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Article

# Refinements of Wilker–Huygens-Type Inequalities via Trigonometric Series

by
Gabriel Bercu
Department of Mathematics and Computer Sciences, “Dunărea de Jos” University of Galaţi, 800201 Galaţi, Romania
Symmetry 2021, 13(8), 1323; https://doi.org/10.3390/sym13081323
Submission received: 23 June 2021 / Revised: 19 July 2021 / Accepted: 21 July 2021 / Published: 22 July 2021
(This article belongs to the Special Issue Symmetry in Nonlinear Functional Analysis and Optimization Theory II)

## Abstract

:
The study of even functions is important from the symmetry theory point of view because their graphs are symmetrical to the $O y$ axis; therefore, it is essential to analyse the properties of even functions for x greater than 0. Since the functions involved in Wilker–Huygens-type inequalities are even, in our approach, we use cosine polynomials expansion method in order to provide new refinements of the above-mentioned inequalities.
MSC:
41A21; 42B05; 26D05; 26D15

## 1. Introduction

The famous Huygens inequality for trigonometric functions states that for any 0 $< x < π 2$ one has
$2 sin x x + tan x x > 3$
while the Wilker inequality asserts that
$sin x x 2 + tan x x > 2 .$
In [1], S.-H. Wu and H. M. Srivastava established the following inequality, which is sometime known as the second Wilker inequality:
$2 x sin x + x tan x > 3 , 0 < x < π 2$
and the following inequality, which is also sometime known as the second Wilker inequality:
$x sin x 2 + x tan x > 2 , 0 < x < π 2 .$
In [2], the inequality (3) is established with another bound
$2 x sin x + x tan x − 3 > 1 60 x 3 sin x ,$
for .
In [3], the inequality (4) is written with another bound
for .
In [4], E. Neumann and J. Sándor proved the following inequality
$3 x sin x + cos x > 4 for 0 < x < π 2 .$
In the paper [2], the inequality (7) is established with another bound
$3 x sin x + cos x − 4 > 1 10 x 3 sin x ,$
for .
In the same work, [4], E. Neumann and J. Sándor also showed the hyperbolic variants of the inequalities (3) and (4)
$2 x sinh x + x tanh x − 3 > 0 , for all x ≠ 0$
and
$x sinh x 2 + x tanh x − 2 > 0 , for all x ≠ 0 .$
In the paper [3], the inequality (10) is written with another bound
for $x > 0$.
The hyperbolic counterpart of the inequality (7) is:
$3 x sinh x + cosh x − 4 > 0 , for all x ≠ 0 .$
These inequalities were extended in different forms in the recent past. We refer to [1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17] and closely related references therein. Some of the recent improvements were obtained using Taylor’s expansion or Padé approximation of the trigonometric functions involved.
In [6], we improved the Huygens and Wilker inequalities using the cosine polynomial method.
The aim of this work is to reformulate the inequalities (3)–(12) using again the cosine polynomial method. The main idea is that the functions involved in the above inequalities are even, so can be expanded in trigonometric series:
$2 x sin x + x tan x − 3 = a 1 + b 1 cos x + c 1 cos 2 x + … ,$
$x sin x 2 + x tan x − 2 = a 2 + b 2 cos x + c 2 cos 2 x + … ,$
$3 x sin x + cos x − 4 = a 3 + b 3 cos x + c 3 cos 2 x + … ,$
$2 x sinh x + x tanh x − 3 = a 4 + b 4 cos x + c 4 cos 2 x + … ,$
$x sinh x 2 + x tanh x − 2 = a 5 + b 5 cos x + c 5 cos 2 x + … ,$
$3 x sinh x + cosh x − 4 = a 6 + b 6 cos x + c 6 cos 2 x + … .$
The above functions can be also expanded as hyperbolic cosine polynomials:
$2 x sinh x + x tanh x − 3 = a 7 + b 7 cosh x + c 7 cosh 2 x + … ,$
$x sinh x 2 + x tanh x − 2 = a 8 + b 8 cosh x + c 8 cosh 2 x + … ,$
$3 x sin x + cos x − 4 = a 9 + b 9 cosh x + c 9 cosh 2 x + … .$
In the following we will present our method for the first function.
We introduce the function $F 1 x$ by
$F 1 x = a 1 + b 1 cos x + c 1 cos 2 x .$
The power series expansion of $2 x sin x + x tan x − 3 − F 1 x$ near 0 is
In order to increase the speed of the function $F 1 x$ approximating $2 x sin x + x tan x − 3$, we vanish the first coefficients as follows:
and we obtain $a 1 = 1 10 , b 1 = − 4 30$ and $c 1 = 1 30 .$
Then, we obtain
or, equivalently,
Using the same algorithm, we find
and

