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

## Abstract

This paper provides improved bounds of inequalities of Mitrinović–Adamović type by using two-parameter functions. It achieves a much better approximation effect than those bounds of prevailing methods. A new method based on the Páde interpolation is used to prove the new bounds, which can also be applied for proving the results of prevailing methods and their much more generalized results.

## 1 Introduction

The following inequalities

$$\biggl( \frac{\sin x}{x} \biggr)^{3} > \cos x, \quad 0< x< \frac{\pi }{2}$$
(1)

are known as Mitrinović–Adamović inequalities (see ). Many references  have discussed the problems related to Eq. (1), such as the following power exponential inequality obtained by Nishizawa in 

$$\biggl( \frac{\sin x}{x} \biggr)^{3} < (\cos x)^{1-2x/ \pi}, \quad 0< x< \frac{\pi }{2}$$
(2)

and similar ones in  as follows

\begin{aligned}& (\cos x)^{1-a x^{2}} < \biggl( \frac{\sin x}{x} \biggr)^{3} < (\cos x)^{1-b x^{2}}, \quad 0< x< \frac{\pi }{2}, \end{aligned}
(3)
\begin{aligned}& (\cos x)^{1-2/15 x^{2}-c x^{4}} < \biggl( \frac{\sin x}{x} \biggr)^{3} < (\cos x)^{1-2/15 x^{2}-d x^{4}}, \quad 0< x< \frac{\pi }{2}, \end{aligned}
(4)

where $$a=2/15$$, $$b=4/\pi ^{2}$$, $$c=19/945$$, and $$d=8(30-\pi ^{2})/(15 \pi ^{4})$$ are the best constants in Eq. (3) and Eq. (4). It is obvious that the bounds of Eq. (3) and Eq. (4) are stronger than those bounds of Eq. (1) and Eq. (2).

Exponential-type bounds can be found in [7, 19, 2426]. In , Bhayo and Sandor obtained the following inequalities

$$\begin{gathered} \frac{1+\cos x}{2- \frac{x^{2}}{6}} < \frac{\sin (x)}{x} < \frac{1+\cos x }{2- \frac{2(4-\pi )}{\pi ^{2}} x^{2}}, \\ \frac{2- ( \frac{x}{\pi } )^{3}+\cos x}{3} < \frac{\sin (x)}{x} < \frac{2- ( \frac{x}{\pi } )^{4}+\cos x}{3}, \end{gathered}$$
(5)

for $$0< x< \frac{\pi }{2}$$, while Zhu  and Yang  proved

$$\biggl( \frac{2}{3}+ \frac{2}{3}\cos ^{p} x \biggr)^{ \frac{3}{p}} < \biggl( \frac{\sin (x)}{x} \biggr)^{3} < \biggl( \frac{2}{3}+ \frac{2}{3}\cos ^{q} x \biggr)^{ \frac{3}{q}}$$
(6)

holds if and only if $$p \leq \frac{4}{5}$$ and $$q \geq \frac{\log 3-\log 2}{\log \pi -\log 2}$$.

Nishizawa  found the following inequalities

\begin{aligned} \biggl( \frac{2}{\pi }+ \frac{\pi -2}{\pi ^{3}} \bigl( \pi ^{2}-4x^{2}\bigr) \biggr)^{\pi ^{3}/(24(\pi -2))} &< \frac{\sin (x)}{x} \\ & < \biggl( \frac{2}{\pi }+ \frac{1}{\pi ^{3}} \bigl(\pi ^{2}-4x^{2}\bigr) \biggr)^{4x^{2}/\pi ^{2}} , \quad 0< x< \frac{\pi }{2}. \end{aligned}
(7)

