# A Note on Jordan, Adamović-Mitrinović, and Cusa Inequalities

Authors: Zhen-Hang Yang; Yu-Ming Chu
Journal: Abstract and Applied Analysis
 (2014)

---

## Abstract

We improve the Jordan, Adamović-Mitrinović, and Cusa inequalities. As applications, several new Shafer-Fink type inequalities for inverse sine function and bivariate means inequalities are established, and a new estimate for sine integral is given.


---

## Body

## 1. Introduction

The classical Jordan inequality [1] is given by
(1)
2
π
<
sin
⁡
x
x
<
1
,
for x
∈
(
0
,
π
/
2
).Some new developments on refinements, generalizations, and applications for the Jordan inequality can be found in [2] and the references therein.In the recent past, the following two-side inequality(2)
(
cos
⁡
⁡
x
)
1
/
3
<
sin
⁡
x
x
<
2
+
cos
⁡
⁡
x
3
(
0
<
x
<
π
2
)
has attracted the attention of many researchers (see, e.g., [2–14]). The left inequality in (2) was obtained by Mitrinović (see [1, p. 238]), while the right one is due to Huygens (see, e.g., [15]) and it is calledCusa inequality [3, 5, 6, 8, 14].In [16], the following open problem was proposed: for each p
>
0, there are greatest value q
=
q
(
p
) and least value r
=
r
(
p
) such that the double inequality
(3)
q
sin
⁡
x
1
+
p
cos
⁡
⁡
x
<
x
<
r
sin
⁡
x
1
+
p
cos
⁡
⁡
x
holds for all x
∈
(
0
,
π
/
2
). This was answered by Carver in [17]. In [1, p. 238, 3.4.15], it was listed that
(4)
(
1
+
p
)
sin
⁡
x
1
+
p
cos
⁡
⁡
x
<
x
<
(
π
/
2
)
sin
⁡
x
1
+
p
cos
⁡
⁡
x
for p
∈
(
0,1
/
2
] and x
∈
[
0
,
π
/
2
]. Wu [18] proved that the double inequality
(5)
(
1
+
p
)
cos
⁡
⁡
x
1
+
p
cos
⁡
⁡
x
<
sin
⁡
x
x
<
1
+
q
1
+
q
cos
⁡
⁡
x
holds for x
∈
(
0
,
π
/
2
), p
∈
[
-
1,2
], and q
∈
[
-
1
/
4
,
∞
). In particular, he obtained that for x
∈
(
0
,
π
/
2
),
(6)
3
cos
⁡
⁡
x
1
+
2
cos
⁡
⁡
x
<
sin
⁡
x
x
<
3
4
-
cos
⁡
⁡
x
.
The first inequality in (6) is equivalent to the Huygen inequality:
(7)
2
sin
⁡
x
x
+
tan
⁡
x
x
>
3
.
Jiang [19] showed that for x
∈
(
0
,
π
/
2
),
(8)
sin
⁡
x
x
>
1
+
2
cos
⁡
⁡
x
2
+
cos
⁡
⁡
x
.
Li and He [20] gave an improvement of (6) as follows:
(9)
7
+
5
cos
⁡
⁡
x
11
+
cos
⁡
⁡
x
<
sin
⁡
x
x
<
9
+
6
cos
⁡
⁡
x
14
+
cos
⁡
⁡
x
.The main purpose of this paper is to give sharp bounds for(
sin
x
)
/
x in terms of the functions H
1
(
cos
⁡
⁡
t
,
p
) and H
2
(
cos
⁡
⁡
t
,
p
), where
(10)
H
1
(
x
,
p
)
=
2
p
+
(
p
+
3
)
x
3
p
+
1
+
2
x
,
x
∈
(
0,1
)
,
(
x
,
p
)
p
∈
(
-
∞
,
-
1
]
∪
[
0
,
∞
)
,
(11)
H
2
(
x
,
p
)
=
3
p
+
1
π
p
2
p
+
(
p
+
3
)
x
(
3
p
+
1
)
+
2
x
,
x
∈
(
0,1
)
,
(
x
,
p
)
(
x
,
p
)
p
∈
(
-
∞
,
-
1
]
∪
(
0
,
∞
)
.The rest of this paper is organized as follows. Several lemmas are given in Section2. Main results and their proofs are given in Section 3, in which Theorem 7 unifies and generalizes Jordan and Cusa inequalities; Theorem 13 shows that Adamović-Mitrinović and Cusa inequalities (2) can be interpolated by H
1
(
cos
⁡
⁡
x
,
p
) for suitable p; Theorem 18 gives a hyperbolic version of Theorem 7. In Section 4, some new Shafer-Fink type inequalities for inverse sine function and several inequalities for bivariate means are presented, and a simpler but more accurate estimate for sine integral is provided.


