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The inequality $\frac{2x}{\pi}\le \sin(x)\le x$ for $0 \le x\le \frac \pi 2$ is well known; it can be proved using calculus.

The second part can be proved for $x\in [0,\pi/2]$ by geometric arguments:

Take unit circle with center origin. Then compare areas of (sector with angle $x$) and (right angled triangle with height $\sin x$).

Q. Can we prove $\frac{2x}{\pi}\le \sin(x)$ for $x\in [0,\pi/2]$ by geometric arguments?

Note: There are proof of first inequality are available using calculus, but I want to know if there is a proof, not based on calculus (Rolle's theorem, or mean value theorem etc.), but with some basic geometric arguments as done in the proof of $\sin(x)\le x$.

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  • $\begingroup$ Does this stumblingrobot.com/2015/12/19/… clarify your question $\endgroup$ – Jitendra Singh Jan 29 at 5:31
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    $\begingroup$ There is a “geometric proof” in this Wikipedia article: en.wikipedia.org/wiki/Jordan%27s_inequality $\endgroup$ – Martin R Jan 29 at 7:45
  • $\begingroup$ @Martin: This is nice proof. +1 $\endgroup$ – Maths Rahul Jan 29 at 8:00
  • $\begingroup$ I have taken the liberty to edit the question title. I have tried to state more precisely what you are looking for. Feel free to revert the change or edit it again if that does not match your intentions. $\endgroup$ – Martin R Jan 29 at 8:45
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The following is taken from Wikipedia: Jordan's inequality where it is attributed to

  • Yuefeng, Feng. “Proof without Words: Jordan's Inequality 2x/π ≤ Sin x ≤ x, 0 ≤ x ≤π/2.” Mathematics Magazine, vol. 69, no. 2, 1996, pp. 126–126. JSTOR, https://www.jstor.org/stable/2690669.

enter image description here

(File: https://commons.wikimedia.org/wiki/File:Jordans_inequality.svg, Attribution: Kmhkmh, CC BY 3.0 https://creativecommons.org/licenses/by/3.0, via Wikimedia Commons.)

The arc from $C$ to $D$ on the unit circle has the length $x$, and the arc from $G$ to $D$ on the circle with radius $\sin(x)$ has the length $\frac \pi 2 \sin(x)$. It follows that $$ x \le \frac \pi 2 \sin(x) \iff \frac 2 \pi x \le \sin(x) \, . $$

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Let $$f(x)=\sin x-\frac{2x}{\pi},x\in[0,\pi/2].$$ Then $$f''(x)=-\sin x\leq 0,x\in[0,\pi/2],$$ which implies $f(x)$ is concave function.

$\textbf{Concavity implies the function is above the secant line.}$

Due to $f(0)=f(\pi/2)=0$, we can get $$f(x)\geq tf(0)+(1-t)f(\pi/2)=0,x\in[0,\pi/2].$$

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    $\begingroup$ From the question: “There are proof of first inequality are available using calculus, but I want to know if there is some geometric arguments.” $\endgroup$ – Martin R Jan 29 at 6:07
  • $\begingroup$ Is "concavity" not geometric arguments? concavity implies the function is above the secant line! $\endgroup$ – Riemann Jan 29 at 6:10
  • $\begingroup$ Perhaps I am interpreting the question differently, I expected something using triangles in the unit circle, similar to the example given in the question. Otherwise we could close the question as a duplicate of e.g. this one math.stackexchange.com/a/980553. $\endgroup$ – Martin R Jan 29 at 7:50
  • $\begingroup$ @Martin: I added something in the note, which will little clarify my question. Let me know if the question (or note) is ok now. $\endgroup$ – Maths Rahul Jan 29 at 8:03

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