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SOME REFINEMENTS OF WELL-KNOWN INEQUALITIES INVOLVING TRIGONOMETRIC
FUNCTIONS
Raouf Chouikha, Christophe Chesneau, Yogesh Bagul
To cite this version:
Raouf Chouikha, Christophe Chesneau, Yogesh Bagul. SOME REFINEMENTS OF WELL-KNOWN
INEQUALITIES INVOLVING TRIGONOMETRIC FUNCTIONS. 2020. �hal-02538194�
INVOLVING TRIGONOMETRIC FUNCTIONS
ABD RAOUF CHOUIKHA, CHRISTOPHE CHESNEAU, AND YOGESH J. BAGUL
Abstract. In this paper, we determine new and sharp inequalities involving trigonometric functions. More specifically, a new general result on the lower bound for log(1−uv),u, v∈(0,1) is proved, allowing to determine sharp lower and upper bounds for the so-called sinc function, i.e., sin(x)/x, lower bounds for cos(x) and upper bounds for (cos(x/3))3. The obtained bounds improve some well-established results. The findings are supported by graphical analy- ses.
2010 Mathematics Subject Classification:26D05, 26D07, 26D20, 33B10.
Key words and phrases:Cusa-Huygens inequality; sinc function; series expan- sion; positive and increasing function.
1. Introduction
Over the last two decades, efforts were made to bound special trigonometric and hyperbolic functions as sharp as possible, with a focus on the sinc function sin(x)/x.
The resulting inequalities find applications in many applied fields, allowing quick evaluations of complex functions involving these trigonometric functions. The lit- erature on the subject is vast and growing fast. We may refer the reader to [1], [2], [3], [4], [5], [6], [7], [8], [10], [11], [12], [13], [14], [15], [16], [17] and [18], and the references therein.
This paper contributes to the subject in the following way. First of all, we prove a general and sharp lower bound result for log(1−uv), u, v∈(0,1). Then, we apply this result to determine polynomial-exponential lower bounds for sin(x)/x and cos(x), also with the use of infinite product series. We prove that they are sharp, improving some recent results of the literature. Also, as intermediate results, some new polynomial-exponential inequalities are set. As an alternative approach, we use these results to conjecture upper bounds for sin(x)/xand (cos(x/3))3. Proofs are given in details by the means of Taylor developments. Again, some recent results in the fields are refined, including the famous Cusa-Huygens inequality[9]. All the findings are supported by the visual checks of appropriate functions.
The rest of the paper is planned as follows. Section 2 investigate the lower bounds. Section 3 is devoted to the upper bounds for sin(x)/x and (cos(x/3))3, with discussions.
2. Lower bounds
This section is devoted to the proof of new lower bounds, involving sharp lower bounds for sin(x)/xand cos(x) as applications. Some graphics support the findings.
1
2.1. Some new general results. The result below proposes a new lower bound for log(1−uv) with u, v ∈(0,1), which will be at the center of the proofs of new sharp lower bounds for sin(x)/xand cos(x).
Proposition 2.1. For any u, v∈(0,1), the following inequality holds:
log(1−uv)> uv(u−1)h
u+ 1 + uv 2
i
+u3log(1−v).
Proof. By virtue of the logarithmic series expansion and, for k≥3, uk < u3, after some algebraic manipulations, we get
log(1−uv) =−
+∞
X
k=1
ukvk
k =−uv−u2v2
2 −
+∞
X
k=3
ukvk k
>−uv−u2v2 2 −u3
+∞
X
k=3
vk
k =−uv−u2v2 2 +u3
"
−
+∞
X
k=1
vk
k +v+v2 2
#
=−uv−u2v2 2 +u3
log(1−v) +v+v2 2
=uv(u−1)h
u+ 1 +uv 2
i
+u3log(1−v).
This ends the proof of Proposition 2.1.
Proposition 2.2. Foru, v∈(0,1) the following inequalities hold:
1−uv >(1−v)u3euv(u−1)[u+1+uv2]>(1−v)u2euv(u−1)>(1−v)u. Proof. The first inequality is Proposition 2.1, by taking the exponential trans- formation. The second one takes the first steps of the proof of Proposition 2.1 in the following sense: Sinceu3< u2, we have
uv(u−1)h
u+ 1 + uv 2
i
+u3log(1−v) =−uv−u2v2 2 −u3
+∞
X
k=3
vk k
>−uv−u2v2 2 −u2
+∞
X
k=3
vk
k =−uv−u2v2 2 +u2
"
−
+∞
X
k=1
vk
k +v+v2 2
#
=uv(u−1) +u2log(1−v).
The desired inequality follows by taking the exponential transformation. The last inequality is follows from [7, Theorem 1-1], showing thatuv(u−1) +u2log(1−v)>
ulog(1−v). This ends the proof of Proposition 2.2.
