Prove $\frac2\pi x \le \sin x \le x$ for $x\in [0,\frac {\pi} 2]$.

I tried to do this in two ways, I'm not sure about CMVT and I have a problem with the other way.

Using Cauchy's MVT:

RHS: $\sin x \le x \implies \frac {\sin x}{x}\le 1$ So define: $f(x)=\sin x, \ g(x)=x$ then from CMVT: $\dfrac {f(\frac {\pi} 2)}{g(\frac {\pi} 2)}=\dfrac {f'(c)} {g'(c)}=\cos c$ and from the fact that $c$ is between $0$ and $\pi/2 \implies \cos c \le 1$.

LHS: In the same manner but here I run into some trouble: $\frac2\pi x \le \sin x\implies \frac {2x}{\pi\sin x}\le 1$ So: $\dfrac {f(\frac {\pi} 2)}{g(\frac {\pi} 2)}=\dfrac {f'(c)} {g'(c)}\implies\frac {1}{\sin {\frac {\pi}{2}}}=\frac {2}{\pi \cos c}$ Here actually $\frac {1}{\sin {\frac {\pi}{2}}}=1$ so it's also $\le 1$

Is it correct to use CMVT like this ?

The other way:

We want to show: $f(x)=\sin x - x < 0$ and $g(x)=\frac {2x}{\pi}-sinx <0 $ by deriving both it's easy to show that the inequality stands for $f$ but for $g$ it isn't so obvious that $g'(x)=\frac {2}{\pi}-\cos x$ is negative. In fact for $x=\frac {\pi} 2$ it's positive. Please help figure this out.

This is the same The sine inequality $\frac2\pi x \le \sin x \le x$ for $0<x<\frac\pi2$ but all the answers there are partial or hints and I want to avoid convexity.

Note: I can't use integrals.


To show that $\sin x\le x$ you can apply the Cauchy mean value theorem. (Note that you want to show the inequality for any $x\in\left[0,\frac{\pi}{2}\right].$ )Consider, as you have done, $f(x)=\sin x$ and $g(x)=x.$ Apply the theorem in the interval $[0,x]$ and you will get the inequality, as a consequence of $\cos c\le 1.$ Indeed, there exists $c\in(0,x)$ such that $$\sin x=g'(c)(f(x)-f(0))=f'(c)(g(x)-g(0))=(\cos c)\cdot x\le x.$$

To show the other inequality consider $f(x)=\sin x-\frac{2}{\pi}x.$ We have that $f(0)=f(\pi/2)=0.$ Since $f$ is continuous and $[0,\pi/2]$ is compact it attains a global minimum. If the minimum is not achieved at the extrema of the interval then it belongs to the open interval $(0,\pi/2).$ Let $c$ be the point where $f$ achieves its global minimum. Then $f''(c)\ge 0,$ but $f''(c)=-\sin c<0$ for any $c\in(0,\pi/2).$ So the minimum value is $f(0)=f(\pi/2)=0,$ from where $0\le f(x)=\sin x-\frac{2}{\pi}x,$ which shows the inequality.

  • $\begingroup$ Thank you. Maybe it should be mentioned that we use Roll for the extrema between the two zeros. $\endgroup$ – GinKin Jun 22 '14 at 11:03
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    $\begingroup$ Rolle's theorem states the existence of some point where $f'$ vanishes, but this point doesn't have to be an extrema ($f(x)=x^3(x-1)(x+1),f(-1)=f(1)=0,f'(0)=0$ and $x=0$ is not a local maximum or minimum.) $\endgroup$ – mfl Jun 22 '14 at 11:25
  • $\begingroup$ I see Rolle's doesn't state that there has to be an extrema although I don't think it's possible to not have any extrema at all between two zeros. In your example there are two extrema on the interval. $\endgroup$ – GinKin Jun 22 '14 at 11:34
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    $\begingroup$ If $f(a)=f(b)$ and $a,b$ are not extrema then necessarily there exists $c,d\in (a,b)$ such that $f(c)\le f(x)\le f(d),\forall x\in [a,b]$ as a consequence of Weierstrass theorem. $\endgroup$ – mfl Jun 22 '14 at 11:39

