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How do I go about showing $0 \leq \frac{2x}{\pi} \leq \sin x $ for $x\in[0,\pi/2]$?

I am completely stuck where to start.

Many thanks.

(I see it is a step in the proof of Jordan's lemma, but I'm not interested in this, and the proofs I find do not explain this actual step, ta).

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    what about $x$? – 2017-01-21
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    This is not true. For example let $x=100$ – 2017-01-21
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    The [Jordan's Lemma](https://en.wikipedia.org/wiki/Jordan's_lemma) says it is true for $x\in[0,\pi,2]$, but OP should have really included it in the question (so I edited the question and added it myself) – 2017-01-21

2 Answers 2

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I assume your inequality is for $x\in[0,\pi/2]$, as shown in Jordan's Lemma.

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

$f(0)=0$, and $f(\pi/2)=0$

$f'(x)=\cos(x)-2/\pi$, which is decreasing from $1-2/\pi>0$ to $-2/\pi<0$

So $f$ is first increasing from $0$, then decreasing to $0$, so $f\ge 0$ on $x\in[0,\pi/2]$

EDIT: more simple, you know that $\sin$ is concave on $[0,\pi/2]$, so its graph on $[0,\pi/2]$ is above the line going through its ends, which are $(0,0)$, and $(\pi/2,1)$

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The derivative of $f(x)=2x-\pi\sin x$, defined over $[0,\pi/2]$, is $f'(x)=2-\pi\cos x$ and the second derivative is $f''(x)=\pi\sin x\ge0$ (vanishing only at $0$), so $f$ is convex over $[0,\pi/2]$.

Since $f(0)=f(\pi/2)=0$, we are done: $f(x)\le0$, for every $x\in[0,\pi/2]$