New Inequalities and Approximations of Cusa–Huygens Type

Abstract

In this paper, we introduce a real parameter in some Cusa–Huygens-type inequalities. Using the concept of stratification, we obtain new inequalities and approximations.

1. Introduction

The well known Cusa’s inequality is

$$\frac{3sinx}{2+cosx}<x,$$

for $x\in \left(0,\pi /2\right)$, [1,2,3]. Huygens gave the first proof of this inequality [4]. Cusa’s inequality is used for obtaining the approximation

$$\frac{3sinx}{2+cosx}\approx x,$$

for $x\in \left(0,\pi /2\right]$. The previous approximation is called Cusa’s approximation [5,6], see also [7,8,9,10].

Extensions and generalisations of the Cusa–Huygens inequality have been considered in many papers [5,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,30,31,32,33,34,35,36,37,38,39,40,41,42,43,44,45,46,47,48,49,50]. In [11], Zhu obtained and proved the following theorems:

Theorem 1.

Let $0<x<\pi /2$. Then,

$$\frac{sinx}{x}-\frac{2+cosx}{3}<-\frac{1}{180}{x}^4{\left(\frac{sinx}{x}\right)}^{2/7}$$

holds with the best constant $-1/180$.

Theorem 2.

Let $0<x<\pi /2$. Then,

$$\frac{sinx}{x}-\frac{2+cosx}{3}<-\left(\frac{2}{3}-\frac{2}{\pi }\right){(sinx-xcosx)}^{({\pi }^2-12)/(3{\pi }^2-{\pi }^3)}$$

holds with the best constant $(2/3-2/\pi )$.

In this paper, we improve the previous two theorems using the concept of stratification [6,51,52,53,54,55,56,57].

In Theorem 5, we give the lower bound for the function $\frac{sinx}{x}-\frac{2+cosx}{3}$ from Theorem 1, as well as a new proof that the constant $-1/180$ is the best possible for the inequality from Theorem 1.

In Section 3.1, we provide a new proof that the constant $(2/3-2/\pi )$ is the best possible for the inequality from Theorem 2.

In Theorems 6 and 11, we give some generalisations of Theorems 1 and 2, respectively.

Additionally, in Theorems 10 and 16 and their corollaries, we give some new approximations of the function $\frac{sinx}{x}-\frac{2+cosx}{3}.$

2. Preliminaries

2.1. Stratification

Let ${\left\{{\phi }_p\left(x\right)\right\}}_{p\in \mathbb{P}}$ be a family of functions that we consider for $x\in \mathbb{S}\subseteq \mathbb{R}$ and $p\in \mathbb{P}\subseteq \mathbb{R}$. The family of functions ${\left\{{\phi }_p\left(x\right)\right\}}_{p\in \mathbb{P}}$ is increasingly stratified on the set $\mathbb{S}$ with respect to the parameter $p\in \mathbb{P}$ if

$$(\forall x\in \mathbb{S})(\forall p_1,p_2\in \mathbb{P})p_1<p_2⟺{\phi }_{p_1}\left(x\right)<{\phi }_{p_2}\left(x\right),$$

i.e., decreasingly stratified on the set $\mathbb{S}$ with respect to the parameter $p\in \mathbb{P}$ if

$$(\forall x\in \mathbb{S})(\forall p_1,p_2\in \mathbb{P})p_1<p_2⟺{\phi }_{p_1}\left(x\right)>{\phi }_{p_2}\left(x\right),$$

see [6,51,52,53,54,55,56,57]. According to [55], for the stratified family of functions ${\left\{{\phi }_p\left(x\right)\right\}}_{p\in \mathbb{P}}$, it is significant to determine, if it exists, a function ${\phi }_{p_0}\left(x\right)$ for which holds

$$\underset{p\in \mathbb{P}}{\inf }\underset{x\in \mathbb{S}}{\sup }\left|{\phi }_p\left(x\right)\right|=\underset{x\in \mathbb{S}}{\sup }\left|{\phi }_{p_0}\left(x\right)\right|.$$

The function ${\phi }_{p_0}\left(x\right)$ is called the minimax approximant of the family ${\left\{{\phi }_p\left(x\right)\right\}}_{p\in \mathbb{P}}$.

The approximation ${\phi }_{p_0}\left(x\right)\approx 0$ is called the minimax approximation with the error $d_0=\underset{x\in \mathbb{S}}{\sup }\left|{\phi }_{p_0}\left(x\right)\right|$.

2.2. A Parametric Method

In this paper, to determine the sign of the functions ${\phi }_p\left(x\right)$, we use the parametric method for proving some analytic inequalities [52] (for applications of the method see also [6,51,56,57]).

The parametric method from [52] is based on the analysis of the function $g:\mathbb{S}⟶\mathbb{P}$, which is defined by

$$g\left(x\right)=p⟺{\phi }_p\left(x\right)=0,$$

(1)

if such a function exists. In [52], based on the monotonicity of the function g, the following theorems are proved.

Theorem 3.

Let ${\left\{{\phi }_p\left(x\right)\right\}}_{p\in \mathbb{P}}$ be a family of functions for $x\in \mathbb{S}=(a,b)\subseteq \mathbb{R}$ and let $\mathbb{P}\subseteq \mathbb{R}$$(\mathbb{P}\ne \varnothing )$, satisfying the following conditions:

1.

the family of functions ${\left\{{\phi }_p\left(x\right)\right\}}_{p\in \mathbb{P}}$ is increasingly (decreasingly) stratified on the interval $(a,b)$;

2.

there exists a continuous monotonically increasing function $g:(a,b)⟶\mathbb{P}$ that satisfies $\left(\right)$;

3.

there exist limits $\underset{x\to a}{lim}g\left(x\right)=A$ and $\underset{x\to b}{lim}g\left(x\right)=B$ in $\overline{\mathbb{R}}$ such that $(A,B)\subseteq \mathbb{P}$.

Then, it holds:

(i)

If $p\le A$, then

$$(\forall x\in (a,b)){\phi }_p\left(x\right)\le {\phi }_A\left(x\right)<0\left((\forall x\in (a,b)){\phi }_p\left(x\right)\ge {\phi }_A\left(x\right)>0\right).$$

(ii)

If $p\in (A,B)$, then the equality ${\phi }_p\left(x\right)=0$ has a unique solution $x_0^{\left(p\right)}\in (a,b)$ and it holds that

$$\left(\forall x\in \left(a,x_0^{\left(p\right)}\right)\right){\phi }_p\left(x\right)>0\left(\left(\forall x\in \left(a,x_0^{\left(p\right)}\right)\right){\phi }_p\left(x\right)<0\right)$$

and

$$\left(\forall x\in \left(x_0^{\left(p\right)},b\right)\right){\phi }_p\left(x\right)<0\left(\left(\forall x\in \left(x_0^{\left(p\right)},b\right)\right){\phi }_p\left(x\right)>0\right).$$

(iii)

If $p\ge B$, then

$$(\forall x\in (a,b)){\phi }_p\left(x\right)\ge {\phi }_B\left(x\right)>0\left((\forall x\in (a,b)){\phi }_p\left(x\right)\le {\phi }_B\left(x\right)<0\right).$$

Theorem 4.

