# A Sharp Double Inequality for Trigonometric Functions and Its Applications

Authors: Zhen-Hang Yang; Yu-Ming Chu; Ying-Qing Song; Yong-Min Li
Journal: Abstract and Applied Analysis
 (2014)

---

## Abstract

We present the best possible parametersp and q such that the double inequality (
2
/
3
)
cos
2
p
(
t
/
2
)
+
1
/
3
1
/
p
<
sin
t
/
t
<
(
2
/
3
)
cos
2
q
(
t
/
2
)
+
1
/
3
1
/
q holds for any t
∈
(
0
,
π
/
2
). As applications, some new analytic inequalities are established.


---

## Body

## 1. Introduction

It is well known that the double inequality(1)
cos
⁡
1
/
3
t
<
sin
t
t
<
2
+
cos
⁡
t
3
holds for any t
∈
(
0
,
π
/
2
). The first inequality in (1) was found by Mitrinović (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
(2)
cos
⁡
2
t
2
<
sin
t
t
holds for any t
∈
(
0
,
π
/
2
). It is easy to verify that cos
⁡
1
/
3
t and cos
⁡
2
(
t
/
2
) cannot be compared on the interval (
0
,
π
/
2
).Neuman and Sándor [6] gave an improvement for the first inequality in (1) as follows:
(3)
cos
⁡
4
/
3
t
2
=
(
1
+
cos
⁡
t
2
)
2
/
3
<
sin
t
t
,
t
∈
(
0
,
π
2
)
.Inequality (3) was also proved by Lv et al. in [10]. In [11, 12], Neuman proved that the inequalities
(4)
cos
⁡
1
/
3
t
<
(
sin
t
t
cos
⁡
t
)
1
/
4
<
(
sin
t
tanh
-
1
(
sin
t
)
)
1
/
2
<
(
t
cos
⁡
t
+
sin
t
2
t
)
1
/
2
<
(
1
+
2
cos
⁡
t
3
)
1
/
2
<
(
1
+
cos
⁡
t
2
)
2
/
3
<
sin
t
t
hold for any t
∈
(
0
,
π
/
2
).For the second inequality in (1), Klén et al. [13] established
(5)
sin
t
t
≤
cos
⁡
3
t
3
≤
2
+
cos
⁡
t
3
for t
∈
(
-
135
/
5
,
135
/
5
).Inequality (5) was improved by Yang [14]. In [15], Yang further proved
(6)
sin
t
t
<
(
2
3
cos
⁡
t
2
+
1
3
)
2
<
cos
⁡
3
t
3
<
2
+
cos
⁡
t
3
,
for t
∈
(
0
,
π
/
2
).Yang [16] proved that the inequalities
(7)
cos
⁡
1
/
3
t
<
cos
⁡
t
3
<
cos
⁡
4
/
3
t
2
<
sin
t
t
<
cos
⁡
3
t
3
<
cos
⁡
16
/
3
t
4
<
e
-
t
2
/
6
<
2
+
cos
⁡
t
3
hold for t
∈
(
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
(8)
(
2
3
+
1
3
cos
⁡
p
t
)
1
/
p
<
sin
t
t
<
(
2
3
+
1
3
cos
⁡
q
t
)
1
/
q
holds for all t
∈
(
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 M
p
(
x
,
ω
) is defined by
(9)
M
p
(
x
,
ω
)
=
(
ω
x
p
+
1
-
ω
)
1
/
p
(
p
≠
0
)
,
M
0
(
x
,
ω
)
=
lim
⁡
p
→
0
M
p
(
x
,
ω
)
=
x
ω
.It is well known thatM
p
(
x
,
ω
) is strictly increasing with respect to p
∈
R for fixed x
>
0 and 0
<
ω
<
1 (see [23]). If 0
<
x
<
1, then it is easy to check that
(10)
M
-
∞
(
x
,
ω
)
=
lim
⁡
p
→
-
∞
M
p
(
x
,
ω
)
=
x
,
M
∞
(
x
,
ω
)
=
lim
⁡
p
→
∞
M
p
(
x
,
ω
)
=
1
.It follows from (2) and (3) together with (6) that
(11)
M
-
∞
(
cos
⁡
2
t
2
,
2
3
)
=
cos
⁡
2
t
2
<
cos
⁡
4
/
3
t
2
=
M
0
(
cos
⁡
2
t
2
,
2
3
)
<
sin
t
t
<
(
2
3
cos
⁡
t
2
+
1
3
)
2
=
M
1
/
2
(
cos
⁡
2
t
2
,
2
3
)
<
2
+
cos
⁡
t
3
=
M
1
(
cos
⁡
2
t
2
,
2
3
)
<
1
=
M
∞
(
cos
⁡
2
t
2
,
2
3
)
,
for t
∈
(
0
,
π
/
2
).The main purpose of this paper is to present the best possible parametersp and q such that the double inequality
(12)
M
p
(
cos
⁡
2
t
2
,
2
3
)
<
sin
t
t
<
M
q
(
cos
⁡
2
t
2
,
2
3
)
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.
Letp
∈
R and the function g
p be defined on (
1
/
2,1
) by
(13)
g
p
(
x
)
=
2
p
x
-
x
1
-
p
+
2
x
p
-
(
2
p
+
1
)
.
