F or \(0 < x < \pi / 2\), we know \[\frac{\sin x}{x} < 1 < \frac{\tan x}{x},\] or equivalently \[\frac{ x}{\tan x} < 1 < \frac{ x}{\sin x}.\] Exploiting these inequalities to the aim to provide other similar Wilker proposed open problems [1]. One of them is \[\left(\frac{\sin x}{x}\right)^2 + \frac{\tan x}{x} > 2.\] Moreover, Wilker asked about the largest constant \(c\) such that \[\left(\frac{\sin x}{x}\right)^2 + \frac{\tan x}{x} > 2 + c x^3 \tan x\,.\] These problems was solved by Sumner et al., [2] who proved in addition \[2 + \frac{16}{\pi^4} x^3 \tan x < \left(\frac{\sin x}{x}\right)^2 + \frac{\tan x}{x} < 2 + \frac{8}{45} x^3 \tan x.\] As we may remark these constants \(\frac{16}{\pi^4}\) and \(\frac{8}{45}\) are the limits at \(0\) and \(\pi/2\) of the function \[x \rightarrow \frac{(\frac{\sin x}{x})^2 + \frac{\tan x}{x} – 2}{x^3 \tan x}.\]
Chen and Cheung [3] proved that this function decreases and provided \[\begin{aligned} &2 + \frac{8 x^3}{45} + \frac{16 x^5}{315} \tan x < \left(\frac{\sin x}{x}\right)^2 + \frac{\tan x}{x} < 2 + \frac{8 x^3}{45} + \left(\frac{2}{\pi}\right)^6\ x^5 \tan x,\\ & 2 + \frac{8 x^3}{45} + \frac{16 x^5}{315} + \frac{104 x^7}{4725} \tan x < \left(\frac{\sin x}{x}\right)^2 + \frac{\tan x}{x} < 2 + \frac{8 x^3}{45} + \frac{16 x^5}{315} + \left(\frac{12}{\pi}\right)^8\ x^7 \tan x. \end{aligned}\] Moreover, all these constants are the best possible constants.
Later in using a natural approach Mortici [4] gave improvements of that inequalities and produced the following for \(0 < x < \pi / 2\); \[2+\frac{1}{\cos x} \left(\frac{8 x^4}{45} – \frac{4 x^6}{105}\right) < \left(\frac{\sin(x)}{x}\right)^2 +\frac{\tan(x)}{x} < 2+\frac{1}{\cos x} \left(\frac{8 x^4}{45} \right) .\]
Male\(\check{{\text s}}\)evic et al., [5] further refined the above inequality for \(0 < x < \pi / 2\) and \(n \in IN\): \[2+ \frac{1}{\cos x}\sum_{k=2}^{2n+1} (-1)^k D_k x^{2k} < \left(\frac{\sin(x)}{x}\right)^2 +\frac{\tan(x)}{x} < 2 + \frac{1}{\cos x}\sum_{k=2}^{2n} (-1)^k D_k x^{2k},\] where \(D_k = \frac{\left(-9 + 3^{2k+2} – 40k- 32k^2\right)}{4 (2k+2)!}.\)
On the other hand, Wu and Srivastava [[6], Lemma 3] proved the following dual inequality for \(0 < x < \pi / 2\); \[\left(\frac{x}{\sin(x)}\right)^2 +\frac{x}{\tan(x)} > 2, \qquad 0 < x < \frac{\pi}{2}\,.\] Mortici [4] improved that result; \[\begin{aligned} &\left(\frac{x}{\sin(x)}\right)^2 +\frac{x}{\tan(x)} > 2+\frac{2 x^4}{45}\,,\\ &\left(\frac{x}{\sin(x)}\right)^2 +\frac{x}{\tan(x)} > \frac{2\sin(x)}{x} + \frac{\tan(x)}{x}. \end{aligned}\] Male\(\check{{\text s}}\)evic et al., [5] further improved the above result for \(0 < x < \pi / 2\) and \(n \in IN\): \[\left(\frac{x}{\sin(x)}\right)^2 +\frac{x}{\tan(x)} > 2 + \sum_{k=2}^{n} \frac{\mid B_{2k}\mid (2k-2) 2^{2k}}{(2k)!} x^{2k}.\]
Theorem 1. ([5], Theorem 4 p.9) For \(0 < x < \frac{\pi}{2}\);
\[\begin{aligned} 2+&\frac{1}{\cos x} \left(\frac{8 x^4}{45} – \frac{4 x^6}{105} + \frac{19 x^8}{4725} – \frac{37 x^{10}}{133650}\right)< \left(\frac{\sin(x)}{x}\right)^2 +\frac{\tan(x)}{x}< 2+\frac{1}{\cos x} \left(\frac{8 x^4}{45} – \frac{4 x^6}{105} + \frac{19 x^8}{4725}\right) . \end{aligned}\]
\[\begin{aligned} 2+&\frac{1}{\cos x} \left(\frac{8 x^4}{45} – \frac{4 x^6}{105} + \frac{19 x^8}{4725} – \frac{37 x^{10}}{133650}+ \frac{283 x^{12}}{20638800} – \frac{3503 x^{14}}{6810604000} \right)\\ &< \left(\frac{\sin(x)}{x}\right)^2 +\frac{\tan(x)}{x} < 2+\frac{1}{\cos x} \left(\frac{8 x^4}{45} – \frac{4 x^6}{105} + \frac{19 x^8}{4725}- \frac{37 x^{10}}{133650}+ \frac{283 x^{12}}{20638800} \right) . \end{aligned}\]
Theorem 2. ([5], Theorem 5 p.11) For \(0 < x < \frac{\pi}{2}\);
\[\begin{aligned} 2+ \frac{2 x^4}{45}+\frac{8 x^6}{945} < \left(\frac{x}{\sin(x)}\right)^2 +\frac{x}{\tan(x)} <\ 2 +\frac{2 x^4}{45} + (\frac{2}{\pi})^6\ \left(-2 +\frac{\pi^2}{4}-\frac{\pi^4}{360}\right) x^6. \end{aligned}\]
\[\begin{aligned} 2+& \frac{2 x^4}{45}+\frac{8 x^6}{945} +\frac{2 x^8}{1575} <\ \left(\frac{x}{\sin(x)}\right)^2 +\frac{x}{\tan(x)} \\ &<2 +\frac{2 x^4}{45} +\frac{8 x^6}{945}+ (\frac{2}{\pi})^8 \left(-2 +\frac{\pi^2}{4}-\frac{\pi^4}{360} – \frac{\pi^6}{7560}\right) x^8. \end{aligned}\]
\[\begin{aligned} 2+ &\frac{2 x^4}{45}+\frac{8 x^6}{945} +\frac{2 x^8}{1575}+\frac{16 x^{10}}{93555} <\ \left(\frac{x}{\sin(x)}\right)^2 +\frac{x}{\tan(x)}\\ &< 2 +\frac{2 x^4}{45} +\frac{8 x^6}{945}+ \frac{2x^{8}}{1575} +(\frac{2}{\pi})^{10} \left(-2 +\frac{\pi^2}{4}-\frac{\pi^4}{360} – \frac{\pi^6}{7560}-\frac{\pi^8}{201600}\right) x^{10}. \end{aligned}\]
In this paper, we aim to refine the inequalities from Theorems 1 and 2.
The first Wilker inequality [1]; \[\left(\frac{\sin(x)}{x}\right)^2 +\frac{\tan(x)}{x} > 2, \qquad x\in (0,\frac{\pi}{2})\] was intensively studied by many authors, see for example, [7-14].
Mortici [4] and Male\(\check{{\text s}}\)evic et al., [5] proved \[2+ \left( {\frac {8}{45}}\,{x}^{4}-{\frac {8}{105}}\,{x}^{6} \right) \left( \frac{1}{\cos x}\right) < \left(\frac{\sin(x)}{x}\right)^2 +\frac{\tan(x)}{x} < 2+ \left( {\frac {8 x^4}{45 \cos(x)}}\right).\] The following result provides bounds permitting to refine some previous ones;
Theorem 3. For \(0 < x < \pi / 2\), the following inequalities holds for \(1 \leq m \leq n\) and \(p \leq n\); \[\begin{aligned} &\left(1 – \sum_{k=m+1}^\infty \frac{2^{2k+1}}{\pi^{2k} (2^{2k}-2^\beta)} x^{2k} – \sum_{k=1}^m \frac{2^{2k} \mid B_{2k}\mid}{(2k)!} x^{2k} + \sum_{k=3}^n a^k x^{2k-2}\right)\\ & \quad \times \left(2 + \sum_{k=1}^p \frac{2^{2k}(2^{2k}-2)\mid B_{2k}\mid}{(2k)!} x^{2k} + \sum_{k=p+1}^n \frac{2^{2k+1}(2^{2k}-2)}{\pi^{2k} (2^{2k}-1)} x^{2k}\right)\\ &< \left(\frac{\sin(x)}{x}\right)^2 +\frac{\tan(x)}{x}\\ &<\left( 1 – \sum_{k=m+1}^\infty \frac{2^{2k+1}}{\pi^{2k} (2^{2k}-1)} x^{2k}+ \sum_{k=1}^m \frac{2^{2k} \mid B_{2k}\mid}{(2k)!} x^{2k} + \sum_{k=3}^\infty a^k x^{2k-2}\right) \\ & \quad \times \left(2 + \sum_{k=1}^p \frac{2^{2k}(2^{2k}-2)\mid B_{2k}\mid}{(2k)!} x^{2k} + 2 \left(\frac{2 x}{\pi}\right)^{2p+2} \frac{1}{1-(\frac{2 x}{\pi})^{2}}\right), \end{aligned}\] where \(B_{2k}\) are the Bernoulli numbers and \(\beta = 2 + \frac{ln(1 – \frac{6}{\pi^2})}{ln 2} = 0.6491…\) is the Alzer constant.
