1 Introduction

Trigonometric inequalities have caused interest of a lot of researchers, they analyzed the Wilker inequality [6–11, 14, 16–19], Jordan inequality [3, 5, 15, 20, 21], Shafer–Fink inequality [12], Becker–Stark inequalities [13], and so on.

Recently, Bercu provided a Padé-approximant-based method and obtained the following inequalities [2].

$$\begin{aligned} &b_{1}(x)= \frac{-7 x^{2} + 60}{3 x^{2} + 60} < \frac{\sin(x)}{x} < \frac{11 x^{4} - 360 x^{2}+ 2520}{60 x^{2}+ 2520}= b_{2}(x),\quad \forall x \in (0,\pi/2); \end{aligned}$$
(1)
$$\begin{aligned} & \begin{aligned}[b] b_{3}(x) &=\frac{17x^{4} - 480x^{2}+ 1080}{2x^{4}+ 60x^{2}+ 1080} < \cos(x) < \frac{3x^{4} - 56x^{2}+ 120}{4x^{2}+ 120} \\ &= b_{4}(x),\quad \forall x \in (0,\pi/2); \end{aligned} \end{aligned}$$
(2)
$$\begin{aligned} & b_{5}(x) < \frac{\tan(x)}{x}< b_{6}(x),\quad \forall x \in (0,1.5701); \end{aligned}$$
(3)
$$\begin{aligned} & \biggl(\frac{x}{\sin(x)}\biggr)^{2}+\frac{x}{\tan(x)} > b_{7}(x),\quad \forall x \in (0,1.5701), \end{aligned}$$
(4)

where \(b_{5}(x)=\frac{-28x^{4} - 600x^{2}+ 7200}{9x^{6}+ 12x^{4} - 3000x^{2}+ 7200}\), \(b_{6}(x)=\frac{22x^{8} - 60x^{6} - 4680x^{4}- 237\mbox{,}600x^{2}+ 2\mbox{,}721\mbox{,}600}{1020x^{6}+ 14\mbox{,}040x^{4}- 1\mbox{,}144\mbox{,}800x^{2}+ 2\mbox{,}721\mbox{,}600}\) and \(b_{7}(x)=\frac{11\mbox{,}220x^{10}-205\mbox{,}560x^{8}-14\mbox{,}256\mbox{,}000x^{6}+512\mbox{,}179\mbox{,}200x^{4}- 3\mbox{,}157\mbox{,}056\mbox{,}000x^{2}+13\mbox{,}716\mbox{,}864\mbox{,}000}{242x^{12}-8580x^{10} +25\mbox{,}560x^{8}-1\mbox{,}080\mbox{,}000x^{6}+103\mbox{,}680\mbox{,}000x^{4}-1\mbox{,}578\mbox{,}528\mbox{,}000x^{2}+6\mbox{,}858\mbox{,}432\mbox{,}000}\).

In this paper, we present a two-point Padé-approximant-based method [1] for refining the rational bounds of several trigonometric inequalities, and also provide a method for proving the refined bounds. By applying the new method to \(\frac{\sin (x)}{x}\) and \(\cos(x)\), we refine the bounds of Eq. (1) ∼ (2), for \(\forall x \in [0,\pi/2]\), see also Theorems 3.1 and 3.2. Applied to \(\frac{\tan (x)}{x}\) and \((\frac{x}{\sin(x)})^{2}+\frac{x}{\tan(x)}\), it not only provides refined two-sided bounds with better approximation effect for Eq. (3) ∼ (4), but also extends the interval \((0,1.5701)\) to the interval \([0,\pi/2]\), see also the theorems and remarks in Sect. 3.

2 Find bounds by using two-point Padé approximant

Given a bounded smooth function \(f(x)\), \(x \in [x_{0}, x_{1}]\), let \(R(x)=\frac{\sum^{n}_{i=0} c_{i} x^{i}}{1+\sum^{m}_{i=1} d_{i} x^{i}}\) be a rational polynomial interpolating derivatives of \(f(x)\) at two points \(x_{0}\) and \(x_{1}\) such that

$$ E^{(i)}(x_{0})=0, \qquad E^{(j)}(x_{1})=0,\quad i=0,1, \ldots,k, j=0,1,\ldots,l, $$
(5)

where \(E(x)=(1+\sum^{m}_{i=1}d_{i} x^{i}) \cdot f(x) -(\sum^{n}_{i=1}c_{i} x^{i})\). There are \(m+n+2\) unknowns in Eq. (5). By selecting suitable values of k and l, we have that Eq. (5) consists of \(m+n+2\) linear equations in the unknown variables \(c_{i}\) and \(d_{j}\), and the interpolation polynomial \(R(x)\) can be determined by solving Eq. (5).