Then the following statements are true:(i)
g
p
(
x
)
<
0 for all x
∈
(
1
/
2,1
) if and only if p
≥
1
/
5;
(ii)
g
p
(
x
)
>
0 for all x
∈
(
1
/
2,1
) if and only if p
≤
p
2, where p
2
=
0.1872
…  is the unique solution of equation
(14)
g
p
(
1
2
)
=
2
1
-
p
-
2
p
-
1
-
p
-
1
=
0
;
(iii)
ifp
2
<
p
<
1
/
5, then there exists x
1
=
x
1
(
p
)
∈
(
1
/
2,1
) such that g
p
(
x
)
<
0 for x
∈
(
1
/
2
,
x
1
) and g
p
(
x
)
>
0 for x
∈
(
x
1
,
1
).Proof.
It follows from (13) and (14) that
(15)
g
0.1872
(
1
2
)
=
0.000141
…
>
0
,
g
0.1873
(
1
2
)
=
-
0.000119
…
<
0
,
∂
g
p
(
x
)
∂
p
=
(
x
1
-
p
+
2
x
p
)
log
⁡
x
-
2
(
1
-
x
)
<
0
,
for x
∈
(
0,1
).
Inequalities (15) lead to the conclusion that the function g
p
(
x
) is strictly decreasing with respect to p
∈
R for fixed x
∈
(
0,1
) and p
2
=
0.1872
…  is the unique solution of (14).
(i) Ifx
∈
(
1
/
2,1
) and p
≥
1
/
5, then from the monotonicity of the function p
→
g
p
(
x
) we clearly see that
(16)
g
p
(
x
)
≤
g
1
/
5
(
x
)
=
2
5
x
-
x
4
/
5
+
2
x
1
/
5
-
7
5
=
-
1
5
(
1
-
x
1
/
5
)
2
×
(
-
2
x
3
/
5
+
x
2
/
5
+
4
x
1
/
5
+
7
)
<
0
.
Ifg
p
(
x
)
<
0 for all x
∈
(
1
/
2,1
), then (13) leads to
(17)
lim
⁡
x
→
1
-
g
p
(
x
)
1
-
x
=
1
-
5
p
≤
0
.
(ii) Ifx
∈
(
1
/
2,1
) and p
≤
0, then the monotonicity of the function p
→
g
p
(
x
) leads to the conclusion that g
p
(
x
)
≥
g
0
(
x
)
=
1
-
x
>
0.
Ifx
∈
(
1
/
2,1
) and 0
<
p
≤
p
2, then (13) and the monotonicity of the function p
→
g
p
(
x
) lead to
(18)
g
p
(
1
2
)
≥
g
p
2
(
1
2
)
=
0
,
g
p
(
1
)
=
0
,
(19)
∂
2
g
p
(
x
)
∂
x
2
=
p
(
p
-
1
)
x
p
-
2
(
2
-
x
1
-
2
p
)
<
0
.
Inequality (19) implies that the function g
p
(
x
) is concave with respect to x on the interval (
1
/
2,1
). Therefore, g
p
(
x
)
>
0 follows from (18) and the concavity of g
p
(
x
).
Ifg
p
(
x
)
>
0 for all x
∈
(
1
/
2,1
), then p
≤
p
2 follows easily from the monotonicity of the function p
→
g
p
(
1
/
2
) and g
p
(
1
/
2
)
≥
0 together with the fact that g
p
2
(
1
/
2
)
=
0.
(iii) Ifx
∈
(
1
/
2,1
) and p
2
<
p
<
1
/
5, then from (13) and (19) together with the monotonicity of the function p
→
g
p
(
1
/
2
) we get
(20)
g
p
(
1
)
=
0
,
g
p
(
1
2
)
<
g
p
2
(
1
2
)
=
0
,
(21)
g
p
′
(
1
)
=
5
p
-
1
<
0
,
(22)
g
p
′
(
1
2
)
=
2
p
-
2
p
+
p
2
p
+
2
p
2
1
-
p
>
2
×
0.1872
-
2
0.1873
+
0.1872
×
2
0.1872
+
2
×
0.1872
×
2
0.8127
=
0.1065
…
>
0
,
and g
p
′
(
x
) is strictly decreasing on (
1
/
2,1
).
It follows from (21) and (22) together with the monotonicity of g
p
′
(
x
) that there exists x
0
=
x
0
(
p
)
∈
(
1
/
2,1
) such that g
p
(
x
) is strictly increasing on (
1
/
2
,
x
0
] and strictly decreasing on [
x
0
,
1
). Therefore, Lemma 1  (
iii
) follows from (20) and the piecewise monotonicity of g
p
(
x
).
Letp
∈
R and the function f
p be defined on (
0
,
π
/
2
) by
(23)
f
p
(
t
)
=
t
-
2
cos
⁡
2
p
(
t
/
2
)
+
1
cos
⁡
t
+
2
cos
⁡
2
p
(
t
/
2
)
sin
t
.
Then elaborated computations lead to(24)
f
p
′
(
t
)
=
4
(
1
-
cos
⁡
2
(
t
/
2
)
)
cos
⁡
2
p
(
t
/
2
)
(
2
cos
⁡
2
p
(
t
/
2
)
+
2
cos
⁡
2
(
t
/
2
)
-
1
)
2
g
p
(
cos
⁡
2
t
2
)
,
where g
p
(
x
) is defined by (13).From Lemma1 and (24) we get the following Lemma 2 immediately.Lemma 2.
