CUBO A Mathematical Journal
Vol.21, No¯ 01, (21–35). April 2019

http: // dx. doi. org/ 10. 4067/ S0719-06462019000100021

Some New Simple Inequalities Involving Exponential,
Trigonometric and Hyperbolic Functions

Yogesh J. Bagul1, Christophe Chesneau2

1Department of Mathematics, K. K. M. College Manwath,

Parbhani(M.S.) - 431505, India

yjbagul@gmail.com
2LMNO, University of Caen Normandie, France

christophe.chesneau@unicaen.fr

ABSTRACT

The prime goal of this paper is to establish sharp lower and upper bounds for useful

functions such as the exponential functions, with a focus on exp(−x2), the trigonometric

functions (cosine and sine) and the hyperbolic functions (cosine and sine). The bounds

obtained for hyperbolic cosine are very sharp. New proofs, refinements as well as new

results are offered. Some graphical and numerical results illustrate the findings.

RESUMEN

El objetivo principal de este art́ıculo es establecer cotas inferiores y superiores precisas

para funciones útiles tales como las funciones exponenciales, con énfasis especial en

exp(−x2), las funciones trigonométricas (coseno y seno) y las funciones hiperbólicas

(coseno y seno). Las cotas obtenidas para el coseno hiperbólico son muy precisas.

Se presentan, tanto nuevas demostraciones y refinamientos, como resultados nuevos.

Algunos resultados numéricos y gráficos ilustran los resultados encontrados.

Keywords and Phrases: Exponential function; trigonometric function; hyperbolic function.

2010 AMS Mathematics Subject Classification: 26D07, 33B10, 33B20.

http://dx.doi.org/10.4067/S0719-06462019000100021


22 Yogesh J. Bagul and Christophe Chesneau CUBO
21, 1 (2019)

1 Introduction

Sharp bounds for useful functions play a central role in many areas of mathematics and theoretical

physics. They aim to provide some properties of functions of interest, possibly complex, by dealing

with more tractable functions (in the context). The literature on the bounds dealing with the

special functions such as e−x
2

, cos(x), sin(x), sinc(x), cosh(x), sinh(x) and tanh(x), is very vast.

Recent developments can be found in [10, 11, 7, 5, 1, 20, 17, 4, 15, 6, 21, 16, 3, 8, 14, 13, 18, 19]

and the references therein. In this paper, we offer new simple tight (lower and upper) bounds

involving these functions, with a high potential of interest for many researchers in mathematics

or theoretical physics. Some proofs of our results are based on the so-called l’Hospital’s rule of

monotonicity, the others used recent results with a new approach. The sharpness of our bounds

are highlighted by some graphics and numerical studies using a global L2 error as benchmark.

The result below shows bounds for e−x
2

defined with the cosine function and well-chosen

constants.

Proposition 1.1. For x ∈ (0, π/2), the best possible constants α and β in the following inequalities

cos(x) − 1 + α

α
6 e−x

2

6
cos(x) − 1 + β

β
(1.1)

are 1/2 and ≈ 1.092663 respectively.

The interest of Proposition 1.1 is the simplicity of the bounds, with very tractable expressions.

It can be useful to evaluate complex functions depending on e−x
2

(Gaussian probability density

function, error function etc.). The bounds of Proposition 1.1 are illustrated in Figure 1. We see

that the lower bound is sharp for small values for x.

0.0 0.5 1.0 1.5

0
.2

0
.4

0
.6

0
.8

1
.0 exp(− x2)

(cos(x) − 1 + α) α
(cos(x) − 1 + β) β



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Some New Simple Inequalities Involving Exponential . . . 23

Figure 1: Graphs of the functions of the bounds (1.1) for x ∈ (0, π/2).

Note: Using exponential and cosine series, Proposition 1.1 can be expressed in terms of

alternating series as follows.

For x ∈ (−π/2, π/2), we have

1

α

∞∑

k=1

(−1)kx2k

(2k)!
6

∞∑

k=1

(−1)kx2k

k!
6

1

β

∞∑

k=1

(−1)kx2k

(2k)!
,

where α and β are as defined above.

Now let us recall that the sinc function is defined by

sinc(x) =

{
sin(x)

x
x 6= 0,

1 x = 0.
(1.2)

It is of importance due to it’s frequent occurrence in Fourier analysis. So the interest of finding

the bounds of this type of functions is increasing. In the next proposition, we give new bounds to

sinc function using hyperbolic tangent.

Proposition 1.2. For x ∈ (0, π/2), we have

(

tanh(x)

x

)δ

<
sin(x)

x
<

(

tanh(x)

x

)η

(1.3)

with the best possible constants δ = 0.839273 and η = 1/2.