Recently, Zhu and Zhang  provided the following improved bounds

\begin{aligned}& Z_{1}(x) \triangleq \lambda _{1} x^{3} \sin x < F(x) < \lambda _{2} x^{3} \sin x \triangleq Z_{2}(x), \quad 0< x< \frac{\pi }{2}, \end{aligned}
(8)
\begin{aligned}& Z_{3}(x) \triangleq \lambda _{3} x^{4} \biggl( \frac{\sin x}{x} \biggr)^{\frac{23}{21}} < F(x) < \lambda _{4} x^{4} \biggl( \frac{\sin x}{x} \biggr)^{\frac{23}{21}} \triangleq Z_{4}(x), \quad 0< x< \frac{\pi }{2}, \end{aligned}
(9)

where

$$F(x) = \biggl( \frac{\sin x}{x} \biggr)^{3} - \cos x,$$

$$\lambda _{1} \approx 0.06593$$, $$\lambda _{2}=1/15$$, $$\lambda _{3}=1/15$$, and $$\lambda _{4}= (2/\pi )^{124/21}$$ are the best constants in Eq. (8) and Eq. (9). Other bounds related to $$\frac{\sin x}{x}$$ can be found in [26, 3056] and the references therein.

This paper aims to present new bounds for $$F(x)$$ by using two functions $$G_{1}$$ and $$G_{2}$$ in the form of two parameters as

\begin{aligned}& G_{1}(x,\alpha )= \frac{64 x^{3} \sin (\alpha x)}{\pi ^{6} \sin ( \frac{\pi \alpha }{2})}, \end{aligned}
(10)
\begin{aligned}& G_{2}(x,\alpha )= \frac{1}{15} x^{4} \biggl( \frac{\sin (\alpha x)}{\alpha x} \biggr)^{ \frac{23}{21 \alpha ^{2}}}. \end{aligned}
(11)

Let $$D_{i}(x)=F(x)-G_{i}(x,\alpha )$$, $$i=1,2$$. It can be verified that

$$\begin{gathered} D_{1}(0)=D_{1}^{\prime }(0)= D_{1}^{\prime \prime }(0) =D_{1}^{\prime \prime \prime }(0)=0, \qquad D_{1}(\pi /2)=0, \\ D_{1}^{(4)}(0)= \frac{(8 \pi ^{6} \sin (\pi \alpha /2)-7680 \alpha )}{5 \pi ^{6} \sin (\pi \alpha /2)} \triangleq g_{1}(\alpha ), \\ D_{1}^{\prime }(\pi /2)= \frac{(\pi ^{4}-96) \sin (\pi \alpha /2)-8 \pi \alpha \cos (\pi \alpha /2)}{\pi ^{4} \sin (\pi \alpha /2)} \triangleq g_{2}(\alpha ), \\ D_{2}(0)=D_{2}^{\prime }(0)=\cdots = D_{2}^{(7)}(0) =0, \\ D_{2}^{(8)}(0)= \frac{736 \alpha ^{2}}{45} - \frac{992}{945} \triangleq g_{3}(\alpha ), \\ D_{2}(\pi /2) = \frac{1920 -\pi ^{7} ( \frac{\sin (\alpha \pi /2)}{\alpha \pi /2} )^{ \frac{23}{21 \alpha ^{2}}}}{240 \pi ^{3}} \triangleq g_{4}(\alpha ), \end{gathered}$$
(12)

where $$g_{i}(\alpha )$$ has a unique root $$\alpha _{i} \in [0,1.5]$$ satisfying

$$\begin{gathered} \alpha _{1} \approx 1.00144, \qquad \alpha _{2} \approx 0.962976, \\ \alpha _{3}= \frac{\sqrt{14{,}973}}{483} \approx 0.253342, \qquad \alpha _{4} \approx 0.264986. \end{gathered}$$
(13)

By using suitable values of $$\alpha _{i}$$, we obtain the following main results, and also provide a new method for proving these results.

### Theorem 1

For $$0< x<\pi /2$$, one has

\begin{aligned} G_{1}(x, \alpha _{2}) < F(x) < G_{1}(x,\alpha _{1}), \end{aligned}
(14)

where $$\alpha _{2}$$ and $$\alpha _{1}$$ determined by Eq. (13) are the maximum and the minimum constants within [0,1.5] satisfying Eq. (14).

### Theorem 2

For $$0< x<\pi /2$$, one has

\begin{aligned} G_{2}(x, \alpha _{4}) < F(x) < G_{2}(x,\alpha _{3}), \end{aligned}
(15)

where $$\alpha _{4}$$ and $$\alpha _{3}$$ determined by Eq. (13) are the maximum and the minimum constants within $$[0,1.5]$$ satisfying Eq. (15).

