a sharp double inequality for trigonometric functions and its applications

S´andor, “On some inequalities involvingtrigonometric and hyperbolic functions with emphasis on the Cusa-Huygens, Wilker, and Huygens inequalities,” Mathemati-cal Inequalities & Applica

Research Article

A Sharp Double Inequality for Trigonometric Functions and

Its Applications

Zhen-Hang Yang,1Yu-Ming Chu,1Ying-Qing Song,1and Yong-Min Li2

1 School of Mathematics and Computation Sciences, Hunan City University, Yiyang 413000, China

2 Department of Mathematics, Huzhou Teachers College, Huzhou 313000, China

Correspondence should be addressed to Yu-Ming Chu; chuyuming2005@126.com

Received 26 April 2014; Accepted 20 June 2014; Published 10 July 2014

Academic Editor: Josip E Peˇcari´c

Copyright © 2014 Zhen-Hang Yang et al This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited

We present the best possible parameters 𝑝 and 𝑞 such that the double inequality ((2/3)cos2𝑝(𝑡/2) + 1/3)1/𝑝 < sin 𝑡/𝑡 < ((2/3)cos2𝑞(𝑡/2) + 1/3)1/𝑞holds for any𝑡 ∈ (0, 𝜋/2) As applications, some new analytic inequalities are established

1 Introduction

It is well known that the double inequality

cos1/3𝑡 < sin𝑡

𝑡 <

2 + cos 𝑡

holds for any𝑡 ∈ (0, 𝜋/2) The first inequality in (1) was

found by Mitrinovi´c (see [1]), while the second inequality

in (1) is due to Huygens (see [2]) and it is called Cusa

inequality Recently, the improvements, refinements, and

generalizations for inequality (1) have attracted the attention

of many mathematicians [3–8]

Qi et al [9] proved that the inequality

cos2𝑡

2 <

sin𝑡

holds for any𝑡 ∈ (0, 𝜋/2) It is easy to verify that cos1/3𝑡 and

cos2(𝑡/2) cannot be compared on the interval (0, 𝜋/2)

Neuman and S´andor [6] gave an improvement for the first

inequality in (1) as follows:

cos4/3𝑡

2 = (

1 + cos 𝑡

2 )

2/3

< sin𝑡𝑡, 𝑡 ∈ (0,𝜋2) (3)

Inequality (3) was also proved by Lv et al in [10] In [11,

12], Neuman proved that the inequalities

cos1/3𝑡 < (sin𝑡

𝑡 cos𝑡)

1/4

< ( sin𝑡 tanh−1(sin 𝑡))

1/2

< (𝑡 cos 𝑡 + sin 𝑡

2𝑡 )

1/2

< (1 + 2 cos 𝑡

1/2

< (1 + cos 𝑡

2 )

2/3

< sin𝑡 𝑡

(4)

hold for any𝑡 ∈ (0, 𝜋/2)

For the second inequality in (1), Kl´en et al [13] established

sin𝑡

𝑡 ≤ cos3

𝑡

3 ≤

2 + cos 𝑡

for𝑡 ∈ (−√135/5, √135/5)

Inequality (5) was improved by Yang [14] In [15], Yang further proved

sin𝑡

𝑡 < (

2

3cos

𝑡

2+

1

3)

2

< cos33𝑡 < 2 + cos 𝑡3 , (6) for𝑡 ∈ (0, 𝜋/2)

http://dx.doi.org/10.1155/2014/592085

Yang [16] proved that the inequalities

cos1/3𝑡 < cos 𝑡

√3 < cos4/3

𝑡

2 <

sin𝑡 𝑡

< cos33𝑡 < cos16/34𝑡 < 𝑒−𝑡2/6< 2 + cos 𝑡3

(7)

hold for𝑡 ∈ (0, 𝜋/2)

Zhu [8] and Yang [17] proved that 𝑝 = 4/5 and 𝑞 =

(log 3−log 2)/(log 𝜋−log 2) = 0.8978 are the best possible

constants such that the double inequality

(23+13cos𝑝𝑡)1/𝑝< sin𝑡𝑡 < (23+13cos𝑞𝑡)1/𝑞 (8)

holds for all𝑡 ∈ (0, 𝜋/2)

More results involving inequality (1) can be found in the

literature [18–22]

Let𝑝 ∈ R, 𝑥 > 0, and 0 < 𝜔 < 1 Then 𝑀𝑝(𝑥, 𝜔) is defined

by

𝑀𝑝(𝑥, 𝜔) = (𝜔𝑥𝑝+ 1 − 𝜔)1/𝑝(𝑝 ̸= 0) ,

𝑀0(𝑥, 𝜔) = lim

𝑝 → 0𝑀𝑝(𝑥, 𝜔) = 𝑥𝜔 (9)

It is well known that𝑀𝑝(𝑥, 𝜔) is strictly increasing with

respect to𝑝 ∈ R for fixed 𝑥 > 0 and 0 < 𝜔 < 1 (see [23]) If

0 < 𝑥 < 1, then it is easy to check that

𝑀−∞(𝑥, 𝜔) = lim

𝑝 → −∞𝑀𝑝(𝑥, 𝜔) = 𝑥,

𝑀∞(𝑥, 𝜔) = lim𝑝 → ∞𝑀𝑝(𝑥, 𝜔) = 1 (10)

It follows from (2) and (3) together with (6) that

𝑀−∞(cos2𝑡

2,

2

3) = cos2

𝑡

2 < cos4/3

𝑡 2

= 𝑀0(cos2𝑡

2,

2

3) <

sin𝑡 𝑡

< (23cos 𝑡

2+

1

3)

2

= 𝑀1/2(cos2𝑡

2,

2

3) <

2 + cos 𝑡 3

= 𝑀1(cos2𝑡

2,

2

3) < 1

= 𝑀∞(cos22𝑡,23) ,

(11)

for𝑡 ∈ (0, 𝜋/2)

The main purpose of this paper is to present the best

possible parameters𝑝 and 𝑞 such that the double inequality

𝑀𝑝(cos2𝑡

2,

2

3) <

sin𝑡

𝑡 < 𝑀𝑞(cos2

𝑡

2,

2

3) (12) holds for all 𝑡 ∈ (0, 𝜋/2) As applications, some new

analytic inequalities are found All numerical computations

are carried out using MATHEMATICA software

2 Lemmas

In order to prove our main results we need several lemmas, which we present in this section

