# Finest bounds for the Sinc trigonometric function and applications

Abd Raouf Chouikha

<figure><img src="image_1.png" /></figure>

## To cite this version:

Abd Raouf Chouikha. Finest bounds for the Sinc trigonometric function and applications. 2026. hal-05456227

HAL Id: hal-05456227

https://hal.science/hal-05456227v1

Preprint submitted on 13 Jan 2026

HAL is a multi-disciplinary open access
archive for the deposit and dissemination of sci-
entific research documents, whether they are pub-
lished or not. The documents may come from
teaching and research institutions in France or
abroad, or from public or private research centers.

L'archive ouverte pluridisciplinaire HAL, est
destinée au dépôt et à la diffusion de documents
scientifiques de niveau recherche, publiés ou non,
émanant des établissements d'enseignement et de
recherche français ou étrangers, des laboratoires
publics ou privés.

<figure><img src="image_2.png" /></figure>
# Finest bounds for the Sinc trigonometric function and applications

Abd Raouf Chouikha *

## Abstract

In this paper we prove for $x \in (0, \frac{\pi}{2})$ the double inequality for the Sinc function

$$ \left( -2 \left( \frac{\tan x}{x} \right)^{-\frac{r}{2}} + 3 \left( \cos x \right)^{\frac{r}{3}} \right)^{\frac{1}{r}} < \frac{\sin x}{x} < \cos x \left( \frac{2}{3} \left( \frac{\tan x}{x} \right)^{\frac{3}{2}r} + \frac{1}{3} \right)^{\frac{1}{r}}, $$

with equality when $r \to 0$.

Similar double inequalities are provided for the Huygens and The Wilker functions as well as for the Mitrinovic-Adamovic function.

**Key Words and phrases:** Trigonometric inequalities; Huygens inequalities, Wilker inequalities, Sinc function ¹.

## 1 Introduction

The obvious inequalities for $x \in (0, \frac{\pi}{2})$, $\cos x < \frac{\sin x}{x} < 1$, have been refined in [12], [11] as

$$ \cos x < \left( \frac{\sin x}{x} \right)^3, \qquad (1) $$

which known as the Mitrinovic-Adamovic Inequality. There are many references refining and generalizing the lower and upper bounds in the double inequality (1), which can be divided in to several different types bounds, among them :

- Polynomial and rational function type bounds, [1, 2, 3, 18, 19, 20, 21, 34].

- Irrational function type bounds, [20]

- Exponential type bounds, [4, 5, 6]

- Functions of $\cos x$ type bounds, [19]

- The bounds of hyperbolic functions, [23].

Other bounds for $\frac{\sin x}{x}$ can be found in [7, 8, 9, 10, 11, 12, 13, 14, 16, 17, 25, 26, 27, 28, 29, 32] and the references therein.

<footnote>0*Universite Paris-Sorbonne, Paris-Nord, villetaneuse, 93430. juju153@yahoo.com.</footnote>

<footnote n="12010">12010 Mathematics Subject Classification : 26D07, 33B10, 26D05, 26D15</footnote>
Moreover, in the past recent years a lot of mathematicians tried to refine and extend inequality (1) by providing finer and more elegant inequalities. Among them [18], [20], [21], [28], [35], [30], [31]. Inequalities concerning the sinc function abound in the literature as well as for the Huygens function

$$ \frac{\sin x}{x} + \frac{\tan x}{2x} > \frac{3}{2} $$

and for the Wilker function

$$ \left(\frac{\sin x}{x}\right)^2 + \frac{\tan x}{2x} > 2. $$

They mainly concern fine bounds of these functions. Researchers have used ingenious tricks and methods to improve these bounds. However, to my knowledge, very little have thought of making these bounds dependent on a special parameter. In [34], [33] one introduced the n-order Mitrinovic-Adamovic inequality

$$ h(n, x) = \left(\frac{\sin x}{x}\right)^n - \frac{\cos x}{3} + \frac{n-3}{3}. $$

One proved the function $h(n, x)$ is increasing with respect to $n$.

The idea that came to mind was not only to try to highlight two functions $f(x, t)$ and $g(x, t)$ depending on a parameter $t$ bounding one of these three functions, but also, for a particular value $t_0$ of this parameter, to observe that $f(x, t_0)$ and $g(x, t_0)$ could coincide with any one of the above functions. In this sense, for example when we have

$$ f(x, t) < \frac{\sin x}{x} < g(x, t), $$

and

$$ \lim_{t \to t_0} f(x, t) = \lim_{t \to t_0} g(x, t) = \frac{\sin x}{x}. $$

In this paper we aim to present new bounds for the Sinc function as well as for the Huygens and the Wilker functions in terms of one parameter functions $f(r, x)$ defined for $r \ge 0$ and for $x \in (0, \frac{\pi}{2})$. These bounds appear to be finer than older

$$ \left(-2 \left(\frac{\tan x}{x}\right)^{-\frac{3r}{2}} + 3 (\cos x)^r\right)^{\frac{1}{3r}} < \left(\frac{\sin x}{x}\right) < \cos x \left(\frac{1}{3} + \frac{2}{3} \left(\frac{\sin x}{x}\right)^{\frac{3r}{2}} (\cos x)^{-\frac{3r}{2}}\right)^{\frac{1}{r}}. \quad (2) $$
To be more precise, notice that when $r$ tends to 0 one prove

$$ 
\begin{aligned}
\lim_{r \to 0} \left( -2 \left( \frac{\tan x}{x} \right)^{-\frac{3r}{2}} + 3 (\cos x)^r \right)^{\frac{1}{3r}} &= \lim_{r \to 0} \cos x \left( \frac{1}{3} + \frac{2}{3} \left( \frac{\sin x}{x} \right)^{\frac{9r}{2}} (\cos x)^{-\frac{3r}{2}} \right)^{\frac{1}{r}} \\
&= \left( \frac{\sin x}{x} \right).
\end{aligned}
$$

That means if we bound $\frac{\sin x}{x}$ by two functions $f(x)$ and $g(x)$ :

$$ f(x) < \left( \frac{\sin x}{x} \right) < g(x), $$

then there exists $r_0$ such that for $r < r_0$ one has

$$ f(x) < \left( -2 \left( \frac{\tan x}{x} \right)^{-\frac{3r}{2}} + 3 (\cos x)^r \right)^{\frac{1}{3r}}, \quad g(x) > \cos x \left( \frac{1}{3} + \frac{2}{3} \left( \frac{\sin x}{x} \right)^{\frac{9r}{2}} (\cos x)^{-\frac{3r}{2}} \right)^{\frac{1}{r}}. $$

Thus (2) appears to be the finest for framing the Sinc function.

In addition we also prove analog bounds for Huygens and Wilker functions as well as for the Mitrinovic-Adamovic function.

## 2 Main results

Next results give bounds for $\frac{\tan x}{x}$ as well for the sinc function $\frac{\sin x}{x}$

**Theorem 1.1** For $x \in (0, \frac{\pi}{2})$ the function $\frac{\tan x}{x}$ satisfies the following double inequality for any $r > 0$

$$ \left( -2 \left( \frac{\tan x}{x} \right)^{-\frac{1}{2}r} + 3 \right)^{\frac{1}{r}} < \frac{\tan x}{x} < \left( \frac{2}{3} \left( \frac{\tan x}{x} \right)^{\frac{3}{2}r} + \frac{1}{3} \right)^{\frac{1}{r}}. $$

Moreover, the equalities are attained when $r$ tends towards 0.