## 2. Lemmas

Lemma 1.
LetH
1 and H
2 be defined by (10) and (11), respectively. Then H
1 and H
2 are, respectively, increasing and decreasing with respect to p on (
-
∞
,
-
1
]
∪
(
0
,
∞
) with the limits
(12)
lim
⁡
p
→
-
∞
H
1
(
x
,
p
)
=
lim
⁡
p
→
∞
H
1
(
x
,
p
)
=
2
+
x
3
,
lim
⁡
p
→
-
∞
H
2
(
x
,
p
)
=
lim
⁡
p
→
∞
H
2
(
x
,
p
)
=
2
+
x
π
.Proof.
From (10) and (11) we have
(13)
∂
H
1
∂
p
=
2
(
x
-
1
)
2
(
3
p
+
1
+
2
x
)
2
>
0
,
∂
H
2
∂
p
=
-
3
x
π
p
2
(
3
p
+
2
x
+
1
)
2
(
p
+
1
)
×
(
(
5
-
2
x
)
p
+
(
2
x
+
1
)
)
.
If p
∈
(
0
,
∞
), then we clearly see that ∂
H
2
/
∂
p
<
0. If ∈
(
-
∞
,
-
1
), then (
5
-
2
x
)
p
+
(
2
x
+
1
)
<
4
(
x
-
1
)
<
0, and then ∂
H
2
/
∂
p
<
0.
Simple computations give (12).Lemma 2.
Letu
1, u
2 be defined on (
0,1
)
×
(
-
∞
,
-
1
]
∪
[
0
,
∞
) by
(14)
u
1
(
x
,
p
)
=
(
2
p
+
(
3
+
p
)
x
)
(
3
p
+
1
+
2
x
)
,
(15)
u
2
(
x
,
p
)
=
2
(
p
+
3
)
x
3
+
8
p
x
2
+
2
p
(
3
p
+
1
)
x
+
3
(
p
+
1
)
2
,
respectively. Then u
1
(
x
,
p
), u
2
(
x
,
p
)
>
0.Proof.
It is not difficult to see thatu
1
(
x
,
p
), u
2
(
x
,
p
)
>
0 for p
∈
[
0
,
∞
). If p
∈
(
-
∞
,
-
1
], then
(16)
2
p
+
(
3
+
p
)
x
<
2
(
x
-
1
)
<
0
,
(
3
p
+
1
+
2
x
)
<
2
(
x
-
1
)
<
0
,
and then u
1
(
x
,
p
)
>
0. It remains to prove that u
2
(
x
,
p
)
>
0 for p
∈
(
-
∞
,
-
1
]. Differentiation leads to
(17)
∂
u
2
∂
p
=
(
12
x
+
6
)
p
+
(
2
x
3
+
8
x
2
+
2
x
+
6
)
.
Hence, ∂
u
2
/
∂
p
<
-
(
12
x
+
6
)
+
(
2
x
3
+
8
x
2
+
2
x
+
6
)
=
2
x
(
x
+
5
)
(
x
-
1
)
<
0, which implies that u
2 is decreasing in p on (
-
∞
,
-
1
), and therefore,
(18)
u
2
(
x
,
p
)
>
u
2
(
x
,
-
1
)
=
4
x
(
x
-
1
)
2
>
0
.
This completes the proof.Lemma 3.
Letu
3 be defined on (
0,1
)
×
(
-
∞
,
-
1
]
∪
[
0
,
∞
) by
(19)
u
3
(
x
,
p
)
=
(
p
+
3
)
2
x
2
+
(
p
+
3
)
(
7
p
+
3
)
x
+
(
-
3
p
3
+
13
p
2
+
21
p
+
9
)
.
Then(i)
u
3
(
x
,
p
)
≥
0 for all x
∈
(
0,1
) if and only if p
∈
(
-
∞
,
p
3
], where p
3
≈
5.6630 is the unique solution of the equation u
3
(
0
,
p
)
=
-
3
p
3
+
13
p
2
+
21
p
+
9
=
0;
(ii)
u
3
(
x
,
p
)
≤
0 if and only if p
∈
[
9
,
∞
);
(iii)
for everyp
∈
(
p
3
,
9
), there exists a unique x
1
∈
(
0,1
) such that u
3
(
x
,
p
)
<
0 for x
∈
(
0
,
x
1
) and u
3
(
x
,
p
)
>
0 for x
∈
(
x
1
,
1
).Proof.
In order to prove the desired results, we need to rewriteu
3
(
x
,
p
) as
(20)
u
3
(
x
,
p
)
=
(
(
p
+
3
)
x
+
7
p
+
3
2
)
2
-
3
(
p
-
9
4
)
(
p
+
1
)
2
.
We clearly see that
(21)
u
3
(
0
,
p
)
=
-
3
p
3
+
13
p
2
+
21
p
+
9
,
u
3
(
1
,
p
)
=
-
3
(
p
+
1
)
2
(
p
-
9
)
.
We claim that there exists uniquep
3
∈
(
5,6
) such that u
3
(
0
,
p
)
>
0 for p
∈
(
-
∞
,
p
3
) and u
3
(
0
,
p
)
<
0 for p
∈
(
p
3
,
∞
). Indeed, we have
(22)
u
3
′
(
0
,
p
)
=
-
9
p
2
+
26
p
+
21
=
-
9
(
p
-
13
-
358
9
)
(
p
-
13
+
358
9
)
,
which implies that u
3
(
0
,
p
) is increasing on (
(
13
-
358
)
/
9
,
(
13
+
358
)
/
9
) and decreasing on (
-
∞
,
(
13
-
358
)
/
9
)
∪
(
(
13
+
358
)
/
9
,
∞
). Since
(23)
u
3
(
0
,
13
-
358
9
)
=
13952
243
-
716
243
358
≈
1.6652
>
0
,
u
3
(
0
,
13
+
358
9
)
=
716
243
358
+
13952
243
>
0
,
u
3
(
0
,
∞
)
=
-
∞
,
there exists unique p
3
∈
(
(
13
+
358
)
/
9
,
∞
) such that u
3
(
0
,
p
3
)
=
0 and u
3
(
0
,
p
)
>
0 for p
∈
(
-
∞
,
p
3
) and u
3
(
0
,
p
)
<
0 for p
∈
(
p
3
,
∞
). An easy calculation reveals that p
3
≈
5.6630.
(i) Now we prove the necessary and sufficient condition foru
3
(
x
,
p
)
≥
0 for all x
∈
(
0,1
). Since u
3
(
x
,
-
3
)
=
144
>
0, we assume that p
≠
-
3. Denote the minimum point of u
3
(
x
,
p
) by x
0. Then x
0
=
-
(
7
p
+
3
)
/
(
2
(
p
+
3
)
). And then, due to ∂
2
u
3
/
∂
x
2
>
0, u
3
(
x
,
p
)
≥
0 for all x
∈
(
0,1
) if and only if at least one of the following cases occur.
Case  1. Consider that x
0
=
-
(
7
p
+
3
)
/
(
2
(
p
+
3
)
)
≥
1, u
3
(
1
,
p
)
≥
0. It is derived that p
∈
(
-
3
,
-
1
].
Case  2. Consider that x
0
=
-
(
7
p
+
3
)
/
(
2
(
p
+
3
)
)
≤
0, u
3
(
0
,
p
)
≥
0. It implies that p
∈
(
-
∞
,
-
3
)
,
∪
[
-
3
/
7
,
p
3
].
Case  3. Consider that x
0
=
-
(
7
p
+
3
)
/
(
2
(
p
+
3
)
)
∈
(
0,1
), u
3
(
x
0
,
p
)
=
-
3
(
p
-
9
/
4
)
(
p
+
1
)
2
≥
0. It yields p
∈
(
-
1
,
-
3
/
7
).
To sum up,u
3
(
x
,
p
)
≥
0 for all x
∈
(
0,1
) if and only if p
∈
(
-
∞
,
p
3
].
(ii) It is clear thatu
3
(
x
,
p
)
≤
0 if and only if u
3
(
0
,
p
)
≤
0 and u
3
(
1
,
p
)
≤
0. Solving the inequalities for p leads to p
≥
9.
(iii) In the case ofp
∈
(
p
3
,
9
), we clearly see that u
3
(
0
,
p
)
<
0, u
3
(
1
,
p
)
>
0, and x
0
=
-
(
7
p
+
3
)
/
(
2
(
p
+
3
)
)
<
0. This implies that there exists a unique x
1
∈
(
0,1
) such that u
3
(
x
,
p
)
<
0 for x
∈
(
0
,
x
1
) and u
3
(
x
,
p
)
>
0 for x
∈
(
x
1
,
1
).
This completes the proof.Now let us consider the sign of functiong defined on (
0
,
π
/
2
)
×
(
-
∞
,
-
1
]
∪
[
0
,
∞
) by
(24)
g
(
t
,
p
)
=
t
-
(
+
2
p
(
3
p
+
1
)
cos
⁡
⁡
t
+
3
(
p
+
1
)
2
)
-
1
(
(
2
p
+
(
p
+
3
)
cos
⁡
⁡
t
)
(
3
p
+
1
+
2
cos
⁡
⁡
t
)
)
(
(
(
2
ee
)
)
×
(
2
(
p
+
3
)
cos
⁡
3
t
+
8
p
cos
⁡
2
t
(
(
(
2
+
ew
)
)
+
2
p
(
3
p
+
1
)
cos
⁡
⁡
t
+
3
(
p
+
1
)
2
)
-
1
sin
⁡
t
)
=
t
-
u
1
(
cos
⁡
⁡
t
,
p
)
u
2
(
cos
⁡
⁡
t
,
p
)
sin
⁡
t
,
where u
1
(
x
,
p
) and u
2
(
x
,
p
) are defined by (14) and (15), respectively. We have the following.Lemma 4.
Letg be defined on (
0
,
π
/
2
)
×
(
-
∞
,
-
1
]
∪
[
0
,
∞
) by (24). Then(i)
g
(
t
,
p
)
<
0 for all t
∈
(
0
,
π
/
2
) if and only if p
∈
(
-
∞
,
-
1
]
∪
[
9
,
∞
);
(ii)
g
(
t
,
p
)
>
0 for all t
∈
(
0
,
π
/
2
) if and only if p
∈
[
0
,
p
1
], where
(25)
p
1
=
2
6
π
+
1
+
3
π
-
2
12
-
3
π
≈
6.3433
;
(iii)
in the case ofp
∈
(
p
1
,
9
), there exists a unique t
0
∈
(
0
,
π
/
2
) such that g
(
t
,
p
)
>
0 for t
∈
(
0
,
t
0
) and g
(
t
,
p
)
<
0 for t
∈
(
t
0
,
π
/
2
).Proof.
We first give two limit relations as follows:(26)
lim
⁡
t
→
0
+
g
(
t
,
p
)
t
5
=
-
1
45
p
-
9
p
+
1
if
p
≠
-
1
,
g
(
π
2
-
,
p
)
=
-
12
-
3
π
6
(
p
+
1
)
2
(
p
-
p
1
)
(
p
-
p
2
)
if
p
≠
-
1
,
where
(27)
p
1
=
2
6
π
+
1
+
3
π
-
2
12
-
3
π
≈
6.3433
,
p
2
=
-
2
6
π
+
1
-
3
π
+
2
12
-
3
π
<
0
.
In fact, if p
≠
-
1, then making use of power series we get
(28)
g
(
t
,
p
)
=
-
1
45
p
-
9
p
+
1
t
5
+
o
(
t
5
)
,
which implies the first relation. Direct computations yield the second one.
Differentiatingg
(
t
,
p
) with respect to t leads to
(29)
∂
g
∂
t
=
1
-
u
1
(
cos
⁡
⁡
t
,
p
)
u
2
(
cos
⁡
⁡
t
,
p
)
cos
⁡
⁡
t
+
(
sin
2
t
)
d
d
x
u
1
(
x
,
p
)
u
2
(
x
,
p
)
|
x
=
cos
⁡
⁡
t
=
4
(
1
-
x
)
(
1
-
x
2
)
u
2
2
(
x
,
p
)
×
h
(
x
,
p
)
,
where u
1
(
x
,
p
) and u
2
(
x
,
p
) are defined by (14) and (15), respectively, and
(30)
h
(
x
,
p
)
=
(
x
+
3
p
+
1
2
)
×
u
3
(
x
,
p
)
;
here u
3
(
x
,
p
) is defined by (19) and x
=
cos
⁡
⁡
t
∈
(
0,1
).
(i) We now prove thatg
(
t
,
p
)
≤
0 for all t
∈
(
0
,
π
/
2
) if and only if p
∈
(
-
∞
,
-
1
]
∪
[
9
,
∞
). The necessity easily follows from the inequalities lim
⁡
t
→
0
+
t
-
5
g
(
t
,
p
)
≤
0 and g
(
π
/
2
-
,
p
)
≤
0 if p
≠
-
1 and g
(
t
,
-
1
)
=
t
-
tan
t
<
0 together with the relation (26).
Next we prove the sufficiency. Ifp
∈
[
9
,
∞
), then by Lemma 3  u
3
(
x
,
p
)
≤
0, and then h
(
x
,
p
)
≤
0. This indicates that g is decreasing in t on (
0
,
π
/
2
), and therefore, we get g
(
t
,
p
)
<
g
(
0
+
,
p
)
=
0. If p
∈
(
-
∞
,
-
1
], then u
3
(
x
,
p
)
≥
0 and x
+
(
3
p
+
1
)
/
2
<
x
-
1
<
0, which yields h
(
x
,
p
)
≤
0. This also yields that g is decreasing in t on (
0
,
π
/
2
), and so g
(
t
,
p
)
<
g
(
0
+
,
p
)
=
0.
(ii) Similarly, we can prove thatg
(
t
,
p
)
>
0 for all t
∈
(
0
,
π
/
2
) if and only if p
∈
[
0
,
p
1
]. If g
(
t
,
p
)
>
0 for all t
∈
(
0
,
π
/
2
), then we have lim
⁡
t
→
0
+
t
-
5
g
(
t
,
p
)
≥
0 and g
(
π
/
2
-
,
p
)
≥
0, which together with (26) and p
∈
(
-
∞
,
-
1
]
∪
[
0
,
∞
) lead to p
∈
[
0
,
p
1
].
In order to prove the sufficiency, we distinguish two cases.
In the case ofp
∈
[
0
,
p
3
], by Lemma 3 we have u
3
(
x
,
p
)
≥
0, which implies that g is increasing in t on (
0
,
π
/
2
), and so, g
(
t
,
p
)
>
g
(
0
+
,
p
)
=
0.
In the case ofp
∈
(
p
3
,
p
1
], from Lemma 3 there is a unique x
1
∈
(
0,1
) such that u
3
(
x
,
p
)
<
0 for x
∈
(
0
,
x
1
) and u
3
(
x
,
p
)
>
0 for x
∈
(
x
1
,
1
). This in conjunction with (30) and (29) shows that g is decreasing in t on (
arccos
⁡
x
1
,
π
/
2
) and increasing on (
0
,
arccos
⁡
x
1
), and consequently, we have
(31)
g
(
t
,
p
)
>
g
(
0
+
,
p
)
=
for
t
∈
(
0
,
arccos
⁡
x
1
)
,
g
(
t
,
p
)
>
g
(
π
2
+
,
p
)
=
-
12
-
3
π
6
(
p
+
1
)
2
(
p
-
p
1
)
(
p
-
p
2
)
≥
0
12
-
3
π
6
(
p
+
1
)
2
for
t
∈
(
arccos
⁡
x
1
,
π
2
)
,
which proves the sufficiency.
(iii) In the case whenp
∈
(
p
1
,
9
), we have seen that g is decreasing in t on (
arccos
⁡
x
1
,
π
/
2
) and increasing on (
0
,
arccos
⁡
x
1
) and g
(
t
,
p
)
>
0 for t
∈
(
0
,
arccos
⁡
x
1
), but
(32)
g
(
π
2
-
,
p
)
=
-
12
-
3
π
6
(
p
+
1
)
2
(
p
-
p
1
)
(
p
-
p
2
)
<
0
.
Thus, there is a unique t
0
∈
(
arccos
⁡
x
1
,
π
/
2
) such that g
(
t
,
p
)
>
0 for t
∈
(
0
,
t
0
) and g
(
t
,
p
)
<
0 for t
∈
(
t
0
,
π
/
2
).
The whole proof is complete.We next observe the functionf defined on (
0
,
π
/
2
)
×
(
-
∞
,
-
1
]
∪
[
0
,
∞
) by
(33)
f
(
t
,
p
)
=
ln
⁡
⁡
sin
⁡
t
t
-
ln
⁡
⁡
H
1
(
cos
⁡
⁡
t
,
p
)
=
ln
⁡
⁡
sin
⁡
t
t
-
ln
⁡
⁡
2
p
+
(
p
+
3
)
cos
⁡
⁡
t
(
3
p
+
1
)
+
2
cos
⁡
⁡
t
.
Differentiation yields that
(34)
∂
f
∂
t
=
cos
⁡
⁡
t
sin
⁡
t
-
1
t
-
2
sin
⁡
t
3
p
+
1
+
2
cos
⁡
⁡
t
+
(
p
+
3
)
sin
⁡
t
2
p
+
(
p
+
3
)
cos
⁡
⁡
t
=
(
(
2
(
p
+
3
)
cos
⁡
3
t
+
(
3
p
2
+
14
p
+
3
)
cos
⁡
2
t
+
3
(
p
+
1
)
2
(
sin
2
t
)
+
2
p
(
3
p
+
1
)
cos
⁡
t
)
e
×
(
(
sin
t
)
(
2
p

