证明: $\dps{0\leq x\leq \frac{\pi}{2}}$ 时, $\dps{\sin x\leq x-\frac{1}{3\pi}x^3}$.
证明: 由例 4.3.19, $$\bex \sin x<x-\frac{x^3}{6}+\frac{x^5}{120} =x-\frac{x^3}{3\pi}\sex{\frac{\pi}{2}-\frac{\pi x^2}{40}}; \eex$$ 再据 $$\beex \bea \frac{\pi}{2}-\frac{\pi x^2}{40}\geq 1&\lra \frac{\pi x^2}{40}\leq \frac{\pi}{2}-1\\ &\lra \pi x^2\leq 20(\pi-2)\\ &\lra x^2\leq 20\sex{1-\frac{2}{\pi}}\\ &\ra x^2\leq \sex{\frac{\pi}{2}}^2\leq 4<\frac{20}{3}=20\sex{1-\frac{2}{3}}<20\sex{1-\frac{2}{\pi}} \eea \eeex$$ 即知结论成立.