Proof. First we may write obviously \[\left(\frac{\sin(x)}{x}\right)^2 \left(1 +\frac{2x}{\sin(2x)}\right) = \left(\frac{\sin(x)}{x}\right)^2 +\frac{\tan(x)}{x}.\]
Lemma 1. For \(0 < x < \pi 2\), the following inequalities holds for any integer \(m \geq 5\); \[\left(\frac{\sin(x)}{x}\right)^2 < 1 – \sum_{k=m+1}^\infty \frac{2^{2k+1}}{\pi^{2k} (2^{2k}-1)} x^{2k}+ \sum_{k=1}^m \frac{2^{2k} \mid B_{2k}\mid}{(2k)!} x^{2k} + \sum_{k=3}^\infty a^k x^{2k-2},\] \[1 – \sum_{k=m+1}^\infty \frac{2^{2k+1}}{\pi^{2k} (2^{2k}-2)} x^{2k} – \sum_{k=1}^m \frac{2^{2k} \mid B_{2k}\mid}{(2k)!} x^{2k} + \sum_{k=3}^n a^k x^{2k-2} < \left(\frac{\sin(x)}{x}\right)^2.\]
Lemma 2. For \(0 < x < \pi / 2\), consider the function \[f(x) = (\sin x)^2 – x^3\cot x -\frac{x^6}{15}+\frac{x^8}{945}\,,\] then \(f(x)\) can be expressed as power series \[f(x)= \sum_{k\geq 5} a^k x^{2k},\] with \[a_{{k}}=\frac {{2}^{2\,k-2} \left| {B}_{ 2\,k-2} \right| }{ \left( 2\,k-2 \right) !}+{\frac { \left( -1 \right) ^{k+1 }{2}^{2\,k-1}}{ \left( 2\,k \right) !}} = \frac{{2}^{2\,k-2}}{ \left( 2 \,k-2 \right) ! } \left( \left| { B}_{ 2\,k-2} \right| +{\frac { \left( -1 \right) ^{k+1}}{ \left( 2\,k-1 \right) k}} \right),\] where \(B_{2k}\) are the Bernoulli numbers. Moreover, the coefficients \(a_k, k\geq 5\) are all positive: \[a_5 = \frac{1}{255150}, \ a_6 = \frac{2}{15436575}, \ a_7 = \frac{103}{8300667375},\ldots\,.\]
Proof. The following series expansions can be found in [15-17] \[\cot x = \frac{1}{x} – \sum_{k=1}^\infty \frac{2^{2k} \mid B_{2k}\mid}{(2k)!} x^{2k-1}, \qquad x \in (0,\pi),\] and \[\sin^2 x = \sum_{k=1}^\infty (-1)^{k+1}\frac{2^{2k-1}}{(2k)!} x^{2k},\qquad x \in \left(0,\frac{\pi}{2}\right)\,.\] \[\begin{aligned} (\sin x)^2 – x^3\cot x =& \left({\frac {1}{15}}{x}^{6}-{\frac {1}{945}}{x}^{8}+{\frac {1}{2835}}{x}^{ 10}+{\frac {8}{467775}}{x}^{12}+{\frac {206}{91216125}}{x}^{14}\right.\\&\left.+{ \frac {139}{638512875}}{x}^{16}+{\frac {10861}{488462349375}}{x}^{18}+ {\frac {438628}{194896477400625}}{x}^{20}+O \left( {x}^{22} \right) \right). \end{aligned}\] To prove the positivity of the coefficients \(a_k,\quad k\) even we will use the following inequality for Bernoulli numbers established by D’aniello [18]: \[\frac{2 (2k)!}{\pi^{2k} (2^{2k}-1)} < \mid B_{2k}\mid < \frac{2 (2k)!}{\pi^{2k} (2^{2k}-2)}.\] For any odd value of \(k\), we have \[a_{{k}}=\frac{{2}^{2\,k-2}}{\ \left( 2\,k-2 \right) ! } \left( \left| {B}_{ 2\,k-2} \right| +{\frac { \left( -1 \right) ^{k+1}}{ \left( 2\,k-1 \right) k}} \right) = \frac{{2}^{2\,k-2}}{\ \left( 2\,k-2 \right) ! } \left( \left| { B}_{ 2\,k-2} \right| +{\frac { 1}{ \left( 2\,k-1 \right) k}} \right) >0.\] Consider the even case, then \[a_{{k}}=\frac{{2}^{2\,k-2}}{\ \left( 2\,k-2 \right) ! } \left( \left| { B}_{ 2\,k-2} \right| +{\frac { -1}{ \left( 2\,k-1 \right) k}} \right).\] By definition of Bernoulli numbers \[S_n(p) = \sum_{k=0}^{n-1} k^p = \frac{1}{p+1} \sum_{k=0}^{p} \left (^{p+1}_k\right) B_k n^{p+1-k} .\] Then for \(k\) even \[a_{{k}}=\frac{{2}^{2\,k-2}}{\ \left( 2\,k-2 \right) ! } \left( \left| { B}_{ 2\,k-2} \right| -{\frac { 1}{ \left( 2\,k-1 \right) k}} \right) > \frac{{2}^{2\,k-2}}{\ \left( 2\,k-2 \right) ! } \left( \frac{2 (2k-2)!}{\pi^{2k-2} (2^{2k-2}-2)} -{\frac { 1}{ \left( 2\,k-1 \right) k}} \right)\,,\] and \[a_{{k}} < \frac{{2}^{2\,k-2}}{\ \left( 2\,k-2 \right) ! } \left( \frac{2 (2k-2)!}{\pi^{2k-2} (2^{2k-2}-1)} -{\frac { 1}{ \left( 2\,k-1 \right) k}} \right).\] Therefore \[\begin{aligned} \frac {{2}^{2\,k-2} \frac{2 (2k-2)!}{\pi^{2k-2} \left(2^{2k-2}-1\right)} }{ \left( 2\,k-2 \right) !}+{\frac { \left( -1 \right) ^{k+1 }{2}^{2\,k-1}}{ \left( 2\,k \right) !}} &< a_{{k}} = \frac {{2}^{2\,k-2} \left| {B}_{ 2\,k-2} \right| }{ \left( 2\,k-2 \right) !}+{\frac { \left( -1 \right) ^{k+1 }{2}^{2\,k-1}}{ \left( 2\,k \right) !}} \\ &< \frac {{2}^{2\,k-2} \frac{2 (2k-2)!}{\pi^{2k-2} \left(2^{2k-2}-2\right)} }{ \left( 2\,k-2 \right) !}+{\frac { \left( -1 \right) ^{k+1 }{2}^{2\,k-1}}{ \left( 2\,k \right) !}}\,, \end{aligned}\] and \[\sum_{k\geq 5} \left(\frac{{2}^{2k-1}}{2\pi^{2k-2} \left(2^{2k-2}-1\right)} + {\frac {\left(-1\right)^{k+1} 2^{2k-1}}{ \left( 2\,k \right)! }} \right) x^{2k} < f(x).\] Thus it implies the left hand, since for any integer \(n\geq 5\), we have \[\begin{aligned} \cos x &\leq \cos x + \sin x\ \left(\frac{x^3}{15}-\frac{x^5}{945}\right) + \sin x \sum_{5\leq k\leq n} a_k x^{2k-3}\\ &\leq \cos x + \sin x\ \left(\frac{x^3}{15}-\frac{x^5}{945}\right) + \sin x \sum_{5\leq k\leq \infty} a_k x^{2k-3}\\ & = \left(\frac{\sin x}{x}\right)^3. \end{aligned}\]