We give two examples. Without loss of generality, let \(\Gamma=[0,\pi/2]\).

Example 1

Let \(f_{1}(x)=\sin(x)\). By setting \(n_{1}=13\), \(m_{1}=0\), \(n_{2}=11\), and \(m_{2}=0\) and introducing the following constraints

$$ f_{1}^{(i)}(0)=R_{j}^{(i)}(0),\qquad f_{1}(\pi/2)=R_{j}(\pi/2),\quad j=1,2, i=0,1,\ldots,14-2j, $$
(6)

we obtain that

$$ R_{1}(x)= \beta_{1}(x)+ \alpha_{1} \cdot x^{13}, \qquad R_{2}(x)= \beta_{2}(x)- \alpha_{2} \cdot x^{11}, $$
(7)

where \(\alpha_{1}=\frac{\pi^{11}-440 \pi^{9}+126\mbox{,}720 \pi^{7}-21\mbox{,}288\mbox{,}960 \pi^{5}+1\mbox{,}703\mbox{,}116\mbox{,}800 \pi^{3}-40\mbox{,}874\mbox{,}803\mbox{,}200 \pi+81\mbox{,}749\mbox{,}606\mbox{,}400}{9\mbox{,}979\mbox{,}200 \pi^{13}}\), \(\beta_{1}(x)=t-\frac{t^{3}}{6}+ \frac{t^{5}}{120} - \frac{t^{7}}{5040} + \frac{t^{9}}{362\mbox{,}880}-\frac{x^{11}}{39\mbox{,}916\mbox{,}800}\), \(\alpha_{2}=\frac{\pi^{9}-288 \pi^{7}+48\mbox{,}384 \pi^{5}-3\mbox{,}870\mbox{,}720 \pi^{3} +92\mbox{,}897\mbox{,}280 \pi-185\mbox{,}794\mbox{,}560}{90\mbox{,}720 \pi^{11}} \), \(\beta_{2}(x)=t-\frac{t^{3}}{6}+ \frac{t^{5}}{120} - \frac{t^{7}}{5040} + \frac{t^{9}}{362\mbox{,}880}\). It can be verified that \(R_{j}(x) \geq 0, \forall x \in \Gamma, j=1,2\). From Eq. (6), \(\forall x \in \Gamma\), there exists \(\xi_{j}(x) \in \Gamma\) such that [4]

$$ f_{1}(x) - R_{j}(x) = \frac{f_{1}^{(16-2j)}(\xi_{j}(x))}{(16-2j)!} \cdot (x-\pi/2) \cdot x^{15-2j}, \quad x \in \Gamma, j=1,2. $$
(8)

Note that \(f_{1}^{(14)}(x) = -\sin(x) \leq 0\) and \(f_{1}^{(12)}(x) = \sin(x) \geq 0\), \(\forall x \in \Gamma\). Combining with Eq. (8), one obtains that

$$ 0 \leq R_{1}(x) \leq \sin(x) \leq R_{2}(x), \quad x \in \Gamma. $$
(9)

Example 2

Let \(f_{2}(x)=\cos(x)\). By setting \(n_{3}=12\), \(m_{3}=0\), \(n_{4}=10\), and \(m_{4}=0\) and introducing the following constraints

$$ f_{2}^{(i)}(0)=R_{j}^{(i)}(0),\qquad f_{2}(\pi/2)=R_{j}(\pi/2),\quad j=3,4, i=0,1,\ldots,17-2j, $$
(10)

we obtain that

$$ R_{3}(x)= \beta_{3}(x)+ \alpha_{3} \cdot x^{12} , \qquad R_{4}(x)= \beta_{4}(x)- \alpha_{4} \cdot x^{10}, $$
(11)

