How can I show that $\left|\sin \frac{s}{2}\right| \geq \frac{|s|}{\pi}, s \in [- \pi , \pi]$?

Question

How can I show that $$\left|\sin \frac{s}{2}\right| \geq \frac{|s|}{\pi}$$ $s \in [- \pi , \pi]$, using that $\psi : x \mapsto \sin x$ is a concave function on $[0 , \pi]$?

By definition of concave function, $$ \psi(t \, x + (1 - t) \, y) \geq t \, \psi(x) + (1 - t) \, \psi(y) $$ for each $x , y \in [0 , \pi]$ and for all $t \in [0 , 1]$. My thought was obtain that using just the definition but I think it is not possible but I am not sure. I have drown that and it's obvious but I want to prove that analytically. Can you help me, please? Thank you very much.

What is the relationship between the graph of $y=\sin \frac x2$ between $0\le x\le \pi$, and the line joining the endpoints of that portion of the graph? What is the equation of that line? — Greg Martin, Feb 15 '17 at 18:12

Michael Rozenberg · Answer 1 · 2017-02-15T19:03:33.983

1

Let $f(x)=\sin\frac{x}{2}$.

Thus, $f$ is a concave function on $[0,\pi]$ and we are done

because graph of $y=\frac{x}{\pi}$ is under graph of $f$

edited Feb 15 '17 at 19:03

answered Feb 15 '17 at 18:15

Michael Rozenberg

203,855

I knew that. I would like to obtain the inequality. Can you be more explicit? – joseabp91 Feb 15 '17 at 18:27
How do you deduce graph of $y = \frac{x}{\pi}$ is under graph of $f$? – joseabp91 Feb 15 '17 at 21:39
@joseabp91 A graph of $f$ and a graph of $y=\frac{x}{\pi}$ have on $[0,\pi]$ two common points $(0,0)$ and $(\pi,1)$. But $f$ is a concave function. Thus, the graph of $y=\frac{x}{\pi}$ located under the graph of $f$. – Michael Rozenberg Feb 15 '17 at 21:54
Yes, but do you use the definition to obtain the statement? – joseabp91 Feb 15 '17 at 21:56
Do you think that I can obtain that with my definition? – joseabp91 Feb 15 '17 at 21:59
I have drown it one day ago. I want to prove that analytically. – joseabp91 Feb 15 '17 at 22:01
@joseabp91 They are the same! Meaning of your definition is: the line $y=\frac{x}{\pi}$ located under the graph of $f$ – Michael Rozenberg Feb 15 '17 at 22:03
1

For all $t\in[0,1]$ the segment $t(0,0)+(1-t)(\pi,1)$ it's segment on the line $y=\frac{x}{\pi}$ and he located under graph $f$. – Michael Rozenberg Feb 15 '17 at 22:11

How can I show that $\left|\sin \frac{s}{2}\right| \geq \frac{|s|}{\pi}, s \in [- \pi , \pi]$?

1 Answers1