For any $x \in (0,\frac{\pi}{2})$, consider the expression

$$\frac{\sin x - \sin 0}{x - 0} - \frac{\sin\frac{\pi}{2} - \sin x}{\frac{\pi}{2}- x} = \frac{\sin x}{x} - \frac{1-\sin x}{\frac{\pi}{2} - x}\tag{*1} $$ Apply MVT on for the first term on $[0,x]$ and the second term on $[x,\frac{\pi}{2}]$, we can find two numbers $y, z$ such that

$$0 < y < x < z < \frac{\pi}{2}\quad\text{ and }\quad \frac{\sin x}{x} = \cos y \;\land\; \frac{1-\sin x}{\frac{\pi}{2} - x} = \cos z$$ Since $\cos t$ is strictly decreasing on $[0,\frac{\pi}{2}]$, we have $\cos y > \cos z$ and hence $$\begin{align}\frac{\sin x}{x} - \frac{1-\sin x}{\frac{\pi}{2} - x} > 0 &\iff \left(\frac{\pi}{2} - x \right)\sin x - x \left(1 - \sin x\right) > 0\\ &\iff \sin x > \frac{2x}{\pi}\end{align}$$ You may wonder how I arrive the expression in $(*1)$. Geometrically,

  • $\displaystyle\;\frac{\sin x}{x}\;$ is the slope of $\sin x$ over $[0,x]$.
  • $\displaystyle\;\frac{1-\sin x}{\frac{\pi}{2} - x}\;$ is the slope over $[x,\frac{\pi}{2}]$.

This proof works because the slope $\cos x$ is decreasing on $[0,\frac{\pi}{2}]$. This is sort of equivalent to $\sin''(x) = -\sin x < 0$. In certain sense, this is really a proof with convexity hiding under the carpet.


The inequality $\sin x \leq x$ for $x \in [0, \pi/2]$ has many proofs (one of them being geometrical and is used in proof of $\lim_{x \to 0}(\sin x)/x = 1$). Another approach is to show that that $f(x) = x - \sin x$ is increasing in $[0, \pi/2]$ because its derivative $f'(x) = 1 - \cos x \geq 0$. And then we have $f(x) \geq f(0) = 0$ for $x \in [0, \pi/2]$ and we are done. But this is based on derivative of $\sin x$ which is based on $\lim_{x \to 0}(\sin x)/x = 1$ so that the geometrical proof referred above is the more fundamental one.

For the other inequality we need to show that $\dfrac{\sin x}{x} \geq \dfrac{2}{\pi} = \dfrac{\sin (\pi/2)}{\pi/2}$. It thus follows that we need to show that $g(x) = (\sin x)/x, g(0) = 1$ is decreasing in $[0, \pi/2]$. This is easy again because $g'(x) = \dfrac{x\cos x - \sin x}{x^{2}} \leq 0$ if $x \leq \tan x$. Again the proof of $x \leq \tan x$ is geometrical (used in proof of proof of $\lim_{x \to 0}(\sin x)/x = 1$).

The geometrical proof is available at MSE as well as many other places online (including my post).

  • $\begingroup$ I can't imagine I would be able to use geometrical proofs for a test. Are there other ways to show that $x \leq \tan x$ ? $\endgroup$ – GinKin Jun 22 '14 at 10:52
  • $\begingroup$ It can be proved with in the same manner as mfl's answer. $\endgroup$ – GinKin Jun 22 '14 at 11:09
  • $\begingroup$ @GinKin: I really feel that a proof should be valid in order to be used in exam/test. In fact proofs based on derivatives can be used to prove $x\leq \tan x$, but for reasons mentioned in answer it is circular and hence invalid. $\endgroup$ – Paramanand Singh Jun 22 '14 at 13:13

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