Approximating trigonometric functions by using exponential inequalities

Abstract

In this paper, some exponential inequalities are derived from the inequalities containing trigonometric functions. Numerical examples show that one can achieve much tighter bounds than those of prevailing methods, which are presented by Cusa, Huygens, Chen and Sándor.

Some new estimates of precision of Cusa-Huygens and Huygens approximations

“…Let us emphasize that by Theorem 8 we gave some bounds of tangent function by use of rational functions which can be applied to other parts of Theory of analytical inequalities. Lastly, let us notice that the proofs of the considered inequalities can be also obtained by applying some methods and algorithms presented in papers [16], [17], [18], [19]- [26], [31]- [34] and in dissertation [35]. One automatic theorem prover related to some classes of inequalities, such as those presented in this paper, is currently being developed by our project team [36].…”

Some new estimates of precision of Cusa-Huygens and Huygens approximations

“…Let us emphasize that with the Theorem 8 were given some bounds of tangent function by use of rational functions which can be applied to other parts of Theory of analytical inequalities. Lastly, let us notice that the proofs of the considered inequalities can be also obtained by application of methods and algorithms presented in papers [16], [17], [18], [19]- [26], [31]- [34] and in dissertation [35].…”

Improved bounds of Mitrinović–Adamović-type inequalities by using two-parameter functions

“…are known as Mitrinović-Adamović inequalities (see [1][2][3]). Many references [4][5][6][7][8][9][10][11][12][13][14][15][16][17][18][19][20][21][22][23] have discussed the problems related to Eq. ( 1), such as the following power exponential inequality obtained by Nishizawa in [4] sin x x…”

“…λ 1 ≈ 0.06593, λ 2 = 1/15, λ 3 = 1/15, and λ 4 = (2/π) 124/21 are the best constants in Eq. (8) and Eq. (9).…”

“…Let D i (x) = F(x) -G i (x, α), i = 1, 2. It can be verified that D 1 (0) = D 1 (0) = D 1 (0) = D 1 (0) = 0, D 1 (π/2) = 0, D (4) 1 (0) = (8π 6 sin(πα/2) -7680α) 5π 6 sin(πα/2) g 1 (α), D 1 (π/2) = (π 4 -96) sin(πα/2) -8πα cos(πα/2) π 4 sin(πα/2) g 2 (α), (7) 2 (0) = 0, D (8) 2 (0) = 736α 2 45 -992 945 g 3 (α),…”