Let ${\left\{{\phi }_p\left(x\right)\right\}}_{p\in \mathbb{P}}$ be a family of functions for $x\in \mathbb{S}=(a,b)\subseteq \mathbb{R}$ and let $\mathbb{P}\subseteq \mathbb{R}$$(\mathbb{P}\ne \varnothing )$, satisfying the following conditions:

1.

the family of functions ${\left\{{\phi }_p\left(x\right)\right\}}_{p\in \mathbb{P}}$ is increasingly (decreasingly) stratified on the interval $(a,b)$;

2.

there exists a continuous monotonically decreasing function $g:(a,b)⟶\mathbb{P}$ that satisfies (1);

3.

there exist limits $\underset{x\to b}{lim}g\left(x\right)=A$ and $\underset{x\to a}{lim}g\left(x\right)=B$ in $\overline{\mathbb{R}}$ such that $(A,B)\subseteq \mathbb{P}$.

Then, it holds:

(i)

If $p\le A$, then

$$(\forall x\in (a,b)){\phi }_p\left(x\right)\le {\phi }_A\left(x\right)<0\left((\forall x\in (a,b)){\phi }_p\left(x\right)\ge {\phi }_A\left(x\right)>0\right).$$

(ii)

If $p\in (A,B)$, then the equality ${\phi }_p\left(x\right)=0$ has a unique solution $x_0^{\left(p\right)}\in (a,b)$ and it holds that

$$\left(\forall x\in \left(a,x_0^{\left(p\right)}\right)\right){\phi }_p\left(x\right)<0\left(\left(\forall x\in \left(a,x_0^{\left(p\right)}\right)\right){\phi }_p\left(x\right)>0\right)$$

and

$$\left(\forall x\in \left(x_0^{\left(p\right)},b\right)\right){\phi }_p\left(x\right)>0\left(\left(\forall x\in \left(x_0^{\left(p\right)},b\right)\right){\phi }_p\left(x\right)<0\right).$$

(iii)

If $p\ge B$, then

$$(\forall x\in (a,b)){\phi }_p\left(x\right)\ge {\phi }_B\left(x\right)>0\left((\forall x\in (a,b)){\phi }_p\left(x\right)\le {\phi }_B\left(x\right)<0\right).$$

Let us note that in [52], the case when a function g has exactly one local minimum on the observed interval $\mathbb{S}=(a,b)\subseteq \mathbb{R}$ was also considered.

2.3. A Method for Proving Mixed Trigonometric Polynomial Inequalities

We use the method for proving mixed trigonometric polynomial (MTP) inequalities from [51,58,59]. A function

$$f\left(x\right)=\sum _{i=1}^n{\alpha }_i{x}^{p_i}{cos}^{q_i}x{sin}^{r_i}x,$$

where ${\alpha }_i\in \mathbb{R}\setminus \left\{0\right\},p_i,q_i,r_i\in {\mathbb{N}}_0$ and $n\in \mathbb{N}$, for $x\in \mathbb{R}$, we call an MTP function. An inequality $f\left(x\right)>0$ for $x\in \mathbb{S}\subseteq \mathbb{R}$, we call an MTP inequality.

The method for proving MTP inequalities from [51,58,59] is based on the following:

(1)

Transformation of a given MTP function $f\left(x\right)$ in the following form:

$$f\left(x\right)=\sum _{i=1}^m{\beta }_i{x}^{s_i}{\mathrm{t}\mathrm{r}\mathrm{i}\mathrm{g}}_i\left(k_{i}x\right),$$

where ${\beta }_i\in \mathbb{R}\setminus \left\{0\right\},s_i,k_i\in {\mathbb{N}}_0$, $m\in \mathbb{N}$, ${\mathrm{t}\mathrm{r}\mathrm{i}\mathrm{g}}_i=cos$ or ${\mathrm{t}\mathrm{r}\mathrm{i}\mathrm{g}}_i=sin$, see Table 1 [51].

(2)

Approximation of each function ${\mathrm{t}\mathrm{r}\mathrm{i}\mathrm{g}}_i\left(k_{i}x\right)$ by a Taylor polynomial using the following estimates:

$$(\ast )\left\{\begin{array}{cc}cos\left(k_{i}x\right)>T_{4{\ell }_i+2}^{cos,0}\left(k_{i}x\right),& \mathrm{if}{\beta }_i>0,\\ cos\left(k_{i}x\right)<T_{4{\ell }_i}^{cos,0}\left(k_{i}x\right),& \mathrm{if}{\beta }_i<0,\\ sin\left(k_{i}x\right)>T_{4{\ell }_i+3}^{sin,0}\left(k_{i}x\right),& \mathrm{if}{\beta }_i>0,\\ sin\left(k_{i}x\right)<T_{4{\ell }_i+1}^{sin,0}\left(k_{i}x\right),& \mathrm{if}{\beta }_i<0\end{array}\right.$$

where $T_n^{\varphi ,a}$ denote the Taylor expansion of order n of a function $\varphi$ in a neighbourhood of the point a (${\ell }_i\in {\mathbb{N}}_0$), see Lemmas 1.1 and 1.2 from [58].

By applying $(\ast )$, we determine a polynomial

$$P\left(x\right)=\sum _{i=1}^m{\beta }_i{x}^{s_i}{T}_{{\vartheta }_i}^{{\mathrm{t}\mathrm{r}\mathrm{i}\mathrm{g}}_i,a}\left(k_{i}x\right),$$

(2)

where

$${\vartheta }_i=\left\{\begin{array}{cccc}4{\ell }_i+2,& \mathrm{i}\mathrm{f}{\mathrm{t}\mathrm{r}\mathrm{i}\mathrm{g}}_i=cos\hfill & \mathrm{a}\mathrm{n}\mathrm{d}& {\beta }_i>0,\hfill \\ 4{\ell }_i,& \mathrm{i}\mathrm{f}{\mathrm{t}\mathrm{r}\mathrm{i}\mathrm{g}}_i=cos\hfill & \mathrm{a}\mathrm{n}\mathrm{d}& {\beta }_i<0,\hfill \\ 4{\ell }_i+3,& \mathrm{i}\mathrm{f}{\mathrm{t}\mathrm{r}\mathrm{i}\mathrm{g}}_i=sin\hfill & \mathrm{a}\mathrm{n}\mathrm{d}& {\beta }_i>0,\hfill \\ 4{\ell }_i+1,& \mathrm{i}\mathrm{f}{\mathrm{t}\mathrm{r}\mathrm{i}\mathrm{g}}_i=sin\hfill & \mathrm{a}\mathrm{n}\mathrm{d}& {\beta }_i<0\hfill \end{array}\right.$$

for which it holds that $P\left(x\right)$ is downward polynomial approximation of the function $f\left(x\right)$ in the sense

$$f\left(x\right)>P\left(x\right)$$

for $x\in \mathbb{S}$ and $x>0$.

(3)

Searching for the values ${\ell }_1,{\ell }_2,\dots ,{\ell }_m$ (${\ell }_i\in {\mathbb{N}}_0$), if there exist, such that $P\left(x\right)>0$ for $x\in \mathbb{S}$. If there exist values ${\ell }_1,{\ell }_2,\dots ,{\ell }_m$ such that $P\left(x\right)>0$ for $x\in \mathbb{S}$, then we have proof that

$$f\left(x\right)>0$$

for $x\in \mathbb{S}$ and $x>0$.