## 2 Lemmas

Given a function $$g(x)$$, let $$\varphi _{g,n}(x)$$ and $$\psi _{g,n}(x)$$ be the polynomials of degree n satisfying

$$\begin{gathered} g(0)^{(i)}=\varphi _{g,n}^{(i)}(0), \quad i=0,1,\ldots ,n-1,\quad \text{and}\quad g\biggl( \frac{\pi}{2}\biggr)= \varphi _{g,n}\biggl(\frac{\pi}{2}\biggr), \\ g(0)^{(i)}=\psi _{g,n}^{(i)}(0) \quad \text{and}\quad g^{(j)}\biggl(\frac{\pi}{2}\biggr)= \psi ^{(j)}_{g,n} \biggl(\frac{\pi}{2}\biggr), \quad i=0,1,\ldots ,n-2, j=0,1. \end{gathered}$$
(16)

Let $$D_{g,n}(x)=g(x)-\varphi _{g,n}(x)$$ and $$E_{g,n}(x)=g(x)-\psi _{g,n}(x)$$. From Theorem 3.5.1 on page 67, Sect. 3.5 of , combining with Eq. (12), for $$\forall x \in [0,\pi /2]$$, there exists $$\bar{\xi}_{i}(x) \in [0,\pi /2]$$, $$i=1,2$$, such that

$$\begin{gathered} D_{g,n}(x)= \frac{D_{g,n}^{(n+1)}(\bar{\xi }_{1}(x))}{(n+1)!} x^{n} (x-\pi /2), \quad x \in [0,\pi /2], \\ E_{g,n}(x)= \frac{D_{g,n}^{(n+1)}(\bar{\xi }_{2}(x))}{(n+1)!} x^{n-1} (x-\pi /2)^{2}, \quad x \in [0,\pi /2]. \end{gathered}$$
(17)

We take $$g_{1,\beta}(x)=\cos (\beta x)$$ and $$g_{2,\beta}(x)=\sin (\beta x)$$, $$\beta \in [0,4]$$ as examples, for finding their bounding polynomials in the form of $$\varphi _{g,n}(x)$$ and $$\psi _{g,n}(x)$$. It can be verified that

$$\begin{gathered} \varphi _{g_{1,\beta},2k}(x)=1 + \sum _{i=1}^{k-1}{ (-1)^{i} \frac{(\beta x)^{2i}}{(2i)!}} + \lambda _{1,2k} x^{2k}, \\ \varphi _{g_{2,\beta},2k-1}(x)= \sum_{i=1}^{k-1}{ (-1)^{i+1} \frac{(\beta x)^{2i-1}}{(2i-1)!}} + \lambda _{2,2k-1} x^{2k-1}, \\ \psi _{g_{1,\beta},2k}(x)=1 + \sum_{i=1}^{k-2}{ (-1)^{i} \frac{(\beta x)^{2i}}{(2i)!}}+ \gamma _{1,2k} x^{2k-1} + \gamma _{2,2k} x^{2k}, \\ \psi _{g_{2,\beta},2k-1}(x)= \sum_{i=1}^{k-1}{ (-1)^{i+1} \frac{(\beta x)^{2i-1}}{(2i-1)!}} + \gamma _{3,2k-1} x^{2k-2} + \gamma _{4,2k-1} x^{2k-1}, \end{gathered}$$
(18)