Lemma 1 Let 𝑝 ∈ R and the function 𝑔𝑝be defined on(1/2, 1)

by

𝑔𝑝(𝑥) = 2𝑝𝑥 − 𝑥1−𝑝+ 2𝑥𝑝− (2𝑝 + 1) (13)

Then the following statements are true:

(i)𝑔𝑝(𝑥) < 0 for all 𝑥 ∈ (1/2, 1) if and only if 𝑝 ≥ 1/5;

(ii)𝑔𝑝(𝑥) > 0 for all 𝑥 ∈ (1/2, 1) if and only if 𝑝 ≤

𝑝2, where𝑝2 = 0.1872 is the unique solution of

equation

𝑔𝑝(12) = 21−𝑝− 2𝑝−1− 𝑝 − 1 = 0; (14)

(iii) if𝑝2< 𝑝 < 1/5, then there exists 𝑥1= 𝑥1(𝑝) ∈ (1/2, 1)

such that𝑔𝑝(𝑥) < 0 for 𝑥 ∈ (1/2, 𝑥1) and 𝑔𝑝(𝑥) > 0

for𝑥 ∈ (𝑥1, 1).

Proof It follows from (13) and (14) that

𝑔0.1872(1

2) = 0.000141 > 0,

𝑔0.1873(12) = − 0.000119 < 0,

𝜕𝑔𝑝(𝑥)

𝜕𝑝 = (𝑥1−𝑝+ 2𝑥𝑝) log 𝑥 − 2 (1 − 𝑥) < 0,

(15)

for𝑥 ∈ (0, 1)

Inequalities (15) lead to the conclusion that the function

𝑔𝑝(𝑥) is strictly decreasing with respect to 𝑝 ∈ R for fixed

𝑥 ∈ (0, 1) and 𝑝2= 0.1872 is the unique solution of (14) (i) If 𝑥 ∈ (1/2, 1) and 𝑝 ≥ 1/5, then from the monotonicity of the function𝑝 → 𝑔𝑝(𝑥) we clearly see that

𝑔𝑝(𝑥) ≤ 𝑔1/5(𝑥) =25𝑥 − 𝑥4/5+ 2𝑥1/5−75

= −1

5(1 − 𝑥1/5)

2

× (−2𝑥3/5+ 𝑥2/5+ 4𝑥1/5+ 7) < 0

(16)

If𝑔𝑝(𝑥) < 0 for all 𝑥 ∈ (1/2, 1), then (13) leads to

lim

𝑥 → 1 −

𝑔𝑝(𝑥)

1 − 𝑥 = 1 − 5𝑝 ≤ 0. (17) (ii) If𝑥 ∈ (1/2, 1) and 𝑝 ≤ 0, then the monotonicity of the function𝑝 → 𝑔𝑝(𝑥) leads to the conclusion that 𝑔𝑝(𝑥) ≥

𝑔0(𝑥) = 1 − 𝑥 > 0

If 𝑥 ∈ (1/2, 1) and 0 < 𝑝 ≤ 𝑝2, then (13) and the monotonicity of the function𝑝 → 𝑔𝑝(𝑥) lead to

𝑔𝑝(12) ≥ 𝑔𝑝2(12) = 0, 𝑔𝑝(1) = 0, (18)

𝜕2𝑔𝑝(𝑥)

𝜕𝑥2 = 𝑝 (𝑝 − 1) 𝑥𝑝−2(2 − 𝑥1−2𝑝) < 0 (19)

Inequality (19) implies that the function𝑔𝑝(𝑥) is concave

with respect to𝑥 on the interval (1/2, 1) Therefore, 𝑔𝑝(𝑥) > 0

follows from (18) and the concavity of𝑔𝑝(𝑥)

If𝑔𝑝(𝑥) > 0 for all 𝑥 ∈ (1/2, 1), then 𝑝 ≤ 𝑝2follows

easily from the monotonicity of the function𝑝 → 𝑔𝑝(1/2)

and𝑔𝑝(1/2) ≥ 0 together with the fact that 𝑔𝑝2(1/2) = 0

(iii) If 𝑥 ∈ (1/2, 1) and 𝑝2 < 𝑝 < 1/5, then from (13)

and (19) together with the monotonicity of the function𝑝 →

𝑔𝑝(1/2) we get

𝑔𝑝(1) = 0, 𝑔𝑝(1

2) < 𝑔𝑝2(1

2) = 0, (20)

𝑔𝑝󸀠(1) = 5𝑝 − 1 < 0, (21)

𝑔𝑝󸀠(12) = 2𝑝 − 2𝑝+ 𝑝2𝑝+ 2𝑝21−𝑝

> 2 × 0.1872 − 20.1873 + 0.1872 × 20.1872 + 2 × 0.1872 × 20.8127

= 0.1065 > 0,

(22)

and𝑔󸀠

𝑝(𝑥) is strictly decreasing on (1/2, 1)

It follows from (21) and (22) together with the

monotonic-ity of𝑔󸀠

𝑝(𝑥) that there exists 𝑥0 = 𝑥0(𝑝) ∈ (1/2, 1) such that

𝑔𝑝(𝑥) is strictly increasing on (1/2, 𝑥0] and strictly decreasing

on[𝑥0, 1) Therefore,Lemma 1(iii) follows from (20) and the

piecewise monotonicity of𝑔𝑝(𝑥)

Let𝑝 ∈ R and the function 𝑓𝑝be defined on(0, 𝜋/2) by

𝑓𝑝(𝑡) = 𝑡 − 2cos2𝑝(𝑡/2) + 1

cos𝑡 + 2cos2𝑝(𝑡/2)sin𝑡. (23) Then elaborated computations lead to

𝑓𝑝󸀠(𝑡) = 4 (1 − cos

2(𝑡/2)) cos2𝑝(𝑡/2) (2cos2𝑝(𝑡/2) + 2cos2(𝑡/2) − 1)2𝑔𝑝(cos

2𝑡

2) , (24) where𝑔𝑝(𝑥) is defined by (13)

FromLemma 1and (24) we get the followingLemma 2

immediately

Lemma 2 Let 𝑝 ∈ R and 𝑓𝑝be defined on (0, 𝜋/2) by (23).

