Hindawi Publishing Corporation Abstract and Applied Analysis Volume 2014, Article ID 592085, pages http://dx.doi.org/10.1155/2014/592085 Research Article A Sharp Double Inequality for Trigonometric Functions and Its Applications Zhen-Hang Yang,1 Yu-Ming Chu,1 Ying-Qing Song,1 and Yong-Min Li2 School of Mathematics and Computation Sciences, Hunan City University, Yiyang 413000, China Department of Mathematics, Huzhou Teachers College, Huzhou 313000, China Correspondence should be addressed to Yu-Ming Chu; chuyuming2005@126.com Received 26 April 2014; Accepted 20 June 2014; Published 10 July 2014 Academic Editor: Josip E Peˇcari´c Copyright © 2014 Zhen-Hang Yang et al This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited We present the best possible parameters 𝑝 and 𝑞 such that the double inequality ((2/3)cos2𝑝 (𝑡/2) + 1/3) 2𝑞 ((2/3)cos (𝑡/2) + 1/3) 1/𝑞 Inequality (3) was also proved by Lv et al in [10] In [11, 12], Neuman proved that the inequalities It is well known that the double inequality cos sin 𝑡 + cos 𝑡 𝑡< < 𝑡 (1) holds for any 𝑡 ∈ (0, 𝜋/2) The first inequality in (1) was found by Mitrinovi´c (see [1]), while the second inequality in (1) is due to Huygens (see [2]) and it is called Cusa inequality Recently, the improvements, refinements, and generalizations for inequality (1) have attracted the attention of many mathematicians [3–8] Qi et al [9] proved that the inequality cos2 𝑡 sin 𝑡 < 𝑡 𝑡 + cos 𝑡 2/3 sin 𝑡 ) < =( , 2 𝑡 cos1/3 𝑡 < ( 𝜋 𝑡 ∈ (0, ) (3) 1/2 1/4 sin 𝑡 sin 𝑡 ) cos 𝑡) < ( 𝑡 tanh−1 (sin 𝑡) for all 𝑥 ∈ (1/2, 1) if and only if 𝑝 ≤ 𝑝2 , where 𝑝2 = 0.1872 is the unique solution of equation 𝑔𝑝 ( ) = 21−𝑝 − 2𝑝−1 − 𝑝 − = 0; (14) (iii) if 𝑝2 < 𝑝 < 1/5, then there exists 𝑥1 = 𝑥1 (𝑝) ∈ (1/2, 1) such that 𝑔𝑝 (𝑥) < for 𝑥 ∈ (1/2, 𝑥1 ) and 𝑔𝑝 (𝑥) > for 𝑥 ∈ (𝑥1 , 1) Proof It follows from (13) and (14) that 𝑔0.1872 ( ) = 0.000141 > 0, 𝑔0.1873 ( ) = − 0.000119 < 0, 𝜕𝑝 (15) = (𝑥1−𝑝 + 2𝑥𝑝 ) log 𝑥 − (1 − 𝑥) < 0, 𝑔𝑝 (𝑥) ≤ 𝑔1/5 (𝑥) = 𝑥 − 𝑥4/5 + 2𝑥1/5 − 5 = − (1 − 𝑥1/5 ) (11) (16) × (−2𝑥3/5 + 𝑥2/5 + 4𝑥1/5 + 7) < If 𝑔𝑝 (𝑥) < for all 𝑥 ∈ (1/2, 1), then (13) leads