## Journal of Inequalities and Applications

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# Sharp Cusa and Becker-Stark inequalities

Journal of Inequalities and Applications20112011:136

DOI: 10.1186/1029-242X-2011-136

Accepted: 7 December 2011

Published: 7 December 2011

## Abstract

We determine the best possible constants θ,ϑ,α and β such that the inequalities

${\left(\frac{2+cosx}{3}\right)}^{\theta }<\frac{sinx}{x}<\phantom{\rule{2.77695pt}{0ex}}{\left(\frac{2+cosx}{3}\right)}^{\vartheta }$

and

${\left(\frac{{\pi }^{2}}{{\pi }^{2}-4{x}^{2}}\right)}^{\alpha }<\frac{tanx}{x}<{\left(\frac{{\pi }^{2}}{{\pi }^{2}-4{x}^{2}}\right)}^{\beta }$

are valid for 0 < × < π/ 2. Our results sharpen inequalities presented by Cusa, Becker and Stark.

Mathematics Subject Classification (2000): 26D05.

### Keywords

Inequalities trigonometric functions

## 1. Introduction

For 0 < × < π/ 2, it is known in the literature that
$\frac{sinx}{x}<\frac{2+cosx}{3}.$
(1)

Inequality (1) was first mentioned by the German philosopher and theologian Nicolaus de Cusa (1401-1464), by a geometrical method. A rigorous proof of inequality (1) was given by Huygens [1], who used (1) to estimate the number π. The inequality is now known as Cusa's inequality [25]. Further interesting historical facts about the inequality (1) can be found in [2].

It is the first aim of present paper to establish sharp Cusa's inequality.

Theorem 1. For 0 < × < π/ 2,
${\left(\frac{2+cosx}{3}\right)}^{\theta }<\frac{sinx}{x}<{\left(\frac{2+cosx}{3}\right)}^{\vartheta }$
(2)
with the best possible constants
$\theta =\frac{ln\left(\pi ∕2\right)}{ln\left(3∕2\right)}=1.11373998\dots \phantom{\rule{1em}{0ex}}\mathsf{\text{and}}\phantom{\rule{1em}{0ex}}\vartheta =1.$
Becker and Stark [6] obtained the inequalities
$\frac{8}{{\pi }^{2}-4{x}^{2}}<\frac{tanx}{x}<\frac{{\pi }^{2}}{{\pi }^{2}-4{x}^{2}}\phantom{\rule{1em}{0ex}}\left(0
(3)

The constant 8 and π2 are the best possible.

Zhu and Hua [7] established a general refinement of the Becker-Stark inequalities by using the power series expansion of the tangent function via Bernoulli numbers and the property of a function involving Riemann's zeta one. Zhu [8] extended the tangent function to Bessel functions.

It is the second aim of present paper to establish sharp Becker-Stark inequality.

Theorem 2. For 0 < × < π/ 2,
${\left(\frac{{\pi }^{2}}{{\pi }^{2}-4{x}^{2}}\right)}^{\alpha }<\frac{tanx}{x}<{\left(\frac{{\pi }^{2}}{{\pi }^{2}-4{x}^{2}}\right)}^{\beta }$
(4)
with the best possible constants
$\alpha =\frac{{\pi }^{2}}{12}=0.822467033\dots \phantom{\rule{1em}{0ex}}\mathsf{\text{and}}\phantom{\rule{1em}{0ex}}\beta =1.$

Remark 1. There is no strict comparison between the two lower bounds$\frac{8}{{\pi }^{2}-4{x}^{2}}$ and

${\left(\frac{{\pi }^{2}}{{\pi }^{2}-4{x}^{2}}\right)}^{{\pi }^{2}∕12}$in (3) and (4).

The following lemma is needed in our present investigation.

Lemma 1 ([911]). Let -< a < b < ∞, and f, g : [a, b] → be continuous on [a, b] and differentiable in (a, b). Suppose g' ≠ 0 on (a; b). If f'(x)/g' (x) is increasing (decreasing) on (a, b), then so are
$\left[f\left(x\right)-f\left(a\right)\right]∕\left[g\left(x\right)-g\left(a\right)\right]\phantom{\rule{1em}{0ex}}\mathsf{\text{and}}\phantom{\rule{1em}{0ex}}\left[f\left(x\right)-f\left(b\right)\right]∕\left[g\left(x\right)-g\left(b\right)\right].$

If f'(x) = g'(x) is strictly monotone, then the monotonicity in the conclusion is also strict.