Then

(i)𝑓𝑝(𝑡) is strictly decreasing on (0, 𝜋/2) if and only if 𝑝 ≥

1/5;

(ii)𝑓𝑝(𝑡) is strictly increasing on (0, 𝜋/2) if and only if 𝑝 ≤

𝑝2, where𝑝2= 0.1872 is the unique solution of (14);

(iii) if𝑝2 < 𝑝 < 1/5, then there exists 𝑡1 = 𝑡1(𝑝) ∈

(0, 𝜋/2) such that 𝑓𝑝(𝑡) is strictly increasing on (0, 𝑡1]

and strictly decreasing on[𝑡1, 𝜋/2).

Lemma 3 Let 𝑝 ∈ R and 𝑓𝑝be defined on (0, 𝜋/2) by (23).

Then

(i)𝑓𝑝(𝑡) < 0 for all 𝑡 ∈ (0, 𝜋/2) if and only if 𝑝 ≥ 1/5;

(ii)𝑓𝑝(𝑡) > 0 for all 𝑡 ∈ (0, 𝜋/2) if and only if 𝑝 ≤ 𝑝1 = log(𝜋 − 2)/ log 2 = 0.1910 ;

(iii) if𝑝1< 𝑝 < 1/5, then there exists 𝑡0 = 𝑡0(𝑝) ∈ (0, 𝜋/2)

such that𝑓𝑝(𝑡) > 0 for 𝑡 ∈ (0, 𝑡0) and 𝑓𝑝(𝑡) < 0 for

𝑡 ∈ (𝑡0, 𝜋/2).

Proof (i) If𝑡 ∈ (0, 𝜋/2) and 𝑝 ≥ 1/5, then from (23) and

Lemma 2(i) we clearly see that

𝑓𝑝(𝑡) < 𝑓𝑝(0+) = 0 (25)

If𝑓𝑝(𝑡) < 0 for all 𝑡 ∈ (0, 𝜋/2), then (23) leads to

0 ≥ lim

𝑡 → 0 +

𝑓𝑝(𝑡)

𝑡5 = lim

𝑡 → 0 +

(1/180) (1 − 5𝑝) 𝑡5+ 𝑜 (𝑡5)

𝑡5

= 1 − 5𝑝180

(26)

(ii) If𝑓𝑝(𝑡) > 0 for all 𝑡 ∈ (0, 𝜋/2), then from (23) we get

0 ≤ 𝑓𝑝(𝜋2−) = 𝜋 − 2 − 22 𝑝 (27) Inequality (27) leads to the conclusion that𝑝 ≤ log(𝜋 − 2)/ log 2

If𝑡 ∈ (0, 𝜋/2) and 𝑝 ≤ 𝑝1 = log(𝜋 − 2)/ log 2, then we divide the proof into two cases

Case 1 Consider𝑝 ≤ 𝑝2, where𝑝2is the unique solution of (14) Then fromLemma 2(ii) and (23) we clearly see that

𝑓𝑝(𝑡) > 𝑓𝑝(0+) = 0 (28)

Case 2 Consider𝑝2 < 𝑝 ≤ 𝑝1 Then (23) andLemma 2(iii) lead to

𝑓𝑝(0+) = 0,

𝑓𝑝(𝜋 2

−

) = 𝜋 − 2 − 2𝑝

𝜋 − 2 − 2𝑝 1

2 = 0,

(29)

and there exists𝑡1= 𝑡1(𝑝) such that 𝑓𝑝(𝑡) is strictly increasing

on (0, 𝑡1] and strictly decreasing on [𝑡1, 𝜋/2) Therefore,

𝑓𝑝(𝑡) > 0 for all 𝑡 ∈ (0, 𝜋/2) follows from (29) and the piecewise monotonicity of𝑓𝑝(𝑡)

(iii) If𝑝1 < 𝑝 < 1/5, then 𝑝2 < 𝑝 < 1/5 It follows from (23) andLemma 2(iii) that

𝑓𝑝(0+) = 0,

𝑓𝑝(𝜋 2

−

) = 𝜋 − 2 − 2𝑝

2 <

𝜋 − 2 − 2𝑝 1

2 = 0,

(30)

and there exists 𝑡1 = 𝑡1(𝑝) such that 𝑓𝑝(𝑡) is strictly increasing on (0, 𝑡1] and strictly decreasing on [𝑡1, 𝜋/2) Therefore,Lemma 3(iii) follows from (30) and the piecewise monotonicity of𝑓𝑝(𝑡)

Let𝑝 ∈ R and 𝐹𝑝be defined on(0, 𝜋/2) by

𝐹𝑝(𝑡) = logsin𝑡

𝑡 −

1

𝑝log(

2

3cos2𝑝

𝑡

2 +

1

3) (𝑝 ̸= 0) , (31)

𝐹0(𝑡) = lim

𝑝 → 0𝐹𝑝(𝑡) = logsin𝑡

𝑡 −

4

3log(cos

𝑡

2) (32)

Then elaborated computations give

𝐹𝑝󸀠(𝑡) = cos𝑡 + 2cos2𝑝(𝑡/2)

𝑡 (1 + 2cos2𝑝(𝑡/2)) sin 𝑡𝑓𝑝(𝑡) , (33) where𝑓𝑝(𝑡) is defined by (23)

FromLemma 3and (33) we getLemma 4immediately

Lemma 4 Let 𝑝 ∈ R and 𝐹𝑝be defined on (0, 𝜋/2) by (31)

and (32) Then

(i)𝐹𝑝(𝑡) is strictly decreasing on (0, 𝜋/2) if and only if 𝑝 ≥

1/5;

(ii)𝐹𝑝(𝑡) is strictly increasing on (0, 𝜋/2) if and only if 𝑝 ≤

𝑝1= log(𝜋 − 2)/ log 2 = 0.1910 ;

(iii) if𝑝1 < 𝑝 < 1/5, then there exists 𝑡0 = 𝑡0(𝑝) ∈

(0, 𝜋/2) such that 𝐹𝑝(𝑡) is strictly increasing on (0, 𝑡0]

and strictly decreasing on[𝑡0, 𝜋/2).

Lemma 5 Let 𝑝 ∈ R and 𝐹𝑝be defined on (0, 𝜋/2) by (31)

and (32) Then the following statements are true:

(i) if𝐹𝑝(𝑡) < 0 for all 𝑡 ∈ (0, 𝜋/2), then 𝑝 ≥ 1/5;

(ii) if𝐹𝑝(𝑡) > 0 for all 𝑡 ∈ (0, 𝜋/2), then 𝑝 ≤ 𝑝0, where

𝑝0= 0.1941 is the unique solution of the equation

𝑝 log2

𝜋− log (1 + 21−𝑝) + log 3 = 0, (34)

on the interval (0.1, ∞).

