Some New Inequalities of Jordan Type for Sine

Abstract

The authors find some new inequalities of Jordan type for the sine function. These newly established inequalities are of new form and are applied to deduce some known results.

1. Introduction

For 𝑥∈(0,𝜋/2], we have sin𝑥𝑥≥2𝜋. (1) The inequality is sharp with equality if and only if 𝑥=𝜋/2. This inequality is known in the literature as Jordan’s inequality for the sine function. See [1, page 33] and other references cited in the first page of [2].

In [3, 4], it was obtained that 1𝜋³𝜋²−4𝑥²+2𝜋≤sin𝑥𝑥≤2𝜋+𝜋−2𝜋³𝜋²−4𝑥², (2) for 𝑥∈(0,𝜋/2]. The equalities hold if and only if 𝑥=𝜋/2. This refines Jordan’s inequality (1).

Motivated by [3], it was established in [5] that 12𝜋⁵𝜋⁴−16𝑥⁴+2𝜋≤sin𝑥𝑥≤2𝜋+𝜋−2𝜋⁵𝜋⁴−16𝑥⁴, (3) for 𝑥∈(0,𝜋/2]. The equalities are valid if and only if 𝑥=𝜋/2. This also refines Jordan’s inequality (1). Also, see the double inequality (3.10) in the survey article [2, page 17].

In recent years, the above inequalities have been refined, extended, generalized, and applied by many mathematicians in a large amount of papers. See, for example, [3–19]. For a systematic review on this topic, please refer to the expository paper [2].

The aim of this paper is to further refine and generalize these inequalities of Jordan type for the sine function.

Our main results may be stated as in the following theorems.

Theorem 1. If 𝑛≥0 and 𝑚≥2 are integers, then 2^𝑚+2(2𝑚+𝑛𝜋)𝜋^𝑚+1𝜋2^𝑚−𝑥^𝑚𝑛𝜋exp𝑥−2+2𝜋≤sin𝑥𝑥≤2𝜋+2^𝑚(𝜋−2)𝜋^𝑚+1𝜋2^𝑚−𝑥^𝑚𝑛𝜋exp𝑥−2, (4) on (0,𝜋/2].

Theorem 2. Suppose that 𝑔 is a 3-time differentiable function on [0,𝜋/2]. If the function 𝑔 satisfies 𝑔(𝜋/2)≠𝑔(0) and 𝑔^′(𝑥)>0,3𝑔^′′(𝑥)+𝑥𝑔^′′′(𝑥)>0,2𝑔^′(𝑥)≤2𝑥𝑔^′′(𝑥)+𝑥²𝑔^′′′(𝑥), (5) on (0,𝜋/2], then 4𝜋²𝑔^′()𝑔𝜋𝜋/22(𝑥+2−𝑔)𝜋≤sin𝑥𝑥≤2𝜋+𝜋−2𝜋𝑔()(0𝑔𝜋𝜋/2−𝑔)2−𝑔(𝑥). (6)

Remark 3. Taking 𝑛=0 in Theorem 1 yields 2^𝑚+1𝑚𝜋^𝑚+1𝜋2^𝑚−𝑥^𝑚+2𝜋≤sin𝑥𝑥≤2𝜋+2^𝑚(𝜋−2)𝜋^𝑚+1𝜋2^𝑚−𝑥^𝑚, (7) on (0,𝜋/2] for 𝑚≥2. The equalities in (7) are valid if and only if 𝑥=𝜋/2.
Putting 𝑚=2,4 in (7) results in (2) and (3), respectively. This means that Theorem 1 generalizes the inequalities (2) and (3).

Remark 4. Let the function 𝑔(𝑥) in Theorem 2 be 𝑥^𝑚 for 𝑚≥2. A straightforward computation gives 𝑔𝜋2=𝜋2^𝑚(0),𝑔≠0=𝑔_′(𝑥)=𝑚𝑥^𝑚−1>0,3𝑔^′′(𝑥)+𝑥𝑔^′′′(𝑥)=3𝑚(𝑚−1)𝑥^𝑚−2+𝑥𝑚(𝑚−1)(𝑚−2)𝑥^𝑚−2>0,2𝑔^′(𝑥)−2𝑥𝑔^′′(𝑥)+𝑥²𝑔^′′′(𝑥)()4=−𝑚𝑚−1)(𝑚−2≤0,𝜋²𝑔^′()𝑔𝜋𝜋/22(𝑥=2−𝑔)_𝑚+1𝑚𝜋^𝑚+1𝜋2^𝑚−𝑥^𝑚+2𝜋,𝜋−2𝜋𝑔()(0𝑔𝜋𝜋/2−𝑔)2(𝑥=2−𝑔)𝜋+2^𝑚(𝜋−2)𝜋^𝑚+1𝜋2^𝑚−𝑥^𝑚. (8) This implies inequality (7). Hence, Theorem 2 generalizes Theorem 1.

