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Given $x\in \left(0; \frac\pi2\right)$. Prove that $$\sin x>\frac{2x}{\pi}$$

This is my try:

Let $y=\sin x-\frac{2x}\pi\implies y ' = \cos x - \frac2\pi\implies y ''=-\sin x <0; \forall x\in \left(0; \frac\pi2\right)$ $\implies \left\{\begin{matrix} y '(x)> y '(\frac\pi2) = -\frac2\pi\\ y' (x) < y'(0) =1-\frac2\pi\end{matrix}\right.$

$\bullet\quad y ' < 0 \implies y (x) > y (\frac\pi2)=0$

$\bullet\quad y' >0 \implies y(x) > y(0) =0$

But it went wrong.

mja
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    If you wish to proceed with $y=\sin x-\frac{2x}{\pi}$, then recognizing that $y'=\cos x-2\pi$ is decreasing on $[0,\pi/2]$ implies that $y$ is concave there. Inasmuch as $y(0)=y(\pi/2)=0$, then $y\ge 0$ and you're done! – Mark Viola Aug 18 '15 at 03:42

3 Answers3

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Consider the function , $f(x)=\frac{\sin x}{x}$ in $(0,\pi/2)$.

Then , $$f'(x)=\frac{x\cos x-\sin x}{x^2}<0$$[Using the relation $x<\tan x$ in $(0,\pi/2)$]

So, $f$ is monotone decreasing in $(0,\pi/2)$.

So, $x<\pi/2\implies f(x)>f(\pi/2)\implies \sin x>\frac{2x}{\pi}$

Empty
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Alternative solution: Let $y = \sin x$. Then $y''=-\sin x <0$ for $0<x<\pi /2$. So $y$ is strictly concave on that interval. Note that $y(0)=0$ and $y(\pi/2)=1$. Therefore, $y$ lies strictly above the line segment joining $(0,0)$ and $(\pi/2, 1)$. Therefore $y > 2x/ \pi$ for $0<x<\pi /2$.

suncup224
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enter image description here

Now Slope of $$\displaystyle \bf{OB} = m_{0B} = \frac{\sin x-0}{x-0} = \frac{\sin x}{x}.$$

and Slope of $$\displaystyle \bf{OA} = m_{0A} = \frac{1-0}{\frac{\pi}{2}-0} = \frac{2}{\pi}.$$

So Here in above graph $$\displaystyle \bf{m_{OB}>m_{OA}}$$

So we get $$\displaystyle \bf{\frac{\sin x}{x}>\frac{2}{\pi}}$$ for all $$\displaystyle \bf{x\in \left(0,\frac{\pi}{2}\right)}$$

juantheron
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