\[\begin{aligned} (\sin x)^2 – x^3 \cot x &= (\sin x)^2 – x^3 \left(\frac{1}{x} – \sum_{k=1}^\infty \frac{2^{2k} \mid B_{2k}\mid}{(2k)!} x^{2k-1}\right) \\ &=(\sin x)^2 -x^2 + \sum_{k=1}^\infty \frac{2^{2k} \mid B_{2k}\mid}{(2k)!} x^{2k+2} \\ &< (\sin x)^2 -x^2 + \sum _{k=1}^{ \infty }{\frac {{2}^{2\,k+1}{x}^{2\,k+2}}{{\pi }^{2\,k} \left( {2}^{2 \,k}-2 \right) }}. \end{aligned}\] In the other hand we know that for \(k > 1\); \[(2k)! > \sqrt{4\pi k}\ \left(\frac{2k}{e}\right)^{2k}\ e^{\frac{1}{24k+1}}.\] It implies \[\begin{aligned} \left( \frac{2 (2k-2)!}{\pi^{2k-2} (2^{2k-2}-1)} -{\frac { 1}{ \left( 2\,k-1 \right) k}} \right)& > \left( \frac{2 \sqrt{4\pi (k-1)} (\frac{2k-2}{ e})^{2k-2} e^{\frac{1}{48k-23}}}{\pi^{2k-2} (2^{2k-2}-1)} -{\frac { 1}{ \left( 2\,k-1 \right) k}} \right)\\ &>\left( \frac{2 \sqrt{4\pi (k-1)} (\frac{2k-2}{\pi e})^{2k-2} }{ (2^{2k-2}-1)} -{\frac { 1}{ \left( 2\,k-1 \right) k}} \right)\\ &> \left( \frac{2 \sqrt{4\pi (k-1)} (\frac{2k-2}{\pi e})^{2k-2} }{ 2^{2k-2}} -{\frac { 1}{ \left( 2\,k-1 \right) k}} \right)\\ &> \left( 2 \sqrt{4\pi (k-1)} (\frac{k-1}{\pi e})^{2k-2} -{\frac { 1}{ \left( 2\,k-1 \right) k}} \right). \end{aligned}\] Thanks to Maple, we may easily verify that the last expression is non negative as soon as \(k> 6\). This means that \(a_k\) are non negative. ◻
By this Lemma, we deduce \[\frac{f(x)}{x^2} = \left(\frac{\sin(x)}{x}\right)^2 – x \cot x – \frac{x^4}{15} + \frac{x^6}{945} = \sum_{k\geq 5} a^k x^{2k-2},\] where \[a_{{k}}= \frac{{2}^{2\,k-2}}{ \left( 2 \,k-2 \right) ! } \left( \left| { B}_{ 2\,k-2} \right| +{\frac { \left( -1 \right) ^{k+1}}{ \left( 2\,k-1 \right) k}} \right).\] Thus for any integer \(n \geq 5\) \[\left(\frac{\sin(x)}{x}\right)^2 = x \cot x + \frac{x^4}{15} – \frac{x^6}{945} + \sum_{k\geq 5} a^k x^{2k-2} \geq x \cot x + \frac{x^4}{15} – \frac{x^6}{945} + \sum_{k=5}^n a^k x^{2k-2}\,,\] since coefficients \(a_k > 0.\)
Therefore, since by [[17], p.145], for \(x \in (0,\pi)\), \[\begin{aligned} \cot x &= \frac{1}{x} – \frac{x}{3} – \frac{x^3}{45} – \frac{2 x^5}{945} -…\\ &= \frac{1}{x} – \sum_{k=1}^\infty \frac{2^{2k} \mid B_{2k}\mid}{(2k)!} x^{2k-1}\\ &=\frac{1}{x} – \sum_{k=1}^{n-1} \frac{2^{2k} \mid B_{2k}\mid}{(2k)!} x^{2k-1} – \sum_{k=n}^\infty \frac{2^{2k} \mid B_{2k}\mid}{(2k)!} x^{2k-1}, \end{aligned}\] and by [18] \[\label{eq1} \frac{2 (2k)!}{\pi^{2k} (2^{2k}-1)} < \mid B_{2k}\mid < \frac{2 (2k)!}{\pi^{2k} (2^{2k}-2)}. \tag{1}\] Then we deduce the inequalities \[\sum_{k=n+1}^\infty \frac{-2^{2k+1}}{\pi^{2k} (2^{2k}-2)} x^{2k-1} < \cot x – \frac{1}{x} + \sum_{k=1}^n \frac{2^{2k} \mid B_{2k}\mid}{(2k)!} x^{2k-1} < \sum_{k=n+1}^\infty \frac{-2^{2k+1}}{\pi^{2k} (2^{2k}-1)} x^{2k-1}.\] Notice that Alzer [19] provides a further result, \[\label{eq2} \frac{2 (2k)!}{\pi^{2k} (2^{2k}-1)} < \mid B_{2k}\mid < \frac{2^{2k+1} (2k)!}{\pi^{2k} (2^{2k}-2^\beta)} < \frac{2^{2k+1} (2k)!}{\pi^{2k} (2^{2k}-2)}\,, \tag{2}\] where \(\beta = 2 + \frac{ln(1 – \frac{6}{\pi^2})}{ln 2} = 0.6491…\) is the best possible constant in the sense that it can not be replaced respectively by any bigger and smaller constant in the double inequality.
We then obtain better inequalities for \(\cot x\), \[\sum_{k=n+1}^\infty \frac{-2^{2k+1}}{\pi^{2k} (2^{2k}-2^\beta)} x^{2k-1} < \cot x – \frac{1}{x} + \sum_{k=1}^n \frac{2^{2k} \mid B_{2k}\mid}{(2k)!} x^{2k-1}.\] On the other hand, it follows by the same way for any integer \(m \geq 5\), \[\left(\frac{\sin(x)}{x}\right)^2 < 1 – \sum_{k=m+1}^\infty \frac{2^{2k+1}}{\pi^{2k} (2^{2k}-1)} x^{2k}- \sum_{k=1}^m \frac{2^{2k} \mid B_{2k}\mid}{(2k)!} x^{2k} + \sum_{k=3}^\infty a^k x^{2k-2},\] implies \[1 – \sum_{k=m+1}^\infty \frac{2^{2k+1}}{\pi^{2k} (2^{2k}-2^\beta)} x^{2k} – \sum_{k=1}^m \frac{2^{2k} \mid B_{2k}\mid}{(2k)!} x^{2k} + \sum_{k=3}^n a^k x^{2k-2} < \left(\frac{\sin(x)}{x}\right)^2.\] Let us consider now expansions trigonometric functions with power series. We will use the Taylor expansions of \(\sin(x),\) \[\sin(x) = x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+….+(-1)^{k-1}\frac{x^{2k-1}}{(2k-1)!}+(-1)^{k}\frac{\sin \theta x}{(2k+1)!} x^{2k+1}\,,\] where \(0 < \theta < 1\).
It is easy to remark that \[1-\frac{x^2}{3!}+\frac{x^4}{5!}-\frac{x^6}{7!} < \frac{\sin x}{x} < 1-\frac{x^2}{3!}+\frac{x^4}{5!}-\frac{x^6}{7!}+\frac{x^8}{9!} \,,\] for \(0 < x < \frac{\pi}{2}\). We then deduce bounds for \(\left(\frac{\sin x}{x}\right)^2\), \[\sum_{k=1}^{2p} (-1)^{k+1}\frac{2^{2k+1}}{(2k)!} x^{2k-2} < \left(\frac{\sin x}{x}\right)^2 < \sum_{k=1}^{2p+1} (-1)^{k+1}\frac{2^{2k+1}}{(2k)!} x^{2k-2}.\] On the other hand, we know that [[17] p.145] \[\frac{1}{\sin x} = \frac{1}{x} +\frac{x}{6} +\frac{7 x^3}{360}+… = \frac{1}{x} + \sum_{k=1}^\infty \frac{(2^{2k}-2)\mid B_{2k}\mid}{(2k)!} x^{2k-1}.\] We then derive the following inequalities for any \(n \geq 1\).