2.2. Lower bounds for sin(x)/x. The result below presents a new sharp bound for sin(x)/xinvolving the exponential function.
Proposition 2.3. Forx∈(0, π), we have the following inequalities:
sin(x) x >
1−x2
π2 π
6 945
ex
2hπ4
945−16+x22π2 945−901i
>
1−x2
π2 π
4 90
ex
2
π2 90−16
>
1−x2
π2 π
2 6
.
Proof. By using the infinite product expression of sin(x)/x, takingu= 1/k2and v=x2/π2in Proposition 2.2, and using the following well-known results on the zeta function, i.e., ζ(k) =P+∞
n=11/nk: ζ(2) =π2/6, ζ(4) = π4/90 andζ(6) =π6/945, we get
sin(x)
x =
+∞
Y
k=1
1− x2
π2k2
>
+∞
Y
k=1
"
1−x2
π2 k16
e x
2
π2k2(k12−1)hk12+1+ x2
2k2π2
i#
=
1−x2 π2
P+∞k=1k16
e
x2 π2
P+∞
k=1
h1
k6−1
k2+x2
2π2(k16−1
k4)i
=
1−x2 π2
π
6 945
ex2
hπ4
945−16+x22
π2 945−901i
.
The second inequality is due to Proposition 2.2, with similar lines of proof. The last inequality follows from [7, Proposition 2-4]. This ends the proof of Proposition
2.3.
Figure 1 illustrates the two first inequalities of Proposition 2.3 by plotting the two following functions forx∈(0, π):
A(x) =sin(x)
x −
1−x2
π2 π
6 945
ex2
hπ4
945−16+x22
π2 945−901i
and
B(x) =
1−x2 π2
π
6 945
ex
2h
π4
945−16+x22
π2 945−901i
−
1−x2 π2
π
4 90
ex
2
π2 90−16
.
0.0 0.5 1.0 1.5 2.0 2.5 3.0
0e+002e−044e−046e−04
A(x)
0.0 0.5 1.0 1.5 2.0 2.5 3.0
0.0000.0010.0020.003
B(x)
, Figure 1. Plots for A(x) andB(x) forx∈(0, π), respectively.
As expected, we see thatA(x) andB(x) are positive, with very small variations, attesting the sharpness of the obtained bounds.
2.3. Lower bounds forcos(x). The result below presents a new sharp bound for cos(x) involving the exponential function.
Proposition 2.4. Forx∈(0, π/2), we have the following inequalities:
cos(x)>
1−4x2
π2 π
6 960
ex2
hπ4
240−12+x2
π2 120−121i
>
1−4x2
π2 π
4 96
ex
2
π2 24−12
>
1−4x2
π2 π
2 8
.
Proof. By using the infinite product expression of cos(x), takingu= 1/(2k−1)2 and v = 4x2/π2 in Proposition 2.2, and using the following well-known results:
P+∞
k=11/(2k−1)2 = π2/8, P+∞
k=11/(2k−1)4 = π4/96 and P+∞
k=11/(2k−1)6 = π6/960 we get
cos(x) =
+∞
Y
k=1
1− 4x2 π2(2k−1)2
>
+∞
Y
k=1
"
1−4x2
π2
(2k−1)61 e
4x2 π2 (2k−1)2
1
(2k−1)2−1h 1
(2k−1)2+1+π2 (2k−1)22x2 i#
=
1−4x2 π2
P+∞
k=1 1 (2k−1)6
e
4x2 π2
P+∞
k=1
h 1
(2k−1)6−(2k−1)21 +2x2
π2
1
(2k−1)6−(2k−1)41 i
=
1−4x2 π2
π
6 960
ex2
hπ4 240−12+x2
π2 120−121i
.
The second inequality is also derived to Proposition 2.2, with similar mathematical arguments. The last one is a consequence of [7, Proposition 2-4]. This ends the
proof of Proposition 2.4.
As a matter of fact, Proposition 2.4 improves [2, Proposition 4], i.e., for any x∈(0, π/2),
cos(x)>
1−4x2
π2 π
2 8
. (2.1)
Figure 2 illustrates the two first inequalities of Proposition 2.4 by plotting the two following functions forx∈(0, π/2):
C(x) = cos(x)−
1−4x2 π2
π
6 960
ex2
hπ4
240−12+x2
π2 120−121i
and
D(x) =
1−4x2 π2
π
6 960
e4x
2h
π4
960−18+2x2
π2 960−961i
−
1−4x2 π2
π
4 96
ex
2
π2 24−12
.
0.0 0.5 1.0 1.5
0e+004e−058e−05
C(x)
0.0 0.5 1.0 1.5
0e+004e−048e−04
D(x)
, Figure 2. Plots forC(x) andD(x) forx∈(0, π/2), respectively.