to 𝑡 = 𝑀1 (cos2 , ) < lim− 𝑡 = 𝑀∞ (cos2 , ) , 𝑥→1 for 𝑡 ∈ (0, 𝜋/2) The main purpose of this paper is to present the best possible parameters 𝑝 and 𝑞 such that the double inequality 𝑡 𝑡 sin 𝑡 < 𝑀𝑞 (cos2 , ) 𝑀𝑝 (cos2 , ) < 𝑡 (13) for 𝑥 ∈ (0, 1) Inequalities (15) lead to the conclusion that the function 𝑔𝑝 (𝑥) is strictly decreasing with respect to 𝑝 ∈ R for fixed 𝑥 ∈ (0, 1) and 𝑝2 = 0.1872 is the unique solution of (14) (i) If 𝑥 ∈ (1/2, 1) and 𝑝 ≥ 1/5, then from the monotonicity of the function 𝑝 → 𝑔𝑝 (𝑥) we clearly see that 𝑝→∞ 𝑡 < ( cos + ) 3 𝑔𝑝 (𝑥) = 2𝑝𝑥 − 𝑥1−𝑝 + 2𝑥𝑝 − (2𝑝 + 1) 𝜕𝑔𝑝 (𝑥) 𝑀−∞ (𝑥, 𝜔) = lim 𝑀𝑝 (𝑥, 𝜔) = 𝑥, 𝑀∞ (𝑥, 𝜔) = lim 𝑀𝑝 (𝑥, 𝜔) = In order to prove our main results we need several lemmas, which we present in this section (12) holds for all 𝑡 ∈ (0, 𝜋/2) As applications, some new analytic inequalities are found All numerical computations are carried out using MATHEMATICA software 𝑔𝑝 (𝑥) 1−𝑥 = − 5𝑝 ≤ (17) (ii) If 𝑥 ∈ (1/2, 1) and 𝑝 ≤ 0, then the monotonicity of the function 𝑝 → 𝑔𝑝 (𝑥) leads to the conclusion that 𝑔𝑝 (𝑥) ≥ 𝑔0 (𝑥) = − 𝑥 > If 𝑥 ∈ (1/2, 1) and < 𝑝 ≤ 𝑝2 , then (13) and the monotonicity of the function 𝑝 → 𝑔𝑝 (𝑥) lead to 1 𝑔𝑝 ( ) ≥ 𝑔𝑝2 ( ) = 0, 2 𝜕2 𝑔𝑝 (𝑥) 𝜕𝑥2 𝑔𝑝 (1) = 0, = 𝑝 (𝑝 − 1) 𝑥𝑝−2 (2 − 𝑥1−2𝑝 ) < (18) (19) Abstract and Applied Analysis Inequality (19) implies that the function 𝑔𝑝 (𝑥) is concave with respect to 𝑥 on the interval (1/2, 1) Therefore, 𝑔𝑝 (𝑥) > follows from (18) and the concavity of 𝑔𝑝 (𝑥) If 𝑔𝑝 (𝑥) > for all 𝑥 ∈ (1/2, 1), then 𝑝 ≤ 𝑝2 follows easily from the monotonicity of the function 𝑝 → 𝑔𝑝 (1/2) and 𝑔𝑝 (1/2) ≥ together with the fact that 𝑔𝑝2 (1/2) = (iii) If 𝑥 ∈ (1/2, 1) and 𝑝2 < 𝑝 < 1/5, then from (13) and (19) together with the monotonicity of the function 𝑝 → 𝑔𝑝 (1/2) we get 1 𝑔𝑝 ( ) < 𝑔𝑝2 ( ) = 0, 2 𝑔𝑝 (1) = 0, (20) 𝑔𝑝 (1) = 5𝑝 − < 0, (21) > × 0.1872 − 20.1873 (22) + × 0.1872 × = 0.1065 > 0, and 𝑔𝑝 (𝑥) is strictly decreasing on (1/2, 1) It follows from (21) and (22) together with the monotonicity of 𝑔𝑝 (𝑥) that there exists 𝑥0 = 𝑥0 (𝑝) ∈ (1/2, 1) such that 𝑔𝑝 (𝑥) is strictly increasing on (1/2, 𝑥0 ] and