A Sharp Double Inequality for Trigonometric Functions and Its Applications

and Applied Analysis 3 Inequality (19) implies that the function g p (x) is concave with respect to x on the interval (1/2, 1).Therefore, g p (x) > 0 follows from (18) and the concavity of g p (x). If g p (x) > 0 for all x ∈ (1/2, 1), then p ≤ p 2 follows easily from the monotonicity of the function p → g p (1/2) and g p (1/2) ≥ 0 together with the fact that g p 2 (1/2) = 0. (iii) If x ∈ (1/2, 1) and p 2 < p < 1/5, then from (13) and (19) together with the monotonicity of the function p → g p (1/2) we get g p (1) = 0, g p ( 1 2 ) < g p 2 ( 1 2 ) = 0, (20) g 󸀠 p (1) = 5p − 1 < 0, (21) g 󸀠 p ( 1 2 ) = 2p − 2 p + p2 p + 2p2 1−p > 2 × 0.1872 − 2 0.1873

The main purpose of this paper is to present the best possible parameters  and  such that the double inequality holds for all  ∈ (0, /2).As applications, some new analytic inequalities are found.All numerical computations are carried out using MATHEMATICA software.

Lemmas
In order to prove our main results we need several lemmas, which we present in this section.
Inequalities (15) lead to the conclusion that the function   () is strictly decreasing with respect to  ∈ R for fixed  ∈ (0, 1) and  2 = 0.1872 . . . is the unique solution of (14).
Let  ∈ R and the function   be defined on (0, /2) by Then elaborated computations lead to where   () is defined by (13).
From Lemma 1 and ( 24) we get the following Lemma 2 immediately.
Let  ∈ R and   be defined on (0, /2) by Then elaborated computations give where   () is defined by (23).
From Lemma 3 and (33) we get Lemma 4 immediately.
From Theorem 7 we get Corollaries 8 and 9 as follows.
It follows from Lemma 3 that we get Theorem 11 immediately.
Therefore, we conjecture that inequality (62) holds for all  ∈ (0, /2) if and only if  ≥ 3/20 or  < 0. We leave it to the readers for further discussion.
i) The double inequalities hold if and only if  ≥ 1/5, and all inequalities in (70) are reversed if and only if  ≤  1 .