where \(\alpha_{3}=\frac{\pi^{10}-360 \pi^{8}+80\mbox{,}640 \pi^{6}-9\mbox{,}676\mbox{,}800 \pi^{4}+464\mbox{,}486\mbox{,}400 \pi^{2}-3\mbox{,}715\mbox{,}891\mbox{,}200}{907\mbox{,}200 \pi^{12}}\), \(\beta_{3}(x)=1-\frac{x^{2}}{2} +\frac{x^{4}}{24} -\frac{x^{6}}{720} +\frac{x^{8}}{40\mbox{,}320} -\frac{x^{10}}{3\mbox{,}628\mbox{,}800}\), \(\alpha_{4}=\frac{10\mbox{,}321\mbox{,}920-1\mbox{,}290\mbox{,}240 \pi^{2}+26\mbox{,}880 \pi^{4}-224 \pi^{6}+\pi^{8}}{10\mbox{,}080 \pi^{10}} \), \(\beta_{4}(x)=1-\frac{x^{2}}{2} +\frac{x^{4}}{24} -\frac{x^{6}}{720} +\frac{x^{8}}{40\mbox{,}320} \). It can be verified that \(R_{j}(x) \geq 0, \forall x \in \Gamma, j=3,4\). From Eq. (10), \(\forall x \in \Gamma\), there exists \(\xi_{j}(x) \in \Gamma, j=3,4\), such that [4]

$$ f_{2}(x) - R_{j}(x) = \frac{f_{2}^{(19-2j)}(\xi_{j}(x))}{(19-2j)!} \cdot (x-\pi/2) \cdot x^{18-2j}, \quad x \in \Gamma, j=3,4. $$
(12)

Note that \(f_{2}^{(13)}(x) = -\sin(x) \leq 0\) and \(f_{2}^{(11)}(x) = \sin(x) \geq 0\), \(\forall x \in \Gamma\). Combining with Eq. (12), one obtains that

$$ 0 \leq R_{3}(x) \leq \cos(x) \leq R_{4}(x), \quad x \in \Gamma. $$
(13)

3 Main results

The main results are as follows.

Theorem 3.1

For all \(\forall x \in \Gamma=[0,\pi/2]\), we have that

$$\begin{aligned}{} [b] c_{1}(x) &= \frac{60\mbox{,}480-9240 x^{2}+364 x^{4}-5 x^{6} }{840 (72+x^{2})} \leq \frac{\sin(x)}{x} \\ &\leq \frac{ (166\mbox{,}320-22\mbox{,}260 x^{2}+551 x^{4}) }{15 (11\mbox{,}088+364 x^{2}+5 x^{4})} =c_{2}(x). \end{aligned}$$
(14)

Proof

Eq. (14) is equivalent to

$$ \textstyle\begin{cases} (60\mbox{,}480-9240 x^{2}+364 x^{4}-5 x^{6}) x - 840 (72+x^{2}) \sin(x) \leq 0, \\ (166\mbox{,}320-22\mbox{,}260 x^{2}+551 x^{4}) x - 15 (11\mbox{,}088+364 x^{2}+5 x^{4}) \sin(x) \geq 0, \end{cases}\displaystyle \forall x \in \Gamma. $$
(15)

It is well known that \(\forall x \in \Gamma\),

$$\begin{aligned} \beta_{1}(x)&=t-\frac{t^{3}}{6}+ \frac{t^{5}}{120} - \frac{t^{7}}{5040} + \frac{t^{9}}{362\mbox{,}880}-\frac{x^{11}}{39\mbox{,}916\mbox{,}800} \\ &\leq \sin(x) \leq \beta_{1}(x) + \frac{x^{13}}{6\mbox{,}227\mbox{,}020\mbox{,}800}. \end{aligned}$$
(16)

Combining with Eq. (16), we have that

$$\begin{aligned} &\bigl(60\mbox{,}480-9240 x^{2}+364 x^{4}-5 x^{6}\bigr) x - 840 \bigl(72+x^{2}\bigr) \sin(x) \\ &\quad \leq \bigl(60\mbox{,}480-9240 x^{2}+364 x^{4}-5 x^{6}\bigr) x - 840 \bigl(72+x^{2}\bigr) \beta_{1}(x) \\ &\quad =\frac{x^{11} \cdot (-38+x^{2})}{39\mbox{,}916\mbox{,}800} \leq 0, \quad \forall x \in \Gamma, \\ & \bigl(166\mbox{,}320-22\mbox{,}260 x^{2}+551 x^{4}\bigr) x - 15 \bigl(11\mbox{,}088+364 x^{2}+5 x^{4}\bigr) \sin(x) \\ &\quad \geq \bigl(166\mbox{,}320-22\mbox{,}260 x^{2}+551 x^{4}\bigr) x\\ &\qquad {} - 15 \bigl(11\mbox{,}088+364 x^{2}+5 x^{4}\bigr) \biggl( \beta_{1}(x) + \frac{x^{13}}{6\mbox{,}227\mbox{,}020\mbox{,}800}\biggr) \\ &\quad = \frac{x^{11}}{6\mbox{,}227\mbox{,}020\mbox{,}800} \bigl(1\mbox{,}661\mbox{,}088-40\mbox{,}104 x^{2}+ 416 x^{4}- 5 x^{6}\bigr) \\ &\quad \geq \frac{x^{11}}{6\mbox{,}227\mbox{,}020\mbox{,}800} \bigl(1\mbox{,}661\mbox{,}088-40\mbox{,}104 \cdot 2^{2} - 5 \cdot 2^{6}\bigr) \geq 0,\quad \forall x \in \Gamma, \end{aligned}$$