## 2. Main Results

Using the Fourier trigonometric series method we can establish our main theorems, which are refined and simple forms of the inequalities (3)–(12).
Theorem 1.
(Wilker–Huygens-type inequalities)
(i)
The following inequality
holds for all $0 < x < π 2$.
(ii)
The following inequality
holds for all $0 < x < π 2$.
(iii)
The following inequality
holds for all $0 < x < π 2$.
Theorem 2.
(Wilker–Huygens-type inequalities for hyperbolic functions)
(i)
For all $x ≠ 0$, one has
(ii)
For all $x ≠ 0$, one has
(iii)
For all $x ≠ 0$, one has
Theorem 3.
(Mixed type of Wilker–Huygens inequalities)
(i)
For all $x ≠ 0$, one has
(ii)
For all x, $0 < x < 1.50618$, one has
(iii)
For all $x ≠ 0$, one has

## 3. The Proofs of the Theorems

We first prove two lemmas.
Lemma 1.
(i) For every $x ≥ 0$, one has
$2 sinh x ≥ sin 2 x .$
(ii)
For every , one has
$11 2 sin x + 45 x + 123 2 x cos x − 112 sin x cos 2 x > 0 .$
(iii)
For every , one has
$4 ( 1 − cos x ) 2 > x 3 sin x .$
Proof.
(i) We consider the function
The derivative of the function g is
Then g is increasing on . As $g 0 = 0$, we find that $g ≥ 0$ on .
(ii) We define the function
We can rearrange p as follows
For , we have
It follows that $p > 0$ on .
(iii) We introduce the function
An alternate form of h is
Using the formula
$sin 3 x = 3 sin x − 4 sin 3 x ,$
we have
The Adamović and Mitrinović inequality (see, e.g., ([8] p. 238)) asserts that
$( cos x ) 1 3 < sin x x$
holds for every .
Therefore, we obtain that $h ( x ) > 0$ for every . □
Lemma 2.
For every $x ≠ 0$, one has
$4 ( 1 − cos x ) 2 < x 3 sinh x .$
Proof.
We define the even function
$r ( x ) = x 3 sinh x − 2 cos 2 x + 8 cos x − 6 , x > 0 .$
We have
$r ′ ( x ) = x 3 cosh x − 8 sin x + 4 sin 2 x + 3 x 2 sinh x ,$
$r ( 2 ) x = − 8 cos x + 8 cos 2 x + 6 x 2 cosh x + ( 6 x + x 3 ) sinh x ,$
$r ( 3 ) ( x ) = x ( 18 + x 2 ) cosh x + 8 ( sin x − 2 sin 2 x ) + ( 6 + 9 x 2 ) sinh x ,$
$r ( 4 ) ( x ) = 8 ( cos x − 4 cos 2 x ) + 12 ( 2 + x 2 ) cosh x + x ( 36 + x 2 ) sinh x ,$
$r ( 5 ) ( x ) = x ( 60 + x 2 ) cosh x − 8 ( sin x − 8 sin 2 x ) + 15 ( 4 + x 2 ) sinh x$
and
From Lemma 1, (i), we deduce that $cosh x ≥ 5 − cos 2 x 4$ for all $x ∈ R$.
Then,
or, equivalently,
Therefore, $r 5$ is strictly increasing on . Since $r 5 0 = 0$, it follows that $r 5 x > 0$ for all $x > 0$. Continuing the algorithm, we finally find that $r > 0$ on . □
Proof of Theorem 1.
(i) Due to the form of the inequality (13), if the inequality (13) holds for $0 < x < π 2$, then it holds for $− π 2 < x < 0$.
Therefore, we can consider $x > 0$.
The inequality (13) takes the following equivalent form:
We introduce the function
The derivative of the function $f 1$ is
The function
has the derivative
Since $f 2 ′ > 0$ on , it follows that $f 2$ is strictly increasing on . As $f 2 0 = 0$ we obtain that $f 2 > 0$ on .
Then, $f 1 ′ > 0$ on . Using the same arguments, we finally find that $f 1 > 0$ on .
(ii) The functions involved in the inequality (14) are even functions, so it is sufficient to prove for .
We write the inequality (14) as follows:
We define the function
Elementary calculations reveal that
The function
has the derivative
According to the second part of the Lemma 1, we have
$11 2 sin x + 45 x + 123 2 x cos x − 112 sin x cos 2 x > 0$
on . We also have
on .
We obtain that $f 4 ′ > 0$ on , then $f 4$ is strictly increasing on .
As $f 4 0 = 0$, we prove that $f 4 > 0$ on .
Therefore, $f 3 2 > 0$ on . Using the same arguments, we finally find that $f 3 > 0$ on .
(iii) We can assume that .
We rewrite the inequality (15) as follows:
The function
has the derivative
The function $f 5 ′$ is $> 0$ on , hence $f 5$ is strictly increasing on . Since $f 5 0 = 0$, we find that $f 5 > 0$ on .
