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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.

Martin Sleziak
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Paul
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  • https://math.stackexchange.com/questions/213382/prove-an-inequality-with-a-sin-function-sinx-frac2-pi-x-for-0x-fr/213398?s=1|127.1530#213398 – Guy Fsone Feb 12 '18 at 12:28

4 Answers4

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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) :

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

Joel Cohen
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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?

Vishal Gupta
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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).

john
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  • I am struggling with this. I need more details of proof. – Paul May 31 '13 at 08:36
  • 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. – john May 31 '13 at 08:39
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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 $y = x$

mechanician
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