which is just Eq. (15). So we have completed the proof of Eq. (14). □

Theorem 3.2

For all \(\forall x \in [0,\pi/2]\), we have that

$$ \begin{aligned}[b] c_{3}(x)&= \frac{ 20\mbox{,}160-9720 x^{2}+ 660 x^{4}-13 x^{6}}{360 (x^{2}+56)} \leq \cos(x) \\ & \leq \frac{15\mbox{,}120-6900 x^{2}+313 x^{4}}{15\mbox{,}120+660 x^{2}+13 x^{4}} =c_{4}(x). \end{aligned} $$
(17)

Proof

Eq. (17) is equivalent to

$$ \textstyle\begin{cases} (20\mbox{,}160-9720 x^{2}+ 660 x^{4}-13 x^{6})- 360 (x^{2}+56) \cos(x) \leq 0, \\ (15\mbox{,}120-6900 x^{2}+313 x^{4}) - (15\mbox{,}120+660 x^{2}+13 x^{4}) \cos(x) \geq 0, \end{cases}\displaystyle \forall x \in \Gamma. $$
(18)

It is well known that

$$ \begin{aligned}[b] \beta_{3}(x)&=1-\frac{x^{2}}{2} + \frac{x^{4}}{24} -\frac{x^{6}}{720} +\frac{x^{8}}{40\mbox{,}320} -\frac{x^{10}}{3\mbox{,}628\mbox{,}800} \leq \cos(x) \\ & \leq 1-\frac{x^{2}}{2} +\frac{x^{4}}{24} -\frac{x^{6}}{720} + \frac{x^{8}}{40\mbox{,}320} = \beta_{4}(x), \quad \forall x \in \Gamma. \end{aligned} $$
(19)

Combining with Eq. (19), we have that

$$ \textstyle\begin{cases} (20\mbox{,}160-9720 x^{2}+ 660 x^{4}-13 x^{6})- 360 (x^{2}+56) \cos(x) \\ \quad \leq (20\mbox{,}160-9720 x^{2}+ 660 x^{4}-13 x^{6})- 360 (x^{2}+56) \beta_{3}(x) \\ \quad = \frac{x^{10}}{3\mbox{,}628\mbox{,}800} (-34+x^{2}) \leq 0, \quad \forall x \in [0,\pi/2], \\ (15\mbox{,}120-6900 x^{2}+313 x^{4}) - (15\mbox{,}120+660 x^{2}+13 x^{4}) \cos(x) \\ \quad \geq (15\mbox{,}120-6900 x^{2}+313 x^{4}) - (15\mbox{,}120+660 x^{2}+13 x^{4}) \beta_{4}(x) \\ \quad =\frac{x^{10}}{40\mbox{,}320} (68 -13 x^{2}) \geq 0, \quad \forall x \in [0,\pi/2]. \end{cases} $$
(20)

