Show that $ x< \sin(\frac{\pi x}2)\forall x\in (0,1)$

Question

I don't know how to prove this inequality, some help would be appreciated, thank you!

$x< \sin(\frac{\pi x}2)\quad\quad \forall x \in (0,1)$

Look at the function $f(x) = \sin(\frac{\pi}{2}x) - x$ and use the first derivative test to find out when this function is increasing and decreasing. This should get you started. — MathNewbie, Nov 25 '16 at 05:40
Some questions which contain rather similar inequality: Mean Value Theorem: $\frac{2}{\pi}<\frac{\sin x}{x}<1$, The sine inequality $\frac2\pi x \le \sin x \le x$ for $0<x<\frac\pi2$ — Martin Sleziak, Nov 30 '16 at 05:08
Your question was put on hold, the message above (and possibly comments) should give an explanation why. (In particular, this link might be useful.) You might try to edit your question to address these issues. Note that the next edit puts your post in the review queue, where users can vote whether to reopen it or leave it closed. (Therefore it would be good to avoid minor edits and improve your question as much as possible with the next edit.) — Martin Sleziak, Nov 30 '16 at 13:40

score 3 · Accepted Answer · answered Nov 25 '16 at 05:59

Let we have to prove $$t<\sin \left(\frac{\pi t}{2}\right)\;\forall t \in (0,1)$$

Now put $\displaystyle \frac{\pi t}{2} = x\Rightarrow t = \frac{2x}{\pi}$

So we have to prove $$\frac{2x}{\pi}<\sin x\;\forall x \in \left(0,\frac{\pi}{2}\right)$$

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 $$\displaystyle \frac{\sin x}{x}>\frac{2}{\pi}\Rightarrow \frac{2x}{\pi}<\sin x\;\forall x\in \left(0,\frac{\pi}{2}\right)$$

Nice! And special thanks for the picture! So basically, the solution boils down to the fact that the graph of $y=\sin x$ is concave down on $(0,\pi)$, so the graph lies above any secant line (within this interval). — zipirovich, Nov 25 '16 at 06:11

score -3 · Answer 2 · edited Nov 25 '16 at 05:48

-3

For $x\in(0,1)$ the function $\sin( \frac\pi2x) $ is an increasing function and we have $$1< \frac\pi2x$$ so $1< \sin (\frac\pi2x) $

edited Nov 25 '16 at 05:48

AlgorithmsX

answered Nov 25 '16 at 05:42

Ali

This is not even a sketch of a valid proof. There are plenty of increasing functions that start at $0$ and end at 1 that are not less than $x$ at very point in $(0,1)$. All you have perhaps shown is that $\sin x < \sin(x\pi/2)$ in that interval. – Mark Fischler Nov 25 '16 at 06:01

2 Answers2