Lemma 3. For \(0 < x < \pi / 2\), the following inequalities holds for any integer \(p \geq 1\), \[\begin{aligned} 2 +& \sum_{k=1}^p \frac{2^{2k}(2^{2k}-2)\mid B_{2k}\mid}{(2k)!} x^{2k} + \sum_{k=p+1}^n \frac{2^{2k+1}(2^{2k}-2)}{\pi^{2k} (2^{2k}-1)} x^{2k}\\ & < 1 + \frac{2x}{\sin 2x} \\ &< 2 + \sum_{k=1}^p \frac{2^{2k}(2^{2k}-2)\mid B_{2k}\mid}{(2k)!} x^{2k} + 2 \left(\frac{2 x}{\pi}\right)^{2p+2} \frac{1}{1-(\frac{2 x}{\pi})^{2}}. \end{aligned}\]
Indeed, by [[17] p.146] one has \[1 + \frac{2x}{\sin 2x} = 2 + \sum_{k=1}^\infty \frac{2^{2k}(2^{2k}-2)\mid B_{2k}\mid}{(2k)!} x^{2k} > 2 + \sum_{k=1}^n \frac{2^{2k}(2^{2k}-2)\mid B_{2k}\mid}{(2k)!} x^{2k}.\] Then it follows by (1) \[\begin{aligned} 2 +& \sum_{k=1}^p \frac{2^{2k}(2^{2k}-2)\mid B_{2k}\mid}{(2k)!} x^{2k} + \sum_{k=p+1}^n \frac{2^{2k+1}(2^{2k}-2)}{\pi^{2k} (2^{2k}-1)} x^{2k}\\ &< 1 + \frac{2x}{\sin 2x} \\ &< 2 + \sum_{k=1}^p \frac{2^{2k}(2^{2k}-2)\mid B_{2k}\mid}{(2k)!} x^{2k} + 2\sum_{k=p+1}^\infty \frac{2^{2k}}{\pi^{2k} } x^{2k} \\ &=2 + \sum_{k=1}^p \frac{2^{2k}(2^{2k}-2)\mid B_{2k}\mid}{(2k)!} x^{2k} + 2 \left(\frac{2 x}{\pi}\right)^{2p+2} \frac{1}{1-(\frac{2 x}{\pi})^{2}}. \end{aligned}\] We may also prove the following frame \[\frac{2\left(\frac{2 x}{\pi}\right)^{2n+2}}{1-(\frac{2x}{\pi})^2} – 2 \sum_{k=1}^n \frac{1}{2^k-1} \left(\frac{2 x}{\pi}\right)^{2k} < 1 + \frac{2x}{\sin 2x} < \frac{2}{1-(\frac{2x}{\pi})^2} .\] Finally, we get a lower bound for the product \[\begin{aligned} \left(\frac{\sin(x)}{x}\right)^2 \left(1 + \frac{2x}{\sin 2x}\right) >& \left(1 – \sum_{k=m+1}^\infty \frac{2^{2k+1}}{\pi^{2k} (2^{2k}-2^\beta)} x^{2k} – \sum_{k=1}^m \frac{2^{2k} \mid B_{2k}\mid}{(2k)!} x^{2k} + \sum_{k=3}^n a^k x^{2k-2}\right)\\ &\times\left(2 + \sum_{k=1}^p \frac{2^{2k}(2^{2k}-2)\mid B_{2k}\mid}{(2k)!} x^{2k} + \sum_{k=p+1}^n \frac{2^{2k+1}(2^{2k}-2)}{\pi^{2k} (2^{2k}-1)} x^{2k}\right)\\ &= \phi_{m,n,p}(x). \end{aligned}\] The upper bound is \[\begin{aligned} \left(\frac{\sin(x)}{x}\right)^2 \left(1 + \frac{2x}{\sin 2x}\right) <& \left( 1 – \sum_{k=m+1}^\infty \frac{2^{2k+1}}{\pi^{2k} (2^{2k}-1)} x^{2k}- \sum_{k=1}^m \frac{2^{2k} \mid B_{2k}\mid}{(2k)!} x^{2k} + \sum_{k=3}^\infty a^k x^{2k-2}\right)\\ &\times \left(2 + \sum_{k=1}^p \frac{2^{2k}(2^{2k}-2)\mid B_{2k}\mid}{(2k)!} x^{2k} + 2 \left(\frac{2 x}{\pi}\right)^{2p+2} \frac{1}{1-(\frac{2 x}{\pi})^{2}}\right)\\ &= \psi_{m,p}(x). \end{aligned}\] Thus the following inequalities hold for \(0 < x < \pi / 2\) and for integers \(1 \leq m ; \ 3 \leq n; \ 1 \leq p \leq n\), \[\phi_{m,n,p}(x) < \left(\frac{\sin(x)}{x}\right)^2 +\frac{\tan(x)}{x} < \psi_{m,p}(x).\] Theorem 3 is then proved. ◻
Let \(0 < x < \pi / 2\). By Theorem 3 we are able to precise the lower bound of the Wilker inequality in putting different values of \(n,p\).
Example 1. Taking \(n=3\) and \(p=m=2\), we have \[\phi_{2,4,2} = 2+ \frac{8 x^4}{45} < \left(\frac{\sin(x)}{x}\right)^2 +\frac{\tan(x)}{x}.\] Then we find again a result of [7] since \[\frac{1}{ \cos x } (\frac{8 x^4}{45}-\frac{4 x^6}{105}) > \frac{\frac{8 x^4}{45}-\frac{4 x^6}{105}}{1- \frac{x^2}{2!}+\frac{x^4}{4!}} > \frac{8 x^4}{45}.\]
Example 2. Taking \(n=4\) and \(p=m=3\), we find \[\phi_{3,4,3} = 2+ \frac{8 x^4}{45} + \frac{16 x^6}{315} < \left(\frac{\sin(x)}{x}\right)^2 +\frac{\tan(x)}{x}.\]
Example 3. Taking \(n=5\) and \(p=m=3\), we have \[\phi_{3,5,3} = 2+ \frac{8 x^4}{45} + \frac{16 x^6}{315} + \frac{104 x^8}{4725} < \left(\frac{\sin(x)}{x}\right)^2 +\frac{\tan(x)}{x}.\]
Example 4. Taking \(n=6\) and \(p=m=5\), we have \[\phi_{5,6,5} = 2+ \frac{8 x^4}{45} + \frac{16 x^6}{315} + \frac{104 x^8}{4725} + \frac{592 x^{10}}{66825} < \left(\frac{\sin(x)}{x}\right)^2 +\frac{\tan(x)}{x} .\]
This permits to find again Theorem 1(i) [[8], p.9]. Indeed, since (see Lemma 4 below) for \(0 < x < \pi / 2\), \[\cos x > 1 – \frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}+\frac{x^8}{8!}-\frac{x^{10}}{10!} +\frac{x^{12}}{12!}-\frac{x^{14}}{14!},\] we then deduce \[(\cos x) \left( \frac{8 x^4}{45} + \frac{16 x^6}{315} + \frac{104 x^8}{4725} + \frac{592 x^{10}}{66825}\right) > {\frac {8}{45}}\,{x}^{4}-{\frac {4}{105}}\,{x}^{6}+{\frac {19}{4725}}\,{x}^{8}-{\frac {37}{133650}}\,{x}^{10}.\]
Example 5. Taking \(n=7\) and \(p=m=4\), we obtain \[\phi_{4,7,4} = 2+ \frac{8 x^4}{45} + \frac{16 x^6}{315} + \frac{104 x^8}{4725} + \frac{592 x^{10}}{66825} + \frac{152912 x^{12}}{42567525}< \left(\frac{\sin(x)}{x}\right)^2 +\frac{\tan(x)}{x}.\]
By the same way, using again the lower bound of \(\cos x\), we find a result of [[8], p.10], \[\begin{aligned} (\cos x)& \left(2+ \frac{8 x^4}{45} + \frac{16 x^6}{315} + \frac{104 x^8}{4725} + \frac{592 x^{10}}{66825} + \frac{152912 x^{12}}{42567525}\right)\\ &> {\frac {8}{45}}\,{x}^{4}-{\frac {4}{105}}\,{x}^{6}+{\frac {19}{4725}}\,{x}^{8}-{\frac {37}{133650}}\,{x}^{10} + {\frac {283}{20638800}}\,{x}^{12}-{\frac {3503}{6810804000}}\,{x}^{14}. \end{aligned}\] In the sequel we will find upper and lower bounds of Wilker inequalities which appear to be finer than known previous. Consider at first;
Lemma 4. For \(0 < x < \pi / 2\), the following inequalities holds for any integer \(p \geq 1\), \[\begin{aligned} &\sum_{k=1}^{2p} (-1)^{k+1}\frac{2^{2k+1}}{(2k)!} x^{2k-2} < \left(\frac{\sin x}{x}\right)^2 < \sum_{k=1}^{2p+1} (-1)^{k+1}\frac{2^{2k+1}}{(2k)!} x^{2k-2},\\ &\sum_{k=1}^{2p+1} (-1)^{k}\frac{1}{(2k)!} x^{2k} < \cos x < \sum_{k=1}^{2p} (-1)^{k}\frac{1}{(2k)!} x^{2k}. \end{aligned}\]
Let us consider expansions trigonometric functions with power series. We will use the Taylor expansions of \(\sin x,\ \cos x\), \[\begin{aligned} &\sin x = x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+….+(-1)^{k-1}\frac{x^{2k-1}}{(2k-1)!}+(-1)^{k}\frac{\sin \theta x}{(2k+1)!} x^{2k+1},\\ &\cos x = 1-\frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}+….+(-1)^{k}\frac{x^{2k}}{(2k)!}+(-1)^{k+1}\frac{\cos \theta x}{(2k+2)!} x^{2k+2}, \end{aligned}\] where \(0 < \theta < 1\).