+

3
cos
⁡
t

+

p
cos
⁡
t
)
(
3
p

+

2
cos
⁡
t

+

1
)
)
-
1
)

-

1
t
=
1
sin
t
u
2
(
cos
⁡
t
,
p
)
u
1
(
cos
⁡
t
,
p
)
-
1
t
=
1
t
sin
t
u
2
(
cost
,
p
)
u
1
(
cos
⁡
t
,
p
)
×
g
(
t
,
p
)
,
where u
1
(
x
,
p
), u
2
(
x
,
p
), and g
(
t
,
p
) are defined by (14), (15), and (24), respectively. From Lemmas 2 and 4 the following assertion is immediate.Lemma 5.
Letf be the function defined on (
0
,
π
/
2
)
×
(
-
∞
,
-
1
]
∪
[
0
,
∞
) by (33). Then(i)
f is decreasing in t on (
0
,
π
/
2
) if and only if p
∈
(
-
∞
,
-
1
]
∪
[
9
,
∞
);
(ii)
f is increasing in t on (
0
,
π
/
2
) if and only if p
∈
[
0
,
p
1
], where p
1
≈
6.3433 is given by (25);
(iii)
in the case whenp
∈
(
p
1
,
9
), there is a unique t
0
∈
(
0
,
π
/
2
) such that f is increasing in t on (
0
,
t
0
) and decreasing on (
t
0
,
π
/
2
).Lastly, for later use, we also give the following.Lemma 6.
LetH
1 be defined on (
0,1
)
×
(
-
∞
,
-
1
]
∪
[
0
,
∞
) by (10). Then H
1
(
x
3
,
p
)
≥
x if and only if p
∈
(
-
∞
,
-
1
]
∪
[
1
,
∞
), and H
1
(
x
3
,
p
)
≤
x if and only if p
=
0.Proof.
Forp
∈
(
-
∞
,
∞
), we define
(35)
u
4
(
x
,
p
)
=
2
x
2
+
(
1
-
p
)
x
-
2
p
=
2
(
x
-
p
-
1
4
)
2
-
1
7
(
14
p
+
p
2
+
1
)
.
Then u
4
(
x
,
p
)
≥
0 holds for all x
∈
(
0,1
) if and only if p
∈
(
-
∞
,
0
].
In fact,u
4
(
x
,
p
)
≥
0 if and only if at least one case of the following occurs.
Case  1. Consider that (
p
-
1
)
/
4
≥
1, u
4
(
1
,
p
)
=
3
-
3
p
≥
0. It is impossible.
Case  2. Consider that (
p
-
1
)
/
4
≤
0, u
4
(
0
,
p
)
=
-
2
p
≥
0. It indicates p
∈
(
-
∞
,
0
].
Case  3. Consider that 0
<
(
p
-
1
)
/
4
<
1, u
4
(
(
p
-
1
)
/
4
,
p
)
≥
0. It is impossible.
In the same way, we can prove thatu
4
(
x
,
p
)
≤
0 holds for all x
∈
(
0,1
) if and only if p
∈
[
1
,
∞
).
We now prove thatH
1
(
x
3
,
p
)
≥
x if and only if p
∈
(
-
∞
,
-
1
]
∪
[
1
,
∞
). Factoring yields
(36)
H
1
(
x
3
,
p
)
-
x
=
-
2
(
x
-
1
)
2
2
x
2
+
(
1
-
p
)
x
-
2
p
3
p
+
2
x
3
+
1
=
-
(
x
-
1
)
2
u
4
(
x
,
p
)
3
p
+
2
x
3
+
1
.
If p
∈
(
-
∞
,
-
1
], then 3
p
+
2
x
3
+
1
<
0, and then, H
1
(
x
3
,
p
)
≥
x if and only if u
4
(
x
,
p
)
≥
0, which is equivalent to p
∈
(
-
∞
,
-
1
]
∩
(
-
∞
,
0
]
=
(
-
∞
,
-
1
]. If p
∈
[
0
,
∞
), then 3
p
+
2
x
3
+
1
>
0, and then, H
1
(
x
3
,
p
)
≥
x if and only if u
4
(
x
,
p
)
≤
0, which is equivalent to p
∈
[
0
,
∞
)
∩
[
1
,
∞
)
=
[
1
,
∞
). Consequently, H
1
(
x
3
,
p
)
≥
x if and only if p
∈
(
-
∞
,
-
1
]
∪
[
1
,
∞
).
Next we show thatH
1
(
x
3
,
p
)
≤
x if and only if p
=
0. In fact, if p
∈
(
-
∞
,
-
1
], then H
1
(
x
3
,
p
)
≤
x if and only if u
4
(
x
,
p
)
≤
0, which yields p
∈
[
1
,
∞
). It is clearly a contradiction. If p
∈
[
0
,
∞
), then the statement in question if and only if u
4
(
x
,
p
)
≥
0, which leads to p
∈
[
0
,
∞
)
∩
(
-
∞
,
0
]
=
{
0
}. Thus the proof is complete.