## 2. Proofs of Theorems 1 and 2

Proof of Theorem [1]. Consider the function f(x) defined by
$\begin{array}{c}F\left(x\right)=\frac{\mathsf{\text{ln}}\left(\frac{sinx}{x}\right)}{\mathsf{\text{ln}}\left(\frac{2+cosx}{3}\right)},\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}0
For 0 < x < π/2, let
${F}_{1}\left(x\right)=ln\left(\frac{sinx}{x}\right)\phantom{\rule{1em}{0ex}}\mathsf{\text{and}}\phantom{\rule{1em}{0ex}}{F}_{2}\left(x\right)=ln\left(\frac{2+cosx}{3}\right).$
Then,
$\frac{{{F}^{\prime }}_{1}\left(x\right)}{{{F}^{\prime }}_{2}\left(x\right)}=\frac{-2xcosx-x{cos}^{2}x+2sinx+sinxcosx}{x{sin}^{2}x}=\frac{{F}_{3}\left(x\right)}{{F}_{4}\left(x\right)},$
where
${F}_{3}\left(x\right)=-2xcosx-x{cos}^{2}x+2sinx+sinxcosx\phantom{\rule{1em}{0ex}}\mathsf{\text{and}}\phantom{\rule{1em}{0ex}}{F}_{4}\left(x\right)=x{sin}^{2}x.$
Differentiating with respect to x yields
$\frac{{{F}^{\prime }}_{3}\left(x\right)}{{{F}^{\prime }}_{4}\left(x\right)}=\frac{2x+2xcosx-sinx}{sinx+2xcosx}\triangleq {F}_{5}\left(x\right).$
Elementary calculations reveal that
${F}_{5}^{\prime }\left(x\right)=\frac{2{F}_{6}\left(x\right)}{2xsin\left(2x\right)+4{x}^{2}{cos}^{2}x+{sin}^{2}x},$
where
${F}_{6}\left(x\right)=sin\left(2x\right)+\left(2{x}^{2}+1\right)sinx-2x-xcosx.$
By using the power series expansions of sine and cosine functions, we find that
${F}_{6}\left(x\right)={x}^{3}-\frac{1}{10}{x}^{5}-\frac{19}{2520}{x}^{7}+2\sum _{n=4}^{\infty }{\left(-1\right)}^{n}{u}_{n}\left(x\right),$
where
${u}_{n}\left(x\right)=\frac{{4}^{n}-4{n}^{2}-3n}{\left(2n+1\right)!}{x}^{2n+1}.$
Elementary calculations reveal that, for 0 < × < π/ 2 and n ≥ 4,
$\begin{array}{cc}\hfill \frac{{u}_{n+1}\left(x\right)}{{u}_{n}\left(x\right)}& =\frac{{x}^{2}}{2}\frac{{2}^{2n+2}-4{n}^{2}-11n-7}{\left(n+1\right)\left(2n+3\right)\left({4}^{n}-4{n}^{2}-3n\right)}\hfill \\ <\frac{1}{2}{\left(\frac{\pi }{2}\right)}^{2}\frac{{2}^{2n+2}-4{n}^{2}-11n-7}{\left(n+1\right)\left(2n+3\right)\left({4}^{n}-4{n}^{2}-3n\right)}\hfill \\ =\frac{{\pi }^{2}}{8\left(n+1\right)}\frac{{4}^{n+1}-4{n}^{2}-11n-7}{\left(2n+3\right)\left({4}^{n}-4{n}^{2}-3n\right)}\hfill \\ <\frac{{\pi }^{2}}{8\left(n+1\right)}<1.\hfill \end{array}$
Hence, for fixed x (0, π/ 2), the sequence n u n (x) is strictly decreasing with regard to n ≥ 4. Hence, for 0 < × < π/2,
${F}_{6}\left(x\right)={x}^{3}-\frac{1}{10}{x}^{5}-\frac{19}{2520}{x}^{7}>0\phantom{\rule{1em}{0ex}}\left(0

and therefore, the functions F5(x) and $\frac{{{F}^{\prime }}_{3}\left(x\right)}{{{F}^{\prime }}_{4}\left(x\right)}$are both strictly increasing on (0, π/2).

By Lemma 1, the function
$\frac{{{F}^{\prime }}_{1}\left(x\right)}{{{F}^{\prime }}_{2}\left(x\right)}=\frac{{F}_{3}\left(x\right)}{{F}_{4}\left(x\right)}=\frac{{F}_{3}\left(x\right)-{F}_{3}\left(0\right)}{{F}_{4}\left(x\right)-{F}_{4}\left(0\right)}$
is strictly increasing on (0, π/2). By Lemma 1, the function
$F\left(x\right)=\frac{{F}_{1}\left(x\right)}{{F}_{2}\left(x\right)}=\frac{{F}_{1}\left(x\right)-{F}_{1}\left(0\right)}{{F}_{2}\left(x\right)-F\left(0\right)}$
is strictly increasing on (0, π/2), and we have
$1=F\left(0\right)

By rearranging terms in the last expression, Theorem 1 follows.

