ON GENERALIZATIONS AND REFINEMENTS OF JORDAN
TYPE INEQUALITY
SHAN-HE WU
Abstract. In this paper, we give some refinements and generalizations of
Jordan’s inequality, we also improve some known results.
1. Introduction and Lemmas
The following inequality is well-known
Assume 0 < | x | ≤ π2 , then
2
sin x
≤
< 1.
(1)
π
x
It is called Jordan’s inequality (see [1]). Jordan’s inequality is an important
inequality in calculus and trigonometry, It has broad applications in theory of limit,
There are many generalizations and refinements that had been published in recent
years (see [2] [3] [4] [5]).
In this paper, we give some new generalizations and refinements of Jordan’s
Inequality. Firstly, we prove following Lemmas.
αx
Lemma 1. Assume 0 < β < α, then the function ϕ(x) = sin
sin βx is decreasing on
π
tan αx
π
interval (0, α ), and φ(x) = tan βx is increasing on interval (0, 2α
).
Proof. We note that ϕ is a differentiable function with
α cos αx sin βx − β cos βx sin αx
π
, x ∈ (0, ).
α
sin2 βx
Let g(x) = α cos αx sin βx − β cos βx sin αx, then g 0 (x) = (β 2 − α2 ) sin αx sin βx.
π
π
, we have g 0 (x) < 0, x ∈ (0, α
).
Since 0 < β < α, 0 < x < α
π
Consequently, g is decreasing on interval (0, α ), and then we have
π
g(x) < g(0) = 0 for any x ∈ (0, α
),
π
0
we deduce that ϕ (x) < 0 for any x ∈ (0, α
), it shows that ϕ is decreasing on
π
interval (0, α ).
We note that φ is also a differentiable function with
ϕ0 (x) =
1
π
sec2 αx csc2 βx(α sin 2βx − β sin 2αx), x ∈ (0,
).
2
2α
Let u(x) = α sin 2βx − β sin 2αx, then u0 (x) = 2αβ (cos 2βx − cos 2αx).
π
π
Since 0 < β < α, 0 < x < 2α
, we have u0 (x) > 0, x ∈ (0, 2α
).
φ0 (x) =
2000 Mathematics Subject Classification. 26D15, 26D10, 39B62, 52A40.
Key words and phrases. Jordan’s inequality, refinement, generalization, application, convex
function, concave function.
This paper was typeset using AMS-LATEX.
1
2
SH.-H. WU
π
Consequently, u is increasing on interval (0, 2α
), and then we have
π
u(x) > u(0) = 0 for any x ∈ (0, 2α ),
π
we deduce that φ0 (x) > 0 for any x ∈ (0, 2α
), it shows that φ is increasing on
π
interval (0, 2α ).
The proof is complete.
Lemma 2. Assume 0 < β < α, 0 < |αx| ≤
π
2,
then
x πβ sin αx sin βx ) ≤
<
< |x|.
2 ≤ (sin βx)/(α sin
π
2α α β π
Proof. Case 1, when 0 < x ≤ 2α
. By Lemma 1, we see that ϕ(x) =
π
decreasing on interval (0, 2α ).
Since
α sin αx
βx
α
π
πβ
sin αx
= lim+ ·
·
= , ϕ(
) = csc
,
lim+
αx
sin βx
β
2α
2α
x→0 β
x→0 sin βx
(2)
sin αx
sin βx
is
thus
α
sin αx
πβ
>
≥ csc
β
sin βx
2α
π
for any 0 < β < α and 0 < x ≤ 2α .
Since 0 < x ≤
π
2α ,
(3)
π
0 < β < π/(2( 2α
)), by inequality (3), we obtain
π
π
( )/x ≥ (sin
β)/(sin xβ),
2α
2α
that is
2x
πβ
≤ (sin βx)/(α sin
).
π
2α
Using Jordan’s inequality, we have
sin βx
< x.
β
(4)
(5)
Combining inequalities (3),(4),(5), we get
0<
2x
πβ
sin αx
sin βx
)≤
<
< x,
≤ (sin βx)/(α sin
π
2α
α
β
(6)
that is
x πβ sin αx sin βx 2 ≤ (sin βx)/(α sin
) ≤
<
< |x|.
π
2α α β Case 2, when −
π
2α
≤ x < 0, then 0 < − x ≤
π
2α .
By the result of Case 1, we have
−x πβ sin(− αx) sin(− βx) 2
≤ (sin(−βx))/(α sin
) ≤
<
< | −x | .
π 2α α
β
that is
x πβ sin αx sin βx 2 ≤ (sin βx)/(α sin
) ≤
<
< |x|.
π
2α α β The proof is complete.
ON GENERALIZATIONS AND REFINEMENTS OF JORDAN TYPE INEQUALITY
Lemma 3. The function f (x) = ln sinx x is concave on interval (0,
x
ln tan
is convex on interval (0, π2 ).
x
π
2 ),
3
and g(x) =
Proof. f is a second order differentiable function with
π
csc2 x
( sin2 x − x2 ), x ∈ (0, ).