In the following propositions, the inequalities presented are somewhat Cusa-Huygen’s type

[13, 18]. Proposition 1.3 below provides bounds for the sinc function using e−x
2

or hyperbolic

cosine.

Proposition 1.3. For x ∈ (0, π/2), the inequalities

(

2 + e−x
2

3

)a

<
sin(x)

x
<

(

2 + e−x
2

3

)b

(1.4)

and

(

3

2 + cosh(x)

)c

<
sin(x)

x
<

(

3

2 + cosh(x)

)d

(1.5)

are true with the best possible constants a ≈ 1.240827, b = 1/2, c ≈ 1.108171 and d = 1.



24 Yogesh J. Bagul and Christophe Chesneau CUBO
21, 1 (2019)

In view of Propositions 1.2 and 1.3, it is natural to address the following question: Which

bounds for sinc are the best ? We provide the answer by doing a numerical study. We investigate

the global L2 error defined by

e(u) =

∫π/2

0

(

sin x

x
− u(x)

)2

dx,

where u(x) denotes bound (lower or upper) in (1.3), (1.4) and (1.5). The results are summarized

in Table 1.

Table 1: Global L2 errors e(u) for sinc(x) and the functions u(x) in the bounds of (1.3), (1.4)

and (1.5) for x ∈ (0, π/2).

Inequality (1.3)

u(x) lower upper

e(u) ≈ 0.001421437 ≈ 0.003648618

Inequality (1.4)

u(x) lower upper

e(u) ≈ 0.006242974 ≈ 0.008628254

Inequality (1.5)

u(x) lower upper

e(u) ≈ 6.53313 × 10−5 ≈ 0.0001542441

It follows from Table 1 that the bounds (1.5) are more sharp. This sharpness is illustrated in

Figure 2.

0.0 0.5 1.0 1.5

0
.7

0
.8

0
.9

1
.0 sin(x) x

(3 (2 + cosh(x)))c

(3 (2 + cosh(x)))d

Figure 2: Graphs of the functions of the bounds (1.5) for x ∈ (0, π/2).



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Some New Simple Inequalities Involving Exponential . . . 25

The next result provides bounds for x/ sinh(x) using cosine function.

Proposition 1.4. If x ∈ (0, π/2) then we have

(

2 + cos(x)

3

)m

<
x

sinh(x)
<

(

2 + cos(x)

3

)n

(1.6)

with the constants m ≈ 1.014227 and n ≈ 0.928648.

The obtained bounds are illustrated in Figure 3.

0.0 0.5 1.0 1.5

0
.7

0
0
.7

5
0
.8

0
0
.8

5
0
.9

0
0
.9

5
1
.0

0

x sinh(x)
((2 + cos(x)) 3)m
((2 + cos(x)) 3)n

Figure 3: Graphs of the functions of the bounds (1.6) for x ∈ (0, π/2).

Note: The inequality
2 + cos(x)

3
<

x

sinh(x)

is more sharp version of left inequality of (1.6). It is appeared in [19, Theorem 6].

Proposition 1.5 below presents sharp bounds for sinh(x)/x using hyperbolic cosine.

Proposition 1.5. For x ∈ (0, π/2) one has

(

2 + cosh(x)

3

)p

<
sinh(x)

x
<

(

2 + cosh(x)

3

)q

(1.7)

with the constants p ≈ 0.928648 and q ≈ 1.009155.

The bounds are illustrated in Figure 4.



26 Yogesh J. Bagul and Christophe Chesneau CUBO
21, 1 (2019)

0.0 0.5 1.0 1.5

1
.0

1
.1

1
.2

1
.3

1
.4

sinh(x) x
((2 + cosh(x)) 3)p
((2 + cosh(x)) 3)q

Figure 4: Graphs of the functions of the bounds (1.7) for x ∈ (0, π/2).

Note: The hyperbolic Cusa-Huygen’s inequality[16]

sinh(x)

x
<

2 + cosh(x)

3

is however more sharp than right inequality of (1.7).

The rest of the study is devoted to new bounds for cosh(x), with discussion. A well-known

upper bound for cosh(x) is given by ex
2/2. This result was recently completed by Yogesh Bagul[3,

Theorem 2.1] who finds a sharp lower bound, i.e.

eax
2

< cosh(x) < ex
2/2, x ∈ (0, 1), (1.8)

with the best possible constants a ≈ 0.433781 and 1/2. We now aim to refine the inequalities of

(1.8) in Proposition 1.6 below.