We observe thatC(x) andD(x) are positive, with very small variations, attesting the sharpness of the obtained bounds.
3. Upper bounds
Here, we derive some new sharps upper bounds for sin(x)/x and (cos(x/3))3, which naturally appear in many inequality involving trigonometric functions.
3.1. Upper bounds for sin(x)/x. In [12], the authors proved the inequality of Cusa-Huygens: forx∈(0, π/2),
sin(x)
x < 2 + cos(x)
3 .
(3.1)
That is, is natural to address the following question: Based on (3.1), can we use lower bounds for cos(x) to derive upper bounds for sin(x)/x?
As a first remark, by using the simple but improvable lower bounds in (2.1), we provide a first answer to the question: Is the following inequality true ? For any x∈(0, π/2),
00 2 + cos(x) 3 > 2
3+1 3
1−4x2
π2 π
2 8
>sin(x) x
00.
The answer is negative because the functionh(x) = 3 sin(x)/x−2− 1−4x2/π2π2/8
has not a constant sign. Indeed, as countered example, we haveh(0.3)≈0.0001>0 andh(0.8)≈ −0.00034<0.
However, based on Proposition 2.4, the following chain of inequalities is true, providing a new upper bound for sin(x)/x.
Proposition 3.1. Forx∈(0, π/2), we have 2 + cos(x)
3 > 2 3+1
3
1−4x2 π2
π
6 960
ex
2h
π4
240−12+x2
π2 120−121i
> 2 3+1
3
1−4x2 π2
π
4 96
ex
2π2 24−12
>sin(x) x .
Proof. We only need to prove the last right inequalities, the others follow from Proposition 2.4. The proof is based on the analytical study of the following function:
k(x) = 3sin(x) x −2−
1−4x2
π2 π
4 96
ex
2
π2 24−12
. (3.2)
The desired inequality comes by proving that k(x) is non positive. The Taylor development of sin(x) gives
sin(x) =x−x3 3! +x5
5! −x7
7! +. . .+ (−1)k−1 x2k−1
(2k−1)!+ (−1)k cos(θx) (2k+ 1)!x2k+1, where θ ∈ (0,1). As a direct application, the following inequalities hold forx ∈ (0, π/2):
1−1
6x2+ 1
120x4− 1
5040x6< sin(x)
x < `(x), (3.3)
where
`(x) = 1−1
6x2+ 1
120x4− 1
5040x6+ 1 362880x8
≈1−0.166667x2+ 0.0083333x4−0.00019841x6+ 0.0000027x8. Similarly, we have
ex
2π2 24−12
> m(x), (3.4)
where
m(x) = 1 + π2
24−1 2
x2+1
2 π2
24−1 2
2 x4+1
6 π2
24−1 2
3 x6
≈1−0.088766x2+ 0.0039392x4−0.00011655x6.
Applying again the Taylor decomposition technique, since 4x2/π2<1 andπ4/96>
1, we have
1−4x2
π2 π
4 96
> n(x), (3.5)
where
n(x) = 1− 1
24π2x2+ 1
1152π4− 1 12
x4− 4 3π2
π4 1152− 1
12 π4 96 −2
x6 + 4
3π4 π4
1152− 1 12
π4
96 −2 π4 96−3
x8
≈1−0.411246x2+ 0.001225x4+ 0.00016305x6+ 0.000032799x8.
By putting (3.2), (3.3), (3.4) and (3.5) together, we obtain
k(x)<3 1−0.166667x2+ 0.0083333x4−0.00019841x6+ 0.0000027x8
−2
− 1−0.41124x2+ 0.001225x4+ 0.000163x6+ 0.0000328x8
× 1−0.08876x2+ 0.0039392x4−0.00011655x6 .
After development and a acceptable approximation (with 5 digits), we arrive at k(x)<−0.0000001x12+ 0.000003x10−0.0000654x8+ 0.0010868x6−0.01667x4. This last polynomial is non positive because it has no root in the interval (0, π/2).
This ends the proof of Proposition 3.1.
Thanks to Proposition 3.1, we then obtain a better bound for the Cusa-Huygens inequality.
Also, we may derive from Propositions 2.1 and 3.1 the following new frame for sin(x)/x:
1−4x2
π2 π
6 960
ex
2h
π4
240−12+x2
π2 120−121i
< sin(x) x
<2 3 +1
3
1−4x2 π2
π
6 960
ex
2h
π4
240−12+x2
π2 120−121i
.
Figure 3 provides a graphical illustration of the main finding of Proposition 3.1 by displaying the following function forx∈(0, π/2):
E(x) = sin(x)
x −
2 3 +1
3
1−4x2 π2
π
4 96
ex
2
π2 24−12
.