Proof. (i) If 𝐹𝑝(𝑡) < 0 for all 𝑡 ∈ (0, 𝜋/2), then from (31) and

(32) we have

0 ≥ lim

𝑡 → 0 +

𝐹𝑝(𝑡)

𝑡4 = lim

𝑡 → 0 +

(1/720) (1 − 5𝑝) 𝑡4+ 𝑜 (𝑡4)

𝑡4

= 1 − 5𝑝

720 .

(35)

(ii) We first prove that 𝑝0 = 0.1941 is the unique solution

of (34) on the interval(0.1, ∞) Let 𝑝 ∈ (0.1, ∞) and

𝐻 (𝑝) = 𝑝 log𝜋2 − log (1 + 21−𝑝) + log 3 (36)

Then numerical computations show that

𝐻 (0.1941) = 8.13 × 10−7 > 0,

𝐻 (0.1942) = − 2.52 × 10−7< 0, (37)

𝐻󸀠(𝑝) = log𝜋2 +2 + 2log4𝑝 < log𝜋2 +2 + 2log40.1

= − 2.81 × 10−4< 0

(38)

Inequality (38) implies that𝐻(𝑝) is strictly decreasing on

[0.1, ∞) Therefore, 𝑝0= 0.1941 is the unique solution of

(34) on the interval(0.1, ∞) which follows from (37) and the

monotonicity of𝐻(𝑝)

If𝑝 > 0.1 and 𝐹𝑝(𝑡) > 0 for all 𝑡 ∈ (0, 𝜋/2), then (31) leads to

0 ≤ 𝐹𝑝(𝜋2+) = 1𝑝𝐻 (𝑝) (39) Therefore, 𝑝 ≤ 𝑝0 follows from (39) and 𝐻(𝑝0) =

0 together with the monotonicity of 𝐻(𝑝) on the interval (0.1, ∞)

Lemma 6 Let 𝑝 ∈ R and 𝑥, 𝑐, 𝜔 ∈ (0, 1), and let 𝑀𝑝(𝑥, 𝜔) be

defined by (9) Then the function𝑝 󳨃→ 𝑀𝑝(𝑥, 𝜔)/𝑀𝑝(𝑐, 𝜔) is

strictly decreasing with respect to 𝑝 ∈ R if 𝑥 ∈ (𝑐, 1).

Proof Let𝐻(𝑝, 𝑥) = log 𝑀𝑝(𝑥, 𝜔) − log 𝑀𝑝(𝑐, 𝜔) Then from (9) we get

𝜕𝐻 (𝑝, 𝑥)

𝜕𝑥 =

𝜔𝑥𝑝−1

𝜔𝑥𝑝+ 1 − 𝜔, (40)

𝜕2𝐻 (𝑝, 𝑥)

𝜕𝑝𝜕𝑥 = 𝜔 (1 − 𝜔) 𝑥𝑝−1

(𝜔𝑥𝑝+ 1 − 𝜔)2log𝑥 < 0. (41) Inequality (41) and𝜕2𝐻(𝑝, 𝑥)/𝜕𝑥𝜕𝑝 = 𝜕2𝐻(𝑝, 𝑥)/𝜕𝑝𝜕𝑥 lead

to the conclusion that 𝜕𝐻(𝑝, 𝑥)/𝜕𝑝 is strictly decreasing with respect to 𝑥 ∈ (𝑐, 1) Therefore, 𝜕𝐻(𝑝, 𝑥)/𝜕𝑝 <

𝜕𝐻(𝑝, 𝑥)/𝜕𝑝|𝑥=𝑐 = 0 for 𝑥 ∈ (𝑐, 1), and 𝑀𝑝(𝑥, 𝜔)/𝑀𝑝(𝑐, 𝜔)

is strictly decreasing with respect to𝑝 ∈ R if 𝑥 ∈ (𝑐, 1)

3 Main Results

Theorem 7 Let 𝑀𝑝(𝑥, 𝜔) be defined by (9) Then the double

inequality

𝜆𝑝𝑀𝑝(cos22𝑡,23) < sin𝑡𝑡 < 𝑀𝑝(cos22𝑡,23) (42)

holds for all 𝑡 ∈ (0, 𝜋/2) if and only if 𝑝 ≥ 1/5, and the double

inequality

𝑀𝑝(cos22𝑡,23) < sin𝑡𝑡 < 𝜆𝑝𝑀𝑝(cos22𝑡,23) (43)

holds for all 𝑡 ∈ (0, 𝜋/2) if and only if 𝑝 ≤ 𝑝1, where

𝜆𝑝= 𝜋2(1 + 231−𝑝)

−1/𝑝

(𝑝 ̸= 0) , 𝜆0= 25/3𝜋 , (44)

𝑝1= log(𝜋 − 2)/ log 2 = 0.1910 , and 𝜆𝑝𝑀𝑝(cos2(𝑡/2), 2/3)

is strictly decreasing with respect to 𝑝 ∈ R.

Proof Let𝑝 ∈ R and 𝐹𝑝(𝑡) be defined on (0, 𝜋/2) by (31) and (32) Then

𝐹𝑝(0+) = 0, 𝐹𝑝(𝜋2−) = log 𝜆𝑝 (45)

If𝑝 ≥ 1/5, then inequality (42) follows fromLemma 4 (i) and (45)

If inequality (42) holds for all𝑡 ∈ (0, 𝜋/2), then 𝐹𝑝(𝑡) < 0 for all𝑡 ∈ (0, 𝜋/2) It follows fromLemma 5(i) that 𝑝 ≥ 1/5

If𝑝 ≤ 𝑝1, then inequality (43) follows fromLemma 4 (ii)

and (45)