In the final section of this paper, we will apply Theorem 1 to refine and generalize Yang’s inequality and construct some integral inequalities.

2. A Lemma

In order to prove Theorems 1 and 2, the following lemma is necessary.

Lemma 5. Let 𝑓,𝑔∶[𝑎,𝑏]→ℝ be differentiable on (𝑎,𝑏). If 𝑔^′≠0 and 𝑓^′/𝑔^′ are decreasing on (𝑎,𝑏), then the functions 𝑓(𝑥)−𝑓(𝑏)𝑔(𝑥)(𝑏),𝑓(𝑥)(𝑎)−𝑔−𝑓𝑔(𝑥)(𝑎),−𝑔 (9) are also decreasing on (𝑎,𝑏).

Remark 6. Lemma 5 can be found in many papers such as [10, 12, 13, 20].

3. Proofs of Theorems

We are now in a position to prove our theorems.

Proof of Theorem 1. Let 𝑓₁(𝑥)=sin𝑥𝑥,𝑓₂(𝑥)=−𝑥^𝑚𝑒^𝑛𝑥,𝑓₃(𝑥)=sin𝑥−𝑥cos𝑥,𝑓₄(𝑥)=𝑚𝑥^𝑚+1+𝑛𝑥^𝑚+2𝑒^𝑛𝑥, (10) on (0,𝜋/2]. A direct calculation gives 𝑓^′₁(𝑥)𝑓^′₂(𝑥)=sin𝑥−𝑥cos𝑥𝑚𝑥^𝑚+1+𝑛𝑥^𝑚+2𝑒^𝑛𝑥=𝑓₃(𝑥)𝑓₄(𝑥),𝑓^′₃(𝑥)𝑓^′₄(𝑥)=sin𝑥𝑚(𝑚+1)𝑥^𝑚−1+2𝑛(𝑚+1)𝑥^𝑚+𝑛²𝑥^𝑚+1𝑒^𝑛𝑥,𝑓^′₃(𝑥)𝑓^′₄(𝑥)^′=ℎ_𝑚(𝑥)sec𝑥𝑚(𝑚+1)𝑥^𝑚−1+2𝑛(𝑚+1)𝑥^𝑚+𝑛²𝑥^𝑚+12𝑒^𝑛𝑥, (11) where ℎ_𝑚(𝑥)=𝑚(𝑚+1)𝑥^𝑚−1+2𝑛(𝑚+1)𝑥^𝑚+𝑛²𝑥^𝑚+1−𝑚(𝑚+1)(𝑚−1)𝑥^𝑚−2+3𝑛𝑚(𝑚+1)𝑥^𝑚−1+3𝑛²(𝑚+1)𝑥^𝑚+𝑛³𝑥^𝑚+1tan𝑥. (12) Utilizing tan𝑥>𝑥 on (0,𝜋/2) leads to ℎ_𝑚(𝑥)≤𝑚(𝑚+1)𝑥^𝑚−1+2𝑛(𝑚+1)𝑥^𝑚+𝑛²𝑥^𝑚+1−𝑥𝑚(𝑚+1)(𝑚−1)𝑥^𝑚−2+3𝑛𝑚(𝑚+1)𝑥^𝑚−1+3𝑛²(𝑚+1)𝑥^𝑚+𝑛³𝑥^𝑚+1=−𝑚(𝑚+1)(𝑚−2)𝑥^𝑚−1+𝑛(𝑚+1)(3𝑚−2)𝑥^𝑚+𝑛²(3𝑚−2)𝑥^𝑚+1+𝑛³𝑥^𝑚+2≤0, (13) on (0,𝜋/2) for 𝑚≥2 and 𝑛≥0. As a result, the function 𝑓^′₃(𝑥)/𝑓^′₄(𝑥) is decreasing on (0,𝜋/2). In virtue of Lemma 5, it follows that the functions 𝑓₃(𝑥)𝑓₄(𝑥)=𝑓₃(𝑥)−𝑓₃(0)𝑓₄(𝑥)−𝑓₄(0),𝑓^′₁(𝑥)𝑓^′₂(𝑥),𝐻(𝑥)=𝑓₁(𝑥)−𝑓₁(𝜋/2)𝑓₂(𝑥)−𝑓₂(𝜋/2) (14) are all decreasing on (0,𝜋/2). Since lim_𝑥→0+𝐻(𝑥)=2^𝑚(𝜋−2)𝜋^𝑚+1𝑒^{−𝑛𝜋/2},lim_{()𝑥→𝜋/2−}𝐻(𝑥)=2^𝑚+2(2𝑚+𝑛𝜋)𝜋^𝑚+1𝑒^{−𝑛𝜋/2}, (15) we have 2^𝑚+2(2𝑚+𝑛𝜋)𝜋^𝑚+1𝑒^{−𝑛𝜋/2}(𝑥)≤2≤𝐻^𝑚(𝜋−2)𝜋^𝑚+1𝑒^{−𝑛𝜋/2}, (16) which can be reformulated as the inequality (4). Theorem 1 is thus proved.

