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This question already has an answer here:

There is an exercise on $\sin x$. How could I see that for any $0<x< \frac \pi 2$, $\frac 2 \pi x \le \sin x\le x$?

Thanks for your help.

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marked as duplicate by Guy Fsone, Arnaud D., Parcly Taxel, Moya, Arnaud Mortier Feb 12 '18 at 14:14

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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For $x \in \left[0, \frac{\pi}{2}\right]$, we have $\sin''(x) = -\sin(x) \le 0$. So the sine function is concave on $\left[0, \frac{\pi}{2}\right]$. So the inequality follows from the principle (I suggest drawing the graph to see it clearly) :

$$\textrm{secant} \le \textrm{function} \le \textrm{tangent}$$

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  • $\begingroup$ Brilliant! @OP I suggest after reading this you check out Jensen's inequality. $\endgroup$ – oldrinb May 31 '13 at 8:47
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Consider the function $y=\frac{sin(x)}{x}$. Use calculus techniques to find its range and hence deduce the desired inequality (once you've specified the domain for the inequality).

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  • $\begingroup$ I am struggling with this. I need more details of proof. $\endgroup$ – Paul May 31 '13 at 8:36
  • $\begingroup$ find where the maxima and minima and consider their values and the values (or limiting values) at the endpoints. You can then deduce the range. $\endgroup$ – john May 31 '13 at 8:39
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$\sin(x) -x$ is a decreasing function on $(0,\pi/2)$(look at the derivative) and is equal to zero at zero. Can you fill in the details now?

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Another good way to see it is just to look at plots of $y = \sin(x)$, $y = \frac{2}{\pi}x$ and $y = x$

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