If inequality (43) holds for all 𝑡 ∈ (0, 𝜋/2), then

𝐹𝑝(𝜋/2−) > 𝐹𝑝(𝑡) > 𝐹𝑝(0+) = 0 for all 𝑡 ∈ (0, 𝜋/2) It follows

fromLemma 5(ii) that 𝑝 ≤ 𝑝0, where𝑝0 = 0.1941 is the

unique solution of (34) on the interval(0.1, ∞) We claim that

𝑝 ≤ 𝑝1; otherwise𝑝1 < 𝑝 ≤ 𝑝0 < 1/5, andLemma 4 (iii)

leads to the conclusion that there exists𝑡0 ∈ (0, 𝜋/2) such

that𝐹𝑝(𝑡) > 𝐹𝑝(𝜋/2−) for 𝑡 ∈ [𝑡0, 𝜋/2)

Note that

𝜆𝑝𝑀𝑝(cos2𝑡

2,

2

3) =

2 𝜋

𝑀𝑝(cos2(𝑡/2) , 2/3)

𝑀𝑝(1/2, 2/3) . (46)

It follows from Lemma 6 and (46) that

𝜆𝑝𝑀𝑝(cos2(𝑡/2), 2/3) is strictly decreasing with respect

to𝑝 ∈ R

FromTheorem 7we get Corollaries8and9as follows

Corollary 8 For all 𝑡 ∈ (0, 𝜋/2) one has

2

𝜋 <

2 + cos 𝑡

𝜋 = 𝜆1𝑀1(cos22𝑡,23)

< 𝜆1/2(2

3cos

𝑡

2+

1

3)

2

< 𝜆1/4(2

3cos1/2

𝑡

2+

1

3)

4

< 𝜆1/5(2

3cos2/5

𝑡

2+

1

3)

5

<sin𝑡 𝑡

< (2

3cos2/5

𝑡

2+

1

3)

5

< (2

3cos1/2

𝑡

2+

1

3)

4

< (23cos𝑡

2 +

1

3)

2

< 𝑀1(cos2𝑡

2,

2

3) =

2 + cos 𝑡

3 < 1.

(47)

Corollary 9 For all 𝑡 ∈ (0, 𝜋/2) one has

cos2𝑡

2 = 𝑀−∞(cos2

𝑡

2,

2

3) < 3 (1 + cos 𝑡)

5 + cos 𝑡

= 𝑀−1(cos2𝑡

2,

2

3)

< 9cos2(𝑡/2)

(2 + cos (𝑡/2))2 = 𝑀−1/2(cos

2𝑡

2,

2

3)

< cos4/3𝑡

2 = 𝑀0(cos2

𝑡

2,

2

3)

< (2

3cos1/4

𝑡

2+

1

3)

8

< (2

3cos1/3

𝑡

2+

1

3)

6

< sin𝑡𝑡 < 𝜆1/6(23cos1/3𝑡

2+

1

3)

6

< 𝜆1/8(23cos1/4𝑡

2+

1

3)

8

< 𝜆0cos4/3𝑡

2

< 𝜆−1/2 9cos2(𝑡/2) (2 + cos (𝑡/2))2 < 𝜆−13 (1 + cos 𝑡)

5 + cos 𝑡

< 𝜆−∞cos2𝑡

2 =

4

𝜋cos2

𝑡

2.

(48)

Theorem 10 Let 𝑀𝑝(𝑥, 𝜔) be defined by (9) Then the double

inequality

𝑀𝑝(cos2𝑡

2,

2

3) <

sin𝑡

𝑡 < 𝑀𝑞(cos2

𝑡

2,

2

3) (49)

holds for all 𝑡 ∈ (0, 𝜋/2) if and only if 𝑝 ≤ 𝑝0and 𝑞 ≥ 1/5,

where𝑝0 = 0.1941 is the unique solution of (34) on the

interval (0.1, ∞) Moreover, the inequality

sin𝑡

𝑡 ≤ 𝛼𝑀𝑝 0(cos22𝑡,23) , (50)

if and only if

𝛼 ≥ sin𝑡0

𝑡0𝑀𝑝0(cos2(𝑡0/2) , 2/3) = 1.00004919 , (51)

where𝑡0∈ (0, 𝜋/2) is defined as in Lemma 3 (iii).

Proof Let𝑝 ∈ R and 𝐹𝑝(𝑡) be defined on (0, 𝜋/2) by (31) and (32) Then Lemma 4(iii) leads to the conclusion that

𝐹𝑝0(𝑡) is strictly increasing on (0, 𝑡0] and strictly decreasing

on[𝑡0, 𝜋/2) Note that

𝐹𝑝0(0+) = 𝐹𝑝0(𝜋

2

−

) = 0 (52)

It follows from the piecewise monotonicity of𝐹𝑝0(𝑡) and (52) that

0 < 𝐹𝑝0(𝑡) ≤ 𝐹𝑝0(𝑡0) , (53) for all𝑡 ∈ (0, 𝜋/2) Therefore, sin 𝑡/𝑡 > 𝑀𝑝0(cos2(𝑡/2), 2/3) for all𝑡 ∈ (0, 𝜋/2) follows from the first inequality of (53), while sin𝑡/𝑡 < 𝑀1/5(cos2(𝑡/2), 2/3) for all 𝑡 ∈ (0, 𝜋/2) follows from the second inequality of (42)

Conversely, if the double inequality (49) holds for all𝑡 ∈ (0, 𝜋/2), then we clearly see that the inequalities

𝐹𝑝(𝑡) > 0, 𝐹𝑞(𝑡) < 0 (54) hold for all𝑡 ∈ (0, 𝜋/2) Therefore, 𝑝 ≤ 𝑝0and𝑞 ≥ 1/5 follows fromLemma 5and (54) Moreover, numerical computations show that𝑡0= 1.312 and

𝑒𝐹𝑝0 (𝑡 0 )= 1.00004919 (55) Therefore, the second conclusion ofTheorem 10 follows from (55) and the second inequality of (53)

It follows fromLemma 3that we getTheorem 11 immedi-ately

Theorem 11 The double inequalities

2cos2𝑝(𝑡/2) + cos 𝑡

2cos2𝑝(𝑡/2) + 1 <

sin𝑡

𝑡 <

2cos2𝑞(𝑡/2) + cos 𝑡 2cos2𝑞(𝑡/2) + 1 , 2cos2𝑝2𝑡 < sin𝑡 − sin 𝑡𝑡 − 𝑡 cos 𝑡 < 2cos2𝑞2𝑡