## 3. Main Results

Theorem 7.
Letp
∈
(
-
∞
,
-
1
]
∪
[
0
,
∞
). Then for t
∈
(
0
,
π
/
2
),
(37)
sin
t
t
<
2
p
+
(
p
+
3
)
cos
⁡
⁡
t
(
3
p
+
1
)
+
2
cos
⁡
⁡
t
holds if and only if p
∈
(
-
∞
,
-
1
]
∪
[
9
,
∞
). Moreover, we have
(38)
H
2
(
cos
⁡
t
,
p
)
=
λ
p
2
p
+
(
p
+
3
)
cos
⁡
⁡
t
(
3
p
+
1
)
+
2
cos
⁡
⁡
t
<
sin
⁡
t
t
<
2
p
+
(
p
+
3
)
cos
⁡
⁡
t
(
3
p
+
1
)
+
2
cos
⁡
⁡
t
=
H
1
(
cos
⁡
⁡
t
,
p
)
for p
∈
(
-
∞
,
-
1
]
∪
[
9
,
∞
), where λ
p
=
(
3
p
+
1
)
/
(
π
p
) is the best possible. And the lower and upper bounds in (38) are decreasing and increasing in p on (
-
∞
,
-
1
]
∪
(
0
,
∞
), respectively.Proof.
Clearly, the desired result is equivalent tof
(
t
,
p
)
<
0 if and only if p
∈
(
-
∞
,
-
1
]
∪
[
9
,
∞
), where f
(
t
,
p
) is defined by (33). To this end, we give two limit relations. The first one follows by expanding f
(
t
,
p
) in power series for t. We have
(39)
f
(
t
,
p
)
=
-
1
180
p
-
9
p
+
1
t
4
+
o
(
t
4
)
if
p
≠
-
1
,
which yields
(40)
lim
⁡
t
→
0
+
f
(
t
,
p
)
t
4
=
-
1
180
p
-
9
p
+
1
if
p
≠
-
1
.
The second one is derived by a simple computation; that is,
(41)
f
(
π
2
-
,
p
)
=
ln
⁡
⁡
3
p
+
1
π
p
.
Now we prove that f
(
t
,
p
)
<
0 for all t
∈
(
0
,
π
/
2
) if and only if p
∈
(
-
∞
,
-
1
]
∪
[
9
,
∞
).
The necessity easily follows by solving the simultaneous inequalities:(42)
lim
⁡
t
→
0
+
f
(
t
,
p
)
t
4
=
-
1
180
p
-
9
p
+
1
≤
if
p
≠
-
1
,
f
(
t
,
-
1
)
=
ln
⁡
⁡
sin
⁡
t
t
<
0
,
f
(
π
2
-
,
p
)
=
ln
⁡
⁡
3
p
+
1
π
p
≤
0
,
which implies p
∈
(
-
∞
,
-
1
]
∪
[
9
,
∞
).
The sufficiency is due to Lemma5. In fact, If p
∈
(
-
∞
,
-
1
]
∪
[
9
,
∞
), then by Lemma 5 we see that f is decreasing in t on (
0
,
π
/
2
). Hence, f
(
t
,
p
)
<
f
(
0
+
,
p
)
=
0.
Utilizing the monotonicity off in t on (
0
,
π
/
2
) gives (38). And from Lemma 1 it is seen that the lower and upper bounds in (38) are decreasing and increasing in p on (
-
∞
,
-
1
]
∪
(
0
,
∞
), respectively.
Thus the proof is finished.By Theorem7 and Lemma 1, we have the following interesting chain of inequalities.Corollary 8.
Fort
∈
(
0
,
π
/
2
), one has
(43)
2
π
=
H
2
(
cos
⁡
t
,
-
1
)
<
⋯
<
H
2
(
cos
⁡
t
,
-
∞
)
=
2
+
cos
⁡
t
π
=
H
2
(
cos
⁡
t
,
∞
)
<
⋯
<
H
2
(
cos
⁡
t
,
9
)
<
sin
t
t
<
H
1
(
cos
⁡
t
,
9
)
<
⋯
<
H
1
(
cos
⁡
t
,
∞
)
=
2
+
cos
⁡
t
3
=
H
1
(
cos
⁡
t
,
-
∞
)
⋯
<
H
1
(
cos
⁡
t
,
-
1
)
=
-
1
.Remark 9.
It is clear that our results unify and refine Jordan and Cusa's inequalities and show that the first one in (9) is sharp. Also, Theorem 7 contains other known results, for example, taking p
=
-
3 in (38) we get
(44)
8
π
1
4
-
cos
⁡
t
<
sin
t
t
<
3
1
4
-
cos
⁡
t
,
which contain (6). After a simple transformation, (44) can be written as
(45)
8
π
t
sin
t
+
cos
⁡
t
<
4
<
3
t
sin
t
+
cos
⁡
t
,
where the second inequality in (45) is due to Neuman and Sándor [6, (
2.12
)].Theorem 10.
Letp
∈
(
-
∞
,
-
1
]
∪
[
0
,
∞
). Then for t
∈
(
0
,
π
/
2
)
(46)
2
p
+
(
p
+
3
)
cos
⁡
t
(
3
p
+
1
)
+
2
cos
⁡
t
<
sin
t
t
holds if and only if p
∈
[
0
,
p
0
], where p
0
=
(
π
-
3
)
-
1
≈
7.0625.
Moreover, forp
∈
(
0
,
p
1
], one has
(47)
H
1
(
cos
⁡
t
,
p
)
=
2
p
+
(
p
+
3
)
cos
⁡
t
(
3
p
+
1
)
+
2
cos
⁡
t
<
sin
t
t
<
λ
p
2
p
+
(
p
+
3
)
cos
⁡
t
(
3
p
+
1
)
+
2
cos
⁡
t
=
H
2
(
cos
⁡
t
,
p
)
,
where p
1
≈
6.3433, λ
p
=
(
3
p
+
1
)
/
(
π
p
) is the best possible. And H
1
(
cos
⁡
t
,
p
), H
2
(
cos
⁡
t
,
p
) are decreasing and increasing in p on (
-
∞
,
-
1
]
∪
(
0
,
∞
), respectively.
Forp
∈
(
p
1
,
p
0
] one has
(48)
H
1
(
cos
⁡
t
,
p
)
=
2
p
+
(
p
+
3
)
cos
⁡
t
(
3
p
+
1
)
+
2
cos
⁡
t
<
sin
t
t
<
δ
p
2
p
+
(
p
+
3
)
cos
⁡
t
(
3
p
+
1
)
+
2
cos
⁡
t
=
δ
p
H
1
(
cos
⁡
t
,
p
)
,
where δ
p
=
(
sin
t
0
/
t
0
)
(
(
(
3
p
+
1
)
+
2
cos
⁡
t
0
)
/
(
2
p
+
(
p
+
3
)
cos
⁡
t
0
)
) is the best possible and t
0 is the unique root of the equation
(49)
(
2
p
+
(
3
+
p
)
cos
⁡
t
)
(
3
p
+
1
+
2
cos
⁡
t
)
2
(
p
+
3
)
cos
⁡
3
t
+
8
p
cos
⁡
2
t
+
2
p
(
3
p
+
1
)
cos
⁡
t
+
3
(
p
+
1
)
2
×
sin
t
=
t
on (
0
,
π
/
2
).Proof.
Since the inequality (46) is equivalent to f
(
t
,
p
)
>
0, it suffices to prove that f
(
t
,
p
)
>
0 holds for t
∈
(
0
,
π
/
2
) if and only if p
∈
[
0
,
p
0
].
Similarly, solving the simultaneous inequalitieslim
⁡
t
→
0
t
-
4
f
(
t
,
p
)
≥
0 and f
(
π
/
2
-
,
p
)
≥
0 with p
∈
(
-
∞
,
-
1
]
∪
(
0
,
∞
) yields p
∈
[
0
,
p
0
], which proves the necessity.
Conversely, the conditionp
∈
[
0
,
p
0
] is also sufficient for f
(
t
,
p
)
>
0 to be valid. For this end, we divide the proof into two cases.
Case  1. Consider that p
∈
[
0
,
p
1
]. By Lemma 5 it is seen that f is increasing in t on (
0
,
π
/
2
), which indicates that f
(
t
,
p
)
>
f
(
0
+
,
p
)
=
0.
Case  2. Consider that p
∈
(
p
1
,
p
0
]. By Lemma 5 we see that there is a unique t
0
∈
(
0
,
π
/
2
) such that f is increasing in t on (
0
,
t
0
) and decreasing on (
t
0
,
π
/
2
). It is acquired that
(50)
f
(
t
0
,
p
)
>
f
(
t
,
p
)
>
f
(
0
+
,
p
)
=
0
for
t
∈
(
0
,
t
0
)
,
f
(
t
0
,
p
)
>
f
(
t
,
p
)
>
f
(
π
/
2
-
,
p
)
=
ln
⁡
3
p
+
1
π
p
≥
0
f
(
π
2
-
,
p
)
=
l
3
p
+
1
π
p
≥
0
f
(
p
)
iiiii
for
t
∈
(
t
0
,
π
2
)
;
that is,
(51)
f
(
t
0
,
p
)
≥
f
(
t
,
p
)
>
for
t
∈
(
0
,
π
/
2
)
,
which proves the sufficiency.
In the first case, application of the monotonicity off in t on (
0
,
π
/
2
) leads to (47), and λ
p
=
(
3
p
+
1
)
/
(
π
p
). In the second case, (51) also yields (47), and