**Theorem 1.2** For $x \in (0, \frac{\pi}{2})$ the sinc function satisfies the following double inequality for any $r > 0$

$$ 
\begin{aligned}
\cos x \left( -2 \left( \frac{\tan x}{x} \right)^{-\frac{1}{2}r} + 3 \right)^{\frac{1}{r}} &< \left( -2 \left( \frac{\tan x}{x} \right)^{-\frac{r}{2}} + 3 (\cos x)^{\frac{r}{3}} \right)^{\frac{1}{r}} < \\
\frac{\sin x}{x} &< \left( \frac{1}{3} (\cos x)^r + \frac{2}{3} \left( \frac{\sin x}{x} \right)^{\frac{3r}{2}} (\cos x)^{\frac{1}{2}r} \right)^{\frac{1}{r}}.
\end{aligned}
$$
Moreover, the equalities are attained when $r$ tends towards 0.

We may also derive bounds for the Huygens function

**Theorem 1.3** For $x \in (0, \frac{\pi}{2})$ the following double inequality holds for any $r > 0$

$$ (2 \cos x + 1) \left( 3 - 2 \left( \frac{\tan x}{x} \right)^{-\frac{1}{2}r} \right)^{\frac{1}{r}} < \frac{2 \sin x + \tan x}{x} < (2 \cos x + 1) \left( \frac{2}{3} \left( \frac{\tan x}{x} \right)^{\frac{3}{2}r} + \frac{1}{3} \right)^{\frac{1}{r}}. $$

Moreover, the equalities are attained when $r$ tends towards 0.

By the same way one obtains bounds for the Wilker function

**Theorem 1.4** For $x \in (0, \frac{\pi}{2})$ the following double inequality holds for any $r > 0$

$$ (\cos x)^2 \left( -2 \left( \frac{\tan x}{x} \right)^{-\frac{1}{2}r} + 3 \right)^{\frac{2}{r}} + \left( -2 \left( \frac{\tan x}{x} \right)^{-\frac{1}{2}r} + 3 \right)^{\frac{1}{r}} < \left( \frac{\sin x}{x} \right)^2 + \frac{\tan x}{x} < (\cos x)^2 \left( \frac{2}{3} \left( \frac{\tan x}{x} \right)^{\frac{3}{2}r} + \frac{1}{3} \right)^{\frac{2}{r}} + \left( \frac{2}{3} \left( \frac{\tan x}{x} \right)^{\frac{3}{2}r} + \frac{1}{3} \right)^{\frac{1}{r}}. $$

Moreover, the equalities are attained when $r$ tends towards 0.

**Remark 1.5** (i) - We may prove the following inequality

$$ \left( -2 \left( \frac{\tan x}{x} \right)^{-\frac{3r}{2}} + 3 (\cos x)^r \right)^{\frac{1}{3r}} < \frac{\sin x}{x}. $$

It is less than the one given by Theorem 1.2 because

$$ \left( -2 \left( \frac{\tan x}{x} \right)^{-\frac{3r}{2}} + 3 (\cos x)^r \right)^{\frac{1}{3r}} < \left( -2 \left( \frac{\tan x}{x} \right)^{-\frac{r}{2}} + 3 (\cos x)^{\frac{r}{3}} \right)^{\frac{1}{r}}. $$

(ii) - We may prove that

$$ (\cos x)^2 \left( \frac{2}{3} \left( \frac{\tan x}{x} \right)^{\frac{3}{2}r} + \frac{1}{3} \right)^{\frac{2}{r}} > \cos x \left( \frac{1}{3} + \frac{2}{3} \left( \frac{\sin x}{x} \right)^{3r} (\cos x)^{-\frac{3r}{2}} \right)^{\frac{1}{r}} > \left( \frac{\sin x}{x} \right)^2 $$
which improves slightly Theorem 1.4.

Concerning the Mitrinovic-Adamovic Inequality we prove the following

**Theorem 1.6** For $x \in (0, \frac{\pi}{2})$ the following inequalities hold for any $r > 0$

$$ \left(-2 \left( \frac{1}{\cos x} \left( \frac{\sin x}{x} \right)^3 \right)^{-\frac{r}{2}} + 3\right)^{\frac{1}{r}} < \left(-2 + 3 \left( \frac{1}{\cos x} \left( \frac{\sin x}{x} \right)^3 \right)^{\frac{r}{3}}\right)^{\frac{1}{r}} < \\ \frac{1}{\cos x} \left( \frac{\sin x}{x} \right)^3 < \left( \frac{2}{3} \left( \frac{1}{\cos x} \left( \frac{\sin x}{x} \right)^3 \right)^{\frac{3r}{2}} + \frac{1}{3} \right)^{\frac{1}{r}}. $$

$$ \left(3 - 2 \left(\frac{\sin x}{x}\right)^{-\frac{3r}{2}} \left(\cos x\right)^{\frac{r}{2}}\right)^{\frac{1}{r}} < \frac{1}{\cos x} \left(\frac{\sin x}{x}\right)^3 < \left(\frac{1}{3} + \frac{2}{3} \left(\frac{\sin x}{x}\right)^{\frac{9r}{2}} \left(\cos x\right)^{-\frac{3r}{2}}\right)^{\frac{1}{r}}. $$
Moreover, the equalities are attained when $r$ tends towards 0.

**Remark (i)** - It seems for $x \in (0, \frac{\pi}{2})$ and for any $k \ge 0$ the following double inequality holds :

$$ \left(3 - 2 \left(\frac{\sin x}{x}\right)^{-\frac{kr}{2}} \left(\cos x\right)^{\frac{r}{2}}\right)^{\frac{1}{r}} < \frac{1}{\cos x} \left(\frac{\sin x}{x}\right)^k < \left(\frac{1}{3} + \frac{2}{3} \left(\frac{\sin x}{x}\right)^{\frac{3kr}{2}} \left(\cos x\right)^{-\frac{3r}{2}}\right)^{\frac{1}{r}}. $$

Moreover, the equalities are attained when $r$ tends towards 0. As a corollary we may also derive the double inequality

$$ \left(3 - 2 \left(\cos x\right)^{\frac{r}{2}}\right)^{\frac{1}{r}} < \frac{1}{\cos x} < \left(\frac{1}{3} + \frac{2}{3} \left(\cos x\right)^{-\frac{3r}{2}}\right)^{\frac{1}{r}}. $$

(ii) - The following seems to be true. Then it improves Theorem 1.6

$$ \left(3 - 2 \left(\frac{\sin x}{x}\right)^{-\frac{3r}{2}} \left(\cos x\right)^{\frac{r}{2}}\right)^{\frac{3}{r}} < \left(-2 + 3 \left(\frac{\sin x}{x}\right)^{3r} \left(\cos x\right)^{-r}\right)^{\frac{1}{r}} < \frac{1}{\cos x} \left(\frac{\sin x}{x}\right)^3. $$