(56)

hold for all 𝑡 ∈ (0, 𝜋/2) if and only if 𝑝 ≥ 1/5 and 𝑞 ≤ 𝑝1 =

log(𝜋 − 2)/ log 2 = 0.1910

We clearly see that the function (2cos2𝑝(𝑡/2) +

cos𝑡)/(2cos2𝑝(𝑡/2) + 1) is strictly decreasing with respect

to𝑝 ∈ R for fixed 𝑥 ∈ (0, 𝜋/2) Let 𝑝 = 1/2, 1, 2, ∞ and

𝑞 = 1/6, 0, −1/2, −1, −2, −∞; thenTheorem 11 leads to

the following

Corollary 12 The inequalities

cos𝑡 < 8 cos 𝑡 + cos 2𝑡 + 3

4 cos 𝑡 + cos 2𝑡 + 7 <

2 cos 𝑡 + 1 cos𝑡 + 2

<2 cos (𝑡/2) + cos 𝑡

2 cos (𝑡/2) + 1 <

sin𝑡 𝑡

<cos𝑡 + 2cos1/3(𝑡/2)

2cos1/3(𝑡/2) + 1 <

cos𝑡 + 2 3

<cos(𝑡/2) + cos (3𝑡/2) + 4

2 cos (𝑡/2) + 4 <

cos𝑡cos2(𝑡/2) + 2 cos2(𝑡/2) + 2

<cos𝑡cos4(𝑡/2) + 2

cos4(𝑡/2) + 2 < 1

(57)

hold for all 𝑡 ∈ (0, 𝜋/2).

4 Applications

In this section, we give some applications for our main results

Neuman [24] proved that the Huygens type inequalities

2sin𝑡𝑡 +tan𝑡 𝑡 > sin𝑡𝑡+ 2tan𝑡/2(𝑡/2)

> 2 𝑡 sin𝑡+

𝑡 tan𝑡 > 3, (sin𝑡

𝑡 )

𝑝

+ 2(tan(𝑡/2) 𝑡/2 )

𝑝

> ( 𝑡

sin𝑡)

𝑝

+ 2( 𝑡/2 tan(𝑡/2))

𝑝

(𝑝 > 0) ,

( 𝑡

sin𝑡)

𝑝

+ 2( 𝑡/2 tan(𝑡/2))

𝑝

> 3 (𝑝 ≥ 1)

(58)

hold for all𝑡 ∈ (0, 𝜋/2) Note that

sin𝑡

𝑡 < (>) 𝑀𝑝(cos22𝑡,23) ⇐⇒ ( 𝑡

sin𝑡)

𝑝

+ 2 ( 𝑡/2 tan(𝑡/2)) > (<) 3, sin𝑡

𝑡 > (<) 𝜆𝑝𝑀𝑝(cos22𝑡,23)

⇐⇒ ( 𝑡 sin𝑡)

𝑝

+ 2( 𝑡/2 tan(𝑡/2))

𝑝

< (>) (𝜋2)𝑝+ 2(𝜋4)𝑝,

(59)

if𝑝 > 0, and the second inequalities in (59) are reversed if

𝑝 < 0

From Theorems7and 10together with (59) we get the following

Theorem 13 The double inequality

(𝜋2)𝑝+ 2(𝜋4)𝑝> ( 𝑡

sin𝑡)

𝑝

+ 2(tan(𝑡/2)𝑡/2 )𝑝> 3 (60)

holds for all 𝑡 ∈ (0, 𝜋/2) if and only if 𝑝 ≥ 1/5 or 𝑝 < 0,

and inequality (60) is reversed if and only if0 < 𝑝 ≤ 𝑝1 = log(𝜋 − 2)/ log 2 = 0.1910

Theorem 14 The double inequality

( 𝑡 sin𝑡)

𝑝

+ 2(tan(𝑡/2)𝑡/2 )𝑝> 3 > ( 𝑡

sin𝑡)

𝑞

+ 2(tan(𝑡/2)𝑡/2 )𝑞

(61)

holds for all 𝑡 ∈ (0, 𝜋/2) if and only if 0 < 𝑞 ≤ 𝑝0and𝑝 ≥ 1/5

or 𝑝 < 0, where 𝑝0 = 0.1941 is the unique solution of (34)

on the interval (0.1, ∞).

Neuman [24] also proved that the Wilker type inequality

( 𝑡 sin𝑡)

𝑝

+ ( 𝑡/2 tan(𝑡/2))

2𝑝

> 2 (62)

holds for all𝑡 ∈ (0, 𝜋/2) if 𝑝 ≥ 1

Making use ofTheorem 13and the arithmetic-geometric means inequality

1 + (tan(𝑡/2)𝑡/2 )2𝑝> 2(tan(𝑡/2)𝑡/2 )𝑝, (63)

we getCorollary 15as follows

Corollary 15 The Wilker type inequality (62 ) holds for all𝑡 ∈

(0, 𝜋/2) if 𝑝 ≥ 1/5 or 𝑝 < 0.

In addition, power series expansions show that

( 𝑡 sin𝑡)

𝑝

+ (tan(𝑡/2)𝑡/2 )2𝑝− 2 = 𝑝 (20𝑝 − 3)720 𝑡4+ 𝑜 (𝑡4)

(64)

Therefore, we conjecture that inequality (62) holds for all𝑡 ∈

(0, 𝜋/2) if and only if 𝑝 ≥ 3/20 or 𝑝 < 0 We leave it to the

readers for further discussion.

The Schwab-Borchardt mean 𝑆𝐵(𝑎, 𝑏) [ 25 – 27 ] of two

distinct positive real numbers 𝑎 and 𝑏 is defined by

𝑆𝐵 (𝑎, 𝑏) =

{ { { { {

√𝑏2− 𝑎2

cos−1(𝑎/𝑏), 𝑎 < 𝑏,

√𝑎2− 𝑏2

cosh−1(𝑎/𝑏), 𝑎 > 𝑏,

(65)

where cos−1(𝑥) and cosh−1(𝑥) = log(𝑥 + √𝑥2− 1) are

the inverse cosine and inverse hyperbolic cosine functions,

respectively.