where $$\lambda _{i,n}$$ and $$\gamma _{j,n}$$ satisfy

\begin{aligned}& \lambda _{1,2k} = \Biggl(\cos (\beta \pi /2) - 1 - \sum _{i=1}^{k-1}{ (-1)^{i} \frac{(\beta \pi /2)^{2i}}{(2i)!}} \Biggr) \biggl( \frac{\pi }{2}\biggr)^{-2k}, \\& \lambda _{2,2k-1} = \Biggl( \sin (\beta \pi /2)- \sum _{i=1}^{k-1}{ (-1)^{i+1} \frac{(\beta \pi /2)^{2i-1}}{(2i-1)!}} \Biggr) \biggl( \frac{\pi }{2}\biggr)^{1-2k}, \\& \begin{aligned} \gamma _{1,2k} &= k \cdot 2^{2k} \Biggl(\cos ( \beta \pi /2)- \sum_{i=1}^{k-1}{ (-1)^{i} \frac{(\beta \pi /2)^{2i}}{(2i)!}}-1 \Biggr) { \pi ^{1-2k}} \\ &\quad {} + 2^{2k-2} \Biggl(\beta \sin (\beta \pi /2)+ \sum _{i=1}^{k-1}{ (-1)^{i} \frac{(\beta \pi /2)^{2i-1}}{(2i-1)!}} \Biggr) {\pi ^{2-2k}}, \end{aligned} \\& \begin{aligned} \gamma _{2,2k}&= \Biggl(\cos (\beta \pi /2)- \sum _{i=1}^{k-1}{ (-1)^{i} \frac{(\beta \pi /2)^{2i}}{(2i)!}}-1 \Biggr) (4k-2)\pi ^{-2k} \\ &\quad {} - 2^{2k-1} \Biggl(\beta \sin (\beta \pi /2)+ \sum _{i=1}^{k-1}{ (-1)^{i} \frac{(\beta \pi /2)^{2i-1}}{(2i-1)!}} \Biggr) \pi ^{1-2k}, \end{aligned} \\& \begin{aligned} \gamma _{3,2k-1}&= (2k-1) \Biggl(\sin (\beta \pi /2)- \sum_{i=1}^{k-1}{ (-1)^{i+1} \frac{(\beta \pi /2)^{2i-1}}{(2i-1)!}} \Biggr) 2^{2 k-2} \pi ^{2-2 k} \\ &\quad {} - 2^{2k-3} \pi ^{3-2 k} \Biggl(\beta \cos (\beta \pi /2)- \sum_{i=1}^{k-1}{ (-1)^{i+1} \frac{(\beta \pi /2)^{2i-2}}{(2i-2)!}} \Biggr), \end{aligned} \\& \begin{aligned} \gamma _{4,2k-1}&= -2^{2k} \Biggl( \sin (\beta \pi /2)- \sum_{i=1}^{k-1}{ (-1)^{i+1} \frac{(\beta \pi /2)^{2i-1}}{(2i-1)!}} \Biggr) (k-1) \pi ^{-2 k+1} \\ &\quad {} +2^{2 k-2} \pi ^{2-2 k} \Biggl(\beta \cos ( \beta \pi /2)- \sum_{i=1}^{k-1}{ (-1)^{i+1} \frac{(\beta \pi /2)^{2i-2}}{(2i-2)!}} \Biggr). \end{aligned} \end{aligned}

One has the following lemmas.

### Lemma 1

For $$\forall 0< \beta < 2$$ and $$\forall x \in (0,\pi /2)$$, $$k \geq 1$$, we have

$$\begin{gathered} \varphi _{g_{1,\beta},4k}(x) < \cos (\beta x) \triangleq g_{1,\beta}(x) < \varphi _{g_{1,\beta},4k+2}(x), \\ \varphi _{g_{2,\beta},4k+1}(x) < \sin (\beta x) \triangleq g_{2,\beta}(x) < \varphi _{g_{2,\beta},4k-1}(x), \\ \psi _{g_{1,\beta},4k}(x) > \cos (\beta x) > \psi _{g_{1,\beta},4k+2}(x), \\ \psi _{g_{2,\beta},4k+1}(x) > \sin (\beta x) > \psi _{g_{2,\beta},4k-1}(x). \end{gathered}$$
(19)

### Proof

Let

$$\begin{gathered} H_{1,\beta}(x)= \varphi _{g_{1,\beta},4k}(x) - \cos (\beta x), \qquad H_{2, \beta}(x)= \varphi _{g_{1,\beta},4k+2}(x) - \cos ( \beta x), \\ H_{3,\beta}(x)= \varphi _{g_{2,\beta},4k+1}(x) - \sin (\beta x), \qquad H_{4, \beta}(x)= \varphi _{g_{2,\beta},4k-1}(x) - \sin (\beta x), \\ H_{5,\beta}(x)= \psi _{g_{1,\beta},4k}(x) - \cos (\beta x), \qquad H_{6, \beta}(x)= \psi _{g_{1,\beta},4k+2}(x) - \cos (\beta x), \\ H_{7,\beta}(x)= \psi _{g_{2,\beta},4k+1}(x) - \sin (\beta x), \qquad H_{8, \beta}(x)= \psi _{g_{2,\beta},4k-1}(x) - \sin (\beta x). \end{gathered}$$
(20)