## 3 Proofs

The following lemmas will be useful in the sequel, [22], [36]
**Lemma 3.1** For $x \in (0, \frac{\pi}{2})$ and $p, q, r \ge 0$ the following inequalities
holds

$$
(i) \quad 1 - \frac{px^2}{6} + \left(-\frac{p}{180} + \frac{p^2}{72}\right) x^4 + \left(-\frac{p}{2835} + \frac{p^2}{1080} - \frac{p^3}{1296}\right) x^6 < \left(\frac{\sin x}{x}\right)^p < 1 - \frac{px^2}{6} + \left(-\frac{p}{180} + \frac{p^2}{72}\right) x^4 + \left(-\frac{p}{2835} + \frac{p^2}{1080} - \frac{p^3}{1296}\right) x^6 + \left(-\frac{p}{37800} + \frac{101p^2}{1360800} - \frac{p^3}{12960} + \frac{p^4}{31104}\right) x^8,
$$

$$
(ii) \quad 1 + \frac{qx^2}{3} + \left(\frac{7}{90}q + \frac{q^2}{18}\right)x^4 + \left(\frac{62}{2835}q + \frac{7}{270}q^2 + \frac{1}{162}q^3\right)x^6 < \left(\frac{\tan x}{x}\right)^q <
$$

$$
1 + \frac{qx^2}{3} + \left(\frac{7q}{90} + \frac{q^2}{18}\right) x^4 + \left(\frac{62q}{2835} + \frac{7q^2}{270} + \frac{q^3}{162}\right) x^6 + \left(\frac{3509q^2}{340200} + \frac{127q}{18900} + \frac{7q^3}{1620} + \frac{q^4}{1944}\right) x^8,
$$

$$
(iii) \quad 1 - \frac{rx^2}{2} + \left(-\frac{r}{12} + \frac{r^2}{8}\right) x^4 + \left(\frac{r^2}{24} - \frac{r}{45} - \frac{r^3}{48}\right) x^6 < (\cos x)^r <
$$

$$
1 - \frac{rx^2}{2} + \left(-\frac{r}{12} + \frac{r^2}{8}\right) x^4 + \left(\frac{r^2}{24} - \frac{r}{45} - \frac{r^3}{48}\right) x^6 + \left(\frac{7r^2}{480} - \frac{17r}{2520} - \frac{r^3}{96} + \frac{r^4}{384}\right) x^8.
$$

**Lemma 3.2** (Proposition 1 of [38]) Let the series $f(x) = \sum_{k=1}^{\infty} (-1)^{k+1} A(k)x^{2k}$
converge for $x \in (0, c)$, $c \in IR^+$. Suppose that the following statements are
true:

(i) If $c < 1$, then the sequence $A(k)$, $k \in IN$ is a positive decreasing sequence that converges to 0.

(ii) If $c \ge 1$, then the sequence $A(k)$, $k \in IN$ is a positive sequence, $\lim_{k \to \infty} c^{2k} A(k) = 0$ and $A(k) > c^2 A(k+1)$ for $k \ge 1$. Then, for all $x \in (0, c)$ and for all $n \in N$ and $m \in N$, we have

$$
\sum_{k=1}^{2n} (-1)^{k+1} A(k) x^{2k} < f(x) < \sum_{k=1}^{2n+1} (-1)^{k+1} A(k) x^{2k}.
$$

**Lemma 3.3** Let a positive $C^1$-function $u(x)$ defined for $x \in (a, b)$ such that $u(0) = 1$ then for $r > 0$ the following inequalities hold

$$
(i) \quad \left(-2 \left(u(x)\right)^{-\frac{r}{2}} + 3\right)^{\frac{1}{r}} < \left(-2 + 3 \left(u(x)\right)^{\frac{r}{3}}\right)^{\frac{1}{r}} < u(x) < \left(\frac{2}{3} \left(u(x)\right)^{\frac{3r}{2}} + \frac{1}{3}\right)^{\frac{1}{r}}.
$$

$$
(ii) \quad \left(-2 + 3(u(x))^{\frac{2r}{3}}\right)^{\frac{1}{r}} < (u(x))^2 < \left(\frac{2}{3}(u(x))^{\frac{3r}{2}} + \frac{1}{3}\right)^{\frac{2}{r}} < \left(\frac{2}{3}(u(x))^{3r} + \frac{1}{3}\right)^{\frac{1}{r}}.
$$

$$
(iii) \quad \left(-2 + 3(u(x))^r\right)^{\frac{1}{r}} < (u(x))^3 < \left(\frac{2}{3}(u(x))^{\frac{3r}{2}} + \frac{1}{3}\right)^{\frac{3}{r}} < \left(\frac{2}{3}(u(x))^{\frac{9r}{2}} + \frac{1}{3}\right)^{\frac{1}{r}}.
$$
More generally, for $k > 0$ one gets

$$ (iv) \quad \left(-2 + 3(u(x))^{\frac{kr}{3}}\right)^{\frac{1}{r}} < (u(x))^k < \left(\frac{2}{3}(u(x))^{\frac{3kr}{2}} + \frac{1}{3}\right)^{\frac{1}{r}}. $$

Moreover, all these inequalities become equalities when $r$ tends towards 0.

### Proof

(i) Notice at first

$$ \begin{aligned} \left(-2(u(x))^{-\frac{r}{2}} + 3\right) - \left(-2 + 3(u(x))^{\frac{r}{3}}\right) = & \\ -2u(x)^{-1/2r} + 5 - 3u(x)^{1/3r} &= -\frac{(3v^3 + 6v^2 + 4v + 2)(v-1)^2}{v^3} < 0, \end{aligned} $$

where $v = (u(x))^{\frac{3r}{2}}$. We get also

$$ -2 + 3u(x)^{1/3r} - u(x)^r = -2 + 3v - v^3 = -(v+2)(v-1)^2 < 0, $$

where $v = u(x)^{1/3r}$. As well as

$$ \frac{1}{3} + \frac{2}{3}u(x)^{3/2r} - u(x)^r = \frac{1}{3} + \frac{2}{3}v^3 - v^2 = \frac{1}{3}(2v+1)(v-1)^2 > 0, $$

where $v = u(x)^{1/2r}$.

(ii) Let us compute the following

$$ \begin{aligned} -2 + 3u(x)^{2/3r} - u(x)^{2r} &= -2 + 3v^2 - v^6 = -(v^2 + 2)(v-1)^2(v+1)^2 < 0, \\ \text{where } v &= u(x)^{1/3r}. \end{aligned} $$

$$ \begin{aligned} \left(\frac{2}{3}u(x)^{3/2r} + \frac{1}{3}\right)^2 - u(x)^{2r} &= \frac{4}{9}u(x)^{3r} + \frac{4}{9}u(x)^{3/2r} + \frac{1}{9} - u(x)^{2r} = \\ & \quad \frac{4}{9}v^6 + \frac{4}{9}v^3 + \frac{1}{9} - v^4 = \frac{1}{9}(2v+1)(2v^3+3v^2+1)(v-1)^2 > 0, \\ \text{where } v &= u(x)^{1/2r}. \end{aligned} $$

(iii) By the same way let us compute

$$ -2 + 3u(x)^r - u(x)^{3r} = -2 + 3v - v^3 = -(v+2)(v-1)^2 < 0, $$

where $v = u(x)^r$.

$$ \begin{aligned} \left(\frac{2}{3}u(x)^{3/2r} + \frac{1}{3}\right)^3 - \frac{2}{3}u(x)^{9/2r} - \frac{1}{3} &= \left(\frac{2}{3}v + \frac{1}{3}\right)^3 - \frac{2}{3}v^3 - \frac{1}{3} = \\ & \quad -\frac{10}{27}v^3 + \frac{4}{9}v^2 + \frac{2}{9}v - \frac{8}{27} = -\frac{2}{27}(5v+4)(v-1)^2 < 0, \end{aligned} $$
where $v = u(x)^{3/2}r$.

