1. **State the problems:** - Evaluate $f(x) = 8 \sin x - 4 \cos \frac{x}{2}$ at $x = \frac{\pi}{3}$. - Evaluate $\tan 600^\circ$ without using a calculator. 2. **Evaluate $f\left(\frac{\pi}{3}\right)$:** - Substitute $x = \frac{\pi}{3}$ into the function: $$f\left(\frac{\pi}{3}\right) = 8 \sin \frac{\pi}{3} - 4 \cos \frac{\pi}{6}$$ - Recall important values: $$\sin \frac{\pi}{3} = \frac{\sqrt{3}}{2}, \quad \cos \frac{\pi}{6} = \frac{\sqrt{3}}{2}$$ - Substitute these values: $$f\left(\frac{\pi}{3}\right) = 8 \times \frac{\sqrt{3}}{2} - 4 \times \frac{\sqrt{3}}{2} = 4\sqrt{3} - 2\sqrt{3} = 2\sqrt{3}$$ 3. **Evaluate $\tan 600^\circ$ without a calculator:** - Use the periodicity of tangent: $\tan(\theta) = \tan(\theta - 360^\circ)$ - Calculate: $$600^\circ - 360^\circ = 240^\circ$$ - So, $\tan 600^\circ = \tan 240^\circ$ - $240^\circ$ is in the third quadrant where tangent is positive. - Reference angle: $$240^\circ - 180^\circ = 60^\circ$$ - Therefore: $$\tan 240^\circ = \tan 60^\circ = \sqrt{3}$$ - Since tangent is positive in the third quadrant, $\tan 600^\circ = \sqrt{3}$. **Final answers:** - $f\left(\frac{\pi}{3}\right) = 2\sqrt{3}$ - $\tan 600^\circ = \sqrt{3}$