On Jordan Type Inequalities for Hyperbolic Functions
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Research Article On Jordan Type Inequalities for Hyperbolic Functions ´ M. Visuri, and M. Vuorinen R. Klen, Department of Mathematics, University of Turku, 20014 Turku, Finland Correspondence should be addressed to R. Kl´en, [email protected] Received 28 January 2010; Accepted 29 April 2010 Academic Editor: Andrea Laforgia Copyright q 2010 R. Kl´en 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. This paper deals with some inequalities for trigonometric and hyperbolic functions such as the Jordan inequality and its generalizations. In particular, lower and upper bounds for functions such as sin x/x and x/ sinh x are proved.
1. Introduction During the past several years there has been a great deal of interest in trigonometric inequalities 1–7. The classical Jordan inequality 8, page 31 2 x ≤ sin x ≤ x, π
0 0 on 0, π/2 because x cos x d > 0. g1 x − g2 x dx 6
1.6
In 45, 27 it is proved that for x ∈ 0, π/2 sin x ≥ g1 x. x
1.7
Hence 1 2 cos x/3 x sin x/6 is a better lower bound for sin x/x than 1.2 for x ∈ 0, π/2. 2 Observe that x 2 cos x 2 2cos2 x/2 − 1 ≤ cos , 3 3 2
1.8
which holds true as equality if and only if cosx/2 3 ± 1/4. In conclusion, 1.8 holds for all x ∈ −2π/3, 2π/3. Together with 1.2 we now have cos2
x 1 cos x > cos x, 2 2
1.9
Journal of Inequalities and Applications
3
and by 1.8 cos2
x x sin x < < cos . 2 x 2
1.10
2. Jordan’s Inequality In this section we will find upper and lower bounds for sin x/x by using hyperbolic trigonometric functions. Theorem 2.1. For x ∈ 0, π/2 1 sin x x < < . cosh x x sinh x
2.1
Proof. The lower bound of sin x/x holds true if the function fx sin x cosh x − x is positive on 0, π/2. Since f x 2 cos x sinh x,
2.2
we have f x > 0 for x ∈ 0, π/2 and f x is increasing on 0, π/2. Therefore f x cos x cosh x sin x sinh x − 1 > f 0 0,
2.3
and the function fx is increasing on 0, π/2. Now fx > f0 0 for x ∈ 0, π/2. The upper bound of sin x/x holds true if the function gx x2 −sin x sinh x is positive on 0, π/2. Let us denote hx tan x − tanh x. Since cos x < 1 < cosh x for x ∈ 0, π/2 we have h x cosh−2 x − cos−2 x > 0 and hx > h0 0 for x ∈ 0, π/2. Now g x 2cos x cosh xhx,
2.4
which is positive on 0, π/2, because cos x cosh x > 0 and hx > 0 for x ∈ 0, π/2. Therefore g x 21 − cos x cosh x > g0 0, g x 2x − cos x sinh x − sin x cosh x > g 0 0
2.5
for x ∈ 0, π/2. Now gx > g0 0 for x ∈ 0, π/2. Proof of Theorem 1.1. The upper bound of sin x/x is clear by Theorem 2.1. The lower bound of sin x/x holds true if the function fx sin xsinh2 x − x3 is positive on 0, π/2. Let us assume x ∈ 0, π/2. Since sin x > x − x3 /6 6x − x3 /6 we have fx > 6x − x3 sinh2 x/6 − x3 . We will
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