strictly decreasing on [𝑥0 , 1) Therefore, Lemma (iii) follows from (20) and the piecewise monotonicity of 𝑔𝑝 (𝑥) Let 𝑝 ∈ R and the function 𝑓𝑝 be defined on (0, 𝜋/2) by (23) Then elaborated computations lead to 𝑓𝑝 (𝑡) = 2𝑝 (1 − cos (𝑡/2)) cos (𝑡/2) (2cos2𝑝 (𝑡/2) + 2cos2 𝑓𝑝 (𝑡) < 𝑓𝑝 (0+ ) = (25) If 𝑓𝑝 (𝑡) < for all 𝑡 ∈ (0, 𝜋/2), then (23) leads to ≥ lim+ 𝑓𝑝 (𝑡) 𝑡5 = lim+ (1/180) (1 − 5𝑝) 𝑡5 + 𝑜 (𝑡5 ) 𝑡→0 𝑡5 − 5𝑝 = 180 (ii) If 𝑓𝑝 (𝑡) > for all 𝑡 ∈ (0, 𝜋/2), then from (23) we get (26) 𝜋− 𝜋 − − 2𝑝 (27) )= 2 Inequality (27) leads to the conclusion that 𝑝 ≤ log(𝜋 − 2)/ log If 𝑡 ∈ (0, 𝜋/2) and 𝑝 ≤ 𝑝1 = log(𝜋 − 2)/ log 2, then we divide the proof into two cases ≤ 𝑓𝑝 ( 0.8127 2cos2𝑝 (𝑡/2) + sin 𝑡 𝑓𝑝 (𝑡) = 𝑡 − cos 𝑡 + 2cos2𝑝 (𝑡/2) Proof (i) If 𝑡 ∈ (0, 𝜋/2) and 𝑝 ≥ 1/5, then from (23) and Lemma (i) we clearly see that 𝑡→0 𝑔𝑝 ( ) = 2𝑝 − 2𝑝 + 𝑝2𝑝 + 2𝑝21−𝑝 + 0.1872 × 20.1872 (ii) 𝑓𝑝 (𝑡) > for all 𝑡 ∈ (0, 𝜋/2) if and only if 𝑝 ≤ 𝑝1 = log(𝜋 − 2)/ log = 0.1910 ; (iii) if 𝑝1 < 𝑝 < 1/5, then there exists 𝑡0 = 𝑡0 (𝑝) ∈ (0, 𝜋/2) such that 𝑓𝑝 (𝑡) > for 𝑡 ∈ (0, 𝑡0 ) and 𝑓𝑝 (𝑡) < for 𝑡 ∈ (𝑡0 , 𝜋/2) 𝑡 𝑔𝑝 (cos2 ) , (𝑡/2) − 1) (24) where 𝑔𝑝 (𝑥) is defined by (13) From Lemma and (24) we get the following Lemma immediately Lemma Let 𝑝 ∈ R and 𝑓𝑝 be defined on (0, 𝜋/2) by (23) Then (i) 𝑓𝑝 (𝑡) is strictly decreasing on (0, 𝜋/2) if and only if 𝑝 ≥ 1/5; (ii) 𝑓𝑝 (𝑡) is strictly increasing on (0, 𝜋/2) if and only if 𝑝 ≤ 𝑝2 , where 𝑝2 = 0.1872 is the unique solution of (14); (iii) if 𝑝2 < 𝑝 < 1/5, then there exists 𝑡1 = 𝑡1 (𝑝) ∈ (0, 𝜋/2) such that 𝑓𝑝 (𝑡) is strictly increasing on (0, 𝑡1 ] and strictly decreasing on [𝑡1 , 𝜋/2) Lemma Let 𝑝 ∈ R and 𝑓𝑝 be defined on (0, 𝜋/2) by (23) Then (i) 𝑓𝑝 (𝑡) < for all 𝑡 ∈ (0, 𝜋/2) if and only if 𝑝 ≥ 1/5; Case Consider 𝑝 ≤ 𝑝2 , where 𝑝2 is the unique solution of (14) Then from Lemma (ii) and (23) we clearly see that 𝑓𝑝 (𝑡) > 𝑓𝑝 (0+ ) = (28) Case Consider 𝑝2 < 𝑝 ≤ 𝑝1 Then (23) and Lemma (iii) lead to 𝑓𝑝 (0+ ) = 0, (29) 𝜋− 𝜋 − − 2𝑝 𝜋 − − 2𝑝1 )= ≥ = 0, 2 and there exists 𝑡1 = 𝑡1 (𝑝) such that 𝑓𝑝 (𝑡) is strictly increasing on (0, 𝑡1 ] and strictly decreasing on [𝑡1 , 𝜋/2) Therefore, 𝑓𝑝 (𝑡) > for all 𝑡 ∈ (0, 𝜋/2) follows from (29) and the piecewise monotonicity of 𝑓𝑝 (𝑡) (iii) If 𝑝1 < 𝑝 < 1/5, then 𝑝2 < 𝑝 < 1/5 It follows from (23) and Lemma (iii) that 𝑓𝑝 ( 𝑓𝑝 (0+ ) = 0, (30) 𝜋− 𝜋 − − 2𝑝 𝜋 − − 2𝑝1 )= < = 0, 2 and there exists 𝑡1 = 𝑡1 (𝑝) such that 𝑓𝑝 (𝑡) is strictly increasing on (0, 𝑡1 ] and strictly decreasing on [𝑡1 , 𝜋/2) Therefore, Lemma (iii) follows from (30) and the piecewise monotonicity of 𝑓𝑝 (𝑡) Let 𝑝 ∈ R and 𝐹𝑝 be defined on (0, 𝜋/2) by 𝑓𝑝 ( 𝐹𝑝 (𝑡) = log sin 𝑡 𝑡 − log ( cos2𝑝 + ) 𝑡 𝑝 3 𝐹0 (𝑡) = lim 𝐹𝑝 (𝑡) = log 𝑝→0 (𝑝 ≠ 0) , (31) sin 𝑡 𝑡 − log (cos ) 𝑡 (32) Abstract and Applied Analysis Then elaborated computations give 𝐹𝑝 (𝑡) = cos 𝑡 + 2cos2𝑝 (𝑡/2) 𝑓 (𝑡) , 𝑡 (1 + 2cos2𝑝 (𝑡/2)) sin 𝑡 𝑝 to If 𝑝 > 0.1 and 𝐹𝑝 (𝑡) > for all 𝑡 ∈ (0, 𝜋/2), then (31) leads (33) where 𝑓𝑝 (𝑡) is defined by (23) From Lemma and (33) we get Lemma immediately Lemma Let 𝑝 ∈ R and 𝐹𝑝 be defined on (0, 𝜋/2) by (31) and (32) Then (i) 𝐹𝑝 (𝑡) is strictly decreasing on (0, 𝜋/2) if and only if 𝑝 ≥ 1/5; ≤ 𝐹𝑝 ( 𝜋+ ) = 𝐻 (𝑝) 𝑝 (39) Therefore, 𝑝 ≤ 𝑝0 follows from (39) and 𝐻(𝑝0 ) = together with the monotonicity of 𝐻(𝑝) on the interval (0.1, ∞) Lemma Let 𝑝 ∈ R and 𝑥, 𝑐, 𝜔 ∈ (0, 1), and let 𝑀𝑝 (𝑥, 𝜔) be defined by (9) Then the function 𝑝 → 𝑀𝑝 (𝑥, 𝜔)/𝑀𝑝 (𝑐, 𝜔) is strictly decreasing with respect to 𝑝 ∈ R if 𝑥 ∈ (𝑐, 1) (ii) 𝐹𝑝 (𝑡) is strictly increasing on (0, 𝜋/2) if and only if 𝑝 ≤ 𝑝1 = log(𝜋 − 2)/ log = 0.1910 ; Proof Let 𝐻(𝑝, 𝑥) = log 𝑀𝑝 (𝑥, 𝜔) − log 𝑀𝑝 (𝑐, 𝜔) Then from (9) we get (iii) if 𝑝1 < 𝑝 < 1/5, then there exists 𝑡0 = 𝑡0 (𝑝) ∈ (0, 𝜋/2) such that 𝐹𝑝 (𝑡) is strictly increasing on (0, 𝑡0 ] and strictly decreasing on [𝑡0 , 𝜋/2) 𝜕𝐻 (𝑝, 𝑥) 𝜔𝑥𝑝−1 = , 𝑝 𝜕𝑥 𝜔𝑥 + − 𝜔 (40) 𝜕2 𝐻 (𝑝, 𝑥) 𝜔 (1 − 𝜔) 𝑥𝑝−1 log 𝑥 < = 𝜕𝑝𝜕𝑥 (𝜔𝑥𝑝 + − 𝜔)2 (41) Lemma Let 𝑝 ∈ R and 𝐹𝑝 be defined on (0, 𝜋/2) by (31) and (32) Then the following statements are true: (i) if 𝐹𝑝 (𝑡) < for all 𝑡 ∈ (0, 𝜋/2), then 𝑝 ≥ 1/5; (ii) if 𝐹𝑝 (𝑡) > for all 𝑡 ∈ (0, 𝜋/2), then 𝑝 ≤ 𝑝0 , where 𝑝0 = 0.1941 is the unique solution of the equation 𝑝 log − log (1 + 21−𝑝 ) + log = 0, 𝜋 (34) Main Results on the