Let 𝑏 > 𝑎 > 0, 𝐴(𝑎, 𝑏) = (𝑎 + 𝑏)/2 be the arithmetic

mean of 𝑎 and 𝑏, and 𝑡 = cos−1(𝑎/𝑏) ∈ (0, 𝜋/2) Then simple

computations lead to

sin𝑡

𝑡 =

𝑆𝐵 (𝑎, 𝑏)

𝑏 ,

𝑀𝑝(cos22𝑡,23) = 1𝑏(23𝐴𝑝(𝑎, 𝑏) +𝑏3𝑝)1/𝑝,

2cos2𝑝(𝑡/2) + cos 𝑡

2cos2𝑝(𝑡/2) + 1 =

2𝐴𝑝(𝑎, 𝑏) + 𝑎𝑏𝑝−1 2𝐴𝑝(𝑎, 𝑏) + 𝑏𝑝

(66)

It follows from Theorems 7 , 10 , and 11 together with (66)

that we have the following.

Theorem 16 Let 𝑝1 = 𝑙𝑜𝑔(𝜋 − 2)/𝑙𝑜𝑔2 = 0.1910 , 𝜆𝑝and

𝑝0= 0.1941 be defined as in Theorems 7 and 10 , respectively.

Then for all 𝑏 > 𝑎 > 0, the following statements are true.

(i) The double inequality

𝜆𝑝(23𝐴𝑝(𝑎, 𝑏) +13𝑏𝑝)1/𝑝

< 𝑆𝐵 (𝑎, 𝑏) < (23𝐴𝑝(𝑎, 𝑏) +1

3𝑏𝑝)

holds if and only if 𝑝 ≥ 1/5, and inequality (67) is

reversed if and only if𝑝 ≤ 𝑝1.

(ii) The double inequality

(23𝐴𝑝(𝑎, 𝑏) +13𝑏𝑝)1/𝑝

< 𝑆𝐵 (𝑎, 𝑏) < (23𝐴𝑞(𝑎, 𝑏) +1

3𝑏𝑞)

holds if and only if𝑝 ≤ 𝑝0and 𝑞 ≥ 1/5.

(iii) The double inequality

2𝐴𝑝(𝑎, 𝑏) + 𝑎𝑏𝑝−1

2𝐴𝑝(𝑎, 𝑏) + 𝑏𝑝 𝑏

< 𝑆𝐵 (𝑎, 𝑏) < 2𝐴2𝐴𝑞(𝑎, 𝑏) + 𝑎𝑏𝑞(𝑎, 𝑏) + 𝑏𝑞−1𝑞 𝑏

(69)

holds if and only if 𝑝 ≥ 1/5 and 𝑞 ≤ 𝑝1.

Let𝑏 > 𝑎 > 0, 𝐺(𝑎, 𝑏) = √𝑎𝑏, 𝑄(𝑎, 𝑏) = √(𝑎2+ 𝑏2)/2, 𝑃(𝑎, 𝑏) = (𝑏 − 𝑎)/[2sin−1((𝑏 − 𝑎)/(𝑏 + 𝑎))], 𝑇(𝑎, 𝑏) = (𝑏 − 𝑎)/[2tan−1((𝑏 − 𝑎)/(𝑏 + 𝑎))], and 𝑌(𝑎, 𝑏) = (𝑏 − 𝑎)/[√2tan−1((𝑏 − 𝑎)/√2𝑎𝑏)] be the geometric, quadratic, first Seiffert [28], second Seiffert [29], and Yang [15] means

of 𝑎 and 𝑏, respectively Then it is easy to check that 𝑃(𝑎, 𝑏) = SB(𝐺(𝑎, 𝑏), 𝐴(𝑎, 𝑏)), 𝑇(𝑎, 𝑏) = SB(𝐴(𝑎, 𝑏), 𝑄(𝑎, 𝑏)), and 𝑌(𝑎, 𝑏) = SB(𝐺(𝑎, 𝑏), 𝑄(𝑎, 𝑏)) Therefore, Theorem 16

leads toCorollary 17

Corollary 17 Let 𝑝1 = log(𝜋 − 2)/ log 2 = 0.1910 , 𝜆𝑝

and 𝑝0 = 0.1941 be defined as in Theorems 7 and 10 , respectively Then for all 𝑏 > 𝑎 > 0, the following statements

are true.

(i) The double inequalities

𝜆𝑝[2

3(

𝐺 (𝑎, 𝑏) + 𝐴 (𝑎, 𝑏)

𝑝

+1

3𝐴𝑝(𝑎, 𝑏)]

1/𝑝

< 𝑃 (𝑎, 𝑏)

< [2

3(

𝐺(𝑎, 𝑏) + 𝐴(𝑎, 𝑏)

𝑝

+1

3𝐴𝑝(𝑎, 𝑏)]

1/𝑝

𝜆𝑝[2

3(

𝐴 (𝑎, 𝑏) + 𝑄 (𝑎, 𝑏)

𝑝

+1

3𝑄𝑝(𝑎, 𝑏)]

1/𝑝

< 𝑇 (𝑎, 𝑏)

< [2

3(𝐴 (𝑎, 𝑏) + 𝑄 (𝑎, 𝑏)

𝑝

+1

3𝑄𝑝(𝑎, 𝑏)]

1/𝑝

,

𝜆𝑝[2

3(

𝐺 (𝑎, 𝑏) + 𝑄 (𝑎, 𝑏)

𝑝

+1

3𝑄𝑝(𝑎, 𝑏)]

1/𝑝

< 𝑌 (𝑎, 𝑏)

< [2

3(𝐺 (𝑎, 𝑏) + 𝑄 (𝑎, 𝑏)

𝑝

+1

3𝑄𝑝(𝑎, 𝑏)]

1/𝑝

, (70)

hold if and only if 𝑝 ≥ 1/5, and all inequalities in (70) are

reversed if and only if𝑝 ≤ 𝑝1.

(ii) The double inequalities

[23(𝐺 (𝑎, 𝑏) + 𝐴 (𝑎, 𝑏)

𝑝

+13𝐴𝑝(𝑎, 𝑏)]

1/𝑝

< 𝑃 (𝑎, 𝑏)

< [2

3(𝐺 (𝑎, 𝑏) + 𝐴 (𝑎, 𝑏)

𝑞

+1

3𝐴𝑞(𝑎, 𝑏)]

1/𝑞

,

[2

3(

𝐴 (𝑎, 𝑏) + 𝑄 (𝑎, 𝑏)

𝑝

+1

3𝑄𝑝(𝑎, 𝑏)]

1/𝑝

< 𝑇 (𝑎, 𝑏)

< [23(𝐴 (𝑎, 𝑏) + 𝑄 (𝑎, 𝑏)

𝑞

+13𝑄𝑞(𝑎, 𝑏)]1/𝑞, [23(𝐺 (𝑎, 𝑏) + 𝑄 (𝑎, 𝑏)

𝑝

+13𝑄𝑝(𝑎, 𝑏)]

1/𝑝

< 𝑌 (𝑎, 𝑏)

< [2

3(

𝐺 (𝑎, 𝑏) + 𝑄 (𝑎, 𝑏)

𝑞

+1

3𝑄𝑞(𝑎, 𝑏)]

1/𝑞

(71)

hold if and only if𝑝 ≤ 𝑝0and 𝑞 ≥ 1/5.