(iv) Let us compute

$$2/3 u(x)^{3/2 kr} + 1/3 - u(x)^{kr} = 1/3 + 2/3 v^3 - v^2 = 1/3 (2v+1)(v-1)^2 > 0,$$
where $v = u^{1/2}kr$.

$$-2 + 3u^{1/3kr} - u^{kr} = -2 + 3v - v^3 = -(v+2)(v-1)^2 < 0,$$
where $v = u^{1/3}kr$.

Tun now to the last part of Lemma 3.3. let us consider

$$\ln \left(-2 + 3 \left(u(x)\right)^{\frac{kr}{3}}\right)^{\frac{1}{r}} = \frac{1}{r} \ln \left(-2 + 3 \left(u(x)\right)^{\frac{kr}{3}}\right) = \\ \ln(u)k - \frac{1}{3} \left(\ln(u)\right)^2 k^2 r + O(r^2).$$

We then deduce for any integer $k > 0$

$$\lim_{r \to 0} \left(-2 + 3 \left(u(x)\right)^{\frac{kr}{3}}\right)^{\frac{1}{r}} = u(x)^k.$$

By the same way

$$\ln \left( \frac{2}{3} \left( u(x) \right)^{\frac{3kr}{2}} + \frac{1}{3} \right)^{\frac{1}{r}} = \frac{1}{r} \ln \left( \frac{2}{3} \left( u(x) \right)^{\frac{3kr}{2}} + \frac{1}{3} \right) = \\ \ln(u)k + O(r).$$

Thus for any integer $k > 0$ one has

$$\lim_{r \to 0} \left( \frac{2}{3} \left( u(x) \right)^{\frac{3kr}{2}} + \frac{1}{3} \right)^{\frac{1}{r}} = u(x)^k.$$

## Proof of Theorem 1.1

We will now separately prove the double inequality.

(i) - Let us prove the right one inequality

$$\frac{\tan x}{x} < \left( \frac{1}{3} + \frac{2}{3} \left( \frac{\tan x}{x} \right)^{\frac{3r}{2}} \right)^{\frac{1}{r}}.$$

Since $\frac{\tan x}{x} > 1$ we have now to prove

$$\left(\frac{\tan x}{x}\right)^{r} - \frac{1}{3} - \frac{2}{3}\left(\frac{\tan x}{x}\right)^{\frac{3r}{2}} < 0.$$
It is easy to see that the function

$$Z - \frac{1}{3} - \frac{2}{3} Z^{3/2} = -\frac{1}{3}(2\sqrt{Z} + 1)(\sqrt{Z} - 1)^2 < 0,$$

where $Z = (\frac{\tan x}{x})^r > 1$.

(ii) - The left inequality of Theorem 1.1

$$\left(-2 \left(\frac{\tan x}{x}\right)^{-\frac{1}{2} r} + 3\right)^{\frac{1}{r}} < \frac{\tan x}{x}$$

is equivalent to

$$-2 \left( \frac{\tan x}{x} \right)^{-\frac{1}{2} r} + 3 < \left( \frac{\tan x}{x} \right)^{r}.$$

Notice that if we take $Z = (\frac{\tan x}{x})^{\frac{1}{2}}$ then

$$-2 \left(\frac{\tan x}{x}\right)^{-\frac{1}{2}r} + 3 - \left(\frac{\tan x}{x}\right)^{r} = -\frac{2}{Z} + 3 - Z^2 = -\frac{(Z+2)(Z-1)^2}{Z} < 0,$$

since $Z > 1$. This proves the left inequality.

(iii) - Turn now to the last part of Theorem 1.1

The logarithm yields :

$$\ln \left( -2 \left( \frac{\tan x}{x} \right)^{-\frac{3r}{2}} + 3 (\cos x)^r \right)^{\frac{1}{3r}} = \frac{\ln \left( -2 \left( \frac{\tan x}{x} \right)^{-\frac{3r}{2}} + 3 (\cos x)^r \right)}{3r} = \\ \frac{1}{3r} \left( 3 \ln \left( \frac{\sin x}{x} \right) r \right) + O(r) = \ln \left( \frac{\sin x}{x} \right) + O(r).$$

Thus

$$\lim_{r \to 0} \left( -2 \left( \frac{\tan x}{x} \right)^{-\frac{3r}{2}} + 3 (\cos x)^r \right)^{\frac{1}{3r}} = \frac{\sin x}{x}.$$

By the same way, let us take the logarithm

$$\ln \cos x \left( \frac{1}{3} + \frac{2}{3} \left( \frac{\tan x}{x} \right)^{\frac{3r}{2}} \right)^{\frac{1}{r}} = \frac{1}{r} \ln \cos x \left( \frac{1}{3} + \frac{2}{3} \left( \frac{\tan x}{x} \right)^{\frac{3r}{2}} \right) = \ln \left( \frac{\sin x}{x} \right) + O(r).$$

Thus

$$\lim_{r \to 0} \cos x \left( \frac{1}{3} + \frac{2}{3} \left( \frac{\tan x}{x} \right)^{\frac{3r}{2}} \right)^{\frac{1}{r}} = \frac{\sin x}{x}.$$
**Proof of Theorem 1.2**

Notice that as a corollary of Theorem 1.1 we may deduce the following double inequality since $\frac{\sin x}{x} = \cos x \frac{\tan x}{x}$

$$ \cos x \left( -2 \left( \frac{\tan x}{x} \right)^{-\frac{1}{2}r} + 3 \right)^{\frac{1}{r}} < \frac{\sin x}{x} < \cos x \left( \frac{2}{3} \left( \frac{\tan x}{x} \right)^{\frac{3}{2}r} + \frac{1}{3} \right)^{\frac{1}{r}}. $$

We will prove the following

**Lemma 1.2.1** For $x \in (0, \frac{\pi}{2})$ the following inequality holds

$$ \cos x \left( -2 \left( \frac{\tan x}{x} \right)^{-\frac{1}{2}r} + 3 \right)^{\frac{1}{r}} < \left( -2 \left( \frac{\tan x}{x} \right)^{-\frac{r}{2}} + 3 (\cos x)^{\frac{r}{3}} \right)^{\frac{1}{r}}. $$

Indeed, let us write the difference

$$ \begin{aligned} -2 \left( \frac{\tan x}{x} \right)^{-1/2r} + 3 (\cos x)^{1/3r} - (\cos x)^r \left( -2 \left( \frac{\tan x}{x} \right)^{-1/2r} + 3 \right) &= \\ & 2 \left( \frac{\tan x}{x} \right)^{-1/2r} \left( (\cos x)^r - 1 \right) + 3 (\cos x)^{1/3r} - 3 (\cos x)^r = \\ & 2 \left( \frac{\sin x}{x} \right)^{-1/2r} (\cos x)^{1/2r} \left( (\cos x)^r - 1 \right) + 3 (\cos x)^{1/3r} - 3 (\cos x)^r. \end{aligned} $$

