By Theorem 2, we have
$$ 1+\frac{1}{3p_{0}}\ln \bigl( 1-p_{0}t^{2} \bigr) < \frac{t}{\tan t}< 1+\frac {5}{2}\ln \biggl( 1-\frac{2}{15}t^{2} \biggr) , $$
(4.1)
where \(p_{0}\approx0.13484\).
We denote the lower bounds for \(t/\tan t\) given in the inequalities (4.1), (1.2), (1.3), (1.4), (1.5), and (1.7), respectively, by
$$\begin{aligned}& LY ( t ) =1+\frac{1}{3p_{0}}\ln \bigl( 1-p_{0}t^{2} \bigr) , \qquad ZS ( t ) =\frac{\pi^{2}-4t^{2}}{\pi^{2}+4t^{2}},\qquad Z ( t ) = \biggl( \frac{\pi^{2}-4t^{2}}{\sqrt{\pi^{4}+48t^{4}}} \biggr) ^{\pi^{2}/6}, \\& BS_{1} ( t ) =1-\frac{4}{\pi^{2}}t^{2},\qquad ZH ( t ) =\frac{\pi^{2}-4t^{2}}{\pi^{2}+ ( \pi^{2}/3-4 ) t^{2}}, \qquad BS_{2} ( t ) =\frac{\pi^{2}-4t^{2}-\pi ^{2}t^{2}/3}{\pi^{2}-t^{2}}. \end{aligned}$$
Proposition 1
The comparison inequalities
$$ LY ( t ) >ZH ( t ) >BS_{1} ( t ) >\max \bigl( ZS ( t ) ,Z ( t ) ,BS_{2} ( t ) \bigr) $$
(4.2)
hold for
\(t\in ( 0,\pi/2 ) \). Moreover, \(ZS ( t ) \), \(Z ( t ) \)
and
\(BS_{2} ( t ) \)
are not comparable with each other for all
\(t\in ( 0,\pi/2 ) \).
Proof
(i) We first prove
$$D_{1} ( x ) =LY ( \sqrt{x} ) -ZH ( \sqrt{x} ) =1+\frac{1}{3p_{0}} \ln ( 1-p_{0}x ) -\frac{\pi^{2}-4x}{\pi ^{2}+ ( \pi^{2}/3-4 ) x}>0 $$
for \(x\in ( 0,\pi^{2}/4 ) \). Differentiation yields
$$D_{1}^{\prime} ( x ) =-\frac{ ( 12-\pi^{2} ) ^{2}x}{27 ( 1-xp_{0} ) ( \pi^{2}+ ( \pi^{2}/3-4 ) x ) ^{2}} \biggl( x- \frac{72\pi^{2}-6\pi^{4}-9\pi^{4}p_{0}}{ ( 12-\pi^{2} ) ^{2}} \biggr) , $$
which shows that \(D_{1}\) is increasing on \(( 0,x_{0} ) \) and decreasing on \(( x_{0},\pi^{2}/4 ) \), where
$$x_{0}=\frac{72\pi^{2}-6\pi^{4}-9\pi^{4}p_{0}}{ ( 12-\pi^{2} ) ^{2}}\approx1.75059. $$
Then we conclude that \(D_{1} ( x ) >\min ( D_{1} ( 0 ) ,D_{1} ( \pi^{2}/4 ) ) =0\) with \(D_{1} ( \pi^{2}/4 ) =0\) due to \(p_{0}\) satisfying \(\alpha_{p_{0}} ( \pi/2 ) =1\) shown in Lemma 5.
(ii) The second inequality directly follows from
$$\begin{aligned} ZH ( t ) -BS_{1} ( t ) & =\frac{\pi^{2}-4t^{2}}{\pi^{2}+ ( \pi^{2}/3-4 ) t^{2}}- \biggl( 1- \frac{4}{\pi^{2}}t^{2} \biggr) \\ & =\frac{1}{3}\frac{t^{2} ( 12-\pi^{2} ) ( \pi^{2}-4t^{2} ) }{\pi^{2} ( \pi^{2}+ ( \pi^{2}/3-4 ) t^{2} ) }>0 \end{aligned}$$
for \(t\in ( 0,\pi/2 ) \).