Proposition 1.6. For x ∈ (0, 1), we have

exp

(

3

2

(

1 − e−x
2/3
)

)

6 cosh(x) 6 exp

(

1

2θ

(

1 − e−θx
2
)

)

(1.9)

with θ ≈ 0.272342.

Note: Using the well-known inequality ey > 1+y for y ∈ R, we obtain exp
((

1 − e−θx
2
)

/(2θ)
)

6

ex
2/2. This proves that the upper bound in (1.9) is sharper to the one in (1.8).

Alternative bounds are given in Proposition 1.7 below, with discussion.



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Some New Simple Inequalities Involving Exponential . . . 27

Proposition 1.7. For x ∈ (0, 1), we have

(

1 +
x2

3

)3/2

6 cosh(x) 6

(

1 +
x2

ξ

)ξ/2

(1.10)

with ξ ≈ 3.194528.

Note: Again, using the well-known inequality ey > 1 + y for y ∈ R, we get
(

1 + x2/ξ
)ξ/2

6

ex
2/2. This shows that the upper bound in (1.10) is sharper to the one in (1.8).

We now claim that the bounds obtained in (1.10) are better than those in (1.8) and (1.9).

Numerical results support this claim. Indeed, by considering the global L2 error defined by

e∗(u) =

∫1

0

(cosh(x) − u(x))
2
dx,

where u(x) denotes bound (lower or upper) in (1.8), (1.9) and (1.10), Table 1 indicates that (1.10)

are the best.

Table 2: Global L2 errors e∗(u) for cosh(x) and the functions u(x) in the bounds of (1.8), (1.9)

and (1.10) for x ∈ (0, 1).

Inequality (1.8)

u(x) lower upper

e∗(u) ≈ 0.0001352084 ≈ 0.001139289

Inequality (1.9)

u(x) lower upper

e∗(u) ≈ 1.335929 × 10
−5

≈ 7.004029 × 10−6

Inequality (1.10)

u(x) lower upper

e∗(u) ≈ 9.456552 × 10
−7

≈ 6.895902 × 10−7

The sharpness of the obtained bounds is illustrated in Figures 5 and 6 (for a zoom on the

interval (0.95, 1), where the hierarchy of the bounds is more clear).



28 Yogesh J. Bagul and Christophe Chesneau CUBO
21, 1 (2019)

0.0 0.2 0.4 0.6 0.8 1.0

1
.0

1
.1

1
.2

1
.3

1
.4

1
.5

cosh(x)
(1 + x2 3)(3 2)

(1 + x2 ξ)(ξ 2)

Figura 5: Graphs of the functions of the bounds (1.10) for x ∈ (0, 1).

0.95 0.96 0.97 0.98 0.99 1.00

1
.4

9
1
.5

0
1
.5

1
1
.5

2
1
.5

3
1
.5

4

cosh(x)
(1 + x2 3)(3 2)

(1 + x2 ξ)(ξ 2)

Figura 6: Graphs of the functions of the bounds (1.10) for x ∈ (0.95, 1).

Note: To prove the inequalities (1.5), (1.6) and (1.7), we will simply use the results of [7, 5, 12].

We stress on the fact that it is not difficult to verify that all the results in [5] are also true in (0, π/2)

with the respective best possible constants obtained accordingly (see [12]). Propositions 1.6 and

1.7 will be proved by the techniques of integration on some known results[4, 6]. For proving

Proposition 1.1, Proposition 1.2 and Proposition 1.3, we need the Lemmas presented in the next

section.



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Some New Simple Inequalities Involving Exponential . . . 29

2 Lemmas

The following Lemma is known as l’Hospital’s rule of monotonicity. The details are given in [9]

and [2].

Lemma 2.1. ([2]) Let f, g be two real valued functions which are continuous on [a, b] and differ-

entiable on (a, b), where −∞ < a < b < ∞ and g′(x) 6= 0, for ∀x ∈ (a, b). Let,

A(x) =
f(x) − f(a)

g(x) − g(a)

and

B(x) =
f(x) − f(b)

g(x) − g(b)
.

Then,

I) A(x) and B(x) are increasing on (a, b) if f′/g′ is increasing on (a, b) and

II) A(x) and B(x) are decreasing on (a, b) if f′/g′ is decreasing on (a, b).

The strictness of the monotonicity of A(x) and B(x) depends on the strictness of monotonicity of

f′/g′.

Lemma 2.2. H(x) =
sin(x)−x cos(x)

x2 sin(x)
is strictly positive increasing in (0, π/2).

Proof: H(x) is positive as cos(x) <
sin(x)

x
on (0, π/2).