Thus, we have completed the proof of both Eq. (18) and Eq. (17). □

Theorem 3.3

For all \(\forall x \in \Gamma\), we have that

$$ \begin{aligned}[b] c_{5}(x)&= \frac{ 21(495-60 x^{2}+x^{4})}{10\mbox{,}395-4725 x^{2}+210 x^{4}-x^{6}} \leq \frac{\tan(x)}{x} \\ &\leq \frac{T_{1}(x)}{105 (\pi^{2}-4 x^{2}) \cdot T_{2}(x)} =c_{6}(x), \end{aligned} $$
(21)

where \(T_{1}(x)=(\pi^{6}-840 \pi^{4}+75\mbox{,}600 \pi^{2}-665\mbox{,}280) x^{6} + (210 \pi^{6}+52\mbox{,}920 \pi^{4}-7\mbox{,}620\mbox{,}480 \pi^{2}+69\mbox{,}854\mbox{,}400) x^{4} + (-17\mbox{,}955 \pi^{6}+1\mbox{,}323\mbox{,}000 \pi^{4}+52\mbox{,}390\mbox{,}800 \pi^{2}-628\mbox{,}689\mbox{,}600) x^{2} + (155\mbox{,}925 (\pi^{4}-112 \pi^{2}+1008)) \pi^{2}\) and \(T_{2}(x) = (26 \pi^{4}-2664 \pi^{2}+23\mbox{,}760) x^{4} + (-666 \pi^{4}+73\mbox{,}980 \pi^{2}-665\mbox{,}280) x^{2} + (1485 \pi^{4}-166\mbox{,}320 \pi^{2}+1\mbox{,}496\mbox{,}880)\).

Proof

Eq. (21) is equivalent to

$$ \textstyle\begin{cases} \begin{aligned} H_{5}(x) ={}& 21(495-60 x^{2}+x^{4}) \cdot x \cos(x) \\ &{} - (10\mbox{,}395-4725 x^{2}+210 x^{4}-x^{6}) \cdot \sin(x) \leq 0; \end{aligned} \\ H_{6}(x) =105 (\pi^{2}-4 x^{2}) \cdot T_{2}(x) \cdot \sin(x) -T_{1}(x) \cdot x \cos(x) \leq 0, \end{cases}\displaystyle \forall x \in \Gamma. $$
(22)

It can be verified that

$$ \textstyle\begin{cases} \cos(x) \leq 1-\frac{x^{2}}{2} +\frac{x^{4}}{24} -\frac{x^{6}}{720} +\frac{x^{8}}{40\mbox{,}320} -\frac{x^{10}}{3\mbox{,}628\mbox{,}800} + \frac{x^{12}}{479\mbox{,}001\mbox{,}600}= \beta_{5}(x), \\ \beta_{1}(x)=t-\frac{t^{3}}{6}+ \frac{t^{5}}{120} - \frac{t^{7}}{5040} + \frac{t^{9}}{362\mbox{,}880}-\frac{x^{11}}{39\mbox{,}916\mbox{,}800} \leq \sin(x), \\ 495-60 x^{2}+x^{4}>0,\qquad 10\mbox{,}395-4725 x^{2}+210 x^{4}-x^{6}>0, \end{cases}\displaystyle \forall x \in \Gamma. $$
(23)

Combining with Eq. (23), we have that

$$ \begin{aligned}[b] H_{5}(x) &\leq 21\bigl(495-60 x^{2}+x^{4} \bigr) \cdot x \beta_{5}(x) - \bigl(10\mbox{,}395-4725 x^{2}+210 x^{4}-x^{6}\bigr) \cdot \beta_{1}(x) \\ &=\frac{x^{13}}{159\mbox{,}667\mbox{,}200} \bigl(-915-64 x^{2}+3 x^{4}\bigr) \leq 0, \quad \forall x \in \Gamma. \end{aligned} $$
(24)

Let \(\beta_{6}(x)=T_{1}(x)+105 (\pi^{2}-4 x^{2}) \cdot T_{2}'(x) -840 x \cdot T_{2}(x)\), \(\beta_{7}(x)=105 (\pi^{2}-4 x^{2}) \cdot T_{2}(x)-T_{1}'(x)\). On the other hand, it can be verified that, \(\forall x \in \Gamma\),

$$ \begin{aligned} &H_{6}'(x)= \beta_{6}(x) \cdot \sin(x) + \beta_{7}(x) \cdot \cos(x), \\ &\beta_{6}(x) \leq 0,\qquad \beta_{7}(x) \geq 0,\qquad T_{2}(x) \geq 0,\qquad T_{1}(x) \geq 0, \\ & \begin{aligned} \cos(x) \geq{}& 1-\frac{x^{2}}{2} +\frac{x^{4}}{24} -\frac{x^{6}}{720} + \frac{x^{8}}{40\mbox{,}320} -\frac{x^{10}}{3\mbox{,}628\mbox{,}800} + \frac{x^{12}}{479\mbox{,}001\mbox{,}600} \\ &{}-\frac{x^{14}}{87\mbox{,}178\mbox{,}291\mbox{,}200}= \beta_{8}(x), \end{aligned}\\ &\beta_{9}(x)=t-\frac{t^{3}}{6}+ \frac{t^{5}}{120} - \frac{t^{7}}{5040} + \frac{t^{9}}{362\mbox{,}880}-\frac{x^{11}}{39\mbox{,}916\mbox{,}800}+\frac{x^{13}}{6\mbox{,}227\mbox{,}020\mbox{,}800}\geq \sin(x). \end{aligned} $$
(25)

