Recall that \(x = r \cos \theta\) and \(y = r \sin \theta\). Plugging in the given values, we get:
\(x = -2 \cos \left(-\frac{\pi}{6}\right) = -2 \cos \frac{\pi}{6} = -\sqrt{3}\)
\(y = -2 \sin \left(-\frac{\pi}{6}\right) = -2 \sin \frac{-\pi}{6} = 1\)
So the
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