(iii) The double inequalities

21−𝑝(𝐺 (𝑎, 𝑏) + 𝐴 (𝑎, 𝑏))𝑝𝐴 (𝑎, 𝑏) + 𝐺 (𝑎, 𝑏) 𝐴𝑝(𝑎, 𝑏)

21−𝑝(𝐺 (𝑎, 𝑏) + 𝐴 (𝑎, 𝑏))𝑝+ 𝐴𝑝(𝑎, 𝑏)

< 𝑃 (𝑎, 𝑏)

< 21−𝑞(𝐺 (𝑎, 𝑏) + 𝐴 (𝑎, 𝑏))𝑞𝐴 (𝑎, 𝑏) + 𝐺 (𝑎, 𝑏) 𝐴𝑞(𝑎, 𝑏)

21−𝑞(𝐺 (𝑎, 𝑏) + 𝐴 (𝑎, 𝑏))𝑞+ 𝐴𝑞(𝑎, 𝑏) ,

21−𝑝(𝐴 (𝑎, 𝑏) + 𝑄 (𝑎, 𝑏))𝑝𝑄 (𝑎, 𝑏) + 𝐴 (𝑎, 𝑏) 𝑄𝑝(𝑎, 𝑏)

21−𝑝(𝐴 (𝑎, 𝑏) + 𝑄 (𝑎, 𝑏))𝑝+ 𝑄𝑝(𝑎, 𝑏)

< 𝑇 (𝑎, 𝑏)

< 21−𝑞(𝐴 (𝑎, 𝑏) + 𝑄 (𝑎, 𝑏))𝑞𝑄 (𝑎, 𝑏) + 𝐴 (𝑎, 𝑏) 𝑄𝑞(𝑎, 𝑏)

21−𝑞(𝐴 (𝑎, 𝑏) + 𝑄 (𝑎, 𝑏))𝑞+ 𝑄𝑞(𝑎, 𝑏) ,

21−𝑝(𝐺 (𝑎, 𝑏) + 𝑄 (𝑎, 𝑏))𝑝𝑄 (𝑎, 𝑏) + 𝐺 (𝑎, 𝑏) 𝑄𝑝(𝑎, 𝑏)

21−𝑝(𝐺 (𝑎, 𝑏) + 𝑄 (𝑎, 𝑏))𝑝+ 𝑄𝑝(𝑎, 𝑏)

< 𝑌 (𝑎, 𝑏)

< 21−𝑞(𝐺 (𝑎, 𝑏) + 𝑄 (𝑎, 𝑏))𝑞𝑄 (𝑎, 𝑏) + 𝐺 (𝑎, 𝑏) 𝑄𝑞(𝑎, 𝑏)

21−𝑞(𝐺 (𝑎, 𝑏) + 𝑄 (𝑎, 𝑏))𝑞+ 𝑄𝑞(𝑎, 𝑏)

(72)

hold if and only if 𝑝 ≥ 1/5 and 𝑞 ≤ 𝑝1.

For 𝑥 ∈ (0, 1), the following Shafer-Fink type inequality can

be found in the literature [ 1 , 30 ]:

sin−1𝑥 > 6 (√1 + 𝑥 − √1 − 𝑥)

4 + √1 + 𝑥 + √1 − 𝑥 >

3𝑥

2 + √1 − 𝑥2 (73)

Fink [ 31 ] proved that the double inequality

3𝑥

2 + √1 − 𝑥2 ≤ sin−1𝑥 ≤ 𝜋𝑥

2 + √1 − 𝑥2 (74)

holds for all 𝑥 ∈ [0, 1] It was generalized and improved by Zhu

[ 32 ].

Let𝑡 ∈ (0, 𝜋/2), 𝑥 = sin 𝑡 ∈ (0, 1) Then Theorems7,10,

and11lead toCorollary 18as follows

Corollary 18 Let 𝑝1= log(𝜋 − 2)/ log 2 = 0.1910 , 𝜆𝑝and

𝑝0= 0.1941 be defined as in Theorems 7 and 10 , respectively.

Then for all 𝑥 ∈ (0, 1), the following statements are true.

(i) The double inequality

𝑥

𝑀𝑝((1 + √1 − 𝑥2) /2, 2/3)

< sin−1(𝑥) < 𝑥

𝜆𝑝𝑀𝑝((1 + √1 − 𝑥2) /2, 2/3)

(75)

holds if and only if 𝑝 ≥ 1/5, and inequality (75) is reversed if

and only if𝑝 ≤ 𝑝1.

(ii) The double inequality

𝑥

𝑀𝑝((1 + √1 − 𝑥2) /2, 2/3)

< sin−1(𝑥) < 𝑥

𝑀𝑞((1 + √1 − 𝑥2) /2, 2/3)

(76)

holds if and only if 𝑝 ≥ 1/5 and 𝑞 ≤ 𝑝0.

(iii) The double inequality

21−2𝑝(√1 + 𝑥 + √1 − 𝑥)2𝑝+ 1

21−2𝑝(√1 + 𝑥 + √1 − 𝑥)2𝑝+ √1 − 𝑥2

< sin−1(𝑥)

< 2

1−2𝑞(√1 + 𝑥 + √1 − 𝑥)2𝑞+ 1

21−2𝑞(√1 + 𝑥 + √1 − 𝑥)2𝑞+ √1 − 𝑥2

(77)

holds if and only if𝑝 ≤ 𝑝1and 𝑞 ≥ 1/5.

Conflict of Interests

The authors declare that there is no conflict of interests regarding the publication of this paper

Acknowledgments

This research was supported by the Natural Science Foun-dation of China under Grants 11371125, 61174076, 61374086, and 11171307, the Natural Science Foundation of Zhejiang Province under Grant LY13A010004, and the Natural Science Foundation of Hunan Province under Grant 14JJ2127

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