(iii) The third one is deduced by
$$\begin{aligned}& BS_{1} ( t ) -ZS ( t ) = \biggl( 1-\frac{4}{\pi^{2}}t^{2} \biggr) -\frac{\pi^{2}-4t^{2}}{\pi^{2}+4t^{2}}=\frac{4t^{2} ( \pi^{2}-4t^{2} ) }{\pi^{2} ( \pi^{2}+4t^{2} ) }>0, \\& \frac{BS_{1} ( t ) -Z ( t ) }{\pi^{2}-4t^{2}}=\frac {1}{\pi^{2}}-\frac{ ( \pi^{2}-4t^{2} ) ^{\pi^{2}/6-1}}{ ( \sqrt{\pi^{4}+48t^{4}} ) ^{\pi^{2}/6}}>\frac{1}{\pi^{2}}- \frac { ( \pi^{2} ) ^{\pi^{2}/6-1}}{ ( \sqrt{\pi^{4}} ) ^{\pi^{2}/6} }=0, \\& BS_{1} ( t ) -BS_{2} ( t ) = \biggl( 1- \frac{4}{\pi^{2} }t^{2} \biggr) -\frac{\pi^{2}-4t^{2}-\pi^{2}t^{2}/3}{\pi ^{2}-t^{2}}=\frac {4}{\pi^{2}} \frac{t^{2} ( 12t^{2}+\pi^{4}-3\pi^{2} ) }{ ( \pi^{2}-t^{2} ) }>0, \end{aligned}$$
for \(t\in ( 0,\pi/2 ) \).
(iv) Finally, we prove that \(ZS ( t ) \), \(Z ( t ) \) and \(BS_{2} ( t ) \) are not comparable with each other for all \(t\in ( 0,\pi/2 ) \). Simple computations yield
$$\begin{aligned}& \lim_{t\rightarrow0^{+}}\frac{ZS ( t ) -Z ( t ) }{t^{2}} =\lim_{t\rightarrow0^{+}}t^{-2} \biggl( \frac{\pi^{2}-4t^{2}}{\pi^{2}+4t^{2}}- \biggl( \frac{\pi^{2}-4t^{2}}{\sqrt{\pi^{4}+48t^{4}}} \biggr) ^{\pi^{2}/6} \biggr) =\frac{2}{3}\frac{\pi^{2}-12}{\pi^{2}}< 0, \\& \lim_{t\rightarrow ( \pi/2 ) ^{-}}\frac{ZS ( t ) -Z ( t ) }{\pi^{2}-4t^{2}} =\lim_{t\rightarrow ( \pi/2 ) ^{-}} \biggl( \frac{1}{\pi^{2}+4t^{2}}-\frac{ ( \pi ^{2}-4t^{2} ) ^{\pi^{2}/6-1}}{ ( \sqrt{\pi^{4}+48t^{4}} ) ^{\pi^{2}/6}} \biggr) =\frac{1}{2\pi^{2}}>0, \\& \begin{aligned} \lim_{t\rightarrow0^{+}}\frac{Z ( t ) -BS_{2} ( t ) }{t^{2}} & =\lim_{t\rightarrow0^{+}}t^{-2} \biggl( \biggl( \frac{\pi ^{2}-4t^{2}}{\sqrt{\pi^{4}+48t^{4}}} \biggr) ^{\pi^{2}/6}-\frac{\pi ^{2}-4t^{2}-\pi^{2}t^{2}/3}{\pi^{2}-t^{2}} \biggr)\\ & =-\frac{\pi ^{2}-9}{3\pi ^{2}}< 0, \end{aligned} \\& \begin{aligned} \lim_{t\rightarrow ( \pi/2 ) ^{-}} \bigl[ Z ( t ) -BS_{2} ( t ) \bigr] & = \lim_{t\rightarrow ( \pi/2 ) ^{-}} \biggl( \biggl( \frac{\pi^{2}-4t^{2}}{\sqrt{\pi^{4}+48t^{4}}} \biggr) ^{\pi^{2}/6}-\frac{\pi^{2}-4t^{2}-\pi^{2}t^{2}/3}{\pi^{2}-t^{2}} \biggr)\\ & =\frac{1}{9} \pi^{2}>0, \end{aligned} \\& \begin{aligned} ZS ( t ) -BS_{2} ( t ) & =\frac{\pi^{2}-4t^{2}}{\pi^{2}+4t^{2}}-\frac{\pi^{2}-4t^{2}-\pi^{2}t^{2}/3}{\pi^{2}-t^{2}} \\ & =\frac{1}{3}t^{2}\frac{4 ( \pi^{2}+15 ) }{ ( \pi^{2}-t^{2} ) ( \pi^{2}+4t^{2} ) } \biggl( t^{2}-\frac{\pi ^{2} ( 15-\pi^{2} ) }{4 ( 15+\pi^{2} ) } \biggr) \\ & \left\{ \textstyle\begin{array}{l@{\quad}l}< 0 & \text{if }0< t< \frac{\pi}{2}\sqrt{\frac{15-\pi^{2}}{15+\pi^{2}}},\\ >0 & \text{if }\frac{\pi}{2}\sqrt{\frac{15-\pi^{2}}{15+\pi^{2}}}< t< \frac {\pi }{2}. \end{array}\displaystyle \right . \end{aligned} \end{aligned}$$
This completes the proof. □
Remark 2
From the above proposition we see that the sharp lower bound in (4.1) is superior to those ones given in (1.2), (1.3), (1.4), (1.5), and (1.7).