2
x
2
Using Jordan’s inequality, we get f 00 (x) < 0 for any x ∈ (0,
concave on interval (0, π2 ).
By g is also a second order differentiable function with
f 00 (x) =
π
2 ).
Therefore f is
csc2 2x
π
(sin2 2x − 4x2 cos 2x), x ∈ (0, ).
x2
2
Firstly, suppose π4 ≤ x < π2 , it is easy to see that g 00 (x) > 0.
Secondly, consider the case of 0 < x < π4 .
Now
csc2 2x
csc2 2x cos 2x
g 00 (x) =
(sin2 2x − 4x2 cos 2x) =
(sin 2x tan 2x − 4x2 ).
2
x
x2
Let q(x) = sin 2x tan 2x − 4x2 , we have
g 00 (x) =
q 0 (x) = 2 cos 2x tan 2x + 2 sin 2x sec2 2x − 8x,
π
),
4
Hence q 0 (x) is increasing on interval (0, π4 ), and then we have q 0 (x) > q 0 (0) = 0
for any x ∈ (0, π4 ).
Thus q(x) is increasing on interval (0, π4 ), further we get q(x) > q(0) = 0 for
any x ∈ (0, π4 ).
Therefore g 00 (x) > 0 for any x ∈ (0, π4 ), actually g 00 (x) > 0 for any x ∈ (0, π2 ).
It shows that g is convex on interval (0, π2 ).
q 00 (x) = 4 sec 2x(cos 2x − 1)2 + 8 sin 2x tan 2x sec2 2x > 0,
x ∈ (0,
The proof is complete.
2. Main Results and Their Proof
Theorem 1. Assume 0 < | y | < | x |, 0 < |λ x| ≤
holds
π
2,
then the following inequality
λ πy sin λ x sin λ y
2 ≤ (sin λy)/(x sin
) ≤ <
π
2x
x y
< |λ|.
Proof. By hypothesis, we have
0 < |y| < |x|,
0 < | |x| λ | ≤
π
,
2
Using Lemma 2, we get
λ π y sin(λ | x | ) sin(λ | y | ) 2 ≤ (sin λ |y|)/(|x| sin ) ≤ <
< |λ|.
π
2 x
|x|
|y|
By identical equation
π y πy ),
(sin λ |y|)/(|x| sin ) = (sin λy)/(x sin
2 x
2x
(7)
4
SH.-H. WU
sin(λ | x | ) sin λ x =
x ,
|x|
we immediately obtain inequality (7).
sin(λ | y | ) sin λ y =
y ,
|y|
The proof of Theorem 1 is complete.
From Theorem 1 we get
Corollary 1. Assume 0 < | λx | ≤
π
2,
then the following inequality holds
λ sin λx < |λ|.
2 ≤ π
x (8)
Remark. If we choose λ = 1 in Corollary 1, inequality (8) become Jordan’s
inequality.
Corollary 2. Assume 0 < | y | < | x | ≤
π
2,
then the following inequality holds
πy
sin x
sin y
2
≤ (sin y)/(x sin
)≤
<
< 1.
π
2x
x
y
(9)
Proof. Let λ = 1 in Theorem 1, we have
2
πy sin x sin y
) ≤ <
≤ (sin y)/(x sin
π
2x
x y
for any 0 < | y | < | x | ≤ π2 ,
<1
By inequality (10) and since
sin x
sin y
πy
π sin y
πy
πy
> 0,
> 0, (sin y)/(x sin
)= ·
· (( )/(sin
)) > 0,
x
y
2x
2
y
2x
2x
we obtain inequality (9)
(10)
Remark. The inequality (9) is a refinement of Jordan’s inequality .
Theorem 2. Assume 0 < β < α, 0 < |αx| <
| tan αx | >
π
2,
then the following inequality holds
α
πβ
α
| tan βx | > α | x | > | sin βx | > | sin αx | > csc
| sin βx | . (11)
β
β
2α
Proof. From Lemma 2, we get
α
πβ
| sin βx | > | sin αx | > csc
| sin βx | .
β
2α
Firstly, we prove following inequality
| tan αx | >
α
| tan βx |
β
(12)
(13)
for any 0 < | αx | < π2 .
π
αx
Case 1, when 0 < x ≤ 2α
. By Lemma 1, we see that φ(x) = tan
tan βx is increasing
π
on interval (0, 2α ).
Since
tan αx
α sin αx
βx
cos βx
α
lim
= lim
·
·
·
= ,
αx
sin βx cos αx
β
x→0+ tan βx
x→0+ β
tan αx
α
α
π
,
therefore tan βx > β , it yields that tan αx > β tan βx for any 0 < x < 2α
ON GENERALIZATIONS AND REFINEMENTS OF JORDAN TYPE INEQUALITY
5
By
tan αx = | tan αx | , tan βx = | tan βx | ,
we get inequality (13)
Case 2, when −
have
π
2α
< x < 0, then 0 < − x <
| tan(− αx) | >
π
2α .