Combining Eq. (20) with Eq. (17), for $$x \in (0,\pi /2)$$, there exists $$\xi _{i}(x) \in (0,\pi /2)$$, $$i=1,2,\ldots ,8$$, such that

$$\begin{gathered} H_{1,\beta}(x)= \frac{H_{1,\beta }^{(4k+1)}(\xi _{1}(x))}{(4k+1)!} x^{4k} \biggl(x- \frac{\pi }{2}\biggr) = \frac{\beta ^{4k+1} \sin (\beta \xi _{1}(x))}{(2k+1)!} x^{4k} \biggl(x- \frac{\pi }{2}\biggr), \\ H_{2,\beta}(x)= \frac{H_{2,\beta }^{(4k+3)}(\xi _{2}(x))}{(4k+3)!} x^{4k+2} \biggl(x- \frac{\pi }{2}\biggr) = \frac{- \beta ^{4k+3} \sin (\beta \xi _{2}(x))}{(4k+3)!} x^{4k+2} \biggl(x- \frac{\pi }{2}\biggr), \\ H_{3,\beta}(x)= \frac{H_{3,\beta }^{(4k+2)}(\xi _{3}(x))}{(4k+2)!} x^{4k+1} \biggl(x- \frac{\pi }{2}\biggr) = \frac{\beta ^{4k+2} \sin (\beta \xi _{3}(x))}{(4k+2)!} x^{4k+1} \biggl(x- \frac{\pi }{2}\biggr), \\ H_{4,\beta}(x)= \frac{H_{4,\beta }^{(4k)}(\xi _{4}(x))}{4k!} x^{4k-1} \biggl(x- \frac{\pi }{2}\biggr) = \frac{- \beta ^{4k} \sin (\beta \xi _{4}(x))}{4k!} x^{4k-1} \biggl(x- \frac{\pi }{2}\biggr), \\ H_{5,\beta}(x)= \frac{H_{5,\beta }^{(4k+1)}(\xi _{5}(x))}{(4k+1)!} x^{4k-1} \biggl(x- \frac{\pi }{2}\biggr)^{2} = \frac{\beta ^{4k+1} \sin (\beta \xi _{5}(x))}{(2k+1)!} x^{4k-1} \biggl(x- \frac{\pi }{2}\biggr)^{2}, \\ H_{6,\beta}(x)= \frac{H_{6,\beta }^{(4k+3)}(\xi _{6}(x))}{(4k+3)!} x^{4k+1} \biggl(x- \frac{\pi }{2}\biggr)^{2} = \frac{- \beta ^{4k+3} \sin (\beta \xi _{6}(x))}{(4k+3)!} x^{4k+1} \biggl(x- \frac{\pi }{2}\biggr)^{2}, \\ H_{7,\beta}(x)= \frac{H_{7,\beta }^{(4k+2)}(\xi _{7}(x))}{(4k+2)!} x^{4k} \biggl(x- \frac{\pi }{2}\biggr)^{2} = \frac{\beta ^{4k+2} \sin (\beta \xi _{7}(x))}{(4k+2)!} x^{4k} \biggl(x- \frac{\pi }{2}\biggr)^{2}, \\ H_{8,\beta}(x)= \frac{H_{8,\beta }^{(4k)}(\xi _{8}(x))}{4k!} x^{4k-2} \biggl(x- \frac{\pi }{2}\biggr)^{2} = \frac{- \beta ^{4k} \sin (\beta \xi _{8}(x))}{4k!} x^{4k-2} \biggl(x- \frac{\pi }{2}\biggr)^{2}. \end{gathered}$$
(21)