Proof of Theorem 2. Let 𝑓₁(𝑥)=sin𝑥𝑥,𝑓₂(𝑥)(𝑥),𝑓=−𝑔₃(𝑥)=sin𝑥−𝑥cos𝑥,𝑓₄(𝑥)=𝑥²𝑔^′(𝑥), (17) on (0,𝜋/2]. It is easy to see that 𝑓^′₂(𝑥)=−𝑔^′(𝑥)<0,𝑓^′₂(𝑥)𝑓≠0,_′4(𝑥)𝑓≠0,_′4(𝑥)=2𝑥𝑔^′(𝑥)+𝑥²𝑔^′′(𝑥)=𝑥2𝑔^′(𝑥)+𝑥𝑔^′′(𝑥)>0, (18) on (0,𝜋/2]. Furthermore, we have 𝑓^′₁(𝑥)𝑓^′₂(𝑥)=sin𝑥−𝑥cos𝑥𝑥²𝑔^′(𝑥)=𝑓₃(𝑥)𝑓₄(𝑥),𝑓^′₃(𝑥)𝑓^′₄(𝑥)=sin𝑥2𝑔^′(𝑥)+𝑥𝑔^′′(𝑥),𝑓^′₃(𝑥)𝑓^′₄(𝑥)^′=2𝑔^′(𝑥)+𝑥𝑔^′′(𝑥)−3𝑔^′′(𝑥)+𝑥𝑔^′′′(𝑥)tan𝑥2𝑔^′(𝑥)+𝑥𝑔^′′(𝑥)².sec𝑥 (19)(20) Employing tan𝑥>𝑥 and the conditions in (5), it is not difficult to show that the numerator of [𝑓^′₃(𝑥)/𝑓^′₄(𝑥)]^′ is negative on (0,𝜋/2). This means that the function 𝑓^′₃(𝑥)/𝑓^′₄(𝑥) is decreasing on (0,𝜋/2). Consequently, making use of Lemma 5 consecutively, it is revealed that the functions 𝑓₃(𝑥)𝑓₄(𝑥)=𝑓₃(𝑥)−𝑓₃(0)𝑓₄(𝑥)−𝑓₄(0),𝑓^′₁(𝑥)𝑓^′₂(𝑥),𝐻(𝑥)=𝑓₁(𝑥)−𝑓₁(𝜋/2)𝑓₂(𝑥)−𝑓₂()=()−()𝜋/2sin𝑥/𝑥2/𝜋𝑔(𝜋/2)−𝑔(𝑥) (21) are all decreasing on (0,𝜋/2). Since lim_𝑥→0+𝐻(𝑥)=lim_𝑥→0+𝑓₁(𝑥)−𝑓₁(𝜋/2)𝑓₂(𝑥)−𝑓₂()=()𝜋/21−2/𝜋𝑔()(0)=𝜋/2−𝑔𝜋−2𝜋𝑔()(0,𝜋/2−𝑔)lim_{()𝑥→𝜋/2−}𝐻(𝑥)=lim_{𝑥→(𝜋/2)−}((sin𝑥)/𝑥)−(2/𝜋)𝑔(𝜋/2)−𝑔(𝑥)=lim_{𝑥→(𝜋/2)−}sin𝑥−(2/𝜋)𝑥𝑥𝑔(𝜋/2)−𝑥𝑔(𝑥)=lim_{𝑥→(𝜋/2)−}cos𝑥−(2/𝜋)𝑔(𝜋/2)−𝑔(𝑥)−𝑥𝑔^′(𝑥)=−(2/𝜋)−(𝜋/2)𝑔^′()=4𝜋/2𝜋²𝑔^′(),𝜋/2 (22) from 𝐻(𝜋/2)≤𝐻(𝑥)≤𝐻(0), the inequality (6) follows. The proof of Theorem 2 is complete.

4. Applications of Theorem 1

After proving Theorems 1 and 2, we now start off to apply them to construct some new inequalities.

Let 0≤𝜆≤1 and 𝐴,𝐵>0 with 𝐴+𝐵≤𝜋. Then, cos²(𝜆𝐴)+cos²(𝜆𝐵)−2cos(𝜆𝐴)cos(𝜆𝐵)cos(𝜆𝜋)≥sin²(𝜆𝜋). (23) This inequality is known in the literature as Yang’s inequality. Since paper [16], many mathematicians mistakenly referred this inequality to [21, pages 116–118]. Indeed, the paper we should refer to is [22] or an even earlier paper in Chinese.