Let us make the change $(\cos x) = b^6 r$ the last expression becomes

$$ 2 b^3 \left( \frac{\sin x}{x} \right)^{-1/2r} (b^6 - 1) + 3 b^2 - 3 b^6 $$

which is upper than

$$ 2 (b (b^6 - 1))^3 + 3 b^2 - 3 b^6 = b^2 (b+1) (2 b^4 + 2 b^3 + 4 b^2 + b + 3) (b-1)^2 > 0. $$

Turn now to the proof to Theorem 1.2. We simply need to prove the inequality of the left

Let us consider

$$ v_t(r, x) = \left(\frac{\sin x}{x}\right)^r + 2 \left(\frac{\tan x}{x}\right)^{-\frac{r}{2}} - 3 (\cos x)^{\frac{r}{3}}. $$

By Lemmas 3.1 and 3.2 one have

$$ \left(\frac{\sin x}{x}\right)^r > \left(1 - \frac{1}{6} r x^2 + \left(-\frac{1}{180} r + \frac{1}{72} r^2\right) x^4 + \left(-\frac{1}{2835} r + \frac{1}{1080} r^2 - \frac{1}{1296} r^3\right) x^6 + \right.$$
$$ 
\begin{aligned}
& \left(-\frac{1}{37800} r + \frac{101}{1360800} r^2 - \frac{1}{12960} r^3 + \frac{1}{31104} r^4\right) x^8 + \\
& \left(\frac{1}{233280} r^4 - \frac{61}{8164800} r^3 + \frac{13}{2041200} r^2 - \frac{1}{467775} r - \frac{1}{933120} r^5\right) x^{10} + \\
& \left(\frac{143 r^4}{293932800} - \frac{59 r^3}{81648000} + \frac{7999 r^2}{14145516000} - \frac{691 r}{3831077250} - \frac{r^5}{5598720} + \frac{r^6}{33592320}\right) x^{12}, \\
& 2 \left(\frac{\tan x}{x}\right)^{-\frac{r}{2}} > \left(2-1/3 r x^2+\left(-\frac{7}{90} r+1/36 r^2\right) x^4+\left(-\frac{62}{2835} r+\frac{7}{540} r^2-\frac{1}{648} r^3\right) x^6+\right. \\
& \left. \left(-\frac{127}{18900} r+\frac{3509}{680400} r^2-\frac{7}{6480} r^3+\frac{1}{15552} r^4\right) x^8+\right. \\
& \left(\frac{7}{116640} r^4-\frac{2269}{4082400} r^3+\frac{2011}{1020600} r^2-\frac{146}{66825} r-\frac{1}{466560} r^5\right) x^{10} + \\
& \left(\frac{5567 r^4}{146966400} - \frac{10397 r^3}{40824000} + \frac{5269351 r^2}{7072758000} - \frac{1414477 r}{1915538625} - \frac{7 r^5}{2799360} + \frac{r^6}{16796160}\right) x^{12}, \\
& 3 \left(\cos x\right)^{\frac{r}{3}} < \left(3-1/2 r x^2+\left(-1/12 r+1/24 r^2\right) x^4+\left(-1/45 r+\frac{1}{72} r^2-\frac{1}{432} r^3\right) x^6+\right. \\
& \left. \left(-\frac{17}{2520} r+\frac{7}{1440} r^2-\frac{1}{864} r^3+\frac{1}{10368} r^4\right) x^8+\right. \\
& \left(\frac{1}{15552} r^4-\frac{13}{25920} r^3+\frac{79}{45360} r^2-\frac{31}{14175} r-\frac{1}{311040} r^5\right) x^{10}.
\end{aligned}
$$

By grouping all these inequalities we find after simplifying

$$ v_t(r, x) > \left( \frac{179r^4}{4665600} - \frac{331r^3}{1296000} + \frac{319597r^2}{428652000} - \frac{691r}{935550} - \frac{r^5}{373248} + \frac{r^6}{11197440} \right) x^{12} + \\ \left(-\frac{1}{16200} r^3 + \frac{2}{8505} r^2 \right) x^{10} + \frac{1}{2700} x^8 r^2. $$

Thanks to Maple we may derive for $r > 0$ since $x \in (0, \frac{\pi}{2})$

$$ v_t(r, x) > \frac{r^2}{2700} x^8 > 0. $$

This means

$$ \left(\frac{\sin x}{x}\right)^r > -2 \left(\frac{\tan x}{x}\right)^{-\frac{r}{2}} + 3 (\cos x)^{\frac{r}{3}}. $$

Or equivalently,

$$ \left(-2 \left(\frac{\tan x}{x}\right)^{-\frac{r}{2}} + 3 (\cos x)^{\frac{r}{3}}\right)^{\frac{1}{r}} < \frac{\sin x}{x}. $$
Turn now to the last part of Theorem 1.2. Notice that it follows from
Theorem 1.1

$$
\lim_{r \to 0} \cos x \left( -2 \left( \frac{\tan x}{x} \right)^{-\frac{1}{2}r} + 3 \right)^{\frac{1}{r}} = \cos x \frac{\tan x}{x} = \frac{\sin x}{x}.
$$

On the other hand the equality

$$
\cos x \left( -2 \left( \frac{x}{\sin x} \right)^{3/2r} (\cos x)^{1/2r} + 3 \right)^{r-1} = \left( -2 \left( \frac{\tan x}{x} \right)^{-3/2r} + 3 (\cos x)^r \right)^{r-1}
$$

implies

$$
\lim_{r \to 0} \left( -2 \left( \frac{\tan x}{x} \right)^{-3/2r} + 3 (\cos x)^r \right)^{r-1} = \frac{\sin x}{x}.
$$

**Proofs of Theorems 1.3 and 1.4**

These Theorems are in fact an easy consequence of Theorem 1.1. Indeed,
we may write for Theorem 1.3

$$
\frac{2 \sin x + \tan x}{x} = \frac{2 \cos x \tan x}{x} + \frac{\tan x}{x} = (2 \cos x + 1) \frac{\tan x}{x}.
$$

For Theorem 1.4 we have

$$
\left(\frac{\tan x}{x}\right)^2 + \frac{\tan x}{x} = (\cos x)^2 \left(\frac{\tan x}{x}\right)^2 + \frac{\tan x}{x}.
$$

**Proof of Theorem 1.6**

(i) - Prove at first the left one: Let us consider

$$
u_t(r, x) = \left(\frac{\sin x}{x}\right)^{3r} + 2 \left(\frac{\tan x}{x}\right)^{\frac{-3r}{2}} - 3 (\cos x)^r.
$$