For determining the sign of the polynomial $P\left(x\right)$, we use Sturm’s theorem ([60]; Section 6.4 [61]; [62]), see also [63,64,65,66,67,68].

In this paper, an application of the previously described method for proving MTP inequalities is given in Lemma 2. The other MTP inequalities could be proved analogously.

3. Main Results

3.1. New Results Related to Theorem 1

Based on Theorem 1, we form the family of functions

$${\left\{\phi \left(x\right)\right\}}_{p\in \mathbb{P}},$$

for $\mathbb{P}=\mathbb{R}$, defined on $[0,\pi /2]$ with

$${\phi }_p\left(x\right)=\left\{\begin{array}{ccc}-\frac{sinx}{x}+\frac{2+cosx}{3}-p{x}^4{\left(\frac{sinx}{x}\right)}^{2/7}& ,& x\in (0,\pi /2],\\ 0& ,& x=0.\end{array}\right.$$

Note that Theorem 1 claims that

$${\phi }_A\left(x\right)>0$$

for $A=1/180$ and $x\in (0,\pi /2)$.

Lemma 1.

The family of functions ${\left\{{\phi }_p\left(x\right)\right\}}_{p\in \mathbb{R}}$ is decreasingly stratified on the interval $(0,\pi /2)$.

Proof. 

It holds that $\frac{d{\phi }_p\left(x\right)}{dp}=-x^4{\left(\frac{sinx}{x}\right)}^{2/7}<0$ for $x\in (0,\pi /2)$. □

Let us note that for $x\in (0,\pi /2)$, it holds that

$$\begin{array}{cc}& {\phi }_p\left(x\right)=-\frac{sinx}{x}+\frac{2+cosx}{3}-p{x}^4{\left(\frac{sinx}{x}\right)}^{2/7}=0\hfill \\ \hfill ⟺& p=g\left(x\right)=-\frac{1}{3}\frac{-xcosx+3sinx-2x}{x^5{\left(\frac{sinx}{x}\right)}^{2/7}}.\hfill \end{array}$$

Lemma 2.

The function $g\left(x\right)$ is increasing on the interval $(0,\pi /2)$.

Proof. 

The function $g\left(x\right)$ has the first derivative

$$g^{\prime }\left(x\right)=\frac{f\left(x\right)}{21x^{6}sinx{\left(\frac{sinx}{x}\right)}^{2/7}},$$

where

$$f\left(x\right)=5x^2{cos}^{2}x-41xcosxsinx-4x^{2}cosx-52xsinx-99{cos}^{2}x-7x^2+99$$

is a mixed trigonometric polynomial (MTP) function. Let us examine the sign of the function $f\left(x\right)$ by applying the method for proving MTP inequalities from [58,59].

It holds that

$$f\left(x\right)=-4x^{2}cosx+\frac{5x^2}{2}cos2x-\frac{9x^2}{2}-52xsinx-\frac{41x}{2}sin2x-\frac{99}{2}cos2x+\frac{99}{2}.$$

One downward polynomial approximation of the function $f\left(x\right)$ on the interval $\left(0,\pi /2\right)$ is given by

$$\begin{array}{ccc}\hfill P\left(x\right)& =& -4x^2{T}_{4{\ell }_1}^{cos,0}\left(x\right)+\frac{5}{2}{x}^2{T}_{4{\ell }_2+2}^{cos,0}\left(2x\right)-\frac{9}{2}{x}^2-52x{T}_{4{\ell }_3+1}^{sin,0}\left(x\right)\hfill \\ \hfill & & -\frac{41}{2}x{T}_{4{\ell }_4+1}^{sin,0}\left(2x\right)-\frac{99}{2}{T}_{4{\ell }_5}^{cos,0}\left(2x\right)+\frac{99}{2},\hfill \end{array}$$

where ${\ell }_1,{\ell }_2,\dots ,{\ell }_5\in {\mathbb{N}}_0$.

For example, for $({\ell }_1,{\ell }_2,{\ell }_3,{\ell }_4,{\ell }_5)=(3,2,3,3,3)$ (it is possible to find some other values ${\ell }_1,{\ell }_2,{\ell }_3,{\ell }_4,{\ell }_5$ such that $P\left(x\right)>0$ on the interval $(0,\pi /2)$), it can be proved that

$$\begin{array}{ccc}\hfill P\left(x\right)& =& -4x^2{T}_{12}^{cos,0}\left(x\right)+\frac{5}{2}{x}^2{T}_{10}^{cos,0}\left(2x\right)-\frac{9}{2}{x}^2-52x{T}_{13}^{sin,0}\left(x\right)\hfill \\ \hfill & & -\frac{41}{2}x{T}_{13}^{sin,0}\left(2x\right)-\frac{99}{2}{T}_{12}^{cos,0}\left(2x\right)+\frac{99}{2}\hfill \\ \hfill & =& -\frac{4201}{155675520}{x}^{14}-\frac{31}{415800}{x}^{12}+\frac{17}{25200}{x}^{10}=\phantom{P\left(x\right)}\hfill \\ \hfill & =& x^{10}\left(-\frac{4201}{155675520}{x}^4-\frac{31}{415800}{x}^2+\frac{17}{25200}\right)>0\hfill \end{array}$$

on the interval $(0,\pi /2)$: by applying Sturm’s theorem to the polynomial

$$-\frac{4201}{155675520}{x}^4-\frac{31}{415800}{x}^2+\frac{17}{25200}$$

we conclude that this polynomial does not have zeros on the segment $\left[0,\pi /2\right]$, i.e., the polynomial $P\left(x\right)$ does not have zeros and is positive on the interval $\left(0,\pi /2\right)$. Hence, it holds that

$$f\left(x\right)>0$$

on the interval $\left(0,\pi /2\right)$. Thus,

$$g^{\prime }\left(x\right)>0$$

on the interval $\left(0,\pi /2\right)$. □

Notice that

$$\underset{x\to 0+}{lim}g\left(x\right)=A=\frac{1}{180}=0.00555\dots .$$

Let

$$B=\underset{x\to \pi /2-}{lim}g\left(x\right)=\frac{16\left(\pi -3\right)2^{5/7}}{3{\pi }^{33/7}}=0.00561\dots .$$

With the new constant B, we obtain one improvement of Theorem 1:

Theorem 5.

Let $0<x<\pi /2$. Then it holds

$$-B{x}^4{\left(\frac{sinx}{x}\right)}^{2/7}<\frac{sinx}{x}-\frac{2+cosx}{3}<-A{x}^4{\left(\frac{sinx}{x}\right)}^{2/7}$$

with the best constants $A=\frac{1}{180}$ and $B=\frac{16\left(\pi -3\right)2^{5/7}}{3{\pi }^{33/7}}$.

Based on Lemmas 1 and 2, the first and the second condition of Theorem 3 are satisfied. Since there exist limits $\underset{x\to 0}{lim}g\left(x\right)=A$ and $\underset{x\to \pi /2}{lim}g\left(x\right)=B$, the third condition of Theorem 3 is also satisfied. Based on this, we obtain the following generalisation of Theorem 1.

Theorem 6.

