# A geometric proof for the inequality $\frac{2x}{\pi} \le \sin(x)$

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 $$\sin(x)\le x$$.

• Does this stumblingrobot.com/2015/12/19/… clarify your question – Jitendra Singh Jan 29 at 5:31
• There is a “geometric proof” in this Wikipedia article: en.wikipedia.org/wiki/Jordan%27s_inequality – Martin R Jan 29 at 7:45
• @Martin: This is nice proof. +1 – Maths Rahul Jan 29 at 8:00
• 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. – Martin R Jan 29 at 8:45

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.

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 $$x \le \frac \pi 2 \sin(x) \iff \frac 2 \pi x \le \sin(x) \, .$$
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 $$f(x)\geq tf(0)+(1-t)f(\pi/2)=0,x\in[0,\pi/2].$$