6
$\begingroup$

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.

$\endgroup$
4
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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
  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ 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
5
$\begingroup$

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.

$\endgroup$
2
$\begingroup$

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).

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.