By the result of Case 1, we
α
| tan(− βx) | ,
β
which proves inequality (13).
By the well-known inequality (see [6])
| tan x | > | x | > | sin x |
(14)
π
2,
for any 0 < | x | <
we get
α
α
α
α
| tan βx | > | βx | = α | x | ,
| sin βx | < | βx | = α | x | ,
β
β
β
β
Combining inequalities (12),(13) and above inequalities, we immediately obtain
(11). The proof of Theorem 2 is complete.
Let α = 1 in Theorem 2, we obtain
Corollary 3. Assume 0 < β < 1, 0 < | x | <
π
2,
then the following inequality holds
1
1
| tan βx | > | x | > | sin βx | > | sin x | .
β
β
Remark. The inequality (15) is a refinement of inequality(14).
| tan x | >
Corollary 4. Assume 0 < β < α, 0 < |αx| <
π
2,
then the following inequality holds
α
sin αx
πβ
tan αx
> >
> csc
.
tan βx
β
sin βx
2α
Proof. Case 1, when 0 < x <
sin αx > 0,
π
2α .
(15)
(16)
Then
sin βx > 0,
tan αx > 0,
tan βx > 0,
From Theorem 2, we obtain inequality (16).
π
π
Case 2, when − 2α
< x < 0, then 0 < − x < 2α
, By the result of Case 1, we
get
tan(− αx)
α
sin(− αx)
πβ
> >
> csc
,
tan(− βx)
β
sin(− βx)
2α
that is inequality (16).
Let x = 1 in Corollary 4, we obtain
Corollary 5. Assume 0 < β < α <
π
2,
then the following inequality holds
α
sin α
πβ
>
> csc
.
(17)
β
sin β
2α
πβ −1
2α
−1
= πβ
for any 0 < β < α < π2 .
Remark. Note that csc πβ
> ( πβ
2α = (sin 2α )
2α )
Clearly, inequality (17) sharpens following Garnir’s inequality(see [1])
α
sin α
2α
>
>
.
(18)
β
sin β
πβ
6
SH.-H. WU
Theorem 3. Assume 0 < |xi | <
holds
π
2,
i = 1, 2, · · · , n, then the following inequality
|x1 | + · · · + |xn | n |x1 | + · · · + |xn |
)/(
)) |tan x1 tan x2 · · · tan xn | ≥ x1 x2 · · · xn ((tan
n
n
|x1 | + · · · + |xn |
|x1 | + · · · + |xn | n > |x1 x2 · · · xn | > x1 x2 · · · xn ( (sin
)/(
)) n
n
≥ |sin x1 sin x2 · · · sin xn | .
(19)
Proof. By Lemma 3 and Jensen’s inequality(see [7]), we have
1
tan |x1 |
tan |xn |
|x1 | + · · · + |xn |
|x1 | + · · · + |xn |
( ln
+ · · · + ln
) ≥ ln((tan
)/(
)),
n
|x1 |
|xn |
n
n
sin |x1 |
sin |xn |
|x1 | + · · · + |xn |
|x1 | + · · · + |xn |
1
( ln
+ · · · + ln
) ≤ ln((sin
)/(
)),
n
|x1 |
|xn |
n
n
It is obvious that above inequalities are equivalent to following inequalities
|x1 | + · · · + |xn |
|x1 | + · · · + |xn | n tan |x1 | · · · tan |xn | ≥ x1 x2 · · · xn ( (tan
)/(
)) ,
n
n
|x1 | + · · · + |xn |
|x1 | + · · · + |xn | n sin |x1 | · · · sin |xn | ≤ x1 x2 · · · xn ( (sin
)/(
)) .
n
n
Now
|tan x1 tan x2 · · · tan xn | = tan |x1 | tan |x2 | · · · tan |xn | ,
|sin x1 sin x2 · · · sin xn | = sin |x1 | sin |x2 | · · · sin |xn | ,
( (tan |x1 | + · · · + |xn | )/( |x1 | + · · · + |xn | ) )n > 1,
n
n
( (sin |x1 | + · · · + |xn | )/( |x1 | + · · · + |xn | ) )n < 1.
n
n
Base on the above results, we immediately obtain inequality (19). The proof of
Theorem 3 is complete.
Remark. Inequality (19) involves n variables x1 , x2 , · · · , xn . Clearly, it is the
refinement and generalization of Jordan’s inequality.
ON GENERALIZATIONS AND REFINEMENTS OF JORDAN TYPE INEQUALITY
7
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[4] F. Yuefeng, Jordan’s Inequality. Math. Mag. 69(1996), 126.
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(SH.-H. Wu) Department of Mathematics, Longyan College, Longyan Fujian 364012,
CHINA
E-mail address: wushanhe@yahoo.com.cn
URL: http://rgmia.vu.edu.au/members/Wu2.htm