1. **Problem statement:** Simplify the expression $\cos\left(\frac{\pi}{6}\right) - 2\sin\left(\frac{\pi}{3}\right)$ using trigonometric definitions and values. 2. **Recall exact values:** - $\cos\left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{2}$ - $\sin\left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2}$ 3. **Substitute values:** $$\cos\left(\frac{\pi}{6}\right) - 2\sin\left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2} - 2 \times \frac{\sqrt{3}}{2}$$ 4. **Simplify multiplication:** $$= \frac{\sqrt{3}}{2} - \cancel{2} \times \frac{\sqrt{3}}{\cancel{2}} = \frac{\sqrt{3}}{2} - \sqrt{3}$$ 5. **Express $\sqrt{3}$ as $\frac{2\sqrt{3}}{2}$ to combine terms:** $$= \frac{\sqrt{3}}{2} - \frac{2\sqrt{3}}{2} = \frac{\sqrt{3} - 2\sqrt{3}}{2} = \frac{-\sqrt{3}}{2}$$ 6. **Final simplified answer:** $$\boxed{-\frac{\sqrt{3}}{2}}$$ --- 7. **Problem statement (b):** Differentiate $$y = x \sin^{-1}(2x) + \frac{1}{2} \sqrt{1 - 4x^2}$$ 8. **Recall derivative rules:** - Derivative of $x \sin^{-1}(2x)$ uses product rule: $$\frac{d}{dx}[u v] = u' v + u v'$$ - Derivative of $\sin^{-1}(u)$ is: $$\frac{d}{dx} \sin^{-1}(u) = \frac{u'}{\sqrt{1 - u^2}}$$ - Derivative of $\sqrt{f(x)} = (f(x))^{1/2}$ is: $$\frac{1}{2} (f(x))^{-1/2} f'(x)$$ 9. **Apply product rule:** $$y' = 1 \cdot \sin^{-1}(2x) + x \cdot \frac{2}{\sqrt{1 - (2x)^2}} + \frac{1}{2} \cdot \frac{1}{2} (1 - 4x^2)^{-1/2} (-8x)$$ 10. **Simplify terms:** $$y' = \sin^{-1}(2x) + \frac{2x}{\sqrt{1 - 4x^2}} - \frac{2x}{\sqrt{1 - 4x^2}}$$ 11. **Cancel terms:** $$\frac{2x}{\sqrt{1 - 4x^2}} - \frac{2x}{\sqrt{1 - 4x^2}} = 0$$ 12. **Final derivative:** $$\boxed{y' = \sin^{-1}(2x)}$$