Letp
∈
R and f
p be defined on (
0
,
π
/
2
) by (23). Then(i)
f
p
(
t
) is strictly decreasing on (
0
,
π
/
2
) if and only if p
≥
1
/
5;
(ii)
f
p
(
t
) is strictly increasing on (
0
,
π
/
2
) if and only if p
≤
p
2, where p
2
=
0.1872
… is the unique solution of (14);
(iii)
ifp
2
<
p
<
1
/
5, then there exists t
1
=
t
1
(
p
)
∈
(
0
,
π
/
2
) such that f
p
(
t
) is strictly increasing on (
0
,
t
1
] and strictly decreasing on [
t
1
,
π
/
2
).Lemma 3.
Letp
∈
R and f
p be defined on (
0
,
π
/
2
) by (23). Then(i)
f
p
(
t
)
<
0 for all t
∈
(
0
,
π
/
2
) if and only if p
≥
1
/
5;
(ii)
f
p
(
t
)
>
0 for all t
∈
(
0
,
π
/
2
) if and only if p
≤
p
1
=
log
⁡
(
π
-
2
)
/
log
⁡
2
=
0.1910
…;
(iii)
ifp
1
<
p
<
1
/
5, then there exists t
0
=
t
0
(
p
)
∈
(
0
,
π
/
2
) such that f
p
(
t
)
>
0 for t
∈
(
0
,
t
0
) and f
p
(
t
)
<
0 for t
∈
(
t
0
,
π
/
2
).Proof.
(i) Ift
∈
(
0
,
π
/
2
) and p
≥
1
/
5, then from (23) and Lemma 2  (
i
) we clearly see that
(25)
f
p
(
t
)
<
f
p
(
0
+
)
=
0
.
Iff
p
(
t
)
<
0 for all t
∈
(
0
,
π
/
2
), then (23) leads to
(26)
0
≥
lim
⁡
t
→
0
+
f
p
(
t
)
t
5
=
lim
⁡
t
→
0
+
(
1
/
180
)
(
1
-
5
p
)
t
5
+
o
(
t
5
)
t
5
=
1
-
5
p
180
.
(ii) If f
p
(
t
)
>
0 for all t
∈
(
0
,
π
/
2
), then from (23) we get
(27)
0
≤
f
p
(
π
2
-
)
=
π
-
2
-
2
p
2
.
Inequality (27) leads to the conclusion that p
≤
log
⁡
(
π
-
2
)
/
log
⁡
2.
Ift
∈
(
0
,
π
/
2
) and p
≤
p
1
=
log
⁡
(
π
-
2
)
/
log
⁡
2, then we divide the proof into two cases.
Case  1. Consider p
≤
p
2, where p
2 is the unique solution of (14). Then from Lemma 2  (
ii
) and (23) we clearly see that(28)
f
p
(
t
)
>
f
p
(
0
+
)
=
0
.
Case  2. Consider p
2
<
p
≤
p
1. Then (23) and Lemma 2  (
iii
) lead to
(29)
f
p
(
0
+
)
=
0
,
f
p
(
π
2
-
)
=
π
-
2
-
2
p
2
≥
π
-
2
-
2
p
1
2
=
0
,
and there exists t
1
=
t
1
(
p
) such that f
p
(
t
) is strictly increasing on (
0
,
t
1
] and strictly decreasing on [
t
1
,
π
/
2
). Therefore, f
p
(
t
)
>
0 for all t
∈
(
0
,
π
/
2
) follows from (29) and the piecewise monotonicity of f
p
(
t
).
(iii) Ifp
1
<
p
<
1
/
5, then p
2
<
p
<
1
/
5. It follows from (23) and Lemma 2  (
iii
) that
(30)
f
p
(
0
+
)
=
0
,
f
p
(
π
2
-
)
=
π
-
2
-
2
p
2
<
π
-
2
-
2
p
1
2
=
0
,
and there exists t
1
=
t
1
(
p
) such that f
p
(
t
) is strictly increasing on (
0
,
t
1
] and strictly decreasing on [
t
1
,
π
/
2
). Therefore, Lemma 3  (
iii
) follows from (30) and the piecewise monotonicity of f
p
(
t
).
Letp
∈
R and F
p be defined on (
0
,
π
/
2
) by
(31)
F
p
(
t
)
=
log
⁡
sin
t
t
-
1
p
log
⁡
(
2
3
cos
⁡
2
p
t
2
+
1
3
)
(
p
≠
0
)
,
(32)
F
0
(
t
)
=
lim
⁡
p
→
0
F
p
(
t
)
=
log
⁡
sin
t
t
-
4
3
log
⁡
(
cos
⁡
t
2
)
.
Then elaborated computations give(33)
F
p
′
(
t
)
=
cos
⁡
t
+
2
cos
⁡
2
p
(
t
/
2
)
t
(
1
+
2
cos
⁡
2
p
(
t
/
2
)
)
sin
t
f
p
(
t
)
,
where f
p
(
t
) is defined by (23).From Lemma3 and (33) we get Lemma 4 immediately.Lemma 4.
Letp
∈
R and F
p be defined on (
0
,
π
/
2
) by (31) and (32). Then(i)
F
p
(
t
) is strictly decreasing on (
0
,
π
/
2
) if and only if p
≥
1
/
5;
(ii)
F
p
(
t
) is strictly increasing on (
0
,
π
/
2
) if and only if p
≤
p
1
=
log
⁡
(
π
-
2
)
/
log
⁡
2
=
0.1910
…;
(iii)
ifp
1
<
p
<
1
/
5, then there exists t
0
=
t
0
(
p
)
∈
(
0
,
π
/
2
) such that F
p
(
t
) is strictly increasing on (
0
,
t
0
] and strictly decreasing on [
t
0
,
π
/
2
).Lemma 5.
