, we isolate \( \sin^2(\theta) \):
\[ \sin^2(\theta) = 1 - \frac{3}{4} \]
This simplifies to:
\[ \sin^2(\theta) = \frac{1}{4} \]
Now we take the square root of both sides:
\[ \sin(\theta) = \pm \frac{1}{2} \]
Since \( \theta \) is in Quadrant III, where sine is negative, we choose the negative root:
\[ \sin(\theta) = -\frac{1}{2} \]
Thus, the sine of the angle \( \theta \) is:
\[ \sin(\theta) = -\frac{1}{2} \]
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