It is easy to remark that \[1-\frac{x^2}{3!}+\frac{x^4}{5!}-\frac{x^6}{7!} < \frac{\sin x}{x} < 1-\frac{x^2}{3!}+\frac{x^4}{5!}-\frac{x^6}{7!}+\frac{x^8}{9!} \,,\] for \(0 < x < \frac{\pi}{2}\). We then deduce bounds for \(\left(\frac{\sin x}{x}\right)^2\) \(\cos x\), \[\begin{aligned} &\sum_{k=1}^{2p} (-1)^{k+1}\frac{2^{2k+1}}{(2k)!} x^{2k-2} < \left(\frac{\sin x}{x}\right)^2 < \sum_{k=1}^{2p+1} (-1)^{k+1}\frac{2^{2k+1}}{(2k)!} x^{2k-2}.\\ &\sum_{k=1}^{2p+1} (-1)^{k}\frac{1}{(2k)!} x^{2k} < \cos x < \sum_{k=1}^{2p} (-1)^{k}\frac{1}{(2k)!} x^{2k}. \end{aligned}\] By Lemma 3, we then derive the following which improves Theorem 3.
Theorem 4. For \(0 < x < \pi / 2\) the following inequalities holds for any \(q\geq 1, 1 \leq p \leq n\), \[\begin{aligned} &\left(\sum_{k=1}^{2q} (-1)^{k+1}\frac{2^{2k-1}}{(2k)!} x^{2k-2}\right) \left(2 + \sum_{k=1}^p \frac{2^{2k}(2^{2k}-2)\mid B_{2k}\mid}{(2k)!} x^{2k} + \sum_{k=p+1}^n \frac{2^{2k+1}(2^{2k}-2)}{\pi^{2k} (2^{2k}-1)} x^{2k}\right) \\ &\qquad< \left(\frac{\sin(x)}{x}\right)^2 +\frac{\tan(x)}{x}\\ & \qquad<\left( \sum_{k=1}^{2q+1} (-1)^{k+1}\frac{2^{2k-1}}{(2k)!} x^{2k-2}\right) \left(2 + \sum_{k=1}^p \frac{2^{2k}(2^{2k}-2)\mid B_{2k}\mid}{(2k)!} x^{2k} + 2 \left(\frac{2 x}{\pi}\right)^{2p+2} \frac{1}{1-(\frac{2 x}{\pi})^{2}}\right), \end{aligned}\] where \(B_{2k}\) are the Bernoulli numbers.
Corollary 5. For \(0 < x < \pi / 2\) the following inequalities holds for any \(n, p,\quad 1 \leq p \leq n\), \[\begin{aligned} &\left(\frac{\sin(x)}{x}\right)^2 \left(2 + \sum_{k=1}^p \frac{2^{2k}(2^{2k}-2)\mid B_{2k}\mid}{(2k)!} x^{2k} + \frac{2\left(\frac{2 x}{\pi}\right)^{2n+2}}{1-(\frac{2x}{\pi})^2} – 2 \sum_{k=1}^n \frac{1}{2^k-1} \left(\frac{2 x}{\pi}\right)^{2k}\right)\\ &\qquad< \left(\frac{\sin(x)}{x}\right)^2 +\frac{\tan(x)}{x}\\ &\qquad<\left(\frac{\sin(x)}{x}\right)^2 \times \left(2 + \sum_{k=1}^p \frac{2^{2k}(2^{2k}-2)\mid B_{2k}\mid}{(2k)!} x^{2k} + \frac{2\left(\frac{2 x}{\pi}\right)^{2n+2}}{1-(\frac{2x}{\pi})^2}\right), \end{aligned}\] where \(B_{2k}\) are the Bernoulli numbers.
Corollary 5 means that the following inequalities hold: \[\begin{aligned} &\left(\frac{\sin(x)}{x}\right)^2 \left(2 + \sum_{k=1}^p \frac{2^{2k}(2^{2k}-2)\mid B_{2k}\mid}{(2k)!} x^{2k} – 2 \sum_{k=1}^n \frac{1}{2^k-1} \left(\frac{2 x}{\pi}\right)^{2k}\right)\\ &\qquad<g_n(x)\\ &\qquad= \left(\frac{\sin(x)}{x}\right)^2 \left( 1 – \frac{2\left(\frac{2 x}{\pi}\right)^{2n+2}}{1-(\frac{2x}{\pi})^2}\right) +\frac{\tan(x)}{x}\\ &\qquad< \left(\frac{\sin(x)}{x}\right)^2 \left(2 + \sum_{k=1}^p \frac{2^{2k}(2^{2k}-2)\mid B_{2k}\mid}{(2k)!} x^{2k} \right)\\ &\qquad< \left(\frac{\sin(x)}{x}\right)^2 +\frac{\tan(x)}{x} . \end{aligned}\] The function \(g_n(x)\) is growing as \(n\) increasing. We have \[g_2(x)=\left(\frac{\sin(x)}{x}\right)^2 \left( 1 – \frac{2\left(\frac{2 x}{\pi}\right)^{2}}{1-(\frac{2x}{\pi})^2}\right) +\frac{\tan(x)}{x} < g_n(x) < lim_{n\rightarrow \infty} g_n(x) = \left(\frac{\sin x}{x}\right)^2 +\frac{\tan x}{x}.\] Moreover, we may compute the limit when \(x\) tends to \((\frac{\pi}{2})^-\), \[lim_{x\rightarrow \frac{\pi}{2}} g_n(x) = \frac{2(3+4n)}{\pi^2},\] and \[g_n(x) – \frac{2(3+4n)}{\pi^2} = \,{\frac {-24\,{n}^{2}+24\,n+33+2\,{\pi }^{2}}{{3\pi }^{3}}} (\pi- 2x)+…\,.\] Let \(0 < x < \pi / 2\). Then we have following examples;
Example 6. Taking \(n=3\) and \(p=2\), we find \[\begin{aligned} 2+ \frac{8 x^4}{45} &< g_3(x)=\left(\frac{\sin x}{x}\right)^2 +\frac{\tan x}{x} -\left(\frac{\sin x}{x}\right)^2 \left( \frac{2\left(\frac{2 x}{\pi}\right)^{8}}{1-(\frac{2x}{\pi})^2}\right)\\ & < {\frac {2306 {x}^{10}}{467775}} -{\frac {2 {x}^{8}}{63}} +{\frac {16 {x}^{6}}{ 315}} +{\frac {8 {x}^{4}}{45}}+2 < {\frac {16 {x}^{6}}{315}} +{\frac {8 {x}^{4}}{45}}+2. \end{aligned}\]
Example 7. Taking \(n=4\) and \(p=3\), we find \[\begin{aligned} 2+ \frac{8 x^4}{45} + \frac{16 x^6}{315} &< g_4(x)=\left(\frac{\sin(x)}{x}\right)^2 +\frac{\tan(x)}{x} -\left(\frac{\sin(x)}{x}\right)^2 \left( \frac{2\left(\frac{2 x}{\pi}\right)^{10}}{1-(\frac{2x}{\pi})^2}\right)\\ & <{\frac {61232 {x}^{12}}{30405375}}-{\frac {868 {x}^{10}}{66825}}+{ \frac {104 {x}^{8}}{4725}}+{\frac {16 {x}^{6}}{315}}+{\frac {8 {x}^{4}}{45}}+2 < {\frac {104 {x}^{8}}{4725}}+{\frac {16 {x}^{6}}{315}}+{\frac {8 {x}^{4}}{45}}+2. \end{aligned}\]
Example 8. Taking \(n=5\) and \(p=3\), we obtain \[\begin{aligned} 2+ \frac{8 x^4}{45} + \frac{16 x^6}{315} + \frac{104 x^8}{4725} &< g_5(x)=\left(\frac{\sin(x)}{x}\right)^2 +\frac{\tan(x)}{x} -\left(\frac{\sin(x)}{x}\right)^2 \left( \frac{2\left(\frac{2 x}{\pi}\right)^{12}}{1-(\frac{2x}{\pi})^2}\right)\\ &< {\frac {1566172 {x}^{14}}{1915538625}}-{\frac {480604 {x}^{12}}{91216125}}+{\frac {592 {x}^{10}}{66825}}+{\frac {104 {x}^{8}}{4725}}+{\frac {16 {x}^{6}}{315}}+{\frac {8 {x}^{4}}{45}}+2\\ &<{\frac {592 {x}^{10}}{66825}}+{\frac {104 {x}^{8}}{4725}}+{\frac {16 {x}^{6}}{315}}+{\frac {8 {x}^{4}}{45}}+2. \end{aligned}\] We then improve by another way Theorem 1(i) ([[8], p.9]), since by Lemma 3, \[\begin{aligned} &\left( {\frac {8}{45}}\,{x}^{4}+{\frac {16}{315}}\,{x}^{6}+{\frac { 104}{4725}}\,{x}^{8}+{\frac {592}{66825}}\,{x}^{10} \right) \cos\left( x \right) < {\frac {8}{45}}{x}^{4}-{\frac {4}{105}}{x}^{6}+{\frac {19}{4725}}{x}^ {8}\\ &\left( {\frac {8}{45}}\,{x}^{4}+{\frac {16}{315}}\,{x}^{6}+{\frac { 104}{4725}}\,{x}^{8}+{\frac {592}{66825}}\,{x}^{10}+{\frac {152912}{42567525}}\,{x}^{12} \right) \cos \left( x \right) \\ &\qquad> {\frac {8}{45}}{x}^{4}-{\frac {4}{105}}{x}^{6}+{\frac {19}{4725}}{x}^{8}-{\frac {37}{133650}}{x}^{10}+{\frac {283}{20638800}}{x}^{12}. \end{aligned}\]
