91Ó°ÊÓ

To prove that \(x - \frac{x^3}{6} < \sin x < x\) for all \(x > 0\), we first analyze the function \(f(x) = \sin x - x + \frac{x^3}{6}\) and find its derivative \(f'(x) = \cos x - 1 + \frac{x^2}{2}\). We show that \(f'(x) > 0\) on the interval \((0, \infty)\) by analyzing the properties of the cosine and quadratic functions. Since \(f'(x) > 0\), it means that \(f(x)\) is increasing on the interval \((0, \infty)\), and thus \(f(x) > 0\) for all \(x > 0\), proving the left inequality. For the right inequality, we use the properties of the sine function to show that \(\sin x < x\) for all \(x > 0\). By combining the results, we can conclude that for all \(x > 0\), \(x - \frac{x^3}{6} < \sin x < x\).