(i) If $p\in (-\infty ,A]$, then

$$x\in (0,\pi /2)⟹\frac{sinx}{x}-\frac{2+cosx}{3}<-A{x}^4{\left(\frac{sinx}{x}\right)}^{2/7}\le -p{x}^4{\left(\frac{sinx}{x}\right)}^{2/7}.$$

(ii) If $p\in (A,B)$, then the equation $g\left(x\right)=p$ has a unique solution $x_0^{\left(p\right)}\in (0,\pi /2)$ and it holds that

$$x\in (0,x_0^{\left(p\right)})⟹-p{x}^4{\left(\frac{sinx}{x}\right)}^{2/7}<\frac{sinx}{x}-\frac{2+cosx}{3}$$

and

$$x\in (x_0^{\left(p\right)},\pi /2)⟹\frac{sinx}{x}-\frac{2+cosx}{3}<-p{x}^4{\left(\frac{sinx}{x}\right)}^{2/7}.$$

(iii) If $p\in [B,+\infty )$, then

$$x\in (0,\pi /2)⟹-p{x}^4{\left(\frac{sinx}{x}\right)}^{2/7}\le -B{x}^4{\left(\frac{sinx}{x}\right)}^{2/7}<\frac{sinx}{x}-\frac{2+cosx}{3}.$$

In the following, we will show that the family ${\left\{{\phi }_p\left(x\right)\right\}}_{p\in \mathbb{R}}$ has a unique minimax approximant. For that purpose, we consider the family ${\left\{\frac{d{\phi }_p\left(x\right)}{dx}\right\}}_{p\in \mathbb{R}}$, for $0<x<\pi /2$, which is defined by

$$\frac{d{\phi }_p\left(x\right)}{dx}=-\frac{2}{7}\frac{\left(\frac{7}{6}{x}^{2}sinx+\frac{7}{2}xcosx-\frac{7}{2}sinx\right){\left(\frac{sinx}{x}\right)}^{5/7}+p\left(xcosx+13sinx\right)}{x^2{\left(\frac{sinx}{x}\right)}^{5/7}}.$$

Let us notice that for $x\in (0,\pi /2)$, it holds that

$$\frac{d{\phi }_p\left(x\right)}{dx}=0⟺p=g_1\left(x\right)=-\frac{7}{6}\frac{{\left(\frac{sinx}{x}\right)}^{5/7}\left(x^{2}sinx+3xcosx-3sinx\right)}{x^4\left(xcosx+13sinx\right)}.$$

Lemma 3.

The function $g_1\left(x\right)$ is increasing on the interval $(0,\pi /2)$.

Proof. 

The function $g_1\left(x\right)$ has the first derivative

$$g_1^{\prime }\left(x\right)=\frac{f_1\left(x\right)}{6x^6{\left(\frac{sinx}{x}\right)}^{2/7}\left(26xcosxsinx+\left(x^2-169\right){cos}^{2}x+169\right)}$$

where

$$\begin{array}{ccc}\hfill f_1\left(x\right)& =& \left(24x^3-1089x\right){cos}^{3}x+\left(-5x^4-328x^2+1287\right)sinx{cos}^{2}x+\hfill \\ \hfill & & +\left(-39x^3+1089x\right)cosx+\left(-7x^4+541x^2-1287\right)sinx\hfill \end{array}$$

is an MTP function. By applying the method for proving MTP inequalities analogously as in Lemma 2, it can be proved that

$$f_1\left(x\right)>0$$

on the interval $\left(0,\pi /2\right)$. Thus,

$$g_1^{\prime }\left(x\right)>0$$

on the interval $\left(0,\pi /2\right)$. □

Let us notice that

$$\underset{x\to 0+}{lim}{g}_1\left(x\right)=\frac{1}{180}=A.$$

Let

$$C=\underset{x\to \pi /2-}{lim}{g}_1\left(x\right)=\frac{14·2^{5/7}\left(12-{\pi }^2\right)}{39{\pi }^{33/7}}=0.00568\dots .$$

Theorem 7.

For $p\in (A,C)$, the function ${\phi }_p\left(x\right)$ from the family ${\left\{{\phi }_p\left(x\right)\right\}}_{p\in \mathbb{R}}$ has exactly one minimum at the point $x_p=g_1^{-1}\left(p\right)\in (0,\pi /2)$.

Proof. 

Based on Taylor’s expansion

$$\frac{d{\phi }_p\left(x\right)}{dx}=\left(-\frac{1}{45}+4p\right)x^3+o\left(x^3\right),$$

there exists a right neighbourhood ${\mathcal{U}}_0$ of the point 0 such that

$$p\in (A,C)\wedge x\in {\mathcal{U}}_0⟹\frac{d{\phi }_p\left(x\right)}{dx}<0.$$

Based on Taylor’s expansion

$$\frac{d{\phi }_p\left(x\right)}{dx}=\frac{4}{7}\frac{2^{2/7}}{{\pi }^{2-5/7}}\left(\frac{7{\pi }^2-84}{24}{\left(\frac{2}{\pi }\right)}^{5/7}+\frac{13{\pi }^4}{16}p\right)+o\left(x-\frac{\pi }{2}\right),$$

there exists a left neighbourhood ${\mathcal{V}}_{\pi /2}$ of the point $\pi /2$ such that

$$p\in (A,C)\wedge x\in {\mathcal{V}}_{\pi /2}⟹\frac{d{\phi }_p\left(x\right)}{dx}>0.$$

Since the function ${\phi }_p\left(x\right)$, for $p\in (A,C)$, is decreasing in a right neighbourhood of point 0, increasing in a left neighbourhood of point $\pi /2$ and, based on the monotonicity of function $g_1\left(x\right)$, has exactly one stationary point on the interval $(0,\pi /2)$, it holds that this function has exactly one minimum on interval $(0,\pi /2)$ and that minimum is attained at the point $x_p=g_1^{-1}\left(p\right)\in (0,\pi /2)$. □

Figure 1 illustrates the family ${\left\{\phi \left(x\right)\right\}}_{p\in \mathbb{R}}$ and corresponding functions g and $g_1$.

According to the previous theorem, for functions from the observed family and for values of parameter $p\in [A,C]$, it holds that

$$\underset{x\in [0,\pi /2]}{max}\left|{\phi }_p\left(x\right)\right|=|{\phi }_p\left(x_p\right)|\mathrm{o}\mathrm{r}\underset{x\in [0,\pi /2]}{max}\left|{\phi }_p\left(x\right)\right|=\left|{\phi }_p(\pi /2)\right|,$$

where $x_p$ is the point at which a local minimum of a function ${\phi }_p\left(x\right)$ is attained.

For $p\in \mathbb{R}\setminus [A,C]$, it holds that

$$\underset{x\in [0,\pi /2]}{max}\left|{\phi }_p\left(x\right)\right|=\left|{\phi }_p(\pi /2)\right|.$$

Let us notice that based on the stratification, it holds that

$$p\in (-\infty ,B)⟹\underset{x\in [0,\pi /2]}{max}\left|{\phi }_p\left(x\right)\right|\ge \underset{x\in [0,\pi /2]}{max}\left|{\phi }_B\left(x\right)\right|$$

and

$$p\in (A,+\infty )⟹\underset{x\in [0,\pi /2]}{max}\left|{\phi }_p\left(x\right)\right|\ge \underset{x\in [0,\pi /2]}{max}\left|{\phi }_A\left(x\right)\right|.$$

Therefore, it holds that

$$\underset{p\in \mathbb{R}}{\inf }\underset{x\in [0,\pi /2]}{\sup }\left|{\phi }_p\left(x\right)\right|=\underset{p\in [A,B]}{min}\underset{x\in [0,\pi /2]}{max}\left|{\phi }_p\left(x\right)\right|.$$

In order to determine the minimax approximant, in the following text we will consider two functions.