Letp
∈
R and F
p be defined on (
0
,
π
/
2
) by (31) and (32). Then the following statements are true:(i)
ifF
p
(
t
)
<
0 for all t
∈
(
0
,
π
/
2
), then p
≥
1
/
5;
(ii)
ifF
p
(
t
)
>
0 for all t
∈
(
0
,
π
/
2
), then p
≤
p
0, where p
0
=
0.1941
…  is the unique solution of the equation
(34)
p
log
⁡
2
π
-
log
⁡
(
1
+
2
1
-
p
)
+
log
⁡
3
=
0
,on the interval (
0.1
,
∞
).Proof.
(
i
) If F
p
(
t
)
<
0 for all t
∈
(
0
,
π
/
2
), then from (31) and (32) we have
(35)
0
≥
lim
⁡
t
→
0
+
F
p
(
t
)
t
4
=
lim
⁡
t
→
0
+
(
1
/
720
)
(
1
-
5
p
)
t
4
+
o
(
t
4
)
t
4
=
1
-
5
p
720
.
(
ii
) We first prove that p
0
=
0.1941
…  is the unique solution of (34) on the interval (
0.1
,
∞
). Let p
∈
(
0.1
,
∞
) and
(36)
H
(
p
)
=
p
log
⁡
2
π
-
log
⁡
(
1
+
2
1
-
p
)
+
log
⁡
3
.
Then numerical computations show that(37)
H
(
0.1941
)
=
8.13
…
×
1
0
-
7
>
0
,
H
(
0.1942
)
=
-
2.52
…
×
1
0
-
7
<
0
,
(38)
H
′
(
p
)
=
log
⁡
2
π
+
log
⁡
4
2
+
2
p
<
log
⁡
2
π
+
log
⁡
4
2
+
2
0.1
=
-
2.81
…
×
1
0
-
4
<
0
.
Inequality (38) implies that H
(
p
) is strictly decreasing on [
0.1
,
∞
). Therefore, p
0
=
0.1941
…  is the unique solution of (34) on the interval (
0.1
,
∞
) which follows from (37) and the monotonicity of H
(
p
).
Ifp
>
0.1 and F
p
(
t
)
>
0 for all t
∈
(
0
,
π
/
2
), then (31) leads to
(39)
0
≤
F
p
(
π
2
+
)
=
1
p
H
(
p
)
.
Therefore,p
≤
p
0 follows from (39) and H
(
p
0
)
=
0 together with the monotonicity of H
(
p
) on the interval (
0.1
,
∞
).Lemma 6.
Letp
∈
R and x
,
c
,
ω
∈
(
0,1
), and let M
p
(
x
,
ω
) be defined by (9). Then the function p
↦
M
p
(
x
,
ω
)
/
M
p
(
c
,
ω
) is strictly decreasing with respect to p
∈
R if x
∈
(
c
,
1
).Proof.
LetH
(
p
,
x
)
=
log
⁡
M
p
(
x
,
ω
)
-
log
⁡
M
p
(
c
,
ω
). Then from (9) we get
(40)
∂
H
(
p
,
x
)
∂
x
=
ω
x
p
-
1
ω
x
p
+
1
-
ω
,
(41)
∂
2
H
(
p
,
x
)
∂
p
∂
x
=
ω
(
1
-
ω
)
x
p
-
1
(
ω
x
p
+
1
-
ω
)
2
log
⁡
x
<
0
.
Inequality (41) and ∂
2
H
(
p
,
x
)
/
∂
x
∂
p
=
∂
2
H
(
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 M
p
(
x
,
ω
)
/
M
p
(
c
,
ω
) is strictly decreasing with respect to p
∈
R if x
∈
(
c
,
1
).


## 3. Main Results

Theorem 7.
LetM
p
(
x
,
ω
) be defined by (9). Then the double inequality
(42)
λ
p
M
p
(
cos
⁡
2
t
2
,
2
3
)
<
sin
t
t
<
M
p
(
cos
⁡
2
t
2
,
2
3
)
holds for all t
∈
(
0
,
π
/
2
) if and only if p
≥
1
/
5, and the double inequality
(43)
M
p
(
cos
⁡
2
t
2
,
2
3
)
<
sin
t
t
<
λ
p
M
p
(
cos
⁡
2
t
2
,
2
3
)
holds for all t
∈
(
0
,
π
/
2
) if and only if p
≤
p
1, where
(44)
λ
p
=
2
π
(
1
+
2
1
-
p
3
)
-
1
/
p
(
p
≠
0
)
,
λ
0
=
2
5
/
3
π
,
p
1
=
log
⁡
(
π
-
2
)
/
log
⁡
2
=
0.1910
…, and λ
p
M
p
(
cos
⁡
2
(
t
/
2
)
,
2
/
3
) is strictly decreasing with respect to p
∈
R.Proof.
Letp
∈
R and F
p
(
t
) be defined on (
0
,
π
/
2
) by (31) and (32). Then
(45)
F
p
(
0
+
)
=
0
,
F
p
(
π
2
-
)
=
log
⁡
λ
p
.
Ifp
≥
1
/
5, then inequality (42) follows from Lemma 4  (
i
) and (45).
If inequality (42) holds for all t
∈
(
0
,
π
/
2
), then F
p
(
t
)
<
0 for all t
∈
(
0
,
π
/
2
). It follows from Lemma 5  (
i
) that p
≥
1
/
5.
Ifp
≤
p
1, then inequality (43) follows from Lemma 4  (
ii
) and (45).
