Prove some inequalities of sin x

For all $x > 0$ prove that

$x - \frac{x^3}{6} < \sin x < x.$

From the first exercise of this section on inequalities, we know $\sin x < x$ for all $0 < x < \frac{\pi}{2}$ . But since $\sin x \leq 1$ for all $x$ and $1 < \frac{\pi}{2}$ we have the inequality on the right immediately,

$\sin x < x \qquad \text{for all } x > 0.$

For the inequality on the left let

$f(x) = \sin x - x + \frac{x^3}{6}.$

Then, we’ll consider the first two derivatives of $f$ to show that it is positive for all $x > 0$ .

$\begin{align*} f'(x) &= \cos x - 1 + \frac{x^2}{2} \\ f''(x) &= -\sin x + x = x - \sin x. \end{align*}$

We know from the inequality on the right that $\sin x < x$ for all $x > 0$ . Hence, $f''(x) > 0$ for all $x > 0$ . Therefore, $f'(x)$ is increasing on the positive real axis. Since