In many books and online , the inequality $\sin x<x$ is defined only over the interval $[0, \frac{\pi}{2}]$ . However it is easy to check that the inequality holds good over the interval $[\frac{\pi}{2} , \pi]$ as well. We can check it graphically . In fact it holds good for all positive numbers. Is there a special reason for proving this inequality over the restricted interval of $[0, \frac{\pi}{2}]$ ?

  • $\begingroup$ You probably mean $\left(0, \frac\pi2\right]$. At $x = 0$ we have equality. $\endgroup$ – mechanodroid Jul 16 '18 at 17:08
  • $\begingroup$ @mechanodroid right . I didn’t pay attention to that. Thanks for correcting me. $\endgroup$ – Aditi Jul 16 '18 at 17:15

You are correct. The inequality holds for all positive $x$. Easiest way to see this is probably to compare derivative.

In regards to the restriction of the inequality: It is possible that in your case, you only need the fact that $\sin x<x$ on $[0,\pi/2]$, rather than on the whole positive real line.

Often times, the author will restrict to a smaller interval to emphasize that that is the only necessary place for the inequality to hold in the remainder of the argument.

  • $\begingroup$ Alright ! Thanks for your help :) $\endgroup$ – Aditi Jul 16 '18 at 16:58
  • $\begingroup$ Your welcome :-) $\endgroup$ – ervx Jul 16 '18 at 17:03

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