Example 9. Taking \(n=6\) and \(p=3\), we obtain \[\begin{aligned} 2+ \frac{8 x^4}{45}& + \frac{16 x^6}{315} + \frac{104 x^8}{4725} +{\frac {592}{66825}}\,{x}^{10} < g_6(x)= \left(\frac{\sin(x)}{x}\right)^2 +\frac{\tan(x)}{x} -\left(\frac{\sin(x)}{x}\right)^2 \left( \frac{2\left(\frac{2 x}{\pi}\right)^{14}}{1-(\frac{2x}{\pi})^2}\right)\\ &< {\frac {161934166 {x}^{16}}{488462349375}}-{\frac {123992 {x}^{14}}{58046625}}+{\frac {152912 {x}^{12}}{42567525}}+{\frac {592 {x}^{10}}{66825}}+{\frac {104 {x}^{8}}{4725}}+{\frac {16 {x}^{6}}{315}}+{ \frac {8 {x}^{4}}{45}}+2 \\ &<{\frac {152912 {x}^{12}}{42567525}}+{\frac {592 {x}^{10}}{66825}}+{\frac {104 {x}^{8}}{4725}}+{\frac {16 {x}^{6}}{315}}+{ \frac {8 {x}^{4}}{45}}+2. \end{aligned}\]
The last estimate improves Theorem 1(ii) ( [[8], p.10]).
Wu and Srivastava [[14], Lemma 3] proved the following dual inequality for \(0 < x < \pi / 2\) also called second Wilker inequality \[\left(\frac{x}{\sin(x)}\right)^2 +\frac{x}{\tan(x)} > 2, \qquad 0 < x < \frac{\pi}{2}\,.\] Mortici [4] improved it and gave, \[\begin{aligned} &\left(\frac{x}{\sin(x)}\right)^2 +\frac{x}{\tan(x)} > 2+\frac{2 x^4}{45}.\\ &\left(\frac{x}{\sin(x)}\right)^2 +\frac{x}{\tan(x)} > \frac{2\sin(x)}{x} + \frac{\tan(x)}{x}. \end{aligned}\] Male\(\check{{\text s}}\)evic et al., [[5], Theorem 5] gave an improvement of that and provided the following bounds for \(0 < x < \pi / 2\) and any integer \(m \geq 2\) \[\begin{aligned} 2 + &\sum_{k=2}^m \frac{(2k-2) 2^{2k} \mid B_{2k}\mid}{(2k)!} x^{2k} < \left(\frac{x}{\sin(x)}\right)^2 +\frac{x}{\tan(x)}\\ & < 2 + \sum_{k=2}^{m-1} \frac{(2k-2) 2^{2k} \mid B_{2k}\mid}{(2k)!} x^{2k} + \left(\frac{2x}{\pi}\right)^{2n}\left(\frac{\pi^2}{4} – 2 – 2 + \sum_{k=2}^{m-1} \frac{(2k-2) 2^{2k} \mid B_{2k}\mid}{(2k)!} \left(\frac{\pi}{2}\right)^{2k}\right). \end{aligned}\] It seems that Theorem 6 below improves that result.
Theorem 6. For \(0 < x < \pi / 2\) and any \(p \geq 1\), the following inequalities hold: \[\begin{aligned} & \left(1 + \sum_{k=1}^{2p} (-1)^{k}\frac{(2x)^{2k}}{(2k+1)!}\right) \left(1 + \sum_{k=1}^p \frac{(2k-1) 2^{2k} \mid B_{2k}\mid}{(2k)!} x^{2k}+ \sum_{k=p+1}^\infty \frac{ (2k-1) 2^{2k+1} }{\pi^{2k}(2^{2k}-1)} x^{2k}\right)\\ &\qquad < \left(\frac{x}{\sin(x)}\right)^2 +\frac{x}{\tan(x)}\\ &\qquad< \left(1 + \sum_{k=1}^{2p+1} (-1)^{k}\frac{(2x)^{2k}}{(2k+1)!}\right) \left(1 + \sum_{k=1}^p \frac{(2k-1) 2^{2k} \mid B_{2k}\mid}{(2k)!} x^{2k} + \sum_{k=p+1}^\infty \frac{(2k-1) 2^{2k+1} }{\pi^{2k}(2^{2k}-2^\beta)} x^{2k}\right), \end{aligned}\] where \(B_{2k}\) are the Bernoulli numbers and \(\beta = 2 + \frac{ln(1 – \frac{6}{\pi^2})}{ln 2} = 0.6491…\) is the Alzer constant.
Proof. Recall the following series expansions which can be found in [17] or [16].
\[\cot x = \frac{1}{x} – \sum_{k=1}^\infty \frac{2^{2k} \mid B_{2k}\mid}{(2k)!} x^{2k-1}, \quad \frac{\sin x}{x} = \sum_{k=0}^{\infty} (-1)^{k}\frac{x^{2k}}{(2k+1)!} \,.\] Then we derive \[\left(\frac{x}{\sin x}\right)^2 = – x^2 \ \frac{d\cot x}{dx} = 1 + \sum_{k=1}^\infty \frac{(2k-1) 2^{2k} \mid B_{2k}\mid}{(2k)!} x^{2k}.\] Notice that \[\begin{aligned} \left(\frac{x}{\sin(x)}\right)^2 +\frac{x}{\tan(x)}& = \left(\frac{x}{\sin x}\right)^2 \left(1 + \frac{\sin 2x}{2x}\right) \\ &= \left( 1 + \sum_{k=1}^\infty \frac{(2k-1) 2^{2k} \mid B_{2k}\mid}{(2k)!} x^{2k}\right) \left(1+ \sum_{k=0}^{\infty} (-1)^{k}\frac{ 2^{2k} x^{2k}}{(2k+1)!}\right). \end{aligned}\] By Taylor expansions, we may deduce \[\sum_{k=0}^{2p} (-1)^{k}\frac{x^{2k}}{(2k+1)!} < \frac{\sin x}{x} < \sum_{k=0}^{2p+1} (-1)^{k}\frac{x^{2k}}{(2k+1)!}.\] On the other hand, writing \[\sum_{k=1}^\infty \frac{(2k-1) 2^{2k} \mid B_{2k}\mid}{(2k)!} x^{2k} = \sum_{k=1}^p \frac{(2k-1) 2^{2k} \mid B_{2k}\mid}{(2k)!} x^{2k}+ \sum_{k=p+1}^\infty \frac{(2k-1) 2^{2k} \mid B_{2k}\mid}{(2k)!} x^{2k}.\] Combining with inequalities (2), \[\frac{2 (2k)!}{\pi^{2k} (2^{2k}-1)} < \mid B_{2k}\mid < \frac{2^{2k+1} (2k)!}{\pi^{2k} (2^{2k}-2^\beta)} < \frac{2^{2k+1} (2k)!}{\pi^{2k} (2^{2k}-2)}\,.\] It follows for \(0 < x < \pi / 2\) and any \(p \geq 1\), \[\left(\frac{x}{\sin(x)}\right)^2 +\frac{x}{\tan(x)} > \left(1 + \sum_{k=1}^{2p} (-1)^{k}\frac{(2x)^{2k}}{(2k+1)!}\right) \left(1 + \sum_{k=1}^p \frac{(2k-1) 2^{2k} \mid B_{2k}\mid}{(2k)!} x^{2k}+ \sum_{k=p+1}^\infty \frac{ (2k-1) 2^{2k+1} }{\pi^{2k}(2^{2k}-1)} x^{2k}\right),\] as well as \[\begin{aligned} \left(\frac{x}{\sin(x)}\right)^2 +\frac{x}{\tan(x)} &< \left(1 + \sum_{k=1}^{2p+1} (-1)^{k}\frac{(2x)^{2k}}{(2k+1)!}\right) \left(1 + \sum_{k=1}^p \frac{(2k-1) 2^{2k} \mid B_{2k}\mid}{(2k)!} x^{2k} + \sum_{k=p+1}^\infty \frac{(2k-1) 2^{2k+1} }{\pi^{2k}(2^{2k}-2^\beta)} x^{2k}\right) \\ &< \left(1 + \sum_{k=1}^{2p+1} (-1)^{k}\frac{(2x)^{2k}}{(2k+1)!}\right) \left(1 + \sum_{k=1}^p \frac{(2k-1) 2^{2k} \mid B_{2k}\mid}{(2k)!} x^{2k} + \sum_{k=p+1}^\infty \frac{(2k-1) 2^{2k+1} }{\pi^{2k}(2^{2k}-2)} x^{2k}\right). \end{aligned}\] ◻