(52)
δ
p
=
exp
⁡
f
(
t
0
,
p
)
=
sin
t
0
t
0
(
3
p
+
1
)
+
2
cos
⁡
t
0
2
p
+
(
p
+
3
)
cos
⁡
t
0
.
Thus we complete the proof.Remark 11.
Takingp
=
7 in (46), we get the first inequality in (9).Lettingp
=
p
0
=
(
π
-
3
)
-
1 and solving (49) by mathematical computation software, we find that t
0
≈
1.3055 and δ
p
0
≈
1.0015. Letting p
=
p
1 be defined by (25) yields λ
p
1
=
(
3
p
1
+
1
)
/
(
π
p
1
)
≈
1.0051. By Theorem 10 we get the following.Corollary 12.
Fort
∈
(
0
,
π
/
2
), one has
(53)
2
p
0
+
(
p
0
+
3
)
cos
⁡
t
(
3
p
0
+
1
)
+
2
cos
⁡
t
<
sin
t
t
<
δ
p
0
2
p
0
+
(
p
0
+
3
)
cos
⁡
t
(
3
p
0
+
1
)
+
2
cos
⁡
t
,
2
p
1
+
(
p
1
+
3
)
cos
⁡
t
(
3
p
1
+
1
)
+
2
cos
⁡
t
<
sin
t
t
<
λ
p
1
2
p
1
+
(
p
1
+
3
)
cos
⁡
t
(
3
p
1
+
1
)
+
2
cos
⁡
t
,
where δ
p
0
≈
1.0015 and λ
p
1
≈
1.0051 are the best possible constants.Lettingx
=
cos
⁡
1
/
3
t in Lemma 6 and using Theorems 7 and 10, we obtain a chain of inequalities that interpolates Adamović-Mitrinović and Cusa's inequalities (2) by H
1
(
cos
⁡
⁡
x
,
p
).Theorem 13.
Fort
∈
(
0
,
π
/
2
), the inequalities
(54)
2
p
+
(
p
+
3
)
cos
⁡
t
(
3
p
+
1
)
+
2
cos
⁡
t
<
cos
⁡
1
/
3
t
<
2
q
+
(
q
+
3
)
cos
⁡
t
(
3
q
+
1
)
+
2
cos
⁡
t
<
sin
t
t
<
2
r
+
(
r
+
3
)
cos
⁡
t
(
3
r
+
1
)
+
2
cos
⁡
t
<
2
+
cos
⁡
t
3
<
2
s
+
(
s
+
3
)
cos
⁡
t
(
3
s
+
1
)
+
2
cos
⁡
t
hold if and only if p
=
0, q
∈
[
0
,
p
0
], r
∈
[
9
,
∞
), and s
∈
(
-
∞
,
-
1
], where p
0
=
(
π
-
3
)
-
1.Using the monotonicity off
(
t
,
p
) in t on (
0
,
π
/
4
) given by parts one and two of Lemma 5, we see that
(55)
ln
⁡
(
4
π
3
p
+
2
+
1
(
2
2
+
1
)
p
+
3
)
=
f
(
π
4
,
p
)
≤
f
(
t
2
,
p
)
=
ln
⁡
2
sin
⁡
(
t
/
2
)
t
-
ln
⁡
H
1
(
cos
⁡
t
2
,
p
)
≤
f
(
0
,
p
)
=
0
hold for p
∈
(
-
∞
,
-
1
]
∪
[
9
,
∞
). And then we have
(56)
4
π
3
p
+
2
+
1
(
2
2
+
1
)
p
+
3
H
1
(
cos
⁡
t
2
,
p
)
cos
⁡
t
2
<
sin
t
t
=
H
1
(
cos
⁡
t
2
,
p
)
cos
⁡
t
2
.
It is clear that the right-hand in (56) is increasing in p on (
-
∞
,
-
1
]
∪
[
0
,
∞
), but the monotonicity of left-hand is to be checked. We define
(57)
H
3
(
x
,
p
)
=
4
π
3
p
+
2
+
1
(
2
2
+
1
)
p
+
3
H
1
(
x
,
p
)
,
where x
=
cos
⁡
(
t
/
2
)
∈
[
1
/
2
,
1
]. Logarithmic differentiation leads to
(58)
∂
ln
⁡
H
3
∂
p
=
3
(
3
p
+
2
+
1
)
-
2
2
+
1
(
p
(
2
2
+
1
)
+
3
)
-
3
(
3
p
+
2
x
+
1
)
+
x
+
2
2
p
+
x
(
p
+
3
)
=
-
(
(
6
(
2
2
+
1
)
(
x
-
2
2
)
(
22
-
9
2
7
-
x
)
)
w
×
(
(
3
p
+
2
+
1
)
(
p
(
2
2
+
1
)
+
3
)
(
3
p
+
2
x
+
1
)
w
w
w
×
(
2
p
+
x
(
p
+
3
)
)
(
(
3
p
+
2
+
1
)
(
p
(
2
2
+
1
)
+
3
)
(
3
p
+
2
x
+
1
)
)
-
1
(
6
(
2
2
+
1
)
(
x
-
2
2
)
(
22
-
9
2
7
-
x
)
)
)
(
p
+
1
)
(
p
-
u
5
(
x
)
)
,
where
(59)
u
5
(
x
)
=
(
5
-
2
2
)
x
-
(
2
+
2
)
(
5
2
-
2
)
-
(
2
2
+
1
)
x
.
Since
(60)
u
5
′
(
x
)
=
-
12
(
3
-
2
2
)
(
5
2
-
2
-
(
2
2
+
1
)
x
)
2
<
0
,
we have -
1
=
u
5
(
1
)
<
u
5
(
x
)
<
u
5
(
1
/
2
)
=
-
(
24
2
+
5
)
/
49
≈
-
0.7947. Consequently, ∂
(
ln
⁡
H
3
)
/
∂
p
<
0 for p
∈
(
-
∞
,
-
1
]
∪
[
0
,
∞
).The result can be stated as a theorem.Theorem 14.
Letp
∈
(
-
∞
,
-
1
]
∪
[
0
,
∞
). Then for t
∈
(
0
,
π
/
2
) the inequalities
(61)
σ
p
2
p
cos
⁡
(
t
/
2
)
+
(
p
+
3
)
cos
⁡
2
(
t
/
2
)
(
3
p
+
1
)
+
2
cos
⁡
(
t
/
2
)
<
sin
t
t
<
2
p
cos
⁡
(
t
/
2
)
+
(
p
+
3
)
cos
⁡
2
(
t
/
2
)
(
3
p
+
1
)
+
2
cos
⁡
⁡
(
t
/
2
)
.
hold if and only if p
∈
(
-
∞
,
-
1
]
∪
[
9
,
∞
), where σ
p
=
(
4
/
π
)
(
(
3
p
+
2
+
1
)
/
(
(
2
2
+
1
)
p
+
3
)
) is the best constant. And the right-hand and left-hand in (61) are increasing and decreasing in p, respectively. Inequality (61) is reversed if and only if p
∈
[
0
,
p
1
], where p
1
≈
6.3433 is defined by (25).Puttingp
=
9
,
∞
,
0,1 in Theorem 14 we have the following.Corollary 15.
Fort
∈
(
0
,
π
/
2
) the following inequalities hold:
(62)
2
(
41
2
-
25
)
7
π
2
cos
⁡
2
(
t
/
2
)
+
3
cos
⁡
⁡
(
t
/
2
)
cos
⁡
(
t
/
2
)
+
14
<
sin
t
t
<
3
2
cos
⁡
2
(
t
/
2
)
+
3
cos
⁡
(
t
/
2
)
cos
⁡
(
t
/
2
)
+
14
,
(63)
4
(
2
2
-
1
)
7
cos
⁡
2
(
t
/
2
)
+
2
cos
⁡
(
t
/
2
)
π
<
sin
t
t
<
cos
⁡
2
(
t
/
2
)
+
2
cos
⁡
(
t
/
2
)
3
,
(64)
3
cos
⁡
2
(
t
/
2
)
2
cos
⁡
(
t
/
2
)
+
1
<
sin
t
t
<
4
(
2
+
1
)
π
cos
⁡
2
(
t
/
2
)
2
cos
⁡
(
t
/
2
)
+
1
,
(65)
2
cos
⁡
2
(
t
/
2
)
+
1
cos
⁡
(
t
/
2
)
+
2
<
sin
t
t
<
2
(
3
-
2
)
π
2
cos
⁡
2
(
t
/
2
)
+
1
cos
⁡
(
t
/
2
)
+
2
.Further, letH
4 be defined on [
1
/
2
,
1
]
×
(
-
∞
,
-
1
]
∪
[
0
,
∞
) by
(66)
H
4
(
x
,
p
)
=
H
1
(
2
x
2
-
1
,
p
)
x
H
1
(
x
,
p
)
,
where H
1 is defined by (10). We can show that the monotonicity of H
4 in x for certain fixed p. Differentiation again yields
(67)
∂
ln
⁡
H
4
(
x
,
p
)
∂
x
=
2
1
+
3
p
+
2
x
-
1
x
-
p
+
3
2
p
+
(
p
+
3
)
x
+
4
(
p
+
3
)
x
(
p
-
3
)
+
2
(
p
+
3
)
x
2
-
8
x
4
x
2
+
3
p
-
1
.
It is easy to verify that
(68)
∂
ln
⁡
H
4
(
x
,
9
)
∂
x
=
-
2
(
x
-
1
)
2
(
594
x
2
+
240
x
3
+
8
x
4
+
910
x
+
273
)
x
(
2
x
+
3
)
(
x
+
14
)
(
2
x
2
+
13
)
(
4
x
2
+
1
)
<
0
,
∂
ln
⁡
H
4
(
x
,
∞
)
∂
x
=
-
2
(
1
-
x
)
(
2
x
+
1
)
x
(
x
+
2
)
(
2
x
2
+
1
)
<
0
,
∂
ln
⁡
H
4
(
x
,
1
)
∂
x
=
2
(
1
-
x
)
(
2
x
3
+
8
x
2
+
x
+
1
)
x
(
2
x
-
1
)
(
x
+
2
)
(
2
x
2
+
1
)
>
0
.
Consequently, we have
(69)
1
=
H
1
(
1
,
p
)
H
1
(
1
,
p
)
<
H
1
(
2
x
2
-
1
,
p
)
x
H
1
(
x
,
p
)
<
H
1
(
0
,
p
)
(
1
/
2
)
H
1
(
1
/
2
,
p
)
=
4
p
3
p
+
1
3
p
+
1
+
2
(
2
2
+
1
)
p
+
3
<
H
1
(
0
,
p
)
(
1
/
2
)
H
1
(
(
1
/
2
)
,
p
)
4
p
3
p
+
1
<
for
p
=
9
,
∞
.
It is reversed for p
=
1. From these we can obtain the following.Theorem 16.
