1. **State the problem:** We need to complete the table of values for the function $f(x) = \sin x$ for $x$ values from $0$ to $2\pi$ in increments of $\frac{\pi}{6}$. 2. **Recall the sine values for special angles:** The sine function values for common angles are: - $\sin 0 = 0$ - $\sin \frac{\pi}{6} = \frac{1}{2}$ - $\sin \frac{\pi}{3} = \frac{\sqrt{3}}{2}$ - $\sin \frac{\pi}{2} = 1$ - $\sin \frac{2\pi}{3} = \sin \left(\pi - \frac{\pi}{3}\right) = \sin \frac{\pi}{3} = \frac{\sqrt{3}}{2}$ - $\sin \frac{5\pi}{6} = \sin \left(\pi - \frac{\pi}{6}\right) = \sin \frac{\pi}{6} = \frac{1}{2}$ - $\sin \pi = 0$ 3. **Use symmetry and periodicity for values beyond $\pi$:** - $\sin \frac{7\pi}{6} = \sin \left(\pi + \frac{\pi}{6}\right) = -\sin \frac{\pi}{6} = -\frac{1}{2}$ - $\sin \frac{4\pi}{3} = \sin \left(\pi + \frac{\pi}{3}\right) = -\sin \frac{\pi}{3} = -\frac{\sqrt{3}}{2}$ - $\sin \frac{3\pi}{2} = -1$ - $\sin \frac{5\pi}{3} = \sin \left(2\pi - \frac{\pi}{3}\right) = -\sin \frac{\pi}{3} = -\frac{\sqrt{3}}{2}$ - $\sin \frac{11\pi}{6} = \sin \left(2\pi - \frac{\pi}{6}\right) = -\sin \frac{\pi}{6} = -\frac{1}{2}$ - $\sin 2\pi = 0$ 4. **Fill in the table:** | $x$ | $\sin x$ | |---------------|-------------------| | $0$ | $0$ | | $\frac{\pi}{6}$ | $\frac{1}{2}$ | | $\frac{\pi}{3}$ | $\frac{\sqrt{3}}{2}$ | | $\frac{\pi}{2}$ | $1$ | | $\frac{2\pi}{3}$ | $\frac{\sqrt{3}}{2}$ | | $\frac{5\pi}{6}$ | $\frac{1}{2}$ | | $\pi$ | $0$ | | $\frac{7\pi}{6}$ | $-\frac{1}{2}$ | | $\frac{4\pi}{3}$ | $-\frac{\sqrt{3}}{2}$ | | $\frac{3\pi}{2}$ | $-1$ | | $\frac{5\pi}{3}$ | $-\frac{\sqrt{3}}{2}$ | | $\frac{11\pi}{6}$| $-\frac{1}{2}$ | | $2\pi$ | $0$ | This completes the first chart for $f(x) = \sin x$ from $0$ to $2\pi$ in increments of $\frac{\pi}{6}$.