For $p\in [A,C]$, it holds that ${\phi }_p\left(x_p\right)={\phi }_{g_1\left(x_p\right)}\left(x_p\right)$ and therefore, based on Theorem 7, the function

$$h\left(x\right)={\phi }_{g_1\left(x\right)}\left(x\right):(0,\pi /2)⟶\mathbb{R}$$

determines the set of points in which the functions from the observed family have minimum. The following assertions holds.

Theorem 8.

The function

$$h\left(x\right)=-\frac{sinx}{x}+\frac{2}{3}+\frac{1}{3}cosx+\frac{7}{6}\frac{x^2{sin}^{2}x+3xsinxcosx-3{sin}^{2}x}{x^{2}cosx+13xsinx},$$

is monotonically decreasing for $x\in (0,\pi /2)$.

Proof. 

By applying the method for proving MTP inequalities analogously as in Lemma 2, it can be proved that

$$h^{\prime }\left(x\right)<0$$

on the interval $\left(0,\pi /2\right)$. □

Let

$$\begin{array}{c}\alpha =g_1^{-1}(A+)=0,\hfill \\ \beta =g_1^{-1}\left(B\right)=1.321...,\hfill \\ \gamma =g_1^{-1}(C-)=\pi /2.\hfill \end{array}$$

Note that the functions ${\phi }_A\left(x\right)$, ${\phi }_B\left(x\right)$ and ${\phi }_C\left(x\right)$ have minimum at the points $\alpha ,\beta$ and $\gamma$, respectively.

There exist values

$$\left.\begin{array}{c}h(\alpha +)=0,\hfill \\ h\left(\beta \right)=-0.0000879\dots ,\hfill \\ h(\gamma -)=\frac{7{\pi }^2+104\pi -396}{-156\pi }=-0.000318\dots .\hfill \end{array}\right\}$$

(3)

For $\overline{p}\in [A,C]$, there exist $\overline{x}\in (0,\pi /2)$ such that the function ${\phi }_{\overline{p}}\left(x\right)$ has minimum at the point $\overline{x}$ and, therefore, it holds that $\overline{p}=g_1\left(\overline{x}\right)$. It is evident that ${\phi }_{\overline{p}}(\pi /2)={\phi }_{g_1\left(\overline{x}\right)}(\pi /2)$. For the function

$$H\left(x\right)={\phi }_{g_1\left(x\right)}(\pi /2):(0,\pi /2)⟶\mathbb{R},$$

the following assertion holds.

Theorem 9.

The function

$$H\left(x\right)=-\frac{2}{\pi }+\frac{2}{3}+\frac{7{\pi }^4}{96}·\frac{x^{2}sinx+21xcosx-21sinx}{x^4\left(xcosx+13sinx\right)}{\left(\frac{2}{\pi }\right)}^{\frac{2}{7}}{\left(\frac{sinx}{x}\right)}^{\frac{5}{7}},$$

is monotonically decreasing for $x\in (0,\pi /2)$.

Proof. 

The function $H\left(x\right)$ has the first derivative

$$H^{\prime }\left(x\right)=\frac{2^{2/7}{\pi }^{26/7}}{96}·\frac{f\left(x\right)}{x^6{\left(\frac{sinx}{x}\right)}^{2/7}\left(169+\left(x^2-169\right){cos}^{2}x+26xcosxsinx\right)}$$

where

$$\begin{array}{ccc}\hfill f\left(x\right)& =& \left(-24x^3+1089x\right){cos}^{3}x+\left(5x^4+328x^2-1287\right)sinx{cos}^{2}x+\hfill \\ \hfill & & +\left(39x^3-1089x\right)cosx+\left(7x^4-541x^2+1287\right)sinx\hfill \end{array}$$

is an MTP function. By applying the method for proving MTP inequalities analogously as in Lemma 2, it can be proved that

$$f\left(x\right)<0$$

on the interval $\left(0,\pi /2\right)$. Thus,

$$H^{\prime }\left(x\right)<0$$

on the interval $\left(0,\pi /2\right)$. □

There exist values

$$\left.\begin{array}{c}H(\alpha +)=\frac{-{\pi }^{\frac{33}{7}}{2}^{\frac{2}{7}}+1920\pi -5760}{2880\pi }=0.000318\dots ,\hfill \\ H\left(\beta \right)=0,\hfill \\ H(\gamma -)=\frac{7{\pi }^2+104\pi -396}{-156\pi }=-0.000381\dots .\hfill \end{array}\right\}$$

(4)

Figure 2 illustrates the graphs of the functions $\left|h\right(x\left)\right|$ and $H\left(x\right)$.

The following assertion holds.

Theorem 10.

The equation

$$\left|h\right(x\left)\right|=H\left(x\right)$$

has a unique solution $x_0\in (0,\pi /2)$, which can be numerically determined with $x_0=1.26105\dots$.

For the value of the parameter

$$p_0=g_1\left(x_0\right)=0.0056039\dots$$

the function

$${\phi }_{p_0}\left(x\right)=-\frac{sinx}{x}+\frac{2+cosx}{3}-p_0{x}^4{\left(\frac{sinx}{x}\right)}^{2/7}$$

is the minimax approximant of the family ${\left\{{\phi }_p\left(x\right)\right\}}_{p\in \mathbb{R}}$.

Proof. 

Based on Theorems 8 and 9 and (3) and (4), the equation $\left|h\right(x\left)\right|=H\left(x\right)$ has a unique solution $x_0\in (0,\pi /2)$, which can be numerically determined with $x_0=1.26105\dots$ using the computer algebra system Maple.

We will show that for $p_1\ne p_0$$(p_1\in \mathbb{P})$, it holds that

$$\underset{x\in [0,\pi /2]}{\sup }\left|{\phi }_{p_1}\left(x\right)\right|>\underset{x\in [0,\pi /2]}{\sup }\left|{\phi }_{p_0}\left(x\right)\right|,$$

which is enough to conclude that $\underset{p\in \mathbb{R}}{\inf }\underset{x\in [0,\pi /2]}{\sup }\left|{\phi }_p\left(x\right)\right|=\underset{x\in [0,\pi /2]}{\sup }\left|{\phi }_{p_0}\left(x\right)\right|$.

Based on the stratification of the family, we conclude that if $p_1<p_0$, then

$$\underset{x\in [0,\pi /2]}{\sup }\left|{\phi }_{p_1}\left(x\right)\right|={\phi }_{p_1}(\pi /2)>{\phi }_{p_0}(\pi /2)=\underset{x\in [0,\pi /2]}{\sup }\left|{\phi }_{p_0}\left(x\right)\right|$$

and if $p_1>p_0$, then

$$\underset{x\in [0,\pi /2]}{\sup }\left|{\phi }_{p_1}\left(x\right)\right|>|{\phi }_{p_1}\left(x_{p_0}\right)|>|{\phi }_{p_0}\left(x_{p_0}\right)|=\underset{x\in [0,\pi /2]}{\sup }\left|{\phi }_{p_0}\left(x\right)\right|.$$

Therefore, the function ${\phi }_{p_0}\left(x\right)$ is the minimax approximant of the family ${\left\{{\phi }_p\left(x\right)\right\}}_{p\in \mathbb{R}}$. □

Corollary 1.