Combining Eq. (23) with Eq. (25), we have that

$$ \begin{aligned} H_{6}(x) &\leq 105 \bigl(\pi^{2}-4 x^{2}\bigr) \cdot T_{2}(x) \cdot \beta_{9}(x) -T_{1}(x) \cdot x \beta_{8}(x) \\ &= \frac{x^{13}}{9\mbox{,}153\mbox{,}720\mbox{,}576\mbox{,}000} \beta_{10}(x) \leq 0, \quad \forall x \in \biggl[0,\frac{31 \pi}{64}\biggr], \\ H_{6}'(x) &\geq \beta_{6}(x) \cdot \beta_{1}(x) + \beta_{7}(x) \cdot \beta_{5}(x) \\ &= \frac{x^{12}}{50\mbox{,}295\mbox{,}168\mbox{,}000} \beta_{11}(x) \geq 0, \quad \forall x \in \biggl[\frac{31 \pi}{64},\frac{\pi}{2}\biggr], \end{aligned} $$
(26)

where \(\beta_{10}(x)=(18\mbox{,}063\mbox{,}360 \pi^{6}-8\mbox{,}128\mbox{,}512\mbox{,}000 \pi^{4}+ 643\mbox{,}778\mbox{,}150\mbox{,}400 \pi^{2}- 5\mbox{,}579\mbox{,}410\mbox{,}636\mbox{,}800)+( -634\mbox{,}725 \pi^{6}+305\mbox{,}912\mbox{,}880 \pi^{4}-24\mbox{,}700\mbox{,}198\mbox{,}320 \pi^{2}+ 214\mbox{,}592\mbox{,}716\mbox{,}800) x^{2} +(6069 \pi^{6}-4\mbox{,}639\mbox{,}320 \pi^{4}+411\mbox{,}823\mbox{,}440 \pi^{2}-3\mbox{,}618\mbox{,}457\mbox{,}920) x^{4} + (28 \pi^{6}+52\mbox{,}920 \pi^{4}-5\mbox{,}715\mbox{,}360 \pi^{2}+51\mbox{,}226\mbox{,}560) x^{6} + (\pi^{6}-840 \pi^{4}+75\mbox{,}600 \pi^{2}-665\mbox{,}280) x^{8} \leq 0, \forall x \in [0,\frac{31 \pi}{64}]\), \(\beta_{11}(x)=(-1\mbox{,}290\mbox{,}240 \pi^{6}+580\mbox{,}608\mbox{,}000 \pi^{4}-45\mbox{,}984\mbox{,}153\mbox{,}600 \pi^{2}+ 398\mbox{,}529\mbox{,}331\mbox{,}200)+ (54\mbox{,}405 \pi^{6}-25\mbox{,}552\mbox{,}800 \pi^{4}+2\mbox{,}048\mbox{,}684\mbox{,}400 \pi^{2}- 17\mbox{,}782\mbox{,}934\mbox{,}400) x^{2}+ ( -1404 \pi^{6}+556\mbox{,}920 \pi^{4}-42\mbox{,}366\mbox{,}240 \pi^{2}+365\mbox{,}238\mbox{,}720) x^{4}+ ( 19 \pi^{6}-5040 \pi^{4}+317\mbox{,}520 \pi^{2}-2\mbox{,}661\mbox{,}120) x^{6}\geq 0, \forall x \in [\frac{31 \pi}{64},\frac{\pi}{2}]\). Combining Eq. (26) with \(H_{6}(\pi/2)=0\), we obtain that

$$ H_{6}(x) \leq 0,\quad \forall x \in \biggl[0,\frac{\pi}{2} \biggr]. $$
(27)

Combining Eq. (24) with Eq. (27), we have completed the proof of both Eq. (22) and Eq. (21). □

From Theorems 3.1, 3.2, and 3.3, we directly obtain the following theorem.