If inequality (43) holds for all t
∈
(
0
,
π
/
2
), then F
p
(
π
/
2
-
)
>
F
p
(
t
)
>
F
p
(
0
+
)
=
0 for all t
∈
(
0
,
π
/
2
). It follows from Lemma 5  (
ii
) that p
≤
p
0, where p
0
=
0.1941
…  is the unique solution of (34) on the interval (
0.1
,
∞
). We claim that p
≤
p
1; otherwise p
1
<
p
≤
p
0
<
1
/
5, and Lemma 4  (
iii
) leads to the conclusion that there exists t
0
∈
(
0
,
π
/
2
) such that F
p
(
t
)
>
F
p
(
π
/
2
-
) for t
∈
[
t
0
,
π
/
2
).
Note that(46)
λ
p
M
p
(
cos
⁡
2
t
2
,
2
3
)
=
2
π
M
p
(
cos
⁡
2
(
t
/
2
)
,
2
/
3
)
M
p
(
1
/
2
,
2
/
3
)
.
It follows from Lemma6 and (46) that λ
p
M
p
(
cos
⁡
2
(
t
/
2
)
,
2
/
3
) is strictly decreasing with respect to p
∈
R.From Theorem7 we get Corollaries 8 and 9 as follows.Corollary 8.
For allt
∈
(
0
,
π
/
2
) one has
(47)
2
π
<
2
+
cos
⁡
t
π
=
λ
1
M
1
(
cos
⁡
2
t
2
,
2
3
)
<
λ
1
/
2
(
2
3
cos
⁡
t
2
+
1
3
)
2
<
λ
1
/
4
(
2
3
cos
⁡
1
/
2
t
2
+
1
3
)
4
<
λ
1
/
5
(
2
3
cos
⁡
2
/
5
t
2
+
1
3
)
5
<
sin
t
t
<
(
2
3
cos
⁡
2
/
5
t
2
+
1
3
)
5
<
(
2
3
cos
⁡
1
/
2
t
2
+
1
3
)
4
<
(
2
3
cos
⁡
t
2
+
1
3
)
2
<
M
1
(
cos
⁡
2
t
2
,
2
3
)
=
2
+
cos
⁡
t
3
<
1
.Corollary 9.
For allt
∈
(
0
,
π
/
2
) one has
(48)
cos
⁡
2
t
2
=
M
-
∞
(
cos
⁡
2
t
2
,
2
3
)
<
3
(
1
+
cos
⁡
t
)
5
+
cos
⁡
t
=
M
-
1
(
cos
⁡
2
t
2
,
2
3
)
<
9
cos
⁡
2
(
t
/
2
)
(
2
+
cos
⁡
⁡
(
t
/
2
)
)
2
=
M
-
1
/
2
(
cos
⁡
2
t
2
,
2
3
)
<
cos
⁡
4
/
3
t
2
=
M
0
(
cos
⁡
2
t
2
,
2
3
)
<
(
2
3
cos
⁡
1
/
4
t
2
+
1
3
)
8
<
(
2
3
cos
⁡
1
/
3
t
2
+
1
3
)
6
<
sin
t
t
<
λ
1
/
6
(
2
3
cos
⁡
1
/
3
t
2
+
1
3
)
6
<
λ
1
/
8
(
2
3
cos
⁡
1
/
4
t
2
+
1
3
)
8
<
λ
0
cos
⁡
4
/
3
t
2
<
λ
-
1
/
2
9
cos
⁡
2
(
t
/
2
)
(
2
+
cos
⁡
⁡
(
t
/
2
)
)
2
<
λ
-
1
3
(
1
+
cos
⁡
t
)
5
+
cos
⁡
t
<
λ
-
∞
cos
⁡
2
t
2
=
4
π
cos
⁡
2
t
2
.Theorem 10.
LetM
p
(
x
,
ω
) be defined by (9). Then the double inequality
(49)
M
p
(
cos
⁡
2
t
2
,
2
3
)
<
sin
t
t
<
M
q
(
cos
⁡
2
t
2
,
2
3
)
holds for all t
∈
(
0
,
π
/
2
) if and only if p
≤
p
0 and q
≥
1
/
5, where p
0
=
0.1941
…  is the unique solution of (34) on the interval (
0.1
,
∞
). Moreover, the inequality
(50)
sin
t
t
≤
α
M
p
0
(
cos
⁡
2
t
2
,
2
3
)
,
if and only if
(51)
α
≥
sin
t
0
t
0
M
p
0
(
cos
⁡
2
(
t
0
/
2
)
,
2
/
3
)
=
1.00004919
…
,
where t
0
∈
(
0
,
π
/
2
) is defined as in Lemma 3  (
iii
).Proof.
Letp
∈
R and F
p
(
t
) be defined on (
0
,
π
/
2
) by (31) and (32). Then Lemma 4  (
iii
) leads to the conclusion that F
p
0
(
t
) is strictly increasing on (
0
,
t
0
] and strictly decreasing on [
t
0
,
π
/
2
). Note that
(52)
F
p
0
(
0
+
)
=
F
p
0
(
π
2
-
)
=
0
.
It follows from the piecewise monotonicity ofF
p
0
(
t
) and (52) that
(53)
0
<
F
p
0
(
t
)
≤
F
p
0
(
t
0
)
,
for all t
∈
(
0
,
π
/
2
). Therefore, sin
t
/
t
>
M
p
0
(
cos
⁡
2
(
t
/
2
)
,
2
/
3
) for all t
∈
(
0
,
π
/
2
) follows from the first inequality of (53), while sin
t
/
t
<
M
1
/
5
(
cos
⁡
2
(
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 all t
∈
(
0
,
π
/
2
), then we clearly see that the inequalities
(54)
F
p
(
t
)
>
0
,
F
q
(
t
)
<
0
hold for all t
∈
(
0
,
π
/
2
). Therefore, p
≤
p
0 and q
≥
1
/
5 follows from Lemma 5 and (54). Moreover, numerical computations show that t
0
=
1.312
…  and
(55)
e
F
p
0
(
t
0
)
=
1.00004919
…
.