For $x\in (0,\pi /2)$ and $p_0=0.0056039\dots$, we obtain the minimax approximation

$$\frac{sinx}{x}-\frac{2+cosx}{3}\approx -p_0·x^4{\left(\frac{sinx}{x}\right)}^{2/7},$$

with the approximation error

$$d_0=\underset{x\in [0,\pi /2]}{\sup }|{\phi }_{p_0}\left(x\right)|=5.94877\dots {10}^{-5}.$$

3.2. New Results Related to Theorem 2

Based on Theorem 2, we form the family of functions

$${\left\{\psi \left(x\right)\right\}}_{p\in \mathbb{P}}$$

for $\mathbb{P}=\mathbb{R}$, defined on $[0,\pi /2]$ with

$${\psi }_p\left(x\right)=\left\{\begin{array}{ccc}-\frac{sinx}{x}+\frac{2+cosx}{3}-p{(sinx-xcosx)}^{({\pi }^2-12)/(3{\pi }^2-{\pi }^3)}& ,& x\in (0,\pi /2],\\ 0& ,& x=0.\end{array}\right.$$

Note that Theorem 2 claims that

$${\psi }_A\left(x\right)>0$$

for $A=(2/3-2/\pi )$ and $x\in (0,\pi /2)$.

Lemma 4.

The family of functions ${\left\{{\psi }_p\left(x\right)\right\}}_{p\in \mathbb{R}}$ is decreasingly stratified on the interval $(0,\pi /2)$.

Proof. 

It holds that $\frac{d{\psi }_p\left(x\right)}{dp}=-{\left(sinx-xcosx\right)}^{({\pi }^2-12)/(3{\pi }^2-{\pi }^3)}<0$ for $x\in (0,\pi /2)$. □

Let us note that for $x\in (0,\pi /2)$, it holds that

$$\begin{array}{cc}& {\psi }_p\left(x\right)=-\frac{sinx}{x}+\frac{2+cosx}{3}-p{(sinx-xcosx)}^{({\pi }^2-12)/(3{\pi }^2-{\pi }^3)}=0\hfill \\ \hfill ⟺& p=g\left(x\right)=-\frac{-xcosx+3sinx-2x}{3x{(sinx-xcosx)}^{({\pi }^2-12)/(3{\pi }^2-{\pi }^3)}}.\hfill \end{array}$$

Lemma 5.

The function $g\left(x\right)$ is decreasing on the interval $(0,\pi /2)$.

Proof. 

The function $p=g\left(x\right)$ has the first derivative

$$g^{\prime }\left(x\right)=\frac{1}{3{\pi }^3-9{\pi }^2}·\frac{{(sinx-xcosx)}^{({\pi }^3-4{\pi }^2+12)/(3{\pi }^2-{\pi }^3)}}{x^2}f\left(x\right),$$

where

$$\begin{array}{ccc}\hfill f\left(x\right)& =& (\left(x^2-6\right){\pi }^3+\left(18-2x^2\right){\pi }^2-12x^2)xsinxcosx+2x^2\left({\pi }^2-12\right)xsinx\hfill \\ \hfill & & +(\left(4x^2-3\right){\pi }^3+\left(9-9x^2\right){\pi }^2-36x^2){cos}^{2}x+\left(3-x^2\right){\pi }^3-9{\pi }^2+36x^2.\hfill \end{array}$$

is an MTP function. Since $f\left(0\right)=0$ and $f(\pi /2)=0$, we consider the following two cases:

Case 1.

If $x\in \left(0,\pi /4\right]$, since $f(\pi /4)\ne 0$, by applying the method for proving MTP inequalities analogously as in Lemma 2, it can be proved that

$$f\left(x\right)<0$$

on the interval $\left(0,\pi /4\right]$.

Case 2.

$x\in \left(\pi /4,\pi /2\right)$, we introduce the function

$$f_1\left(t\right)=f(\pi /2-x)$$

for which, by applying the method for proving MTP inequalities analogously as in Lemma 2, it can be proved that

$$f_1\left(t\right)<0$$

for $t\in \left(0,\pi /4\right)$. Therefore,

$$f\left(x\right)<0$$

on the interval $\left(\pi /4,\pi /2\right)$.

Based on previous two cases

$$f\left(x\right)<0$$

on the interval $\left(0,\pi /2\right)$. Thus,

$$g^{\prime }\left(x\right)<0$$

on the interval $\left(0,\pi /2\right)$. □

Notice that

$$\underset{x\to \pi /2-}{lim}g\left(x\right)=A=\frac{2\pi -6}{3\pi }=0.030046\dots .$$

Let

$$B=\underset{x\to 0+}{lim}g\left(x\right)=+\infty .$$

Based on Lemma 4 and Lemma 5, the first and the second condition of Theorem 4 are satisfied. Since there exist limits $\underset{x\to 0}{lim}g\left(x\right)=B$ and $\underset{x\to \pi /2}{lim}g\left(x\right)=A$, the third condition of Theorem 4 is also satisfied. Based on this, we obtain the following generalisation of Theorem 2.

Theorem 11.

(i) If $p\in (-\infty ,A]$, then

$$\begin{array}{ccc}\hfill x\in (0,\pi /2)⟹\frac{sinx}{x}-\frac{2+cosx}{3}& <& -A{(sinx-xcosx)}^{({\pi }^2-12)/(3{\pi }^2-{\pi }^3)}\le \hfill \\ \hfill & \le & -p{(sinx-xcosx)}^{({\pi }^2-12)/(3{\pi }^2-{\pi }^3)}.\hfill \end{array}$$

(ii) If $p\in (A,+\infty )$, then the equation $g\left(x\right)=p$ has a unique solution $x_0^{\left(p\right)}\in (0,\pi /2)$ and it holds that

$$x\in (0,x_0^{\left(p\right)})⟹\frac{sinx}{x}-\frac{2+cosx}{3}<-p{(sinx-xcosx)}^{({\pi }^2-12)/(3{\pi }^2-{\pi }^3)}$$

and

$$x\in (x_0^{\left(p\right)},\pi /2)⟹-p{(sinx-xcosx)}^{({\pi }^2-12)/(3{\pi }^2-{\pi }^3)}<\frac{sinx}{x}-\frac{2+cosx}{3}.$$

In the following, we will show that the family ${\left\{{\psi }_p\left(x\right)\right\}}_{p\in \mathbb{R}}$ has a unique minimax approximant. For that purpose, we consider the family ${\left\{\frac{d{\psi }_p\left(x\right)}{dx}\right\}}_{p\in \mathbb{R}}$, for $0<x<\pi /2$, which is defined by

$$\begin{array}{c}\frac{d{\psi }_p\left(x\right)}{dx}=(-{\pi }^2\left(\pi -3\right)\left(\left(4x^2-3\right){cos}^{2}x+\left(x^2-6\right)xsinxcosx-x^2+3\right)-\hfill \\ -3p{x}^{3}sinx\left({\pi }^2-12\right){\left(sinx-xcosx\right)}^{\frac{{\pi }^2-12}{3{\pi }^2-{\pi }^3}})/\hfill \\ \left(3{\pi }^2(\pi -3)(xcosx-sinx)x^2\right).\hfill \end{array}$$

Let us notice that for $x\in (0,\pi /2)$, it holds that

$$\begin{array}{c}\frac{d{\psi }_p\left(x\right)}{dx}=0⟺p=g_1\left(x\right)=\frac{1}{3}{\pi }^2(\left(\left(\pi -3\right)x^2-3\pi +9\right){sin}^{2}x+\hfill \\ +\left(\left(\pi -3\right)x^2+6\pi -18\right)xsinxcosx+\hfill \\ +\left(9-3\pi \right)x^2{cos}^{2}x)/\hfill \\ (\left({\pi }^2-12\right)x^{3}sinx{\left(sinx-xcosx\right)}^{({\pi }^2-12)/(3{\pi }^2-{\pi }^3)}).\hfill \end{array}$$

For the function $g_1\left(x\right)$, the following theorem holds.