By Lemmas 3.1 and 3.2 it yields

$$
\begin{align*}
\left(\frac{\sin x}{x}\right)^{3r} &> 2 - rx^2 + \left(-\frac{7}{30}r + \frac{1}{4}r^2\right)x^4 + \left(\frac{7}{60}r^2 - \frac{62}{945}r - \frac{1}{24}r^3\right)x^6 + \\
&\quad \left(\frac{3509}{75600}r^2 - \frac{127}{6300}r - \frac{7}{240}r^3 + \frac{1}{192}r^4\right)x^8 + \\
&\quad \left(\frac{2011}{113400}r^2 - \frac{146}{22275}r - \frac{2269}{151200}r^3 + \frac{7}{1440}r^4 - \frac{1}{1920}r^5\right)x^{10} + ,
\end{align*}
$$
$$ 
\begin{aligned}
& 2 \left( \frac{\tan x}{x} \right)^{-\frac{3r}{2}} > 2 - rx^2 + \left( -2\frac{5}{3}r - 1\frac{1}{6}r(-3\frac{2}{3}r - 1) \right) x^4 + \\
& \left( \frac{7}{60}r^2 - \frac{62}{945}r - 1\frac{1}{24}r^3 \right) x^6 + \left( \frac{3509}{75600}r^2 - \frac{127}{6300}r - \frac{7}{240}r^3 + \frac{1}{192}r^4 \right) x^8 + \\
& \left( \frac{2011}{113400}r^2 - \frac{146}{22275}r - \frac{2269}{151200}r^3 + \frac{7}{1440}r^4 - \frac{1}{1920}r^5 \right) x^{10} \\
& 3 (\cos x)^r < \left( 1\frac{8}{3}r^2 - 1\frac{15}{3}r - 1\frac{16}{3}r^3 \right) x^6 + \left( \frac{7}{160}r^2 - \frac{17}{840}r - 1\frac{1}{32}r^3 + \frac{1}{128}r^4 \right) x^8 + \\
& \left( -\frac{13}{960}r^3 + \frac{79}{5040}r^2 - \frac{31}{4725}r + \frac{1}{192}r^4 - \frac{1}{1280}r^5 \right) x^{10} + \\
& \left( -\frac{677}{120960}r^3 + \frac{863}{151200}r^2 - \frac{691}{311850}r + \frac{31}{11520}r^4 - \frac{1}{1536}r^5 + \frac{1}{15360}r^6 \right) x^{12}.
\end{aligned}
$$

By grouping all these inequalities we find after simplifying

$$ u_t(r, x) > \frac{1}{300}x^8r^2 + \frac{2}{945}x^{10}r^2 - \frac{1}{600}x^{10}r^3 + \frac{677}{120960}x^{12}r^3 - \frac{863}{151200}x^{12}r^2 + \\ \frac{691}{311850}x^{12}r - \frac{31}{11520}x^{12}r^4 + \frac{1}{1536}x^{12}r^5 - \frac{1}{15360}x^{12}r^6. $$

Thanks to *Maple* we may derive for $r > 0$ since $x \in (0, \frac{\pi}{2})$

$$ u_t(r, x) > \frac{r^2}{300}x^8 > 0. $$

This means

$$ \left(\frac{\sin x}{x}\right)^{3r} > -2 \left(\frac{\tan x}{x}\right)^{-\frac{3r}{2}} + 3 (\cos x)^r . $$

(ii) - Let us turn now to the right inequality. Let us consider the following function

$$ w_t(r, x) = \left(\frac{\sin x}{x}\right)^{3r} - (\cos x)^r \left(\frac{1}{3} + \frac{2}{3}\left(\frac{\sin x}{x}\right)^{\frac{9r}{2}} (\cos x)^{-\frac{3r}{2}}\right). $$

By Lemmas 3.1 and 3.2 we obtain

$$ 
\begin{aligned}
\left(\frac{\sin x}{x}\right)^r &< \left(1-\frac{1}{6}rx^2+\left(-\frac{1}{180}r+\frac{1}{72}r^2\right)x^4+\left(-\frac{1}{2835}r+\frac{1}{1080}r^2-\frac{1}{1296}r^3\right)x^6+\right. \\
& \left. \left(-\frac{1}{37800}r+\frac{101}{1360800}r^2-\frac{1}{12960}r^3+\frac{1}{31104}r^4\right)x^8+\right. \\
& \left. \left(\frac{1}{233280}r^4-\frac{61}{8164800}r^3+\frac{13}{2041200}r^2-\frac{1}{467775}r-\frac{1}{933120}r^5\right)x^{10}, \right. \\
(\cos x)^r &> \left(1-\frac{1}{2}rx^2+\left(-\frac{1}{12}r+\frac{1}{8}r^2\right)x^4+\left(-\frac{1}{45}r+\frac{1}{24}r^2-\frac{1}{48}r^3\right)x^6+\right.
\end{aligned}
$$
$$ 
\begin{aligned}
& \left(-\frac{17}{2520} r + \frac{7}{480} r^2 - \frac{1}{96} r^3 + \frac{1}{384} r^4\right) x^8 + \\
& \left(\frac{1}{576} r^4 - \frac{13}{2880} r^3 + \frac{79}{15120} r^2 - \frac{31}{14175} r - \frac{1}{3840} r^5\right) x^{10}, \\
& \left(\frac{1}{3} + \frac{2}{3} \left(\frac{\sin x}{x}\right)^{\frac{9r}{2}} (\cos x)^{-\frac{3r}{2}}\right) > \left(1+\frac{1}{15} r x^4 + \frac{4}{189} r x^6 + \left(\frac{1}{150} r + \frac{1}{300} r^2\right) x^8 + \right. \\
& \qquad \left. \left(\frac{68}{31185} r + \frac{2}{945} r^2\right) x^{10} + \left(\frac{42842}{58046625} r + \frac{1}{9000} r^3 + \frac{5969}{5953500} r^2\right) x^{12}, \\
& (\cos x)^r \left(\frac{1}{3} + \frac{2}{3} \left(\frac{\sin x}{x}\right)^{\frac{9r}{2}} (\cos x)^{-\frac{3r}{2}}\right) > \left(1-\frac{1}{2} r x^2 + \left(-\frac{1}{60} r + \frac{1}{8} r^2\right) x^4 + \right. \\
& \qquad \left. \left(\frac{1}{120} r^2 - \frac{1}{945} r - \frac{1}{48} r^3\right) x^6 + \left(-\frac{1}{12600} r + \frac{269}{151200} r^2 - \frac{1}{480} r^3 + \frac{1}{384} r^4\right) x^8 + \right. \\
& \qquad \left. \left(-\frac{1}{155925} r + \frac{173}{226800} r^2 - \frac{229}{302400} r^3 + \frac{1}{2880} r^4 - \frac{1}{3840} r^5\right) x^{10}. \right)
\end{aligned}
$$

By grouping all these inequalities we find and after simplifying

$$ w_t(r, x) < \left( \frac{-59r^3}{3024000} + \frac{7999r^2}{1571724000} - \frac{691r}{1277025750} + \frac{143r^4}{3628800} - \frac{r}{23040} + \frac{r^6}{46080} \right) x^{12} + \left( \frac{1}{1800} r^3 - \frac{2}{2835} r^2 \right) x^{10} - \frac{1}{900} r^2 x^8. $$

Thanks to Maple we may derive for $r > 0$ and for $x \in (0, \frac{\pi}{2})$

$$ w_t(r, x) < -\frac{1}{900} r^2 x^8 < 0. $$

That means

$$ \left(\frac{\sin x}{x}\right)^{3r} - (\cos x)^r \left(\frac{1}{3} + \frac{2}{3} \left(\frac{\sin x}{x}\right)^{\frac{9r}{2}} (\cos x)^{-\frac{3r}{2}}\right) < 0, $$

or equivalently,

$$ \frac{1}{\cos x} \left(\frac{\sin x}{x}\right)^3 < \left(\frac{1}{3} + \frac{2}{3} \left(\frac{\sin x}{x}\right)^{\frac{9r}{2}} (\cos x)^{-\frac{3r}{2}}\right)^{\frac{1}{r}}. $$