Therefore, the second conclusion of Theorem10 follows from (55) and the second inequality of (53).It follows from Lemma3 that we get Theorem 11 immediately.Theorem 11.
The double inequalities(56)
2
cos
⁡
2
p
(
t
/
2
)
+
cos
⁡
t
2
cos
⁡
2
p
(
t
/
2
)
+
1
<
sin
t
t
<
2
cos
⁡
2
q
(
t
/
2
)
+
cos
⁡
t
2
cos
⁡
2
q
(
t
/
2
)
+
1
,
2
cos
⁡
2
p
t
2
<
sin
t
-
t
cos
⁡
t
t
-
sin
t
<
2
cos
⁡
2
q
t
2
hold for all t
∈
(
0
,
π
/
2
) if and only if p
≥
1
/
5 and q
≤
p
1
=
log
⁡
(
π
-
2
)
/
log
⁡
2
=
0.1910
….We clearly see that the function(
2
cos
⁡
2
p
(
t
/
2
)
+
cos
⁡
t
)
/
(
2
cos
⁡
2
p
(
t
/
2
)
+
1
) is strictly decreasing with respect to p
∈
R for fixed x
∈
(
0
,
π
/
2
). Let p
=
1
/
2,1
,
2
,
∞ and q
=
1
/
6
,
0
,
-
1
/
2
,
-
1
,
-
2
,
-
∞; then Theorem 11 leads to the following.Corollary 12.
The inequalities(57)
cos
⁡
t
<
8
cos
⁡
t
+
cos
⁡
2
t
+
3
4
cos
⁡
t
+
cos
⁡
2
t
+
7
<
2
cos
⁡
t
+
1
cos
⁡
t
+
2
<
2
cos
⁡
⁡
(
t
/
2
)
+
cos
⁡
t
2
cos
⁡
⁡
(
t
/
2
)
+
1
<
sin
t
t
<
cos
⁡
t
+
2
cos
⁡
1
/
3
(
t
/
2
)
2
cos
⁡
1
/
3
(
t
/
2
)
+
1
<
cos
⁡
t
+
2
3
<
cos
⁡
⁡
(
t
/
2
)
+
cos
⁡
⁡
(
3
t
/
2
)
+
4
2
cos
⁡
⁡
(
t
/
2
)
+
4
<
cos
⁡
t
cos
⁡
2
(
t
/
2
)
+
2
cos
⁡
2
(
t
/
2
)
+
2
<
cos
⁡
t
cos
⁡
4
(
t
/
2
)
+
2
cos
⁡
4
(
t
/
2
)
+
2
<
1
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
(58)
2
sin
t
t
+
tan
t
t
>
sin
t
t
+
2
tan
(
t
/
2
)
t
/
2
>
2
t
sin
t
+
t
tan
t
>
3
,
(
sin
t
t
)
p
+
2
(
tan
(
t
/
2
)
t
/
2
)
p
>
(
t
sin
t
)
p
+
2
(
t
/
2
tan
(
t
/
2
)
)
p
(
p
>
0
)
,
(
t
sin
t
)
p
+
2
(
t
/
2
tan
(
t
/
2
)
)
p
>
3
(
p
≥
1
)
hold for all t
∈
(
0
,
π
/
2
). Note that
(59)
sin
t
t
<
(
>
)
M
p
(
cos
⁡
2
t
2
,
2
3
)
⟺
(
t
sin
t
)
p
+
2
(
t
/
2
tan
⁡
(
t
/
2
)
)
>
(
<
)
3
,
sin
t
t
>
(
<
)
λ
p
M
p
(
cos
⁡
2
t
2
,
2
3
)
⟺
(
t
sin
t
)
p
+
2
(
t
/
2
tan
⁡
(
t
/
2
)
)
p
<
(
>
)
(
π
2
)
p
+
2
(
π
4
)
p
,
if p
>
0, and the second inequalities in (59) are reversed if p
<
0.From Theorems7 and 10 together with (59) we get the following.Theorem 13.
The double inequality(60)
(
π
2
)
p
+
2
(
π
4
)
p
>
(
t
sin
t
)
p
+
2
(
t
/
2
tan
(
t
/
2
)
)
p
>
3
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 if 0
<
p
≤
p
1
=
log
⁡
(
π
-
2
)
/
log
⁡
2
=
0.1910
….Theorem 14.
The double inequality(61)
(
t
sin
t
)
p
+
2
(
t
/
2
tan
(
t
/
2
)
)
p
>
3
>
(
t
sin
t
)
q
+
2
(
t
/
2
tan
(
t
/
2
)
)
q
holds for all t
∈
(
0
,
π
/
2
) if and only if 0
<
q
≤
p
0 and p
≥
1
/
5 or p
<
0, where p
0
=
0.1941
… is the unique solution of (34) on the interval (
0.1
,
∞
).Neuman [24] also proved that the Wilker type inequality
(62)
(
t
sin
t
)
p
+
(
t
/
2
tan
(
t
/
2
)
)
2
p
>
2
holds for all t
∈
(
0
,
π
/
2
) if p
≥
1.Making use of Theorem13 and the arithmetic-geometric means inequality
(63)
1
+
(
t
/
2
tan
(
t
/
2
)
)
2
p
>
2
(
t
/
2
tan
(
t
/
2
)
)
p
,
we get Corollary 15 as follows.Corollary 15.