Consider,

H(x) =
sin(x) − x cos(x)

x2 sin(x)
=

H1(x)

H2(x)
,

where H1(x) = sin(x) − x cos(x) and H2(x) = x
2 sin(x) are such that H1(0) = 0 and H2(0) = 0. By

differentiating
H′1(x)

H′2(x)
=

sin(x)

x cos(x) + 2 sin(x)
=

H3(x)

H4(x)
,

where H3(x) = sin(x) and H4(x) = x cos(x) + 2 sin(x) with H3(0) = 0 and H4(0) = 0. Again

differentiating we get

H′3(x)

H′4(x)
=

cos(x)

−x sin(x) + 3 cos(x)
=

1

−x tan(x) + 3
.

Now, it is well known that −x tan(x) is decreasing in (0, π/2) and so is −x tan(x) + 3. By Lemma

1, H(x) is a strictly increasing function in (0, π/2).

3 Proofs of the Main Results

This section is devoted to the proofs of our main results.



30 Yogesh J. Bagul and Christophe Chesneau CUBO
21, 1 (2019)

Proof of Proposition 1.1: Clearly, the equalities hold at x = 0. Consider

f(x) =
cos(x) − 1

e−x
2

− 1
=

f1(x)

f2(x)
,

where f1(x) = cos(x) − 1 and f2(x) = e
−x2 − 1 with f1(0) = 0 and f2(0) = 0. By differentiation,

we obtain
f′1(x)

f′2(x)
=

sin(x)ex
2

2x
=

f3(x)

f4(x)
,

where f3(x) = sin(x)e
x2 and f4(x) = 2x with f3(0) = 0 and f4(0) = 0. Again differentiating we get

f′3(x)

f′4(x)
=

ex
2

2
[cos(x) + 2x sin(x)]

=
ex

2

2
F(x),

where F(x) = cos(x) + 2x sin(x). Differentiation gives

F′(x) = 2x cos(x) + sin(x) > 0

in (0, π/2), which implies that F(x) is increasing. Thus
f′
3
(x)

f′
4
(x)

being a product of two positive in-

creasing functions is a positive increasing. By Lemma 2.1, f(x) is also increasing in (0, π/2). So

α = f(0+) = 1/2 and β = f(π/2−) = −1/[e−(π/2)
2

− 1] ≈ 1.092663.

Proof of Proposition 1.2: Let us set

h(x) =
log(sin(x)/x)

log(tanh(x)/x)
=

h1(x)

h2(x)
,

where h1(x) = log(sin(x)/x) and h2(x) = log(tanh(x)/x) with h1(0+) = 0 and h2(0+) = 0.

Differentiating we get

h′1(x)

h′2(x)
=

sin(x) − x cos(x)

x2 sin(x)

x2 tanh(x)

tanh(x) − x sech2(x)
= H(x) J(x),

where H(x) =
sin(x)−x cos(x)

x2 sin(x)
and J(x) =

x2 tanh(x)

tanh(x)−x sech2(x)
. Now set

J(x) =
J1(x)

J2(x)
,

where J1(x) = x
2 tanh(x) and J2(x) = tanh(x) − x sech

2
(x) with J1(0) = 0 and J2(0) = 0. Differ-

entiation gives

J′1(x)

J′2(x)
=

x sech2(x) + 2 tanh(x)

2 sech2(x) tanh(x)

=
1

2

x

tanh(x)
+ cosh2(x),



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Some New Simple Inequalities Involving Exponential . . . 31

which is clearly increasing as both x/ tanh(x) and cosh2(x) are increasing. By Lemma 2.1, J(x) is

also increasing in (0, π/2). Moreover, J(x) is positive as x/ sinh(x) < cosh(x). By Lemma 2.2, H(x)

is strictly positive increasing in (0, π/2). h′1(x)/h
′

2(x), being product of two positive increasing

functions is positive increasing. Again by Lemma 2.1, h(x) is strictly increasing in (0, π/2). So

δ = log(2/π)/ log(2 tanh(π/2)/π) ≈ 0.839273 and η = f(0+) = 1/2, by l’Hospital’s rule. This

completes the assertion.

Proof of Proposition 1.3:

• Proof of (1.4). Let

f(x) =
log (sin(x)/x)

log
(

2 + e−x
2
)

− log 3
=

f1(x)

f2(x)
,

where f1(x) = log (sin(x)/x) and f2(x) = log
(

2 + e−x
2
)

− log 3 such that f1(0+) = 0 and

f2(0) = 0. Differentiation gives

f′1(x)

f′2(x)
=

1

2

(sin(x) − x cos(x))

x2 sin(x)
(2ex

2

+ 1)

=
1

2
H(x) G(x),

where H(x) =
sin(x)−x cos(x)

x2 sin(x)
is strictly positive increasing in (0, π/2) by Lemma 2.2 and

G(x) = 2ex
2

+ 1 is also clearly positive increasing. Therefore H(x) G(x) is strictly increasing.