interval (0.1, ∞) Proof (i) If 𝐹𝑝 (𝑡) < for all 𝑡 ∈ (0, 𝜋/2), then from (31) and (32) we have ≥ lim+ 𝑡→0 𝐹𝑝 (𝑡) 𝑡4 = lim+ Inequality (41) and 𝜕2 𝐻(𝑝, 𝑥)/𝜕𝑥𝜕𝑝 = 𝜕2 𝐻(𝑝, 𝑥)/𝜕𝑝𝜕𝑥 lead to the conclusion that 𝜕𝐻(𝑝, 𝑥)/𝜕𝑝 is strictly decreasing with respect to 𝑥 ∈ (𝑐, 1) Therefore, 𝜕𝐻(𝑝, 𝑥)/𝜕𝑝 < 𝜕𝐻(𝑝, 𝑥)/𝜕𝑝|𝑥=𝑐 = for 𝑥 ∈ (𝑐, 1), and 𝑀𝑝 (𝑥, 𝜔)/𝑀𝑝 (𝑐, 𝜔) is strictly decreasing with respect to 𝑝 ∈ R if 𝑥 ∈ (𝑐, 1) (1/720) (1 − 5𝑝) 𝑡4 + 𝑜 (𝑡4 ) 𝑡4 𝑡→0 − 5𝑝 = 720 (35) (ii) We first prove that 𝑝0 = 0.1941 is the unique solution of (34) on the interval (0.1, ∞) Let 𝑝 ∈ (0.1, ∞) and 𝐻 (𝑝) = 𝑝 log − log (1 + 21−𝑝 ) + log 𝜋 𝐻 (0.1942) = − 2.52 × 10−7 < 0, log log 2 < log + 𝐻 (𝑝) = log + 𝜋 + 2𝑝 𝜋 + 20.1 (37) = − 2.81 × 10 (42) holds for all 𝑡 ∈ (0, 𝜋/2) if and only if 𝑝 ≥ 1/5, and the double inequality 𝑡 𝑡 sin 𝑡 𝑀𝑝 (cos2 , ) < < 𝜆 𝑝 𝑀𝑝 (cos2 , ) 𝑡 (43) holds for all 𝑡 ∈ (0, 𝜋/2) if and only if 𝑝 ≤ 𝑝1 , where −1/𝑝 𝜆𝑝 = + 21−𝑝 ( ) 𝜋 (𝑝 ≠ 0) , 𝜆0 = 25/3 , 𝜋 (44) 𝑝1 = log(𝜋 − 2)/ log = 0.1910 , and 𝜆 𝑝 𝑀𝑝 (cos2 (𝑡/2), 2/3) is strictly decreasing with respect to 𝑝 ∈ R Proof Let 𝑝 ∈ R and 𝐹𝑝 (𝑡) be defined on (0, 𝜋/2) by (31) and (32) Then −4 𝑡 𝑡 sin 𝑡 𝜆 𝑝 𝑀𝑝 (cos2 , ) < < 𝑀𝑝 (cos2 , ) 𝑡 (36) Then numerical computations show that 𝐻 (0.1941) = 8.13 × 10−7 > 0, Theorem Let 𝑀𝑝 (𝑥, 𝜔) be defined by (9) Then the double inequality (38) < Inequality (38) implies that 𝐻(𝑝) is strictly decreasing on [0.1, ∞) Therefore, 𝑝0 = 0.1941 is the unique solution of (34) on the interval (0.1, ∞) which follows from (37) and the monotonicity of 𝐻(𝑝) 𝐹𝑝 (0+ ) = 0, 𝐹𝑝 ( 𝜋− ) = log 𝜆 𝑝 (45) If 𝑝 ≥ 1/5, then inequality (42) follows from Lemma (i) and (45) If inequality (42) holds for all 𝑡 ∈ (0, 𝜋/2), then 𝐹𝑝 (𝑡) < for all 𝑡 ∈ (0, 𝜋/2) It follows from Lemma (i) that 𝑝 ≥ 1/5 Abstract and Applied Analysis If 𝑝 ≤ 𝑝1 , then inequality (43) follows from Lemma (ii) and (45) If inequality (43) holds for all 𝑡 ∈ (0, 𝜋/2), then 𝐹𝑝 (𝜋/2− ) > 𝐹𝑝 (𝑡) > 𝐹𝑝 (0+ ) = for all 𝑡 ∈ (0, 𝜋/2) It follows from Lemma (ii) that 𝑝 ≤ 𝑝0 , where 𝑝0 = 0.1941 is the unique solution of (34) on the interval (0.1, ∞) We claim that 𝑝 ≤ 𝑝1 ; otherwise 𝑝1 < 𝑝 ≤ 𝑝0 < 1/5, and Lemma (iii) leads to the conclusion that there exists 𝑡0 ∈ (0, 𝜋/2) such that 𝐹𝑝 (𝑡) > 𝐹𝑝 (𝜋/2− ) for 𝑡 ∈ [𝑡0 , 𝜋/2) Note that 𝑡 2 𝑀𝑝 (cos (𝑡/2) , 2/3) 𝜆 𝑝 𝑀𝑝 (cos2 , ) = 𝜋 𝑀𝑝 (1/2, 2/3) 𝑡 𝑡 < 𝜆 1/8 ( cos1/4 + ) < 𝜆 cos4/3 3 < 𝜆 −1/2 9cos2 (𝑡/2) (1 + cos 𝑡) < 𝜆 −1 + cos 𝑡 (2 + cos (𝑡/2)) < 𝜆 −∞ cos2 𝑡 𝑡 = cos2 𝜋 (48) Theorem 10 Let 𝑀𝑝 (𝑥, 𝜔) be defined by (9) Then the double inequality (46) 𝑡 𝑡 sin 𝑡 𝑀𝑝 (cos2 , ) < < 𝑀𝑞 (cos2 , ) 𝑡 (49) It follows from Lemma and (46) that 𝜆 𝑝 𝑀𝑝 (cos2 (𝑡/2), 2/3) is strictly decreasing with respect to 𝑝 ∈ R holds for all 𝑡 ∈ (0, 𝜋/2) if and only if 𝑝 ≤ 𝑝0 and 𝑞 ≥ 1/5, where 𝑝0 = 0.1941 is the unique solution of (34) on the interval (0.1, ∞) Moreover, the inequality From Theorem we get Corollaries and as follows 𝑡 sin 𝑡 ≤ 𝛼𝑀𝑝0 (cos2 , ) , 𝑡 (50) sin 𝑡0 = 1.00004919 , 𝑡0 𝑀𝑝0 (cos2 (𝑡0 /2) , 2/3) (51) Corollary For all 𝑡 ∈ (0, 𝜋/2) one has if and only if 𝑡 2 + cos 𝑡 < = 𝜆 𝑀1 (cos2 , ) 𝜋 𝜋 𝑡 < 𝜆 1/2 ( cos + ) 3 𝛼≥ where 𝑡0 ∈ (0, 𝜋/2) is defined as in Lemma (iii) 𝑡 < 𝜆 1/4 ( cos1/2 + ) 3 𝑡 sin 𝑡 < 𝜆 1/5 ( cos2/5 + ) < 3 𝑡 𝑡 𝑡 2 < ( cos2/5 + ) < ( cos1/2 + ) 3 3 𝑡 < ( cos + ) 3 𝑡 2 + cos 𝑡 < 𝑀1 (cos2 , ) = < 3 Corollary For all 𝑡 ∈ (0, 𝜋/2) one has cos2 𝑡 𝑡 (1 + cos 𝑡) = 𝑀−∞ (cos2 , ) < 2 + cos 𝑡 𝑡 = 𝑀−1 (cos2 , ) < 𝑡 9cos2 (𝑡/2) = 𝑀−1/2 (cos2 , ) (2 + cos (𝑡/2))2 < cos4/3 𝑡 𝑡 = 𝑀0 (cos2 , ) 2 𝑡 𝑡 2 < ( cos1/4 + ) < ( cos1/3 + ) 3 3 < 𝑡 sin 𝑡 < 𝜆 1/6 ( cos1/3 + ) 𝑡 3 (47) Proof Let 𝑝 ∈ R and 𝐹𝑝 (𝑡) be defined on (0, 𝜋/2) by (31) and (32) Then Lemma (iii) leads to the conclusion that 𝐹𝑝0 (𝑡) is strictly increasing on (0, 𝑡0 ] and strictly decreasing on [𝑡0 , 𝜋/2) Note that 𝐹𝑝0 (0+ ) = 𝐹𝑝0 ( 𝜋− ) = (52) It follows from the piecewise monotonicity of 𝐹𝑝0 (𝑡) and (52) that < 𝐹𝑝0 (𝑡) ≤ 𝐹𝑝0 (𝑡0 ) , (53) for all 𝑡 ∈ (0, 𝜋/2) Therefore, sin 𝑡/𝑡 > 𝑀𝑝0 (cos2 (𝑡/2), 2/3) for all 𝑡 ∈ (0, 𝜋/2) follows from the first inequality of (53), while sin 𝑡/𝑡 < 𝑀1/5 (cos2 (𝑡/2), 2/3) for all 𝑡 ∈ (0, 𝜋/2) follows from the second inequality of (42) Conversely, if the double inequality (49) holds for all 𝑡 ∈ (0, 𝜋/2), then