The Wilker type inequality (62) holds for all t
∈
(
0
,
π
/
2
) if p
≥
1
/
5 or p
<
0.
In addition, power series expansions show that(64)
(
t
sin
t
)
p
+
(
t
/
2
tan
(
t
/
2
)
)
2
p
-
2
=
p
(
20
p
-
3
)
720
t
4
+
o
(
t
4
)
.
Therefore, we conjecture that inequality (62) holds for all t
∈
(
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 meanS
B
(
a
,
b
) [25–27] of two distinct positive real numbers a and b is defined by
(65)
S
B
(
a
,
b
)
=
{
b
2
-
a
2
cos
⁡
-
1
(
a
/
b
)
,
a
<
b
,
a
2
-
b
2
cosh
⁡
-
1
(
a
/
b
)
,
a
>
b
,
where cos
⁡
-
1
(
x
) and cosh
⁡
-
1
(
x
)
=
log
⁡
(
x
+
x
2
-
1
) are the inverse cosine and inverse hyperbolic cosine functions, respectively.
Letb
>
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
(66)
sin
t
t
=
S
B
(
a
,
b
)
b
,
M
p
(
cos
⁡
2
t
2
,
2
3
)
=
1
b
(
2
3
A
p
(
a
,
b
)
+
b
p
3
)
1
/
p
,
2
cos
⁡
2
p
(
t
/
2
)
+
cos
⁡
t
2
cos
⁡
2
p
(
t
/
2
)
+
1
=
2
A
p
(
a
,
b
)
+
a
b
p
-
1
2
A
p
(
a
,
b
)
+
b
p
.
It follows from Theorems7, 10, and 11 together with (66) that we have the following.Theorem 16.
Letp
1
=
l
o
g
(
π
-
2
)
/
l
o
g
2
=
0.1910
…, λ
p and p
0
=
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(67)
λ
p
(
2
3
A
p
(
a
,
b
)
+
1
3
b
p
)
1
/
p
<
S
B
(
a
,
b
)
<
(
2
3
A
p
(
a
,
b
)
+
1
3
b
p
)
1
/
p
holds if and only if p
≥
1
/
5, and inequality (67) is reversed if and only if p
≤
p
1.
(ii)
The double inequality(68)
(
2
3
A
p
(
a
,
b
)
+
1
3
b
p
)
1
/
p
<
S
B
(
a
,
b
)
<
(
2
3
A
q
(
a
,
b
)
+
1
3
b
q
)
1
/
q
holds if and only if p
≤
p
0 and q
≥
1
/
5.
(iii)
The double inequality(69)
2
A
p
(
a
,
b
)
+
a
b
p
-
1
2
A
p
(
a
,
b
)
+
b
p
b
<
S
B
(
a
,
b
)
<
2
A
q
(
a
,
b
)
+
a
b
q
-
1
2
A
q
(
a
,
b
)
+
b
q
b
holds if and only if p
≥
1
/
5 and q
≤
p
1.Letb
>
a
>
0, G
(
a
,
b
)
=
a
b, Q
(
a
,
b
)
=
(
a
2
+
b
2
)
/
2, P
(
a
,
b
)
=
(
b
-
a
)
/
[
2
sin
-
1
(
(
b
-
a
)
/
(
b
+
a
)
)
], T
(
a
,
b
)
=
(
b
-
a
)
/
[
2
tan
-
1
(
(
b
-
a
)
/
(
b
+
a
)
)
], and Y
(
a
,
b
)
=
(
b
-
a
)
/
[
2
tan
-
1
(
(
b
-
a
)
/
2
a
b
)
] 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 to Corollary 17.Corollary 17.
Letp
1
=
log
⁡
(
π
-
2
)
/
log
⁡
2
=
0.1910
…, λ
p and p
0
=
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(70)
λ
p
[
2
3
(
G
(
a
,
b
)
+
A
(
a
,
b
)
2
)
p
+
1
3
A
p
(
a
,
b
)
]
1
/
p
<
P
(
a
,
b
)
<
[
2
3
(
G
(
a
,
b
)
+
A
(
a
,
b
)
2
)
p
+
1
3
A
p
(
a
,
b
)
]
1
/
p
λ
p
[
2
3
(
A
(
a
,
b
)
+
Q
(
a
,
b
)
2
)
p
+
1
3
Q
p
(
a
,
b
)
]
1
/
p
<
T
(
a
,
b
)
<
[
2
3
(
A
(
a
,
b
)
+
Q
(
a
,
b
)
2
)
p
+
1
3
Q
p
(
a
,
b
)
]
1
/
p
,
λ
p
[
2
3
(
G
(
a
,
b
)
+
Q
(
a
,
b
)
2
)
p
+
1
3
Q
p
(
a
,
b
)
]
1
/
p
<
Y
(
a
,
b
)
<
[
2
3
(
G
(
a
,
b
)
+
Q
(
a
,
b
)
2
)
p
+
1
3
Q
p
(
a
,
b
)
]
1
/
p
,
hold if and only if p
≥
1
/
5, and all inequalities in (70) are reversed if and only if p
≤
p
1.