Theorem 12.

The function $g_1\left(x\right)$ has exactly one minimum on the interval $(0,\pi /2)$ at the point $c=1.29610\dots$.

Proof. 

The function $g_1\left(x\right)$ has the first derivative

$$g_1^{\prime }\left(x\right)=-\frac{1}{3\left({\pi }^2-12\right)}\frac{{\left(sinx-xcosx\right)}^{({\pi }^2-12)/({\pi }^3-3{\pi }^2)}}{x^4{sin}^{2}x}{f}_1\left(x\right)$$

where

$$\begin{array}{ccc}\hfill f_1\left(x\right)& =& \left(\left({\pi }^3-2{\pi }^2-12\right)x^4+\left(-10{\pi }^3+27{\pi }^2+36\right)x^2+9{\pi }^3-27{\pi }^2\right)sinx{cos}^{2}x\hfill \\ \hfill & & +\left(\left(-{\pi }^3+2{\pi }^2+12\right)x^4+\left(7{\pi }^3-18{\pi }^2-36\right)x^2-9{\pi }^3+27{\pi }^2\right)sinx\hfill \\ \hfill & & +\left(\left(4{\pi }^3-9{\pi }^2-36\right)x^2-15{\pi }^3+45{\pi }^2\right)x{cos}^{3}x\hfill \\ \hfill & & +\left(\left(-7{\pi }^3+18{\pi }^2+36\right)x^2+15{\pi }^3-45{\pi }^2\right)xcosx\hfill \end{array}$$

is an MTP function.

We will show that the function $f_1\left(x\right)$ has exactly one zero on the interval $(0,\pi /2)$. Let us notice that the function $f_1\left(x\right)$ changes the sing at the endpoints of the segment $\left[1.29,1.3\right]$. By applying the method for proving MTP inequalities analogously as in Lemma 2, it can be proved that

(1)

$f_1\left(x\right)<0$ on the interval $\left[0,1.29\right]$.

(2)

$f_1\left(x\right)>0$ on the interval $\left[1.3,\pi /2\right]$.

(3)

$f_1^{\prime }\left(x\right)>0$ on the interval $\left[1.29,1.3\right]$.

Based on (1), (2) and (3), according to the method for isolating zeros from [52], the function $f_1\left(x\right)$ has exactly one zero on the interval $\left(0,\pi /2\right)$, i.e., more precisely on the interval $\left[1.29,1.3\right]$, which can be numerically determined with $c=1.29610\dots$ using the computer algebra system Maple. □

Let us notice that

$$\underset{x\to 0+}{lim}{g}_1\left(x\right)=+\infty =B$$

and

$$\underset{x\to \pi /2-}{lim}{g}_1\left(x\right)=A.$$

For $c=1.29610\dots$ the function $g_1\left(x\right)$ has the minimum

$$C=g_1\left(c\right)=0.029171\cdots <A.$$

It can shown that holds.

Lemma 6.

There exists $x_A=1.02795\cdots \in (0,c)$ such that $g_1:(0,x_A)⟶(A,+\infty )$ is bijection.

Figure 3 illustrates the family ${\left\{\psi \left(x\right)\right\}}_{p\in \mathbb{R}}$ and corresponding functions g and $g_1$.

In the following text, we consider the function $g_1$ on the interval $(0,x_A)$, where the function $g_1$ is monotonic.

Theorem 13.

For each $p\in (A,+\infty )$, the function ${\psi }_p\left(x\right)$ from the observed family has exactly one maximum at the point $x_p=g_1^{-1}\left(p\right)\in (0,x_A)$.

Proof. 

For $p\in (A,+\infty )$, it holds the following:

(i)

${\psi }_p$(0) = 0;

(ii)

${\psi }_p(\pi /2)<0$ (based on the stratification and the fact that $\underset{x\to \pi /2-}{lim}{\psi }_A\left(x\right)=0$);

(iii)

on the interval $(0,\pi /2)$, the function ${\psi }_p\left(x\right)$ has exactly one zero (since the function $g\left(x\right)$ is monotonic on the interval $(0,\pi /2)$);

(iv)

the function ${\psi }_p\left(x\right)$ has exactly one stationary point (since the function $g_1\left(x\right)$ is monotonic on the interval $(0,x_A)$).

Based on $\left(i\right)-\left(iv\right)$, the assertion follows. □

According to the previous theorem, for functions from the observed family and for values of the parameter $p\in [A,+\infty )$, it holds that

$$\underset{x\in [0,\pi /2]}{max}\left|{\psi }_p\left(x\right)\right|=|{\psi }_p\left(x_p\right)|\mathrm{o}\mathrm{r}\underset{x\in [0,\pi /2]}{max}\left|{\psi }_p\left(x\right)\right|=\left|{\psi }_p(\pi /2)\right|,$$

where $x_p$ is the point at which a local maximum of a function ${\psi }_p\left(x\right)$ is attained.

For $p<A$, it holds that

$$\underset{x\in [0,\pi /2]}{max}\left|{\psi }_p\left(x\right)\right|\ge \underset{x\in [0,\pi /2]}{max}\left|{\psi }_A\left(x\right)\right|.$$

Therefore, it holds that

$$\underset{p\in \mathbb{R}}{\inf }\underset{x\in [0,\pi /2]}{\sup }\left|{\psi }_p\left(x\right)\right|=\underset{p\in [A,+\infty )}{\inf }\underset{x\in [0,\pi /2]}{max}\left|{\psi }_p\left(x\right)\right|.$$

In order to determine the minimax approximant, in the following text we will consider two functions.

For $p\in (A,+\infty )$, it holds that ${\psi }_p\left(x_p\right)={\psi }_{g_1\left(x_p\right)}\left(x_p\right)$ and therefore, based on Theorem 13, the function

$$h_1\left(x\right)={\psi }_{g_1\left(x\right)}\left(x\right):(0,x_A)⟶\mathbb{R}$$

determines the set of points in which the functions from the observed family have maximum.

The following assertion holds.

Theorem 14.

The function

$$\begin{array}{cc}\hfill h_1\left(x\right)=& (((4{\pi }^3-9{\pi }^2-36)x^2-3{\pi }^3+9{\pi }^2){cos}^{2}x+\hfill \\ \hfill & +(({\pi }^3-2{\pi }^2-12)x^3+(-6{\pi }^3+18{\pi }^2)x)sinxcosx\hfill \\ \hfill & +(2{\pi }^2-24)x^{3}sinx+(-{\pi }^3+36)x^2+3{\pi }^3-9{\pi }^2)/(3\left({\pi }^2-12\right)x^{3}sinx)\hfill \end{array}$$

is monotonically increasing for $x\in (0,x_A)$.

