证明: \dps0≤x≤π2 时, \dpssinx≤x−13πx3.
证明: 由例 4.3.19, \bexsinx<x−x36+x5120=x−x33π\sexπ2−πx240;\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.