Remark 3
Analogously, by comparing the limits at \(t=0\) and \(t=\pi/2\), we find the sharp upper bound in (4.1) is not comparable with those ones given in (1.2), (1.3), (1.4), (1.5), and (1.7). Here we omit all the details.
Remark 4
We claim that the result stated in Theorem 2 is stronger than the inequality (1.8), that is, for \(t\in ( 0,\pi/2 ) \), we have the inequalities
$$ 1-\frac{4}{\pi^{2}}t^{2}< 1+\frac{1}{3p_{0}}\ln \bigl( 1-p_{0}t^{2} \bigr) < \frac{t}{\tan t}< 1+ \frac{5}{2}\ln \biggl( 1-\frac{2}{15}t^{2} \biggr) < 1- \frac{t^{2}}{3} . $$
(4.3)
Indeed, the right hand side for this inequality in (4.3) follows from Corollary 1, while the left hand side one is the inequality connecting the first and third bounds in (4.2).
Remark 5
Lemma 4 tells us that
$$\begin{aligned}& \frac{2}{15} < \frac{3t-3\cos t\sin t-2t\sin^{2}t}{3t^{2} ( t-\sin t\cos t ) }< \frac{4}{3\pi^{2}}\quad\text{for }t \in ( 0,\pi /2 ) , \end{aligned}$$
(4.4)
$$\begin{aligned}& \frac{1}{\pi^{2}} < \frac{3t-3\cos t\sin t-2t\sin^{2}t}{3t^{2} ( t-\sin t\cos t ) }< \frac{4}{3\pi^{2}}\quad\text{for }t \in ( \pi /2,\pi ) . \end{aligned}$$
(4.5)
Then from equation (3.7) we find that for \(f^{\prime} ( t ) <0\) for \(t\in ( 0,\pi ) \) when \(p=1/\pi^{2}\), and so \(f ( t ) < f ( 0^{+} ) =0\). This gives the following inequality:
$$\frac{t\cos t}{\sin t}-1-\frac{\pi^{2}}{3}\ln \biggl( 1-\frac{t^{2}}{\pi ^{2}} \biggr) < 0 $$
for all \(t\in ( 0,\pi ) \), which can be stated as the following proposition.
Proposition 2
For all
\(t\in ( 0,\pi ) \), we have
$$\frac{t}{\tan t}< 1+\frac{\pi^{2}}{3}\ln \biggl( 1-\frac{t^{2}}{\pi^{2}} \biggr) . $$
Remark 6
The inequality
$$\frac{\sin t}{t}< \frac{2+\cos t}{3},\quad t\in \biggl( 0,\frac{\pi }{2} \biggr), $$
is true due to Cusa and Huygens’ paper (see, e.g. [19]), which is now known as Cusa’s inequality (see e.g. [8, 20–23]). Some refinements and generalizations of Cusa’s inequality can be found in [8, 21, 22, 24–29]. Now by letting \(t=x/2\) and simplifying, inequalities (4.4) and (4.5) can be written as
$$\begin{aligned}& \frac{x ( x-\sin x ) }{30} < \frac{2+\cos x}{3}-\frac{\sin x}{x}< \frac{x ( x-\sin x ) }{3\pi^{2}}\quad\text{for }x\in ( 0,\pi ) , \\& \frac{x ( x-\sin x ) }{4\pi^{2}} < \frac{2+\cos x}{3}-\frac {\sin x}{x}< \frac{x ( x-\sin x ) }{3\pi^{2}}\quad\text{for }x\in ( \pi,2\pi ) , \end{aligned}$$
which give stronger versions of Cusa’s inequality.