Corollary 7. For \(0 < x < \pi / 2\) and any \(n \geq p \geq 1\), the following inequalities hold: \[\begin{aligned} a_{p,n}(x)=& \left(1 + \sum_{k=1}^{2p} (-1)^{k}\frac{(2x)^{2k}}{(2k+1)!}\right) \left(1 + \sum_{k=1}^p \frac{(2k-1) 2^{2k} \mid B_{2k}\mid}{(2k)!} x^{2k}+ \sum_{k=p+1}^n \frac{ (2k-1) 2^{2k+1} }{\pi^{2k}(2^{2k}-1)} x^{2k}\right)\\ <& \left(\frac{x}{\sin(x)}\right)^2 +\frac{x}{\tan(x)} < b_{p}(x)= \Big(1 + \sum_{k=1}^{2p+1} (-1)^{k}\frac{(2x)^{2k}}{(2k+1)!}\Big) \left(1 + \sum_{k=1}^p \frac{(2k-1) 2^{2k} \mid B_{2k}\mid}{(2k)!} x^{2k}\right.\\ &\left.+ \frac{2\left(\frac{x}{\pi}\right)^{2p+2}}{(1-\left(\frac{x}{\pi}\right)^{2})^2}\left[(1-2p)\left(\frac{x}{\pi}\right)^{2}+2p+1\right] + \frac{2\left(\frac{x}{2\pi}\right)^{2p+2}}{(1-\left(\frac{x}{2\pi}\right)^{2})^2}\left[(1-2p)\left(\frac{x}{2\pi}\right)^{2}+2p+1\right]\right). \end{aligned}\]
Proof. Let us consider the function \[\frac{1}{1-2^{1-2x}}, \quad x \geq 1.\] For \(1 \leq x\) this function can be majored as;
Lemma 5. For \(1 \leq x\), the following inequality holds: \[\frac{1}{1-2^{1-2x}} \leq (1 + 2^{2-2x}).\]
Indeed, consider the difference \[\alpha(x) = \frac{1}{1-2^{1-2x}} < (1 + 2^{2-2x}).\] Its derivative is \[\frac{d\alpha(x)}{dx} =-2\,{\frac {{2}^{1-2\,x}\ln \left( 2 \right) }{ \left( 1-{2}^{1-2\,x} \right) ^{2}}}+4\,{2}^{1-2\,x}\ln \left( 2 \right) = 2\,{\frac {{2}^{1-2\,x}\ln \left( 2 \right) \left( 1-4.\,{2}^{1-2\,x} +2\, \left( {2}^{1-2\,x} \right) ^{2} \right) }{ \left( -1+{2}^{1-2\,x } \right) ^{2}}}.\] The study suggest us that this function is non positive for \(x \neq 1.\) It is such that \(\alpha(1) =0, \ \lim_{x\rightarrow \infty} \alpha(x) = 0\) and increases between \(1\) and \(\infty.\) Thus, we get for \(1 \leq x,\quad \frac{1}{1-2^{1-2x}} \leq (1 + 4.2^{-2x}).\)
By Lemma 5 we have for any \(k \geq 1\); \[\begin{aligned} \frac{(2k-1) 2^{2k+1} }{\pi^{2k}(2^{2k}-2)} x^{2k} &= 2(2k-1) \frac{1}{1-2^{1-2x}} \left(\frac{x}{\pi}\right)^{2k} \\ &\leq 2(2k-1) (1 + 4. 2^{-2k}) \left(\frac{x}{\pi}\right)^{2k}\\ & = 2(2k-1) \left(\frac{x}{\pi}\right)^{2k} + 8(2k-1) \left(\frac{x}{2\pi}\right)^{2k}. \end{aligned}\] We then deduce the sum, since \(0 < x < \pi / 2\), \[\begin{aligned} \sum_{k=p+1}^\infty \frac{(2k-1) 2^{2k+1} }{\pi^{2k}(2^{2k}-2)} x^{2k} &\leq \sum_{k=p+1}^\infty 2(2k-1) \left(\frac{x}{\pi}\right)^{2k} + 8(2k-1) \left(\frac{x}{2\pi}\right)^{2k}\\ & = \frac{2\left(\frac{x}{\pi}\right)^{2p+2}}{(1-\left(\frac{x}{\pi}\right)^{2})^2}\left[(1-2p)\left(\frac{x}{\pi}\right)^{2}+2p+1\right] + \frac{2\left(\frac{x}{2\pi}\right)^{2p+2}}{(1-\left(\frac{x}{2\pi}\right)^{2})^2}\left[(1-2p)\left(\frac{x}{2\pi}\right)^{2}+2p+1\right]. \end{aligned}\] Replacing in Theorem 6, we get Corollary 7. ◻
Let \(0 < x < \pi / 2\), we have following examples. We will use Maple.
Example 10.
Taking \(n=3\) and \(p=2\), one obtains \[\begin{aligned} a_{2,3}(x)&= 2 +\frac{2 x^4}{45}+\frac{8x^6}{945}\\ & < \left(\frac{x}{\sin(x)}\right)^2 +\frac{x}{\tan(x)}\\ & < b_{2}(x)\\ &= \left( 2-\frac{2x^2}{3}+\frac{2x^4}{15}-{\frac {4x^6}{315}} \right) \left( 1+\frac{x^2}{3}+\frac{{x}^{4}}{15}+{\frac {85}{8}}\,{\frac {{x}^{6}}{{\pi }^{6}}} \right) \\ &< 2 +\frac{2 x^4}{45} +(\frac{85}{4\pi^6}-\frac{4}{315}) x^6\,. \end{aligned}\] The last expression is lower than \[2 +\frac{2 x^4}{45} +\frac{64}{\pi^6}\left(-2+\frac{\pi^2}{4}-\frac{\pi^4}{360}\right) x^6\,,\] since \((\frac{85}{4\pi^6}-\frac{4}{315})-\frac{64}{\pi^6}(-2+\frac{\pi^2}{4}-\frac{\pi^4}{360}) \approx -0.003697.\) Then, this case is finer than Theorem 2(i).
Example 11. Taking \(n=4\) and \(p=3\), one obtains \[\begin{aligned} a_{3,4}(x)=& 2 +\frac{2 x^4}{45}+\frac{8x^6}{945}+\frac{2x^8}{1575}\\ <& \left(\frac{x}{\sin(x)}\right)^2 +\frac{x}{\tan(x)}\\ <& b_{3}(x)\\ =&\Big( 2-\frac{2x^2}{3}+\frac{2x^4}{15}-{\frac {4x^6}{315}} \Big) \left( 1+\frac{x^2}{3}+\frac{x^4}{15}+{\frac {2}{189}}\,{x}^{6 }+{\frac {16\,{x}^{8}{\pi }^{2}-12\,{x}^{10}}{{\pi }^{6} \left( {x}^{2 }-{\pi }^{2} \right) ^{2}}}\right.\\ &\left.+\frac{1}{4}{\frac {16\,{x}^{8}{\pi }^{2}-3\,{x}^ {10}}{{\pi }^{6} \left( {x}^{2}-4\,{\pi }^{2} \right) ^{2}}}-2\,{ \frac {{x}^{8}}{{\pi }^{6} \left( -{x}^{2}+{\pi }^{2} \right) }}+{ \frac {{x}^{8}}{-32\,{\pi }^{8}+8\,{x}^{2}{\pi }^{6}}} \right)\\ <&2+{\frac {2}{45}}\,{x}^{4}+{\frac {8}{945}}\,{x}^{6}+ \left( -{\frac { 34}{14175}}+{\frac {455}{16}}\,{\pi }^{-8} \right) {x}^{8}. \end{aligned}\] The last expression is lower than \[2+{\frac {2}{45}}\,{x}^{4}+{\frac {8}{945}}\,{x}^{6}+ \frac{256}{\pi^8}\left(-2+\frac{\pi^2}{4}-\frac{\pi^4}{360} -\frac{\pi^6}{7560}\right) x^8\,,\] since \(-{\frac {34}{14175}}+{\frac {455}{16}}\,{\pi }^{-8}-\frac{256}{\pi^8}\left(-2+\frac{\pi^2}{4}-\frac{\pi^4}{360} -\frac{\pi^6}{7560}\right) \approx -0.0012804.\) This case is also finer than Theorem 2(ii).