This completes the proof of the Theorem 1. □
Proof of Theorem 2.
(i) We assume that $x > 0$. We have to prove the following inequality:
for all $x > 0$.
We introduce the function
The derivative of the function $f 6$ is
The function ,
has the derivative
Since $f 7 ′ x < 0$ on , it follows that $f 7$ is strictly decreasing on . Since $f 7 0 = 0$, we have $f 7 < 0$ on , then $f 6 ′ < 0$ on .
Hence, $f 6$ is strictly decreasing on .
As $f 6 0 = 0$, we finally obtain that $f 6 < 0$ on .
(ii) We have to prove that
for all $x > 0$.
The function
has the derivatives:
and
To find critical points of the function $f 8 3$, first, we calculate the derivative $f 8 3$:
Solving the equation $f 8 3 x = 0$ yields $x = 0$.
Therefore, the only critical point of the function $f 8 2$ is $x = 0$. Then, we evaluate $f 8 2$ at the critical point and at the endpoint of the domain:
$f 8 2 0 = 0 , lim x → ∞ f 8 2 x = − ∞ .$
Hence, the function $f 8 2$ has a global maximum at $x = 0$: $f 8 2 0 = 0$.
Then, $f 8 ′$ is strictly decreasing on . As $f 8 ′ 0 = 0$, we obtain $f 8 ′ < 0$ on .
It follows that $f 8$ is strictly decreasing on . As $f 8 0 = 0$, we find $f 8 < 0$ on .
(iii) We have to prove that
for $x > 0 .$
We consider the function
The derivative of the function $f 9$ is
$f 9 ′ x = − 48 sinh 6 x 2 .$
Then $f 9 ′ < 0$ on , hence $f 9$ is strictly decreasing on .
As $f 9 0 = 0$, we find that $f 9 < 0$ on .
The proof of the Theorem 2 is complete. □
Proof of Theorem 3.
(i) Since the functions involved in the inequality (19) are even functions, we can assume that $x > 0$.
The inequality (19) takes the equivalent form:
We consider the function
The derivatives of the function $f 10$ is
The function
has the positive roots $x = 0$, $x ≈ 0.85321$. Then, $s < 0$ on and $s > 0$ on . It follows that $f 10 5 x > 0$ on .
If , then $f 10 5 x$ can be rewritten as
The function
has the positive roots $x = 0$, $x ≈ 2.34534$ and $t > 0$ on .
Hence, $f 10 5 x > 0$ on . It follows that $f 10 4$ is strictly increasing on . As $f 10 4 0 = 0$, we find $f 10 4 > 0$ on . Continuing the algorithm we finally obtain that $f 10 > 0$ on .
(ii) We also can assume $x > 0$.
We write the inequality (20) as follows:
The function
has the derivative
The equation $f 11 ′ x = 0$ has the positive roots $x = 0$, $x ≈ 1.35234$. Moreover, $f 11 ′ x < 0$ for . Then, $f 11$ is strictly decreasing on and it is strictly increasing on . Since $f 11 0 = 0$ and , it follows that $f 11 < 0$ on .
(iii) As in the above theorems, we can assume $x > 0$.
We rearrange the inequality (21) as follows:
We introduce the function
Easy computation yields
and
According to the first part of the Lemma 1 we have $2 sinh x ≥ sin 2 x$ for all $x ≥ 0$.
Then,
Therefore, we find that
In the following, we will prove that the function
is positive on .
The derivatives of the function h are
$h ′ x = 8 sin x − 7 sin 2 x + 16 sinh x$
and
Since the function $h 2 > 0$ on it follows that $h ′$ is strictly increasing on . As $h ′ 0 = 0$, we get $h ′ > 0$ on . Continuing the algorithm, we finally obtain that $h > 0$ on .
Hence, we deduce that $f 12 4 > 0$ on . Using the same arguments as above, we finally find that $f 12 > 0$ on .
The proof of the Theorem 3 is complete. □
Remark 1.
(1)   From the Lemma 1, (iii), it follows that
$1 15 ( 1 − cos x ) 2 > 1 60 x 3 sin x , for 0 < x < π 2 ,$
$8 45 ( 1 − cos x ) 2 > 2 45 x 3 sin x , for 0 < x < π 2 ,$
$2 5 ( 1 − cos x ) 2 > 1 10 x 3 sin x , for 0 < x < π 2 .$
Therefore, we also improved the inequalities (5), (6) and ( 8).
(2) From the Lemma 2 and Theorem 3, we find that
for all x, $0 < x < 1.50618$, hence we improved the inequality (11).

