# The sine inequality $\frac2\pi x \le \sin x \le x$ for $0<x<\frac\pi2$ [duplicate]

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

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

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) :

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

• Brilliant! @OP I suggest after reading this you check out Jensen's inequality. – oldrinb May 31 '13 at 8:47

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).

• I am struggling with this. I need more details of proof. – Paul May 31 '13 at 8:36
• 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 8:39

$\sin(x) -x$ 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?

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