(iii) - Turn now to the last part of Theorem 1.6

A limited expansion of the left side with respect to $r$ yields

$$ \left(3 - 2 \left(\frac{\sin x}{x}\right)^{-3/2} (\cos x)^{1/2} r\right)^{r-1} = $$
$$ \frac{(\sin x)^3}{x^3 \cos x} \left( 1 - \frac{3}{4} \left( -3 \ln \left( \frac{\sin x}{x} \right) + \ln (\cos x)^2 \right) r \right) + O(r^2). $$

This means

$$ \lim_{r \to 0} \left( 3 - 2 \left( \frac{\sin x}{x} \right)^{-3/2r} (\cos x)^{1/2r} \right)^{r^{-1}} = \left( \frac{\sin x}{x} \right)^3 \frac{1}{\cos x}. $$

For the right side one gets the limited expansion with respect to $r$

$$ \left( \frac{1}{3} + \frac{2}{3} \left( \frac{\sin x}{x} \right)^{9/2r} (\cos x)^{-3/2r} \right)^{r^{-1}} = \frac{(\sin x)^3}{x^3 \cos x} \left( 1 + \frac{1}{4} \left( -3 \ln \left( \frac{\sin x}{x} \right) + \ln (\cos x)^2 \right) r \right) + O(r^2). $$

This means

$$ \lim_{r \to 0} \left( \frac{1}{3} + \frac{2}{3} \left( \frac{\sin x}{x} \right)^{9/2r} (\cos x)^{-3/2r} \right)^{r^{-1}} = \frac{(\sin x)^3}{x^3 \cos x}. $$

# 4 Links with older inequalities

In this section, we will examine more closely why the bounds given by Theorem 1.2 are the most precise for the Sinc function. As mentioned in the introduction, we will determine the value of $r_0$ for certain cases in order that for $r < r_0$ one has

$$ f(x) < \left( -2 + 3 \left( \frac{\sin x}{x} \right)^{\frac{r}{3}} \right)^{\frac{1}{r}}, \quad g(x) > \left( \frac{2}{3} \left( \frac{\sin x}{x} \right)^{\frac{3r}{2}} + \frac{1}{3} \right)^{\frac{1}{r}}. $$

But first, we need a brief reminder.

One tried to refine and extend inequality (1). The literature on trigonometric inequalities is very extensive. We will only refer for that to one of the most recent paper, Y. Bagul and W. Du, [24].

Following [24] recall that Mortici [39] and Chouikha [30] obtained, respectively, the following double inequalities for $x \in (0, \frac{\pi}{2})$:

$$ \cos x + \frac{x^4}{15} - \frac{23x^6}{1890} < \left( \frac{\sin x}{x} \right)^3 < \cos x + \frac{x^4}{15}, $$ $$ \cos x + x^3 \left( 1 - \frac{x^2}{63} \right) \frac{\sin x}{15} < \left( \frac{\sin x}{x} \right)^3 < \cos x + \frac{x^3 \sin x}{15}. $$
Moreover, W.-D. Jiang [35] established the double inequality for $x \in (0, \pi)$

$$ \cos x + \frac{21}{5} \frac{x^3 \sin x}{63 + x^2} < \left( \frac{\sin x}{x} \right)^3 < \cos x + \frac{\pi^2}{15} \frac{x^3 \sin x}{\pi^2 - x^2}. $$

More recently, Bagul and Du [24] obtain sharp bounds for the Sinc function and refine all the above inequalities. They prove for $x \in (0, \frac{\pi}{2})$

$$ \cos x + \frac{8}{\pi^3} (1 - \cos x)^2 < \left( \frac{\sin x}{x} \right)^3 < \cos x + \frac{4}{15} (1 - \cos x)^2, $$

as well as

$$ \cos x + \left( \frac{4}{15} - \frac{4x^2}{945} \right) (1 - \cos x)^2 < \left( \frac{\sin x}{x} \right)^3 < \cos x + \left( \frac{4}{15} - \frac{x^2}{0\pi^2} \right) (1 - \cos x)^2. $$

For example notice that for $r = r_0 = \frac{1}{792}$ one proves

$$ - \left( -2 \left( \frac{\tan(x)}{x} \right)^{-\frac{1}{528}} + 3 \left( \cos(x) \right)^{\frac{1}{792}} \right)^{792} + \cos(x) + \frac{1}{4} \left( 1 - \cos(x) \right)^2 > 0. $$

This means Theorem 1.2 is sharper than Theorem 2 of [24].

Notice that for $r = r_0 = \frac{1}{1000}$ one proves

$$ - \left( -2 \left( \frac{\tan x}{x} \right)^{-\frac{3}{2000}} + 3 \left( \cos x \right)^{\frac{1}{1000}} \right)^{1000} + \cos x + \frac{4}{15} \left( 1 - \cos x \right)^2 - \frac{1}{945} x^6 < 0. $$

This means Theorem 1.2 is sharper than Theorem 4 of [24].

Notice also for $r = r_0 = \frac{1}{1500}$ one proves

$$ - \left( -2 \left( \frac{\tan x}{x} \right)^{-\frac{1}{1000}} + 3 \left( \cos x \right)^{\frac{1}{1500}} \right)^{1500} + \cos x + \left( \frac{4}{15} - \frac{4}{945} x^2 \right) (1 - \cos x)^2 < 0. $$

This means Theorem 1.2 is sharper than Theorem 6 of [24].

We can therefore similarly apply all the trigonometric inequalities for the sinc function and determine each time the corresponding value of $r_0$, so that Theorem 1.2 appears more refined. This procedure is also valid for the other functions thanks to Theorems 1.3, 1.4, 1.6.
References

[1] Agarwal, R.P., Kim, Y.H., Sen, S.K.: A new refined Jordan's inequality and its application. Math. Inequal. Appl. 12, 255-264 (2009)

[2] Alzer, H., Kwong, M.K.: On Jordan's inequality. Period. Math. Hungar. 77(2), 191-200 (2018)

[3] Chen, X.D., Shi, J.R., Wang, Y.G., Xiang, P.: A new method for sharpening the bounds of several special functions. Results Math. 72(1-2), 695-702 (2017)

[4] Chen, X.D., Ma, J.Y., Li, Y.X.: Approximating trigonometric functions by using exponential inequalities. J. Inequal. Appl. 2019, Paper no. 53 (2019)

[5] Debnath, L., Zhao, C.J.: New strengthened Jordan inequality and its applications. Math. Inequal. Appl. 16(4), 557-560 (2003)

[6] Nishizawa, Y.: Sharpening of Jordan's type and Shafer-Fink type inequalities with exponential approximations. Appl. Math. Comput. 269, 146-154 (2015)

[7] Ozban, A.Y. A new refined form of Jordan inequality and its applications. Appl. Math. Lett. 19, 155-160 (2006)

[8] Qiu, Y.Y., Zhu, L.: The best approximation of the Sinc function by a polynomial of degree n with the square norm. J. Inequal. Appl. 2010, 12 (2010)