(ii) The double inequalities(71)
[
2
3
(
G
(
a
,
b
)
+
A
(
a
,
b
)
2
)
p
+
1
3
A
p
(
a
,
b
)
]
1
/
p
<
P
(
a
,
b
)
<
[
2
3
(
G
(
a
,
b
)
+
A
(
a
,
b
)
2
)
q
+
1
3
A
q
(
a
,
b
)
]
1
/
q
,
[
2
3
(
A
(
a
,
b
)
+
Q
(
a
,
b
)
2
)
p
+
1
3
Q
p
(
a
,
b
)
]
1
/
p
<
T
(
a
,
b
)
<
[
2
3
(
A
(
a
,
b
)
+
Q
(
a
,
b
)
2
)
q
+
1
3
Q
q
(
a
,
b
)
]
1
/
q
,
[
2
3
(
G
(
a
,
b
)
+
Q
(
a
,
b
)
2
)
p
+
1
3
Q
p
(
a
,
b
)
]
1
/
p
<
Y
(
a
,
b
)
<
[
2
3
(
G
(
a
,
b
)
+
Q
(
a
,
b
)
2
)
q
+
1
3
Q
q
(
a
,
b
)
]
1
/
q
hold if and only if p
≤
p
0 and q
≥
1
/
5.
(iii) The double inequalities(72)
2
1
-
p
(
G
(
a
,
b
)
+
A
(
a
,
b
)
)
p
A
(
a
,
b
)
+
G
(
a
,
b
)
A
p
(
a
,
b
)
2
1
-
p
(
G
(
a
,
b
)
+
A
(
a
,
b
)
)
p
+
A
p
(
a
,
b
)
<
P
(
a
,
b
)
<
2
1
-
q
(
G
(
a
,
b
)
+
A
(
a
,
b
)
)
q
A
(
a
,
b
)
+
G
(
a
,
b
)
A
q
(
a
,
b
)
2
1
-
q
(
G
(
a
,
b
)
+
A
(
a
,
b
)
)
q
+
A
q
(
a
,
b
)
,
2
1
-
p
(
A
(
a
,
b
)
+
Q
(
a
,
b
)
)
p
Q
(
a
,
b
)
+
A
(
a
,
b
)
Q
p
(
a
,
b
)
2
1
-
p
(
A
(
a
,
b
)
+
Q
(
a
,
b
)
)
p
+
Q
p
(
a
,
b
)
<
T
(
a
,
b
)
<
2
1
-
q
(
A
(
a
,
b
)
+
Q
(
a
,
b
)
)
q
Q
(
a
,
b
)
+
A
(
a
,
b
)
Q
q
(
a
,
b
)
2
1
-
q
(
A
(
a
,
b
)
+
Q
(
a
,
b
)
)
q
+
Q
q
(
a
,
b
)
,
2
1
-
p
(
G
(
a
,
b
)
+
Q
(
a
,
b
)
)
p
Q
(
a
,
b
)
+
G
(
a
,
b
)
Q
p
(
a
,
b
)
2
1
-
p
(
G
(
a
,
b
)
+
Q
(
a
,
b
)
)
p
+
Q
p
(
a
,
b
)
<
Y
(
a
,
b
)
<
2
1
-
q
(
G
(
a
,
b
)
+
Q
(
a
,
b
)
)
q
Q
(
a
,
b
)
+
G
(
a
,
b
)
Q
q
(
a
,
b
)
2
1
-
q
(
G
(
a
,
b
)
+
Q
(
a
,
b
)
)
q
+
Q
q
(
a
,
b
)
hold if and only if p
≥
1
/
5 and q
≤
p
1.
Forx
∈
(
0,1
), the following Shafer-Fink type inequality can be found in the literature [1, 30]:
(73)
sin
-
1
x
>
6
(
1
+
x
-
1
-
x
)
4
+
1
+
x
+
1
-
x
>
3
x
2
+
1
-
x
2
.
Fink [31] proved that the double inequality
(74)
3
x
2
+
1
-
x
2
≤
sin
-
1
x
≤
π
x
2
+
1
-
x
2
holds for all x
∈
[
0,1
]. It was generalized and improved by Zhu [32].Lett
∈
(
0
,
π
/
2
), x
=
sin
t
∈
(
0,1
). Then Theorems 7, 10, and 11 lead to Corollary 18 as follows.Corollary 18.
Letp
1
=
log
⁡
(
π
-
2
)
/
log
⁡
2
=
0.1910
…, λ
p and p
0
=
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(75)
x
M
p
(
(
1
+
1
-
x
2
)
/
2
,
2
/
3
)
<
sin
-
1
(
x
)
<
x
λ
p
M
p
(
(
1
+
1
-
x
2
)
/
2
,
2
/
3
)
holds if and only if p
≥
1
/
5, and inequality (75) is reversed if and only if p
≤
p
1.
(ii) The double inequality(76)
x
M
p
(
(
1
+
1
-
x
2
)
/
2
,
2
/
3
)
<
sin
-
1
(
x
)
<
x
M
q
(
(
1
+
1
-
x
2
)
/
2
,
2
/
3
)
holds if and only if p
≥
1
/
5 and q
≤
p
0.
(iii) The double inequality(77)
2
1
-
2
p
(
1
+
x
+
1
-
x
)
2
p
+
1
2
1
-
2
p
(
1
+
x
+
1
-
x
)
2
p
+
1
-
x
2
<
sin
-
1
(
x
)
<
2
1
-
2
q
(
1
+
x
+
1
-
x
)
2
q
+
1
2
1
-
2
q
(
1
+
x
+
1
-
x
)
2
q
+
1
-
x
2
holds if and only if p
≤
p
1 and q
≥
1
/
5.



---

*Source: 592085-2014-07-10.xml*