The first application of Theorem 1 is to refine and generalize Yang’s inequality (23) as follows.

Theorem 7. For 𝑘≥2, let 𝐴_𝑖>0 and ∑^𝑘_𝑖=1𝐴_𝑖≤𝜋. If 0≤𝜆≤1, then 𝑅(𝜆)≤_{1≤𝑖<𝑗≤𝑘}𝐻_𝑖𝑗≤𝑇(𝜆), (24) where 𝐻_𝑖𝑗=cos²𝜆𝐴_𝑖+cos²𝜆𝐴_𝑗−2cos𝜆𝐴_𝑖cos𝜆𝐴_𝑗(),𝑇(𝜆)𝑘2()𝜆cos𝜆𝜋=4𝜆+𝜋−221−𝜆^𝑚𝑒^{𝑛𝜋(𝜆−1)/22},𝑅(𝜆)𝑘2=4𝜆+2𝜆2𝑚+𝑛𝜋1−𝜆^𝑚𝑒^{𝑛𝜋(𝜆−1)/22}cos²𝜆𝜋2, (25) and 𝑛≥0 and 𝑚≥2 are integers.

Proof. Substituting 𝑥=𝜆𝜋/2 in the inequality (4) reveals that sin𝜆𝜋2≥𝜆+2𝜆2𝑚+𝑛𝜋1−𝜆^𝑚𝑒^{𝑛𝜋/2(𝜆−1)},sin𝜆𝜋2()𝜆≤𝜆+𝜋−221−𝜆^𝑚𝑒^{𝑛𝜋/2(𝜆−1)}. (26) Using sin²(𝜆𝜋)=4sin²𝜆𝜋2cos²𝜆𝜋2, (27) the inequality sin²(𝜆𝜋)≤𝐻_𝑖𝑗≤4sin²𝜆𝜋2, (28) see either [22], [16, (2.13)], or [2, page 17, (3.4)], becomes 4𝜆+2𝜆2𝑚+𝑛𝜋1−𝜆^𝑚𝑒^{𝑛𝜋/2(𝜆−1)2}cos²𝜆𝜋2≤𝐻_𝑖𝑗()𝜆≤4𝜆+𝜋−221−𝜆^𝑚𝑒^{𝑛𝜋/2(𝜆−1)2}. (29) Finally, taking the sum of the above inequality for all 1≤𝑖<𝑗≤𝑛 results in (24). The required proof is complete.

Corollary 8. Under the conditions of Theorem 7, one has 𝑅₁(𝜆)≤_{1≤𝑖<𝑗≤𝑘}𝐻_𝑖𝑗≤𝑇₁(𝜆), (30) where 𝑇₁(𝜆)=𝑘2𝜋²𝜆²,𝑅₁(𝜆)𝑘2=4𝜆²cos²𝜆𝜋2. (31)

Proof. When 0≤𝜆<1 and 𝑛=0, we have lim_𝑚→∞𝑇(𝜆)=𝑘2𝜋²𝜆²,lim_𝑚→∞𝑅(𝜆)𝑘2=4𝜆²cos²𝜆𝜋2. (32) This implies the required result.

The second application of Theorem 1 is to construct some new integral inequalities for sin𝑥/𝑥.

Theorem 9. For 𝑥∈(0,𝜋/2], if 𝑛≥0 and 𝑚≥2 are integers, then ₀^𝜋/2sin𝑥𝑥2𝑑𝑥+^𝑚+2𝑒^{−𝑛𝜋/2}(2𝑚+𝑛𝜋)𝜋^𝑚+1₀^𝜋/2𝑥^𝑚𝑒^𝑛𝑥2𝑑𝑥≥1+(),2𝑚+𝑛𝜋_0𝜋/2sin𝑥𝑥2𝑑𝑥+^𝑚(𝜋−2)𝑒^{−𝑛𝜋/2}𝜋^𝑚+1₀^𝜋/2𝑥^𝑚𝑒^𝑛𝑥≤𝜋𝑑𝑥2. (33)

Proof. This follows from integrating on all sides of the double inequality (4).

Remark 10. Applying Theorem 9 to 𝑛=0 gives 1𝑚+1+1≤₀^𝜋/2sin𝑥𝑥()𝑑𝑥≤1+𝜋−22𝑚.𝑚+1 (34) Applying Theorem 9 to 𝑛=0 and 𝑚=2 yields 43≤₀^𝜋/2sin𝑥𝑥𝑑𝑥≤𝜋+13. (35) This is a recovery of an inequality established in [3, page 101]. It was also collected in [2, (2.14)]. Such a kind of inequalities can be found in [23].

Acknowledgment

The authors appreciate Professor Dr. Feng Qi (F. Qi) at Tianjin Polytechnic University in China for his kind and valuable contributions to this paper.