[9] Qi, F., Niu, D.W., Guo, B.N.: Refinements, generalizations, and applications of Jordan inequality and related problems. J. Inequal. Appl. 2009, Article ID 271923 (2009). https://doi.org/10.1155/2009/271923

[10] Wu, S.H., Debnath, L.: A new generalized and sharp version of Jordan's inequality and its applications to the improvement of the Yang Le inequality. Appl. Math. Lett. 19, 1378-1384 (2006)

[11] Wu, S.H., Li, S.G.: Sharpened versions of Mitrinovic-Adamovic, Lazarevic and Wilker inequalities for trigonometric and hyperbolic functions. J. Nonlinear Sci. Appl. 9(5), 2688-2696 (2016)

[12] Wu, S.H., Baricz, A.: Generalizations of Mitrinovic, Adamovic and Lazarevic inequalities and their applications. Publ. Math. Debrecen 75(3-4), 447-458 (2009)

[13] Yang, Z.H.: Refinements of a two-sided inequality for trigonometric functions. J. Math. Inequal. 7(4), 601-615 (2013)
[14] Yang, Z.H., Chu, Y.M.: A note on Jordan, Adamovic-Mitrinovic, and Cusa inequalities. Abstr. Appl. Anal. 2014 (2014), Article ID 364076. https://doi.org/10.1155/2014/364076

[15] Yang, Z.H., Chu, Y.M., Song, Y.Q., Li, Y.M.: A sharp double inequality for trigonometric functions and its applications. Abstr. Appl. Anal. 2014, Article ID 592085 (2014). https://doi.org/10.1155/2014/592085

[16] Yang, Z.H.: The sharp inequalities related to Wilker type. Math. Inequal. Appl.17(3), 1015-1026(2014). https://doi.org/10.7153/mia-17-74

[17] Yang, Z.H., Jiang, Y.L., Song, Y.Q., Chu, Y.M.: Sharp inequalities for trigonometric functions, Abstr. Appl. Anal. 2014 (2014), Article ID 601839, 18 pages. https://doi.org/10.1155/2014/601839

[18] Zhu, L.: Sharp inequalities of Mitrinovic-Adamovic type. Rev. R. Acad. Cienc. Exactas FÃs. Nat. Ser. A Mat. RACSAM 113(2), 957-968 (2019). https://doi.org/10.1007/s13398-018-0521-0

[19] Zhu, L.: Some new bounds for Sinc function by simultaneous approximation of the base and exponential functions. Rev. R. Acad. Cienc. Exactas FÃs. Nat. Ser. A Mat. RACSAM 114(2), Paper No. 81, 17 pp. (2020). https://doi.org/10.1007/s13398-020-00811-9

[20] Zhu, L.: New Mitrinovic-Adamovic type inequalities. Rev. R. Acad. Cienc. Exactas FÃs. Nat. Ser. A Mat. RACSAM 114(3), Paper No. 119, 14 pp. (2020). https://doi.org/10.1007/s13398-020-00848-w

[21] Zhu, L.: An unity of Mitrinovic-Adamovic and Cusa-Huygens inequalities and the analogue for hyperbolic functions. Rev. R. Acad. Cienc. Exactas FÃs. Nat. Ser. A Mat. RACSAM 113(4), 3399-3412 (2019). https://doi.org/10.1007/s13398-019-00706-4

[22] M. Abramowitz and I. Stegun Handbook of mathematical functions, Appl. Math. series 55. Washington, 1972.

[23] Bagul, Y.J., Chesneau, C.: Some new simple inequalities involving exponential, trigonometric and hyperbolic functions. Cubo 21(1), 21-35 (2019)

[24] Y. Bagul and W. Du, On New Generalized Mitrinovic-Adamovic Type Inequalities, MDPI Mathematics 2025, 13, 1174, https://doi.org/10.3390/math13071174.

[25] A. Baricz and J. Sandor, Extensions of generalized Wilker inequality to Bessel functions, J. Math. Inequal. 2 (2008), 397-406.
[26] C.-P. Chen and W.-S. Cheung, Sharpness of Wilker and Huygens type inequalities, J. Inequal. Appl., 2012, 2012:72 doi:10.1186/1029-242X 2012-72.

[27] C.-P. Chen and C. Mortici, The Relationship between Huygens and Wilker inequalities and further remarks. Appl. Anal. Discrete Math. 17 (2023), 092-100

[28] C.-P. Chen and B. Malesevic, Sharp inequalities related to the Adamovic Mitrinovic, Cusa, Wilker and Huygens results. Filomat 37:19 (2023), 6319-6334. https://doi.org/10.2298/FIL2319319C

[29] C.-P. Chen and J. Sandor: Inequality chains for Wilker, Huygens and Lazarevic type inequalities, http://ajmaa.org/RGMIA/papers/v15/v15a11.pdf.

[30] Chouikha A.R., New sharp inequalities related to classical trigonometric inequalities, J. of Inequalities and Special Functions, vol 11, n 4, 2021.

[31] Chouikha, A.R. On natural approaches related to classical trigonometric inequalities. Open J. Math. Sci. 2023, 7, 299-320. https://www.doi.org/10.30538/oms2023.0212. [CrossRef]

[32] A.R. Chouikha, New Look of Trigonometric and Hyperbolic Inequalities Vol. 18 No. 4 (2025): Int. J. Open Problems Compt. Math., Vol. 18, No. 4, December 2025. https://ijopcm.icsrs.uk/index.php/journal/article/view/45

[33] A.R. Chouikha and C. Chesneau Contributions to trigonometric 1 parameter inequalities, Functiones et Approximatio Commentarii Mathematici http://dx.doi.org/10.7169/facm/240419-4-3

[34] A.R. Chouikha On the 1-parameter trigonometric and hyperbolic in equalities chains, Chapter from From "Exploring the Benefits of Numerical Simulation and Modelling", https://www.intechopen.com/chapters/1184491

[35] Jiang, W.-D. New sharp inequalities of Mitrinovic-Adamovic type. Appl. Anal. Discrete Math. 2023, 17, 76-91

[36] I. Gradshteyn, I. Ryzhik, Table of Integrals Series and Products, 8th ed. Academic Press, San Diego (2015).

[37] S. Wu, L. Debnath, A generalization of L'Hospital-type rules for monotonicity and its application, Appl. Math. Lett., 22 (2009), 284-290.

[38] B. Malesevic, T. Lutovac, M. Rasajski and C. Mortici, Extensions of the natural approach to refinements and generalizations of
some trigonometric inequalities. Adv. in Diff. Eq. ( 2018) 2018:90
https://doi.org/10.1186/s13662-018-1545-7

[39] C. Mortici, The natural approach of Wilker-Cusa-Huygens inequalities. Math Inequal Appl. 14, 535-541 (2011).

[40] C. Mortici: A subtly analysis of Wilker inequality, Appl. Math. Comput. 2014, 231, 516-520

[41] J. S. Sumner, A. A. Jagers, M. Vowe and J. Anglesio, Inequalities involv-
ing trigonometric functions, Amer. Math. Monthly 98 (1991), 264-267.

[42] L. Zhu, On Wilker-type inequalities, Math. Inequal. Appl. 10 (2007), 727-731.