we clearly see that the inequalities 𝐹𝑝 (𝑡) > 0, 𝐹𝑞 (𝑡) < (54) hold for all 𝑡 ∈ (0, 𝜋/2) Therefore, 𝑝 ≤ 𝑝0 and 𝑞 ≥ 1/5 follows from Lemma and (54) Moreover, numerical computations show that 𝑡0 = 1.312 and 𝑒𝐹𝑝0 (𝑡0 ) = 1.00004919 (55) Therefore, the second conclusion of Theorem 10 follows from (55) and the second inequality of (53) It follows from Lemma that we get Theorem 11 immediately 6 Abstract and Applied Analysis Theorem 11 The double inequalities hold for all 𝑡 ∈ (0, 𝜋/2) Note that 2cos2𝑝 (𝑡/2) + cos 𝑡 sin 𝑡 2cos2𝑞 (𝑡/2) + cos 𝑡 < < , 2cos2𝑝 (𝑡/2) + 𝑡 2cos2𝑞 (𝑡/2) + 2cos2𝑝 𝑡 sin 𝑡 − 𝑡 cos 𝑡 𝑡 < < 2cos2𝑞 𝑡 − sin 𝑡 𝑡 𝑡 𝑝 sin 𝑡 < (>) 𝑀𝑝 (cos2 , ) ⇐⇒ ( ) 𝑡 sin 𝑡 + 2( (56) 𝑡/2 ) > ( () ( ) + 2( ) , if 𝑝 > 0, and the second inequalities in (59) are reversed if 𝑝 < From Theorems and 10 together with (59) we get the following Theorem 13 The double inequality cos 𝑡 < cos 𝑡 + cos 2𝑡 + cos 𝑡 + < cos 𝑡 + cos 2𝑡 + cos 𝑡 + < cos (𝑡/2) + cos 𝑡 sin 𝑡 < cos (𝑡/2) + 𝑡 < cos 𝑡 + 2cos1/3 (𝑡/2) cos 𝑡 + < 2cos1/3 (𝑡/2) + 𝑝 𝜋 𝑝 𝜋 𝑝 𝑡 𝑝 𝑡/2 ( ) + 2( ) > ( ) + 2( ) >3 sin 𝑡 tan(𝑡/2) holds for all 𝑡 ∈ (0, 𝜋/2) if and only if 𝑝 ≥ 1/5 or 𝑝 < 0, and inequality (60) is reversed if and only if < 𝑝 ≤ 𝑝1 = log(𝜋 − 2)/ log = 0.1910 (57) cos (𝑡/2) + cos (3𝑡/2) + cos 𝑡cos2 (𝑡/2) + < < cos (𝑡/2) + cos2 (𝑡/2) + < (60) Theorem 14 The double inequality ( cos 𝑡cos4 (𝑡/2) + 3>( ) + 2( ) sin 𝑡 tan(𝑡/2) sin 𝑡 tan(𝑡/2) (61) holds for all 𝑡 ∈ (0, 𝜋/2) if and only if < 𝑞 ≤ 𝑝0 and 𝑝 ≥ 1/5 or 𝑝 < 0, where 𝑝0 = 0.1941 is the unique solution of (34) on the interval (0.1, ∞) hold for all 𝑡 ∈ (0, 𝜋/2) Neuman [24] also proved that the Wilker type inequality Applications In this section, we give some applications for our main results Neuman [24] proved that the Huygens type inequalities tan (𝑡/2) sin 𝑡 tan 𝑡 sin 𝑡 + > +2 𝑡 𝑡 𝑡 𝑡/2 >( ( 𝑝 𝑡 𝑝 𝑡/2 ) + 2( ) sin 𝑡 tan(𝑡/2) 𝑝 𝑝 𝑡/2 𝑡 ) + 2( ) >3 sin 𝑡 tan(𝑡/2) (𝑝 ≥ 1) (62) holds for all 𝑡 ∈ (0, 𝜋/2) if 𝑝 ≥ Making use of Theorem 13 and the arithmetic-geometric means inequality 2𝑝 𝑝 𝑡/2 𝑡/2 ) > 2( ) , tan(𝑡/2) tan(𝑡/2) (63) we get Corollary 15 as follows (58) (𝑝 > 