Theorem 3.4

We have that

$$\frac{1}{c_{2}(x)^{2}} + \frac{1}{c_{6}(x)} \leq \biggl(\frac{x}{\sin(x)} \biggr)^{2} + \frac{x}{\tan(x)} \leq \frac{1}{c_{1}(x)^{2}} + \frac{1}{c_{5}(x)},\quad \forall x \in [0,\pi/2]. $$

4 Discussion and conclusions

Firstly, we compare the results of \(\frac{\sin(x)}{x}\) between \(b_{i}(x)\) in [2] and \(c_{i}(x)\) in this paper, \(i=1,2\). It can be verified that \(c_{1}(x)-b_{1}(x)=\frac{x^{6} (264-5x^{2})}{840(72+x^{2})(x^{2}+20)} \geq 0\) and \(c_{2}(x)-b_{2}(x)=\frac{ -11 x^{8}}{12(11\mbox{,}088+364 x^{2}+5 x^{4})(x^{2}+42)} \leq 0\), \(\forall x \in [0,\pi/2]\), we have that

$$b_{1}(x) \leq c_{1}(x) \leq \frac{\sin(x)}{x} \leq c_{2}(x) \leq b_{2}(x), \quad \forall x \in [0,\pi/2]. $$

Secondly, we compare the approximation results of \(\cos(x)\) between previous \(b_{i}(x)\) and present \(c_{i}(x)\), \(i=3,4\). It can be verified that \(c_{3}(x)-b_{3}(x)=\frac{x^{8}(270-13 x^{2})}{360 (56+x^{2})(x^{4}+30 x^{2}+540)} \geq 0\) and \(c_{2}(x)-b_{2}(x)=\frac{ -39 x^{8}}{4 (15\mbox{,}120+660 x^{2}+13 x^{4})(x^{2}+30)} \leq 0\), \(\forall x \in [0,\pi/2]\), we have that

$$b_{3}(x) \leq c_{3}(x) \leq \cos(x) \leq c_{4}(x) \leq b_{4}(x), \quad \forall x \in [0,\pi/2]. $$

Thirdly, we compare the approximation results of \(\frac{\tan(x)}{x}\), which also shows that this paper achieves a much better result. It can be verified that \(\forall x \in [0,\pi/2]\),

$$c_{5}(x)-b_{5}(x)=\frac{x^{6}(161 x^{2}-495) (x^{2}-33)}{3(10\mbox{,}395-4725 x^{2}+210 x^{4}-x^{6}) (x^{2}+20) (3 x^{4}-56 x^{2}+120)} \geq 0. $$

However, note that the denominator of \(b_{6}(x)\) is \(T_{3}(x) = 1020x^{6}+ 14\mbox{,}040x^{4}- 1\mbox{,}144\mbox{,}800x^{2}+ 2\mbox{,}721\mbox{,}600 = 30(17 x^{4}-480 x^{2}+1080)(x^{2}+42)\), which has a real root ≈1.5701 within the interval Γ, and we have \(T_{3}(x) >0, \forall x \in [0,1.5701]\). It can be verified that \(c_{6}(x) - b_{6}(x)= \frac{-x^{8} H_{7}(x)}{210 T_{2}(x) T_{3}(x) (\pi^{2}-4 x^{2})}\), where \(H_{7}(x)=378\mbox{,}675 (\pi^{4}-112 \pi^{2}+1008) \pi^{2} + (-64\mbox{,}350 \pi^{6}+5\mbox{,}536\mbox{,}440 \pi^{4}+106\mbox{,}323\mbox{,}840 \pi^{2}-1\mbox{,}526\mbox{,}817\mbox{,}600)x^{2} + (1968 \pi^{6}+50\mbox{,}400 \pi^{4}-25\mbox{,}764\mbox{,}480 \pi^{2}+ 247\mbox{,}484\mbox{,}160) x^{4} + (-8008 \pi^{4}+820\mbox{,}512 \pi^{2}-7\mbox{,}318\mbox{,}080)x^{6}\). By using the Maple software, \(H_{7}(x)\) has six real roots \(\approx -9.16,-4.97 ,-2.76, 2.76, 4.97, 9.16\), and \(H_{7}(x), T_{2}(x), T_{3}(x)> 0, \forall x \in (0,1.5701)\), we have that

$$c_{6}(x) - b_{6}(x) \leq 0,\quad \forall x \in [0,1.5701]. $$