By making use of Lemma 2.1, we conclude that f(x) is strictly increasing in (0, π/2). So

f(0+) < f(x) < f(π/2); x ∈ (0, π/2).

Hence, a = f(π/2) = log(2/π)/[log(2 + e−(π/2)
2

) − log 3] ≈ 1.240827 and b = f(0+) = 1/2

by l’Hospital’s rule.

• Proof of (1.5). Utilizing [5, Theorem 2], [12, Proposition 3] we have

e−kx
2

<
sin(x)

x
< e−x

2/6,

where k =
− log(2/π)

(π/2)2
. After rearrangement, it can be written as

(

sin(x)

x

)6

< e−x
2

<

(

sin(x)

x

)1/k

. (3.1)

By virtue of [7, Theorem 2] we write

(

3

2 + cosh(x)

)γ

< e−x
2

<

(

3

2 + cosh(x)

)6

, (3.2)



32 Yogesh J. Bagul and Christophe Chesneau CUBO
21, 1 (2019)

where γ =
(π/2)2

log[(2+cosh(π/2))/3]
. Combining (3.1) and (3.2), we get

(

3

2 + cosh(x)

)c

<
sin(x)

x
<

(

3

2 + cosh(x)

)

,

where c = kγ =
− log(2/π)

log[(2+cosh(π/2))/3]
≈ 1.108171.

Proof of Proposition 1.4: According to [5, Theorem 3] and [12] we have

e−x
2/6 <

x

sinh(x)
< e−tx

2

, x ∈ (0, π/2),

where t =
− log[π/(2 sinh(π/2))]

(π/2)2
. It is equivalent to

(

x

sinh(x)

)1/t

< e−x
2

<

(

x

sinh(x)

)6

. (3.3)

Similarly, using [7, Theorem 1] we have

(

2 + cos(x)

3

)λ

< e−x
2

<

(

2 + cos(x)

3

)6

, (3.4)

where λ =
−(π/2)2

log(2/3)
. Combining (3.3) and (3.4) we get

(

2 + cos(x)

3

)m

<
x

sinh(x)
<

(

2 + cos(x)

3

)n

,

where m = λ
6
=

−(π/2)2

6 log(2/3)
≈ 1.014227 and n = 6t =

−6 log[π/(2 sinh(π/2))]

(π/2)2
≈ 0.928648.

Proof of Proposition 1.5: The proof follows easily by combining inequalities (3.2) and (3.3)

to get

p =
−6 log[π/(2 sinh(π/2))]

(π/2)2
≈ 0.928648 and q =

(π/2)2

6 log[(2+cosh(π/2))/3]
≈ 1.009155.

Proof of Proposition 1.6: For x = 0 equalities hold obviously. Rearranging [4, Theorem

5], for any t ∈ (0, 1), we have

te−t
2/3 < tanh(t) < te−θt

2

with θ ≈ 0.272342. Therefore by integration, for x ∈ (0, 1), we get

∫x

0

te−t
2/3dt <

∫x

0

tanh(t)dt <

∫x

0

te−θt
2

dt,

which yields
3

2

(

1 − e−x
2/3
)

< log(cosh(x)) <
1

2θ

(

1 − e−θx
2
)

.



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Some New Simple Inequalities Involving Exponential . . . 33

By composing with the exponential function, we get the required result.

Proof of Proposition 1.7: Clearly, the equalities hold at x = 0. Rearranging [6, Theorem

4], for any t ∈ (0, 1), we have
3t

3 + t2
< tanh(t) <

ξt

ξ + t2

with ξ ≈ 3.194528. On integration, for x ∈ (0, 1), we have

∫x

0

3t

3 + t2
dt <

∫x

0

tanh(t)dt <

∫x

0

ξt

ξ + t2
dt

which implies that
3

2
log

(

1 +
x2

3

)

< log(cosh(x)) <
ξ

2
log

(

1 +
x2

ξ

)

.

The desired result follows by composing with the exponential function.

Acknowledgments: We would like to thank the referee for the thorough comments which

have helped the presentation of the paper.



34 Yogesh J. Bagul and Christophe Chesneau CUBO
21, 1 (2019)

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	Introduction
	Lemmas
	Proofs of the Main Results