## 4. Conclusions

The function
$s i n c x = sin x x$
occurs in Fourier analysis and its applications in signal processing. The Fourier transform of the $s i n c$ function is a rectangle, and the Fourier transform of a rectangular pulse is a $s i n c$ function. The $s i n c$ function also appears in analysis of digital-to-analogue conversion.
In our work, Taylor expansion of the error function between the truncated sum of the first terms of the cosine series of the functions involved in Wilker–Huygens-type inequalities and the functions themselves is carried out. Then the best approximation of the functions which improve Wilker–Huygens-type inequalities is obtained.
These new approximations give sharp bounds to the functions $s i n c ( x )$ and
$t a n c ( x ) = tan x x .$
For example, the inequalities
and
are very sharp and interesting for further studies.

## Funding

The APC was funded by “Dunărea de Jos” University of Galaţi, Romania.

Not applicable.

Not applicable.

Not applicable.

## Conflicts of Interest

The author declare no conflict of interest.

## References

1. Wu, S.-H.; Srivastava, H.M. A further refinement of Wilker’s inequality. Int. Trans. Spec. Funct. 2008, 19, 757–765. [Google Scholar] [CrossRef]
2. Jiang, W.D.; Luo, Q.M.; Qi, F. Refinements and Sharpening of some Huygens and Wilker Type Inequalities. Turk. J. Math. 2014, 2, 134–139. [Google Scholar] [CrossRef] [Green Version]
3. Sun, Z.; Zhu, L. On New Wilker-Type Inequalities. ISRN Math. Anal. 2011. [Google Scholar] [CrossRef] [Green Version]
4. Neumann, E.; Sándor, J. On some inequalities involving trigonometric and huperbolic functions with emphases on the Cusa-Huygens, Wilker and Huygens inequalities. Math. Inequal. Appl. 2010, 13, 715–723. [Google Scholar]
5. Bercu, G. Padé approximant related to remarkable inequalities involving trigonometric functions. J. Inequal. Appl. 2016, 1, 99. [Google Scholar] [CrossRef] [Green Version]
6. Bercu, G. Fourier series method related to Wilker-Cusa-Huygens inequalities. Math. Inequal. Appl. 2019, 22, 1091–1098. [Google Scholar] [CrossRef] [Green Version]
7. Boyd, J.P. Chebyshev and Fourier Spectral Methods; Dover: New York, NY, USA, 2000. [Google Scholar]
8. Mitrinović, D.S. Analytic Inequalities; Springer: Berlin, Germany, 1970. [Google Scholar]
9. Mitrinović, D.S.; Pečarić, J.E.; Fink, A.M. Classical and New Inequalities in Analysis; Kluwer Academic Publishers: Drive Norwell, MA, USA, 1993; Volume 61. [Google Scholar]
10. Mortici, C. The natural approach of Wilker-Cusa-Huygens inequalities. Math. Inequal. Appl. 2011, 14, 535–541. [Google Scholar] [CrossRef] [Green Version]
11. Nenezić, M.; Malešević, B.; Mortici, C. Accurate approximations of some expressions involving trigonometric functions. Appl. Math. Comput. 2016, 283, 299–315. [Google Scholar]
12. Wu, S.-H.; Srivastava, H.M. A weighted and exponential generalization of Wilker’s inequality and its applications. Integral Transform. Spec. Funct. 2007, 18, 525–535. [Google Scholar] [CrossRef]
13. Wu, S.-H.; Debnath, L. Wilker-type inequalities for hyperbolic functions. Appl. Math. Lett. 2012, 25, 837–843. [Google Scholar] [CrossRef] [Green Version]
14. Zhu, L. New inequalities of Wilker’s type for hyperbolic functions. AIMS Math. 2019, 5, 376–386. [Google Scholar] [CrossRef]
15. Zhu, L. On Wilker-type inequalities. Math. Inequal. Appl. 2007, 10, 727–731. [Google Scholar] [CrossRef]
16. Agarwal, R.P. Difference Equations and Inequalities: Theory, Methods and Applications, 2nd ed.; Marcel Dekker, Inc.: New York, NY, USA, 2000. [Google Scholar]
17. Bagul, Y.J.; Dhaigude, R.M.; Bhayo, B.A.; Raut, V.M. Wilker and Huygens’ type inequalities for mixed trigonometric-hyperbolic functions. Tbil. Math. J. 2021, 14, 207–220. [Google Scholar]
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Bercu, G. Refinements of Wilker–Huygens-Type Inequalities via Trigonometric Series. Symmetry 2021, 13, 1323. https://doi.org/10.3390/sym13081323

AMA Style

Bercu G. Refinements of Wilker–Huygens-Type Inequalities via Trigonometric Series. Symmetry. 2021; 13(8):1323. https://doi.org/10.3390/sym13081323

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Bercu, Gabriel. 2021. "Refinements of Wilker–Huygens-Type Inequalities via Trigonometric Series" Symmetry 13, no. 8: 1323. https://doi.org/10.3390/sym13081323

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