0.0 0.5 1.0 1.5
0.0000.0100.0200.030
E(x)
, Figure 3. Plots forE(x) forx∈(0, π/2).
We see thatE(x) is positive, as proved analytically.
3.2. Upper bounds for (cos(x/3))3. In [15], the following chain of inequalities is proved. Forx∈(0, π/2),
sin(x) x <
1 3
2 cosx 2
+ 12
<
cosx 3
3
< 2 + cos(x)
3 .
The following result proposes a refinement of this results, by the use of Propo- sition 2.4.
Proposition 3.2. Forx∈(0, π/2) we have
cosx 3
3
<2 3 +1
3
1−4x2 π2
π
6 960
ex
2h
π4
240−12+x2
π2 120−121i
<2 3 +1
3
1−4x2 π2
π
4 96
ex
2π2 24−12
< 2 + cos(x)
3 .
Proof. We follow the lines of the proof of Proposition 3.1, by proving the first left inequality; the others follows from Proposition 2.4. Firstly, the classic Taylor development of cos(x) gives
cosx= 1−x2 2! +x4
4! −x6
6! +. . .+ (−1)kx2k
2k! + (−1)k+1 cos(θx) (2k+ 2)!x2k+2, whereθ∈(0,1). Among others, this implies that
cos(x)<1−x2 2! +x4
4! −x6 6! +x8
8!. That is, we have
cosx
3 3
< q(x), (3.6)
where, doing a standard development and a acceptable approximation, q(x) =
1−(x/3)2
2! +(x/3)4
4! −(x/3)6
6! +(x/3)8 8!
3
≈1−1
6x2+ 7
648x4− 61
174960x6+ 547 88179840x8
≈1−0.166667x2+ 0.010802x4−0.0003486x6+ 0.000006x8. It follows from (3.4) that
ex
2
π2 24−12
> r(x), (3.7)
where
r(x) = 1 + π2
24−1 2
x2+1
2 π2
24−1 2
2 x4+1
6 π2
24−1 2
3 x6
≈1−0.088766x2+ 0.0039392x4−0.00011655x6. We have also
1−4x2
π2 π
4 96
> s(x), (3.8)
where
s(x) = 1− 1
24π2x2+ 1
1152π4− 1 12
x4− 4 3π2
π4 1152− 1
12 π4 96 −2
x6 + 4
3π4 π4
1152− 1 12
π4
96 −2 π4 96−3
x8
≈1−0.411246x2+ 0.001225x4+ 0.0001631x6+ 0.0000328x8.
The relations (3.6), (3.7) and (3.8) allow us to derive an estimate of the difference
t(x) = 3 cosx
3 3
−2−
1−4x2 π2
π
4 96
ex2(π
4 24−12). Indeed, we have
t(x)<3
1−1/6x2+ 7
648x4− 61
174960x6+ 547 88179840x8
−2
− 1 + π2
24−1 2
x2+1
2 π2
24 −1 2
2 x4+1
6 π2
24 −1 2
3 x6
!
×
1− 1
24π2x2+ 1
1152π4− 1 12
x4− 4 3π2
π4 1152− 1
12 π4 96−2
x6 + 4
3π4 π4
1152− 1 12
π4
96−2 π4 96−3
x8
, implying that
t(x)<3 1.0−0.166667x2+ 0.010802x4−0.0003486x6+ 0.000006x8
−2
− 1−0.088766x2+ 0.0039392x4−0.0001166x6
× 1−0.411246x2+ 0.001225x4+ 0.0001631x6+ 0.0000328x8 .
That is, after development and a suitable approximation (with 5 digits), we arrive at
t(x)<−0.00925926x4+ 0.000636532x6−0.00005245x8+ 0.00000241x10. We verify that this polynomial is non positive because it has no root in the interval (0, π/2). This proved the first left inequality, ending the proof of Proposition 3.2.
As illustration of the main result in Proposition 3.2, Figure 4 shows the curve of the following function forx∈(0, π/2):
F(x) = 2 3+1
3
1−4x2 π2
π
6 960
ex
2h
π4
240−12+x2
π2 120−121i
− cosx
3 3
.
0.0 0.5 1.0 1.5
0.0000.0050.0100.015
F(x)
, Figure 4. Plots forF(x) forx∈(0, π/2).
As expected, we see thatF(x) is positive.
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4, Cour des Quesblais 35430 Saint-Pere Email address:chouikha@math.univ-paris13.fr
LMNO, University of Caen-Normandie, Caen, France Email address:christophe.chesneau@unicaen.fr
Department of Mathematics, K. K. M. College, Manwath, Dist : Parbhani(M.S.) - 431505, India
Email address:yjbagul@gmail.com