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Explanation:

Step1: Use the angle - reduction formula

We know that (\sin(x)=\sin(2\pi - x)) when (x) is in the fourth - quadrant. Here, (\frac{11\pi}{6}=2\pi-\frac{\pi}{6}). So, (\sin(\frac{11\pi}{6})=\sin(2\pi - \frac{\pi}{6})). According to the formula (\sin(A - B)=\sin A\cos B-\cos A\sin B), when (A = 2\pi) and (B=\frac{\pi}{6}), since (\sin(2\pi)=0) and (\cos(2\pi)=1), we have (\sin(2\pi-\frac{\pi}{6})=\sin(2\pi)\cos(\frac{\pi}{6})-\cos(2\pi)\sin(\frac{\pi}{6})). Substituting the values, we get (\sin(2\pi-\frac{\pi}{6})=0\times\frac{\sqrt{3}}{2}-1\times\frac{1}{2}). Another way: Using the property (\sin(2k\pi - \alpha)=-\sin\alpha) ((k\in\mathbb{Z})), when (k = 1) and (\alpha=\frac{\pi}{6}), (\sin(\frac{11\pi}{6})=-\sin(\frac{\pi}{6})).

Step2: Evaluate (\sin(\frac{\pi}{6}))

We know that for a right - triangle with an angle of (\frac{\pi}{6}) (or (30^{\circ})), (\sin(\frac{\pi}{6})=\frac{1}{2}).

Answer:

(-\frac{1}{2})