For $$\beta \in (0,2)$$ and $$x \in (0,\pi /2)$$, one has $$\beta \xi _{i}(x) \in (0,\pi )$$ and $$\sin (\beta \xi _{i}(x))>0$$. By combining with Eq. (21), one obtains

$$\begin{gathered} H_{1,\beta}(x)< 0,\qquad H_{2,\beta}(x) >0,\qquad H_{3,\beta}(x)< 0 \quad \text{and}\quad H_{4, \beta}(x) >0, \quad \forall x \in (0,\pi /2), \\ H_{5,\beta}(x)>0,\qquad H_{6,\beta}(x) < 0,\qquad H_{7,\beta}(x)>0 \quad \text{and}\quad H_{8, \beta}(x) < 0, \quad \forall x \in (0,\pi /2). \end{gathered}$$
(22)

Thus, the proof has been completed. □

### Lemma 2

For $$\beta _{1}=3$$ and $$\forall x \in (0,\pi /2)$$, we have

$$\psi _{g_{2,\beta _{1}},11}(x) < \sin (3 x) < \psi _{g_{2,\beta _{1}},13}(x).$$
(23)

### Proof

Let $$H_{9}(x)=\psi _{g_{2,\beta _{1}},13}(x)-\sin (3 x)$$ and $$H_{10}(x)= \psi _{g_{2,\beta _{1}},11}(x) - \sin (3 x)$$. This is equivalent to proving the following inequalities

$$H_{9}(x) > 0 >H_{10}(x), \quad \forall x \in \biggl(0, \frac{\pi }{2}\biggr).$$
(24)

It can be verified that

$$\begin{gathered} H_{9}^{(i)}(0)=0, \quad i=0,1, \ldots ,11,\qquad H_{9}^{(j)}\biggl( \frac{\pi }{2} \biggr)=0, \quad j=0,1, \\ H_{9}^{(12)}(0)= \frac{12!}{\pi ^{12}} \biggl( \frac{2187}{123{,}200} \pi ^{11}- \frac{243}{140} \pi ^{9}+ \frac{ 2916}{35} \pi ^{7}- \frac{10{,}368}{5} \pi ^{5} \\ \hphantom{H_{9}^{(12)}(0)=}{} + 23{,}040 \pi ^{3}- 73{,}728 \pi -53{,}248\biggr) \\ \hphantom{H_{9}^{(12)}(0)}\approx 34{,}694.4 >0, \\ H_{9}^{\prime \prime }\biggl( \frac{\pi }{2}\biggr)= \biggl(- \frac{531{,}441}{12!\cdot 64} \pi ^{9}+ \frac{59{,}049}{9! \cdot 32} \pi ^{7}- \frac{10{,}935}{5! \cdot 224} \pi ^{5}+ \frac{1701}{5!} \pi ^{3} \\ \hphantom{H_{9}^{\prime \prime }\biggl( \frac{\pi }{2}\biggr)=}{}- \frac{405}{2} \pi -9 + \frac{792}{\pi } + \frac{ 624}{ \pi ^{2}}\biggr)\\ \hphantom{H_{9}^{\prime \prime }\biggl( \frac{\pi }{2}\biggr)} \approx 0.017>0, \\ H_{10}^{(i)}(0)=0, \quad i=0,1,\ldots ,9,\qquad H_{10}^{(j)}\biggl( \frac{\pi }{2}\biggr)=0, \quad j=0,1, \\ H_{10}^{(10)}(0)= \frac{10!}{\pi ^{10}} \biggl( - \frac{243}{1120} \pi ^{9}+ \frac{486}{35} \pi ^{7}- \frac{1944}{5} \pi ^{5} \\ \hphantom{H_{10}^{(10)}(0)=}{} + 4608 \pi ^{3}- 15{,}360 \pi - 11{,}264 \biggr)\\ \hphantom{H_{10}^{(10)}(0)} \approx -5850.1< 0, \\ H_{10}^{\prime \prime }\biggl( \frac{\pi }{2}\biggr)= \biggl(-8! \cdot \frac{2187}{64} \pi ^{7} - \frac{6561}{8!} \pi ^{5} + \frac{243}{32} \pi ^{3} \\ \hphantom{H_{10}^{\prime \prime }\biggl( \frac{\pi }{2}\biggr)=}{} -126 \pi -9 + \frac{540}{\pi } + \frac{ 440}{ \pi ^{2}}\biggr) \\ \hphantom{H_{10}^{\prime \prime }\biggl( \frac{\pi }{2}\biggr)}\approx -0.154< 0. \end{gathered}$$
(25)

By combining Eq. (25) with Eq. (21), one obtains Eq. (24). Thus, the proof has been completed. □

### Remark 1

In principle, Eq. (25) can be manually verified. On the other hand, it is very helpful to use Maple software to verify the corresponding equations with much higher efficiency than manual verification.