Proof. 

By applying the method for proving MTP inequalities analogously as in Lemma 2, it can be proved that

$$h_1^{\prime }\left(x\right)>0$$

on the interval $(0,1.03)\supset (0,x_A)$. □

It also holds that

$$\underset{x\to 0+}{lim}{h}_1\left(x\right)=0.$$

(5)

For $\overline{p}\in (A,+\infty )$, there exists $\overline{x}\in (0,x_A)$ such that the function ${\psi }_{\overline{p}}\left(x\right)$ has maximum at the point $\overline{x}$ and, therefore, it holds that $\overline{p}=g_1\left(\overline{x}\right)$. It is evident that ${\psi }_{\overline{p}}(\pi /2)={\psi }_{g_1\left(\overline{x}\right)}(\pi /2)$. For the function

$$H_1\left(x\right)={\psi }_{g_1\left(x\right)}(\pi /2):(0,x_A)⟶\mathbb{R},$$

the following assertion holds.

Theorem 15.

The function

$$\begin{array}{cc}\hfill H_1\left(x\right)=& -\frac{2}{\pi }+\frac{2}{3}-{\pi }^2(((\pi -3)x^2-3\pi +9){sin}^{2}x+((-\pi +3)x^3+(6\pi -18)x)cosxsinx\hfill \\ \hfill & +(-3\pi +9)x^2{cos}^{2}x)/(3({\pi }^2-12)x^{3}sinx{(sinx-xcosx)}^{-\frac{{\pi }^2-12}{{\pi }^2\left(\pi -3\right)}})\hfill \end{array}$$

is monotonically increasing for $x\in (0,x_A)$.

Proof. 

The function $H_1\left(x\right)$ has the first derivative

$$H_1^{\prime }\left(x\right)=\frac{1}{3({\pi }^2-12)}·\frac{{(sinx-xcosx)}^{\frac{{\pi }^2-12}{{\pi }^2(\pi -3)}}}{x^4{sin}^{2}x}f\left(x\right)$$

where

$$\begin{array}{ccc}\hfill f\left(x\right)& =& (\left(4{\pi }^3-9{\pi }^2-36\right)x^3+\left(-15{\pi }^3+45{\pi }^2\right)x){cos}^{3}x+\hfill \\ \hfill & & +(\left({\pi }^3-2{\pi }^2-12\right)x^4+\left(-10{\pi }^3+27{\pi }^2+36\right)x^2+9{\pi }^3-27{\pi }^2)sinx{cos}^{2}x+\hfill \\ \hfill & & +(\left(-7{\pi }^3+18{\pi }^2+36\right)x^3+\left(15{\pi }^3-45{\pi }^2\right)x)cosx+\hfill \\ \hfill & & +(\left(-{\pi }^3+2{\pi }^2+12\right)x^4+\left(7{\pi }^3-18{\pi }^2-36\right)x^2-9{\pi }^3+27{\pi }^2)sinx\hfill \end{array}$$

is an MTP function. By applying the method for proving MTP inequalities analogously as in Lemma 2, it can be proved that

$$f\left(x\right)<0$$

on the interval $(0,1.03)\supset (0,x_A)$. Thus,

$$H_1^{\prime }\left(x\right)>0$$

on the interval $(0,1.03)\supset (0,x_A)$. □

There exist values

$$\left.\begin{array}{c}\underset{x\to 0+}{lim}{H}_1\left(x\right)=-\infty ,\hfill \\ H_1\left(x_A\right)=0.\hfill \end{array}\right\}$$

(6)

Figure 4 illustrates the graphs of the functions $h_1\left(x\right)$ and $|H_1\left(x\right)|$.

The following assertion holds.

Theorem 16.

The equation

$$h_1\left(x\right)=\left|H_1\left(x\right)\right|$$

has a unique solution $x_0\in (0,x_A)$, which can be numerically determined with $x_0=0.97411\dots$.

For the value of the parameter

$$p_0=g_1\left(x_0\right)=0.030456\dots$$

the function

$${\psi }_{p_0}\left(x\right)=-\frac{sinx}{x}+\frac{2+cosx}{3}-p_0{(sinx-xcosx)}^{({\pi }^2-12)/(3{\pi }^2-{\pi }^3)}$$

is the minimax approximant of the family ${\left\{{\psi }_p\left(x\right)\right\}}_{p\in \mathbb{R}}$.

Proof. 

Based on Theorems 14 and 15 and (5) and (6), the equation $h_1\left(x\right)=\left|H_1\left(x\right)\right|$ has a unique solution $x_0\in (0,x_A)$, which can be numerically determined with $x_0=0.97411\dots$ using the computer algebra system Maple.

We will show that for $p_1\ne p_0$$(p_1\in \mathbb{P})$, it holds that

$$\underset{x\in [0,\pi /2]}{\sup }\left|{\psi }_{p_1}\left(x\right)\right|>\underset{x\in [0,\pi /2]}{\sup }\left|{\psi }_{p_0}\left(x\right)\right|,$$

which is enough to conclude that $\underset{p\in \mathbb{R}}{\inf }\underset{\in [0,\pi /2]}{\sup }\left|{\psi }_p\left(x\right)\right|=\underset{x\in [0,\pi /2]}{\sup }\left|{\psi }_{p_0}\left(x\right)\right|$.

Based on the stratification of the family, we conclude that if $p_1>p_0$, then

$$\underset{x\in [0,\pi /2]}{\sup }\left|{\psi }_{p_1}\left(x\right)\right|=|{\psi }_{p_1}(\pi /2)|>|{\psi }_{p_0}(\pi /2)|=\underset{x\in [0,\pi /2]}{\sup }\left|{\psi }_{p_0}\left(x\right)\right|$$

and if $p_1<p_0$, then

$$\underset{x\in [0,\pi /2]}{\sup }\left|{\psi }_{p_1}\left(x\right)\right|\ge {\psi }_{p_1}\left(x_{p_0}\right)>{\psi }_{p_0}\left(x_{p_0}\right)=\underset{x\in [0,\pi /2]}{\sup }\left|{\psi }_{p_0}\left(x\right)\right|.$$

Therefore, the function ${\psi }_{p_0}\left(x\right)$ is the minimax approximant of the family ${\left\{{\psi }_p\left(x\right)\right\}}_{p\in \mathbb{R}}$. □

Corollary 2.

For $x\in (0,\pi /2)$ and $p_0=0.030456\dots$, we obtain the minimax approximation

$$\frac{sinx}{x}-\frac{2+cosx}{3}\approx -p_0·{(sinx-xcosx)}^{({\pi }^2-12)/(3{\pi }^2-{\pi }^3)},$$

with the approximation error

$$d_0=\underset{x\in [0,\pi /2]}{\sup }\left|{\psi }_{p_0}\left(x\right)\right|=4.10100\dots {10}^{-4}.$$

4. Conclusions

In this paper, by using methods from [52], an improvement of Cusa–Huygens-type inequalities from [11] is given, and therefore, new Cusa–Huygens-type inequalities and approximations are obtained. The presented method for determining approximations could be applied to many inequalities in the theory of analytical inequalities [1,2,69,70].