Fort
∈
(
0
,
π
/
2
) the following inequalities hold:
(70)
28
9
π
6
cos
⁡
t
+
9
cos
⁡
t
+
14
<
41
(
2
2
-
25
)
7
π
2
cos
⁡
2
(
t
/
2
)
+
3
cos
⁡
(
t
/
2
)
cos
⁡
(
t
/
2
)
+
14
<
sin
t
t
<
6
cos
⁡
2
(
t
/
2
)
+
9
cos
⁡
(
t
/
2
)
cos
⁡
(
t
/
2
)
+
14
<
6
cos
⁡
t
+
9
cos
⁡
t
+
14
,
2
+
cos
⁡
t
π
<
12
(
2
2
-
1
)
7
π
cos
⁡
2
(
t
/
2
)
+
2
cos
⁡
(
t
/
2
)
3
<
sin
t
t
<
cos
⁡
2
(
t
/
2
)
+
2
cos
⁡
(
t
/
2
)
3
<
2
+
cos
⁡
t
3
,
2
cos
⁡
t
+
1
cos
⁡
t
+
2
<
2
cos
⁡
2
(
t
/
2
)
+
1
cos
⁡
(
t
/
2
)
+
2
<
sin
t
t
<
2
(
3
-
2
)
π
2
cos
⁡
2
(
t
/
2
)
+
1
cos
⁡
(
t
/
2
)
+
2
<
4
π
2
cos
⁡
t
+
1
cos
⁡
t
+
2
.Additionally, Lemma4 implies an optimal two-side inequality.Theorem 17.
Letp
∈
(
-
∞
,
-
1
]
∪
[
0
,
∞
) and let u
1
(
x
,
p
) and u
2
(
x
,
p
) be defined by (14) and (15), respectively. Then for t
∈
(
0
,
π
/
2
) the two-side inequality
(71)
u
2
(
cos
⁡
t
,
p
)
u
1
(
cos
⁡
t
,
p
)
<
sin
t
t
<
u
2
(
cos
⁡
t
,
q
)
u
1
(
cos
⁡
t
,
q
)
holds if and only if p
∈
(
-
∞
,
-
1
]
∪
[
9
,
∞
) and q
∈
[
0
,
p
1
], where p
1
≈
6.3433. And, for x
∈
(
0,1
), the function p
↦
u
2
(
x
,
p
)
/
u
1
(
x
,
p
) is decreasing on (
-
∞
,
-
1
]
∪
[
0
,
∞
).Proof.
Sinceu
1
(
x
,
p
)
,
u
2
(
x
,
p
)
>
0 for p
∈
(
-
∞
,
-
1
]
∪
[
0
,
∞
) and x
∈
(
0,1
) by Lemma 2 and g
(
t
,
p
) defined by (24) can be written as
(72)
g
(
t
,
p
)
=
-
t
u
1
(
cos
⁡
t
,
p
)
u
2
(
cos
⁡
t
,
p
)
(
sin
t
t
-
u
2
(
cos
⁡
t
,
p
)
u
1
(
cos
⁡
t
,
p
)
)
,
it follows from Lemma 4 that (71) holds if and only if p
∈
(
-
∞
,
-
1
]
∪
[
9
,
∞
) and q
∈
[
0
,
p
1
]. It remains to check the monotonicity of u
2
(
cos
⁡
t
,
p
)
/
u
1
(
cos
⁡
t
,
p
) in p. Differentiation yields
(73)
d
d
p
u
2
(
x
,
p
)
u
1
(
x
,
p
)
=
-
6
(
x
+
1
)
(
x
-
1
)
2
×
(
p
+
1
)
(
(
5
+
x
)
p
+
5
x
+
1
)
(
2
p
+
3
x
+
p
x
)
2
(
3
p
+
2
x
+
1
)
2
,
where x
∈
(
0,1
). If p
∈
[
0
,
∞
), then the numerator of the fraction in right-hand above is clearly positive. Consider that (
p
+
1
)
(
(
5
+
x
)
p
+
5
x
+
1
)
>
0. If p
∈
(
-
∞
,
-
1
], then (
p
+
1
)
≤
0 and (
(
5
+
x
)
p
+
5
x
+
1
)
≤
5
(
x
-
1
)
<
0, which yields that the numerator is nonnegative.
This proves the assertion.Similarly, we can obtain a hyperbolic version of Theorems7 and 10Theorem 18.
Letp
∈
(
-
∞
,
-
1
]
∪
[
0
,
∞
). Then for t
∈
(
0
,
∞
)
(74)
2
+
(
1
+
3
p
)
cosh
⁡
t
3
+
p
+
2
p
cosh
⁡
t
<
sinh
⁡
t
t
holds if and only if p
∈
(
-
∞
,
-
1
]
∪
[
1
/
9
,
∞
). It is reversed if and only if p
=
0.Proof.
LetF be the function defined on (
0
,
∞
)
×
(
-
∞
,
-
1
]
∪
[
0
,
∞
) by
(75)
F
(
t
,
p
)
=
3
+
p
+
2
p
cosh
⁡
t
2
+
(
1
+
3
p
)
cosh
⁡
t
sinh
⁡
t
-
t
.
Then the inequalities (74) are equivalent to F
(
t
,
p
)
>
0. Expanding in power series yields
(76)
F
(
t
,
p
)
=
t
5
180
9
p
-
1
p
+
1
+
o
(
t
5
)
,
which implies
(77)
lim
⁡
t
→
0
F
(
t
,
p
)
t
5
=
1
20
p
-
1
/
9
p
+
1
if
p
≠
-
1
,
F
(
t
,
-
1
)
=
sinh
⁡
t
-
t
>
0
.
On the other hand, we have
(78)
lim
⁡
t
→
∞
F
(
t
,
p
)
sinh
⁡
t
=
2
p
1
+
3
p
.
Now we prove desired results.
(i) We first prove thatF
(
t
,
p
)
>
0 holds if and only if p
∈
(
-
∞
,
-
1
]
∪
[
1
/
9
,
∞
).
IfF
(
t
,
p
)
>
0 for all t
>
0, then we have
(79)
lim
⁡
t
→
0
F
(
t
,
p
)
t
5
=
1
20
p
-
1
/
9
p
+
1
≥
0
,
F
(
t
,
-
1
)
=
sinh
⁡
t
-
t
>
0
,
lim
⁡
t
→
∞
F
(
t
,
p
)
sinh
⁡
t
=
2
p
1
+
3
p
≥
0
.
Solving the inequalities yields p
∈
(
-
∞
,
-
1
]
∪
[
1
/
9
,
∞
).
We prove the conditionp
∈
(
-
∞
,
-
1
]
∪
[
1
/
9
,
∞
) is sufficient for F
(
t
,
p
)
>
0 to hold for t
∈
(
0
,
∞
). Differentiation gives
(80)
∂
F
∂
t
=
(
3
p
+
1
+
2
cosh
⁡
t
)
2
p
+
(
p
+
3
)
cosh
⁡
t
cosh
⁡
t
-
3
(
p
+
1
)
2
sinh
⁡
2
t
(
2
p
+
(
p
+
3
)
cosh
⁡
t
)
2
-
1
=
(
x
-
1
)
2
2
p
(
3
p
+
1
)
x
+
(
3
p
2
+
6
p
-
1
)
(
x
+
3
p
x
+
2
)
2
,
where x
=
cosh
⁡
t
∈
(
1
,
∞
).
Due top
∈
(
-
∞
,
-
1
]
∪
[
1
/
9
,
∞
), we see that 2
p
(
3
p
+
1
)
>
0, which yields
(81)
2
p
(
3
p
+
1
)
x
+
(
3
p
2
+
6
p
-
1
)
>
2
p
(
3
p
+
1
)
+
(
3
p
2
+
6
p
-
1
)
=
(
p
+
1
)
(
9
p
-
1
)
≥
0
.
Then∂
F
/
∂
t
>
0; that is, F is increasing in t on (
0
,
∞
). It is obtained that F
(
t
,
p
)
>
F
(
0
,
p
)
=
0, which proves the sufficiency.
(ii) Next we prove that the reverse inequality of (74) holds if and only if p
=
0. The necessity follows from
(82)
lim
⁡
t
→
0
F
(
t
,
p
)
t
5
=
1
20
p
-
1
/
9
p
+
1
≤
0
,
lim
⁡
t
→
∞
F
(
t
,
p
)
sinh
⁡
t
=
2
p
1
+
3
p
≤
0
,
and the assumption p
∈
(
-
∞
,
-
1
]
∪
[
0
,
∞
). We get p
=
0.
Now we proveF
(
t
,
p
)
<
0 when p
=
0. We have
(83)
∂
F
∂
t
=
-
(
x
-
1
)
2
(
x
+
3
p
x
+
2
)
2
<
0
,
where x
=
cosh
⁡
t
∈
(
1
,
∞
), then F
(
t
,
0
)
<
F
(
0,0
)
=
0.
Thus the proof of Theorem18 is complete.Denote(84)
H
5
(
x
,
p
)
=
2
+
(
1
+
3
p
)
x
3
+
p
+
2
p
x
.
It is easy to verify that H
5
(
x
,
p
)
=
H
1
(
x
,
p
-
1
) for p
≠
0. By Lemma 1, we see that H
5 is decreasing in p on (
-
∞
,
-
1
]
∪
[
0
,
∞
). Thus, as a consequence of Theorem 14, we have the following.Corollary 19.
One has(85)
2
+
cosh
⁡
t
3
>
sinh
⁡
t
t
>
H
5
(
cosh
⁡
t
,
1
9
)
>
⋯
>
H
5
(
cosh
⁡
t
,
∞
)
=
3
cosh
⁡
t
2
cosh
⁡
t
+
1
=
H
5
(
cosh
⁡
t
,
-
∞
)
>
⋯
H
5
(
cosh
⁡
t
,
-
1
)
=
1
.Furthermore, note thatH
5
(
x
,
p
)
=
H
1
-
1
(
x
-
1
,
p
) and by Lemma 6 we have the following.Corollary 20.
One has(86)
2
+
cosh
⁡
t
3
>
sinh
⁡
t
t
>
cosh
⁡
1
/
3
t
>
1
+
2
cosh
⁡
t
2
+
cosh
⁡
t
>
H
5
(
cosh
⁡
t
,
p
)
,
where p
∈
(
-
∞
,
-
1
]
∪
(
1
,
∞
).