0) , 2𝑝 𝑡 𝑝 𝑡/2 ) +( ) >2 sin 𝑡 tan(𝑡/2) 1+( 𝑡 𝑡 >2 + > 3, sin 𝑡 tan 𝑡 tan(𝑡/2) 𝑝 sin 𝑡 𝑝 ) + 2( ) ( 𝑡 𝑡/2 ( Corollary 15 The Wilker type inequality (62) holds for all 𝑡 ∈ (0, 𝜋/2) if 𝑝 ≥ 1/5 or 𝑝 < In addition, power series expansions show that ( 2𝑝 𝑝 (20𝑝 − 3) 𝑡/2 𝑡 𝑝 ) +( ) −2= 𝑡 + 𝑜 (𝑡4 ) sin 𝑡 tan(𝑡/2) 720 (64) Abstract and Applied Analysis Therefore, we conjecture that inequality (62) holds for all 𝑡 ∈ (0, 𝜋/2) if and only if 𝑝 ≥ 3/20 or 𝑝 < We leave it to the readers for further discussion The Schwab-Borchardt mean 𝑆𝐵(𝑎, 𝑏) [25–27] of two distinct positive real numbers 𝑎 and 𝑏 is defined by √𝑏2 − 𝑎2 { , 𝑎 < 𝑏, { { { cos−1 (𝑎/𝑏) 𝑆𝐵 (𝑎, 𝑏) = { { { √𝑎2 − 𝑏2 { , 𝑎 > 𝑏, { cosh−1 (𝑎/𝑏) −1 −1 (65) √𝑥2 where cos (𝑥) and cosh (𝑥) = log(𝑥 + − 1) are the inverse cosine and inverse hyperbolic cosine functions, respectively Let 𝑏 > 𝑎 > 0, 𝐴(𝑎, 𝑏) = (𝑎 + 𝑏)/2 be the arithmetic mean of 𝑎 and 𝑏, and 𝑡 = cos−1 (𝑎/𝑏) ∈ (0, 𝜋/2) Then simple computations lead to sin 𝑡 𝑆𝐵 (𝑎, 𝑏) = , 𝑡 𝑏 , (66) 2cos2𝑝 (𝑡/2) + cos 𝑡 2𝐴𝑝 (𝑎, 𝑏) + 𝑎𝑏𝑝−1 = 2cos2𝑝 (𝑡/2) + 2𝐴𝑝 (𝑎, 𝑏) + 𝑏𝑝 It follows from Theorems 7, 10, and 11 together with (66) that we have the following Theorem 16 Let 𝑝1 = 𝑙𝑜𝑔(𝜋 − 2)/𝑙𝑜𝑔2 = 0.1910 , 𝜆 𝑝 and 𝑝0 = 0.1941 be defined as in Theorems and 10, respectively Then for all 𝑏 > 𝑎 > 0, the following statements are true 1/𝑝 𝜆 𝑝 ( 𝐴𝑝 (𝑎, 𝑏) + 𝑏𝑝 ) 3 1/𝑝 (67) holds if and only if 𝑝 ≥ 1/5, and inequality (67) is reversed if and only if 𝑝 ≤ 𝑝1 (ii) The double inequality 1/𝑝 ( 𝐴𝑝 (𝑎, 𝑏) + 𝑏𝑝 ) 3 1/𝑞 < 𝑆𝐵 (𝑎, 𝑏) < ( 𝐴𝑞 (𝑎, 𝑏) + 𝑏𝑞 ) 3 1/𝑝 1/𝑝 𝐺(𝑎, 𝑏) + 𝐴(𝑎, 𝑏) 𝑝 𝑝 𝑎 > 0, 𝐺(𝑎, 𝑏) = √𝑎𝑏, 𝑄(𝑎, 𝑏) = √(𝑎2 + 𝑏2 )/2, 𝑃(𝑎, 𝑏) = (𝑏 − 𝑎)/[2sin−1 ((𝑏 − 𝑎)/(𝑏 + 𝑎))], 𝑇(𝑎, 𝑏) = (𝑏 − 𝑎)/[2tan−1 ((𝑏 − 𝑎)/(𝑏 + 𝑎))], and 𝑌(𝑎, 𝑏) = (𝑏 − 𝑎)/[√2tan−1 ((𝑏 − 𝑎)/√2𝑎𝑏)] be the geometric, quadratic, first Seiffert [28], second Seiffert [29], and Yang [15] means of 𝑎 and 𝑏, respectively Then it is easy to check that 𝑃(𝑎, 𝑏) = SB(𝐺(𝑎, 𝑏), 𝐴(𝑎, 𝑏)), 𝑇(𝑎, 𝑏) = SB(𝐴(𝑎, 𝑏), 𝑄(𝑎, 𝑏)), and 𝑌(𝑎, 𝑏) = SB(𝐺(𝑎, 𝑏), 𝑄(𝑎, 𝑏)) Therefore, Theorem 16 leads to Corollary 17 < 𝑃 (𝑎, 𝑏) 1/𝑞 𝐺 (𝑎, 𝑏) + 𝐴 (𝑎, 𝑏) 𝑞 𝑞 ) + 𝐴 (𝑎, 𝑏)]