Proposition 3
We have
$$\begin{aligned}& \frac{2+\cos x}{3}-\frac{x ( x-\sin x ) }{3\pi^{2}} < \frac {\sin x}{x}< \frac{2+\cos x}{3}-\frac{x ( x-\sin x ) }{30}\quad\textit {for }x\in ( 0,\pi ) , \end{aligned}$$
(4.6)
$$\begin{aligned}& \frac{2+\cos x}{3}-\frac{x ( x-\sin x ) }{3\pi^{2}} < \frac {\sin x}{x}< \frac{2+\cos x}{3}-\frac{x ( x-\sin x ) }{4\pi^{2}}\quad \textit{for }x\in ( \pi,2\pi ) . \end{aligned}$$
(4.7)
Moreover, the two double inequalities are sharp.
Remark 7
In [12], Corollary 12, Yang et al. proved that, for \(t\in ( 0,\pi/2 ) \),
$$\sqrt{\exp \biggl( \frac{t}{\tan t}-1 \biggr) }< \frac{\sin t}{t}< \exp \biggl( \frac{t}{\tan ( t/2 ) }-2 \biggr) . $$
Then by inequalities (3.10) for \(p=4/ ( 3\pi^{2} ) \) and \(q=2/15\), we obtain
$$\biggl( 1-\frac{4t^{2}}{3\pi^{2}} \biggr) ^{\pi^{2}/8}< \sqrt{\exp \biggl( \frac{t}{\tan t}-1 \biggr) }< \frac{\sin t}{t}< \exp^{2} \biggl( \frac{t/2}{\tan ( t/2 ) }-1 \biggr) < \biggl( 1-\frac{t^{2}}{30} \biggr) ^{5} $$
for \(t\in ( 0,\pi/2 ) \). Further, the right hand side inequalities can be improved as follows.
Proposition 4
The inequalities
$$ \begin{aligned}[b]\rho_{r} \bigl( 1-rt^{2} \bigr) ^{1/ ( 6r ) }&< \lambda_{s}\exp \biggl( \frac{st}{\tan ( st ) }-1 \biggr) \\ &< \frac{\sin t}{t}< \exp \biggl( \frac{st}{\tan ( st ) }-1 \biggr) < \bigl( 1-rt^{2} \bigr) ^{1/ ( 6r ) }\end{aligned} $$
(4.8)
hold for
\(t\in ( 0,\pi/2 ) \)
with the best constants
\(s=1/\sqrt {2}\), \(r=1/15\)
and
$$\begin{aligned}& \lambda_{s} =\frac{2}{\pi\exp ( \sqrt{2}\pi\cot ( \sqrt{2}\pi/4 ) -1 ) }\approx0.99801, \\& \rho_{r} =\frac{2}{\pi ( 1-\pi^{2}/60 ) ^{5/2}}\approx 0.99771. \end{aligned}$$
Proof
Let
$$h ( t ) =\frac{st\cos ( st ) }{\sin ( st ) }-1-\ln\frac{\sin t}{t}, $$
where \(s=1/\sqrt{2}\). Differentiation yields
$$h^{\prime} ( t ) =\frac{1}{t}-\frac{\cos t}{\sin t}+s \frac{\cos st}{\sin st}-s^{2}\frac{t}{\sin^{2}st}. $$
Expanding in power series leads to
$$h^{\prime} ( t ) =\sum_{n=1}^{\infty} \bigl( 1-2ns^{2n} \bigr) \frac{2^{2n}}{ ( 2n ) !}|B_{2n}|t^{2n-1}= \sum_{n=1}^{\infty } \bigl( 2^{n}-2n \bigr) \frac{2^{n}}{ ( 2n-1 ) !}|B_{2n}|t^{2n-1}>0. $$
This indicates that \(h ( \pi/2 ) >h ( t ) >h ( 0^{+} ) =0\) for \(t\in ( 0,\pi/2 ) \), which proves the second and third inequalities of (4.8). Considering the limit
$$\lim_{t\rightarrow0}\frac{h ( t ) }{t^{2}}=\lim_{t\rightarrow 0} \frac{\frac{st\cos ( st ) }{\sin ( st ) }-1-\ln \frac{\sin t}{t}}{t^{2}}=\frac{1}{3} \biggl( \frac{1}{2}-s^{2} \biggr) , $$
it is seen that \(s=1/\sqrt{2}\) and \(\lambda_{s}\) are the best possible constants.
The first and fourth ones are derived from the decreasing property of \(f ( st ) \equiv f ( u ) \) for \(u\in ( 0,s\pi/2 ) \subset ( 0,\pi/2 ) \) proved in Theorem 1 for \(p=r/s^{2}=2/15\), and then \(r=1/15\) and \(\rho_{r}\) are also the best. This completes the proof. □