## 4. Applications

In this section, we give some applications of our results.


### 4.1. Shafer-Fink Type Inequalities

In [1, p. 247, 3.4
.
31], it was listed that the inequality
(87)
arcsin
x
>
6
(
x
+
1
-
1
-
x
)
4
+
x
+
1
+
1
-
x
>
3
x
2
+
1
-
x
2
holds for x
∈
(
0,1
), which is due to Shafer [21]. Fink [22] proved that the double inequality
(88)
3
x
2
+
1
-
x
2
≤
arcsin
x
≤
π
x
2
+
1
-
x
2
is true for x
∈
[
0,1
]. There has been some improvements and generalizations of Shafer-Fink inequality (see [23]). Letting sin
t
=
x in Theorems 7, 10, 13, 14, 16 and 17 we can obtain corresponding Shafer-Fink type inequalities, which clearly contain many known results. For example, Theorems 7 and 10 can be changed into the following.Proposition 21.
Forx
∈
(
0,1
), the two-side inequality
(89)
x
H
1
(
1
-
x
2
,
p
)
=
x
(
3
p
+
1
)
+
2
1
-
x
2
2
p
+
(
p
+
3
)
1
-
x
2
<
arcsin
x
<
π
p
3
p
+
1
x
(
3
p
+
1
)
+
2
1
-
x
2
2
p
+
(
p
+
3
)
1
-
x
2
=
x
H
2
(
1
-
x
2
,
p
)
holds if and only if p
∈
(
-
∞
,
-
1
]
∪
[
9
,
∞
), where π
p
/
(
3
p
+
1
) is the best possible. And, the lower and upper bounds in (89) are decreasing and increasing in p on (
-
∞
,
-
1
]
∪
(
0
,
∞
), respectively.
Inequality (89) is reversed if p
∈
[
0
,
p
1
], where p
1
≈
6.3433 is defined by (25).Lettingsin
t
=
x, then cos
⁡
⁡
(
t
/
2
)
=
(
1
/
2
)
(
1
+
x
+
1
-
x
). Theorem 14 can be restated as follows.Proposition 22.
Forx
∈
(
0,1
), the two-side inequality
(90)
2
(
3
p
+
1
)
(
1
+
x
-
1
-
x
)
+
2
x
4
p
+
(
p
+
3
)
(
1
+
x
+
1
-
x
)
<
arcsin
⁡
x
<
2
σ
p
(
3
p
+
1
)
(
1
+
x
-
1
-
x
)
+
2
x
4
p
+
(
p
+
3
)
(
1
+
x
+
1
-
x
)
holds if and only if p
∈
(
-
∞
,
-
1
]
∪
[
9
,
∞
), where σ
p
=
(
4
/
π
)
(
(
3
p
+
2
+
1
)
/
(
(
2
2
+
1
)
p
+
3
)
) is the best constant. And, the lower and upper bounds in (90) are decreasing and increasing in p on (
-
∞
,
-
1
]
∪
(
0
,
∞
), respectively.
Inequality (90) is reversed if p
∈
[
0
,
p
1
], where p
1
≈
6.3433 is defined by (25).As another example, Theorem16 can be rewritten as follows.Proposition 23.
Forx
∈
(
0,1
), all the following chains of inequalities hold:
(91)
x
3
1
-
x
2
+
14
2
1
-
x
2
+
3
<
1
3
x
+
14
(
x
+
1
-
1
-
x
)
3
+
x
+
1
+
1
-
x
<
arcsin
x
<
(
41
2

+

25
)
π
782
x

+

14
(
x
+
1

-

1
-
x
)
3

+

x
+
1

+

1
-
x
<
3
π
x
28
1
-
x
2
+
14
2
1
-
x
2
+
3
,
(92)
3
x
2
+
1
-
x
2
<
6
(
x
+
1
-
1
-
x
)
4
+
x
+
1
+
1
-
x
<
arcsin
x
<
(
1
+
2
2
)
π
12
6
(
x
+
1
-
1
-
x
)
4
+
x
+
1
+
1
-
x
<
π
x
2
+
1
-
x
2
,
(93)
π
x
4
1
-
x
2
+
2
2
1
-
x
2
+
1
<
(
2
+
3
)
π
14
x
+
2
(
x
+
1
-
1
-
x
)
1
+
x
+
1
+
1
-
x
<
arcsin
x
<
x
+
2
(
x
+
1
-
1
-
x
)
1
+
x
+
1
+
1
-
x
<
x
1
-
x
2
+
2
2
1
-
x
2
+
1
.Remark 24.
Inequalities (92) are due to Zhu [23].


### 4.2. Inequalities for Certain Means

Fora, b
>
0 with a
≠
b, the first and second Seiffert means [24, 25]; Nueman-Sándor means [26] are defined by
(94)
P
=
P
(
a
,
b
)
=
a
-
b
2
arcsin
⁡
(
(
a
-
b
)
/
(
a
+
b
)
)
,
T
=
T
(
a
,
b
)
=
a
-
b
2
arctan
⁡
(
(
a
-
b
)
/
(
a
+
b
)
)
,
NS
=
NS
(
a
,
b
)
=
a
-
b
2
arcsinh
⁡
(
(
a
-
b
)
/
(
a
+
b
)
)
,
respectively. More new means can be found in [27]. We also denote the logarithmic mean, arithmetic mean, geometric mean, and quadratic mean of a and b by L, A, G, and Q. There has been some inequalities for these means; we quote [7, 26–36]. Now we establish some new ones involving these means.Letx
=
arcsin
⁡
(
(
b
-
a
)
/
(
a
+
b
)
), arctan
(
(
b
-
a
)
/
(
a
+
b
)
). Then (
sin
x
)
/
x
=
P
/
A, cos
⁡
x
=
G
/
A; (
sin
x
)
/
x
=
T
/
Q, cos
⁡
x
=
A
/
Q. And then Theorems 7, 10, 13, 14, 16 and 17 can be stated as equivalent ones involving means P, A, G, and T, Q. For example, from Theorems 7 and 6 we have the following.Proposition 25.
Fora
,
b
>
0 with a
≠
b, both the two-side inequalities
(95)
2
p
A
+
(
p
+
3
)
G
(
3
p
+
1
)
A
+
2
G
A
<
P
<
A
2
q
A
+
(
q
+
3
)
G
(
3
q
+
1
)
A
+
2
G
,
2
p
Q
+
(
p
+
3
)
A
(
3
p
+
1
)
Q
+
2
A
Q
<
T
<
Q
2
q
Q
+
(
q
+
3
)
A
(
3
q
+
1
)
Q
+
2
A
hold if and only if p
∈
[
0
,
p
0
] and q
∈
(
-
∞
,
-
1
]
∪
[
9
,
∞
), where p
0
=
(
π
-
3
)
-
1
≈
7.0625.Making changes of variablesx
=
arctanh
⁡
(
(
b
-
a
)
/
(
a
+
b
)
), arcsinh
⁡
(
(
b
-
a
)
/
(
a
+
b
)
) yield (
sinh
⁡
x
)
/
x
=
L
/
G, cosh
⁡
x
=
A
/
G; (
sinh
⁡
x
)
/
x
=
NS
/
A, cosh
⁡
x
=
Q
/
A, respectively. And then, Theorem 18 can be equivalently written as follows.Proposition 26.
Fora, b
>
0 with a
≠
b, both the inequalities
(96)
2
G
+
(
1
+
3
p
)
A
(
3
+
p
)
G
+
2
p
A
G
<
L
,
2
A
+
(
1
+
3
p
)
Q
(
3
+
p
)
A
+
2
p
Q
A
<
NS
hold if and only if p
∈
(
-
∞
,
-
1
]
∪
[
1
/
9
,
∞
). They are reversed if and only if p
=
0.


### 4.3. The Estimate for the Sine Integral

For the estimations for the sine integral defined by(97)
Si
(
x
)
=
∫
0
x
‍
sin
t
t
d
t
,
there has been some results (see [37–39]). By our results we can obtain many estimates for Si
⁡
(
x
). Here we give a simpler but more accurate one.Proposition 27.
Forx
∈
(
0
,
π
/
2
], we have
(98)
4
2
-
2
7
π
(
x
+
sin
x
+
8
sin
x
2
)
<
Si
⁡

(
x
)
<
1
6
(
x
+
sin
x
+
8
sin
x
2
)
.Proof.
By (63) we see that the inequalities
(99)
4
(
2
2
-
1
)
7
cos
⁡
2
(
t
/
2
)
+
2
cos
⁡
(
t
/
2
)
π
<
sin
t
t
<
cos
⁡
2
(
t
/
2
)
+
2
cos
⁡
(
t
/
2
)
3
hold for t
∈
[
0
,
π
/
2
]. Integrating both sides over [
0
,
x
] and simple calculation yield (98).Remark 28.
By (98) we have
(100)
1.3682
≈
2
2
-
1
7
π
(
π
+
8
2
+
2
)
<
∫
0
π
/
2
‍
sin
t
t
d
t
<
1
12
(
π
+
8
2
+
2
)
≈
1.3713
.