Proof of Theorem 2. Consider the function f(x) defined by
$\begin{array}{ll}\hfill f\left(x\right)& =\frac{ln\left(\frac{tanx}{x}\right)}{ln\left(\frac{{\pi }^{2}}{{\pi }^{2}-4{x}^{2}}\right)},\phantom{\rule{1em}{0ex}}0
For 0 < x < π/2, let
${f}_{1}\left(x\right)=ln\left(\frac{tanx}{x}\right)\phantom{\rule{1em}{0ex}}\mathsf{\text{and}}\phantom{\rule{1em}{0ex}}{f}_{2}\left(x\right)=ln\left(\frac{{\pi }^{2}}{{\pi }^{2}-4{x}^{2}}\right).$
Then,
$\frac{{{f}^{\prime }}_{1}\left(x\right)}{{{f}^{\prime }}_{2}\left(x\right)}=\frac{\left({\pi }^{2}-4{x}^{2}\right)\left(2x-sin\left(2x\right)\right)}{8{x}^{2}sin\left(2x\right)}\triangleq g\left(x\right).$
Elementary calculations reveal that
$4{x}^{3}{sin}^{2}\left(2x\right){g}^{\prime }\left(x\right)=-\left({\pi }^{2}+4{x}^{2}\right)xsin\left(2x\right)-2\left({\pi }^{2}-4{x}^{2}\right){x}^{2}cos\left(2x\right)+{\pi }^{2}{sin}^{2}\left(2x\right)\triangleq h\left(x\right).$
Motivated by the investigations in [12], we are in a position to prove h(x) > 0 for x (0, π/2).Let
$H\left(x\right)=\left\{\begin{array}{cc}\lambda ,\hfill & x=0,\hfill \\ \frac{h\left(x\right)}{{x}^{6}{\left(\frac{\pi }{2}-x\right)}^{2}}\hfill & 0
Where λ and μ are constants determined with limits:
$\begin{array}{cc}\hfill \lambda & =\underset{x\to {0}^{+}}{lim}\frac{h\left(x\right)}{{x}^{6}{\left(\frac{\pi }{2}-x\right)}^{2}}=\frac{224{\pi }^{2}-1920}{45{\pi }^{2}}=0.654740609...,\hfill \\ \hfill \mu & =\underset{t\to {\left(\pi ∕2\right)}^{-}}{lim}\frac{h\left(x\right)}{{x}^{6}{\left(\frac{\pi }{2}-x\right)}^{2}}=\frac{128}{{\pi }^{4}}=1.31404572....\hfill \end{array}$
Using Maple, we determine Taylor approximation for the function H(x) by the polynomial of the first order:
${P}_{1}\left(x\right)=\frac{32\left(7{\pi }^{2}-60\right)}{45{\pi }^{2}}+\frac{128\left(7{\pi }^{2}-60\right)}{45{\pi }^{3}}x,$
which has a bound of absolute error
${\epsilon }_{1}=\frac{-1920-1920{\pi }^{2}+224{\pi }^{4}}{15{\pi }^{4}}=0.650176097...$
for values x [0,π/2]. It is true that
$H\left(x\right)-\left({P}_{1}\left(x\right)-{\mathit{ℰ}}_{1}\right)\ge 0,\phantom{\rule{1em}{0ex}}{P}_{1}\left(x\right)-{\mathit{ℰ}}_{1}=\frac{64\left(60{\pi }^{2}+90-7{\pi }^{4}\right)}{45{\pi }^{4}}+\frac{128\left(7{\pi }^{2}-60\right)}{45{\pi }^{3}}x>0$
for x [0, π/ 2]. Hence, for x [0, π/ 2], it is true that H (x) > 0 and, therefore, h (x) > 0 and g'(x) > 0 for x [0, π/ 2]. Therefore, the function $\frac{{{f}^{\prime }}_{1}\left(x\right)}{{{f}^{\prime }}_{2}\left(x\right)}$is strictly increasing on. (0, π/ 2).By Lemma 1, the function
$f\left(x\right)=\frac{{f}_{1}\left(x\right)}{{f}_{2}\left(x\right)}$
is strictly increasing on (0, π/ 2), and we have
$\frac{{\pi }^{2}}{12}=f\left(0\right)

By rearranging terms in the last expression, Theorem 2 follows.

## Declarations

### Acknowledgements

Research is supported in part by the Research Grants Council of the Hong Kong SAR, Project No. HKU7016/07P.

## Authors’ Affiliations

(1)
School of Mathematics and Informatics, Henan Polytechnic, University
(2)
Department of Mathematics, the University of Hong Kong

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