Example 12. Taking \(n=5\) and \(p=4\), one obtains (details omitted) \[\begin{aligned} a_{3,4}(x)&= 2 +\frac{2 x^4}{45}+\frac{8x^6}{945}+\frac{2x^8}{1575}+\frac{16 x^{10}}{93555}\\ & < \left(\frac{x}{\sin(x)}\right)^2 +\frac{x}{\tan(x)} \\ &< b_{3}(x)\\ & <2+{\frac {2}{45}}\,{x}^{4}+{\frac {8}{945}}\,{x}^{6}+{\frac {2}{1575}} \,{x}^{8}+ \left( -{\frac {4}{18711}}+{\frac {2313}{64}}\,{\pi }^{-10} \right) {x}^{10}. \end{aligned}\] The last expression is lower than \[2 +\frac{2 x^4}{45} +\frac{8 x^6}{945}+ \frac{2x^{8}}{1575} +(\frac{2}{\pi})^{10} \left(-2 +\frac{\pi^2}{4}-\frac{\pi^4}{360} – \frac{\pi^6}{7560}-\frac{\pi^8}{201600}\right) x^{10}\,,\] since \(\left( -{\frac {4}{18711}}+{\frac {2313}{64}}\right) – \left(-2 +\frac{\pi^2}{4}-\frac{\pi^4}{360} – \frac{\pi^6}{7560}-\frac{\pi^8}{201600}\right) \approx -0.00007469.\) This case is also finer than Theorem 2(iii).
Examples 1-5 and Examples 6-9 above suggest us
that Corollary 7 is finer than Theorem 4 of [5]. Therefore, we may ask if the
following is valid for \(0 < x < \pi /
2\) and \(n\geq 2\):
(i)\[2+ \frac{1}{\cos x}\sum_{k=2}^{2n+1}
(-1)^k D_k x^{2k} < g_2(x) = \left(\frac{\sin(x)}{x}\right)^2 \left(
1 – \frac{2\left(\frac{2
x}{\pi}\right)^{6}}{1-(\frac{2x}{\pi})^2}\right)
+\frac{\tan(x)}{x}.\] Recall that for \(n \geq 2\), \[g_2(x) < g_n(x) =
\left(\frac{\sin(x)}{x}\right)^2 \left( 1 – \frac{2\left(\frac{2
x}{\pi}\right)^{2n+2}}{1-(\frac{2x}{\pi})^2}\right) +\frac{\tan(x)}{x}
< \left(\frac{\sin(x)}{x}\right)^2 +\frac{\tan(x)}{x}.\]
(ii)\[2+ \frac{1}{\cos
x}\sum_{k=2}^{2n+1} (-1)^k D_k x^{2k} <
\left(\frac{\sin(x)}{x}\right)^2 \left(2 + \sum_{k=1}^p
\frac{2^{2k}(2^{2k}-2)\mid B_{2k}\mid}{(2k)!} x^{2k} +
\frac{2\left(\frac{2 x}{\pi}\right)^{2n+2}}{1-(\frac{2x}{\pi})^2} – 2
\sum_{k=1}^n \frac{1}{2^k-1} \left(\frac{2
x}{\pi}\right)^{2k}\right)\,,\] and
(iii) \[\left(\frac{\sin(x)}{x}\right)^2 \left(2 +
\sum_{k=1}^p \frac{2^{2k}(2^{2k}-2)\mid B_{2k}\mid}{(2k)!} x^{2k} +
\frac{2\left(\frac{2 x}{\pi}\right)^{2n+2}}{1-(\frac{2x}{\pi})^2}\right)
< 2 + \frac{1}{\cos x}\sum_{k=2}^{2n} (-1)^k D_k x^{2k},\]
where \(D_k = \frac{\left(-9 + 3^{2k+2} –
40k- 32k^2\right)}{4 (2k+2)!}.\)
Examples 10-12 suggest us that Corollary 7 is finer than Theorem 5 of [5] and we may naturally ask if more generally the following inequality \[b_m(x) < 2 + \sum_{k=2}^{m-1} \frac{\mid B_{2k}\mid (2k-2) 4^k}{(2k)!}x^{2k} +\left(\frac{2x}{\pi}\right)^{2n}\left(\frac{\pi^2}{4}-2 – \sum_{k=2}^{m-1} \frac{\mid B_{2k}\mid (2k-2) 4^k}{(2k)!}(\frac{\pi}{2})^{2k}\right),\] holds for any \(m \geq 2\), where \[\begin{aligned} b_{p}(x)=& \Big(1 + \sum_{k=1}^{2p+1} (-1)^{k}\frac{(2x)^{2k}}{(2k+1)!}\Big) \Big(1 + \sum_{k=1}^p \frac{(2k-1) 2^{2k} \mid B_{2k}\mid}{(2k)!} x^{2k}\\ &+ \frac{2\left(\frac{x}{\pi}\right)^{2p+2}}{(1-\left(\frac{x}{\pi}\right)^{2})^2}\left[(1-2p)\left(\frac{x}{\pi}\right)^{2}+2p+1\right] + \frac{2\left(\frac{x}{2\pi}\right)^{2p+2}}{(1-\left(\frac{x}{2\pi}\right)^{2})^2}\left[(1-2p)\left(\frac{x}{2\pi}\right)^{2}+2p+1\right]\Big). \end{aligned}\]
Sumner, J. S., Jagers, A. A., Vowe, M., & Anglesio, J. (1991). Inequalities involving trigonometric functions. American Mathematical Monthly, 98(3), 264-267.
Chen, C. P., & Cheung, W. S. (2012). Sharpness of Wilker and Huygens type inequalities. Journal of Inequalities and Applications, 2012, 72, https://doi.org/10.1186/1029-242X-2012-72
Mortici, C. (2011). The natural approach of Wilker-Cusa-Huygens inequalities. Mathematical Inequalities & Applications, 14(3), 535-541.
Male\(\check{{\text s}}\)evic, B., Lutovac, T., Ra\(\check{{\text s}}\)ajski, M., & Mortici, C. (2018). Extensions of the natural approach to refinements and generalizations of some trigonometric inequalities. Advances in Difference Equations, 2018, 90, https://doi.org/10.1186/s13662-018-1545-7
Wu, S. H., & Srivastava, H. M. (2007). A weighted and exponential generalization of Wilker’s inequality and its applications. Integral Transforms and Special Functions, 18(8), 529-535.
Baricz, A., & Sandor, J. (2008). Extensions of the generalized Wilker inequality to Bessel functions. Journal of Mathematical Inequalities, 2(3), 397-406.
Bercu, G. (2016). Padé approximant related to remarkable inequalities involving trigonometric functions. Journal of Inequalities and Applications, 2016, 99, https://doi.org/10.1186/s13660-016-1044-x
Chen, C. P., & Male\(\check{{\text s}}\)evic, B. (2020). Sharp inequalities related to the Adamovic-Mitrinovic, Cusa, Wilker and Huygens results. https://www.researchgate.net/publication/339569555
Chen, C. P., & Paris, R. B. (2020). On the Wilker and Huygens-type inequalities. Journal of Mathematical Inequalities, 14(3), 685-705.
Ra\(\check{{\text s}}\)ajski, M., Lutovac, T., & Male\(\check{{\text s}}\)evic, B. (2018). Sharpening and generalizations of Shafer-Fink and Wilker type inequalities: a new approach. Journal of Nonlinear Sciences & Applications, 11(7), 885–893.
Male\(\check{{\text s}}\)evic, B., Ra\(\check{{\text s}}\)ajski, M., & Lutovac, T. (2019). Double–Sided Taylor’s Approximations and Their Applications in Theory of Analytic Inequalities. Differential and Integral Inequalities, 151, 569-582.
Gradshteyn, I. S., & Ryzhik, I. M. (2015). Table of Integrals, Series, and Products, 8th edn., edited by D. Zwillinger.
Jeffrey, A., & Dai, H. H. (2008). Handbook of Mathematical Formulas and Integrals. Elsevier.