## 4.1. Shafer-Fink Type Inequalities

In [1, p. 247, 3.4
.
31], it was listed that the inequality
(87)
arcsin
x
>
6
(
x
+
1
-
1
-
x
)
4
+
x
+
1
+
1
-
x
>
3
x
2
+
1
-
x
2
holds for x
∈
(
0,1
), which is due to Shafer [21]. Fink [22] proved that the double inequality
(88)
3
x
2
+
1
-
x
2
≤
arcsin
x
≤
π
x
2
+
1
-
x
2
is true for x
∈
[
0,1
]. There has been some improvements and generalizations of Shafer-Fink inequality (see [23]). Letting sin
t
=
x in Theorems 7, 10, 13, 14, 16 and 17 we can obtain corresponding Shafer-Fink type inequalities, which clearly contain many known results. For example, Theorems 7 and 10 can be changed into the following.Proposition 21.
Forx
∈
(
0,1
), the two-side inequality
(89)
x
H
1
(
1
-
x
2
,
p
)
=
x
(
3
p
+
1
)
+
2
1
-
x
2
2
p
+
(
p
+
3
)
1
-
x
2
<
arcsin
x
<
π
p
3
p
+
1
x
(
3
p
+
1
)
+
2
1
-
x
2
2
p
+
(
p
+
3
)
1
-
x
2
=
x
H
2
(
1
-
x
2
,
p
)
holds if and only if p
∈
(
-
∞
,
-
1
]
∪
[
9
,
∞
), where π
p
/
(
3
p
+
1
) is the best possible. And, the lower and upper bounds in (89) are decreasing and increasing in p on (
-
∞
,
-
1
]
∪
(
0
,
∞
), respectively.
Inequality (89) is reversed if p
∈
[
0
,
p
1
], where p
1
≈
6.3433 is defined by (25).Lettingsin
t
=
x, then cos
⁡
⁡
(
t
/
2
)
=
(
1
/
2
)
(
1
+
x
+
1
-
x
). Theorem 14 can be restated as follows.Proposition 22.
Forx
∈
(
0,1
), the two-side inequality
(90)
2
(
3
p
+
1
)
(
1
+
x
-
1
-
x
)
+
2
x
4
p
+
(
p
+
3
)
(
1
+
x
+
1
-
x
)
<
arcsin
⁡
x
<
2
σ
p
(
3
p
+
1
)
(
1
+
x
-
1
-
x
)
+
2
x
4
p
+
(
p
+
3
)
(
1
+
x
+
1
-
x
)
holds if and only if p
∈
(
-
∞
,
-
1
]
∪
[
9
,
∞
), where σ
p
=
(
4
/
π
)
(
(
3
p
+
2
+
1
)
/
(
(
2
2
+
1
)
p
+
3
)
) is the best constant. And, the lower and upper bounds in (90) are decreasing and increasing in p on (
-
∞
,
-
1
]
∪
(
0
,
∞
), respectively.
Inequality (90) is reversed if p
∈
[
0
,
p
1
], where p
1
≈
6.3433 is defined by (25).As another example, Theorem16 can be rewritten as follows.Proposition 23.
Forx
∈
(
0,1
), all the following chains of inequalities hold:
(91)
x
3
1
-
x
2
+
14
2
1
-
x
2
+
3
<
1
3
x
+
14
(
x
+
1
-
1
-
x
)
3
+
x
+
1
+
1
-
x
<
arcsin
x
<
(
41
2

+

25
)
π
782
x

+

14
(
x
+
1

-

1
-
x
)
3

+

x
+
1

+

1
-
x
<
3
π
x
28
1
-
x
2
+
14
2
1
-
x
2
+
3
,
(92)
3
x
2
+
1
-
x
2
<
6
(
x
+
1
-
1
-
x
)
4
+
x
+
1
+
1
-
x
<
arcsin
x
<
(
1
+
2
2
)
π
12
6
(
x
+
1
-
1
-
x
)
4
+
x
+
1
+
1
-
x
<
π
x
2
+
1
-
x
2
,
(93)
π
x
4
1
-
x
2
+
2
2
1
-
x
2
+
1
<
(
2
+
3
)
π
14
x
+
2
(
x
+
1
-
1
-
x
)
1
+
x
+
1
+
1
-
x
<
arcsin
x
<
x
+
2
(
x
+
1
-
1
-
x
)
1
+
x
+
1
+
1
-
x
<
x
1
-
x
2
+
2
2
1
-
x
2
+
1
.Remark 24.
Inequalities (92) are due to Zhu [23].


## 4.2. Inequalities for Certain Means

Fora, b
>
0 with a
≠
b, the first and second Seiffert means [24, 25]; Nueman-Sándor means [26] are defined by
(94)
P
=
P
(
a
,
b
)
=
a
-
b
2
arcsin
⁡
(
(
a
-
b
)
/
(
a
+
b
)
)
,
T
=
T
(
a
,
b
)
=
a
-
b
2
arctan
⁡
(
(
a
-
b
)
/
(
a
+
b
)
)
,
NS
=
NS
(
a
,
b
)
=
a
-
b
2
arcsinh
⁡
(
(
a
-
b
)
/
(
a
+
b
)
)
,
respectively. More new means can be found in [27]. We also denote the logarithmic mean, arithmetic mean, geometric mean, and quadratic mean of a and b by L, A, G, and Q. There has been some inequalities for these means; we quote [7, 26–36]. Now we establish some new ones involving these means.Letx
=
arcsin
⁡
(
(
b
-
a
)
/
(
a
+
b
)
), arctan
(
(
b
-
a
)
/
(
a
+
b
)
). Then (
sin
x
)
/
x
=
P
/
A, cos
⁡
x
=
G
/
A; (
sin
x
)
/
x
=
T
/
Q, cos
⁡
x
=
A
/
Q. And then Theorems 7, 10, 13, 14, 16 and 17 can be stated as equivalent ones involving means P, A, G, and T, Q. For example, from Theorems 7 and 6 we have the following.Proposition 25.
Fora
,
b
>
0 with a
≠
b, both the two-side inequalities
(95)
2
p
A
+
(
p
+
3
)
G
(
3
p
+
1
)
A
+
2
G
A
<
P
<
A
2
q
A
+
(
q
+
3
)
G
(
3
q
+
1
)
A
+
2
G
,
2
p
Q
+
(
p
+
3
)
A
(
3
p
+
1
)
Q
+
2
A
Q
<
T
<
Q
2
q
Q
+
(
q
+
3
)
A
(
3
q
+
1
)
Q
+
2
A
hold if and only if p
∈
[
0
,
p
0
] and q
∈
(
-
∞
,
-
1
]
∪
[
9
,
∞
), where p
0
=
(
π
-
3
)
-
1
≈
7.0625.Making changes of variablesx
=
arctanh
⁡
(
(
b
-
a
)
/
(
a
+
b
)
), arcsinh
⁡
(
(
b
-
a
)
/
(
a
+
b
)
) yield (
sinh
⁡
x
)
/
x
=
L
/
G, cosh
⁡
x
=
A
/
G; (
sinh
⁡
x
)
/
x
=
NS
/
A, cosh
⁡
x
=
Q
/
A, respectively. And then, Theorem 18 can be equivalently written as follows.Proposition 26.
Fora, b
>
0 with a
≠
b, both the inequalities
(96)
2
G
+
(
1
+
3
p
)
A
(
3
+
p
)
G
+
2
p
A
G
<
L
,
2
A
+
(
1
+
3
p
)
Q
(
3
+
p
)
A
+
2
p
Q
A
<
NS
hold if and only if p
∈
(
-
∞
,
-
1
]
∪
[
1
/
9
,
∞
). They are reversed if and only if p
=
0.


## 4.3. The Estimate for the Sine Integral

For the estimations for the sine integral defined by(97)
Si
(
x
)
=
∫
0
x
‍
sin
t
t
d
t
,
there has been some results (see [37–39]). By our results we can obtain many estimates for Si
⁡
(
x
). Here we give a simpler but more accurate one.Proposition 27.
Forx
∈
(
0
,
π
/
2
], we have
(98)
4
2
-
2
7
π
(
x
+
sin
x
+
8
sin
x
2
)
<
Si
⁡

(
x
)
<
1
6
(
x
+
sin
x
+
8
sin
x
2
)
.Proof.
By (63) we see that the inequalities
(99)
4
(
2
2
-
1
)
7
cos
⁡
2
(
t
/
2
)
+
2
cos
⁡
(
t
/
2
)
π
<
sin
t
t
<
cos
⁡
2
(
t
/
2
)
+
2
cos
⁡
(
t
/
2
)
3
hold for t
∈
[
0
,
π
/
2
]. Integrating both sides over [
0
,
x
] and simple calculation yield (98).Remark 28.
By (98) we have
(100)
1.3682
≈
2
2
-
1
7
π
(
π
+
8
2
+
2
)
<
∫
0
π
/
2
‍
sin
t
t
d
t
<
1
12
(
π
+
8
2
+
2
)
≈
1.3713
.



---

*Source: 364076-2014-03-31.xml*