### Lemma 3

Let $$\beta _{1}=3$$, $$\beta _{2}=3-\alpha _{3}$$, $$\beta _{3}=3+\alpha _{3}$$, $$\beta _{4}=3-\alpha _{4}$$, $$\beta _{5}=3+\alpha _{4}$$. For $$\beta =\beta _{i}$$, $$i=1,2,\ldots ,5$$, and $$\forall x \in (0,\pi /2)$$, we have

$$\begin{gathered} \varphi _{g_{1,\beta},20}(x) < \cos (\beta x) < \varphi _{g_{1,\beta},18}(x), \\ \varphi _{g_{2,\beta},17}(x) < \sin (\beta x) < \varphi _{g_{2,\beta},15}(x). \end{gathered}$$
(26)

### Proof

Equation (26) is equivalent to the following inequalities, where $$i=1,2,\ldots ,5$$.

$$\begin{gathered} \bar{H}_{1,\beta _{i}}(x)= \varphi _{g_{1,\beta},20}(x) - \cos ( \beta x) < 0 , \\ \bar{H}_{2,\beta _{i}}(x) = \varphi _{g_{1,\beta},18}(x) - \cos ( \beta x) >0, \\ \bar{H}_{3,\beta _{i}}(x)=\varphi _{g_{2,\beta},17}(x) - \sin (\beta x) < 0 , \\ \bar{H}_{4,\beta _{i}}(x)=\varphi _{g_{2,\beta},15}(x) - \sin (\beta x) >0. \end{gathered}$$
(27)

First, by using Maple software, it can be verified that

\begin{aligned}& \begin{gathered} H_{1}^{(i)}(0)=0, \quad i=0,1,\ldots ,19, \\ H_{1}^{(20)}(0)= \frac{20!}{\pi ^{20}} \biggl( 1{,}048{,}576 \biggl( \cos \biggl( \frac{\beta \pi }{2}\biggr)-1\biggr) \\ \hphantom{H_{1}^{(20)}(0)=}{} + 131{,}072 (\beta \pi )^{2}- \frac{8192 (\beta \pi )^{4}}{3}+ \frac{1024 (\beta \pi )^{6}}{45}- \frac{32 (\beta \pi )^{8}}{315} + \frac{ 512 (\beta \pi )^{10}}{5\cdot 9!} \\ \hphantom{H_{1}^{(20)}(0)=}{} - \frac{ 64 (\beta \pi )^{12}}{3 \cdot 11!} + \frac{ 32 (\beta \pi )^{14}}{7 \cdot 13!} - \frac{ (\beta \pi )^{16}}{15!} + \frac{ 2 (\beta \pi )^{18}}{9\cdot 17!} - \frac{ (\beta \pi )^{20}}{20 \cdot 19!} \biggr), \\ H_{1}\biggl( \frac{\pi }{2}\biggr)=0, \\ H_{1}^{\prime }\biggl( \frac{\pi }{2}\biggr)= \frac{1}{65{,}536 \pi \cdot 18!} \biggl(18! \cdot \biggl(2{,}621{,}440 \biggl(\cos \biggl( \frac{\beta \pi }{2}\biggr)-1\biggr) \\ \hphantom{H_{1}^{\prime }\biggl( \frac{\pi }{2}\biggr)=}{} +65{,}536 \alpha \pi \sin \biggl( \frac{\beta \pi }{2}\biggr) \biggr)+ 18! \cdot 294{,}912 (\beta \pi )^{2} -16! \cdot 1{,}671{,}168 (\beta \pi )^{4} \\ \hphantom{H_{1}^{\prime }\biggl( \frac{\pi }{2}\biggr)=}{} +14! \cdot 2{,}924{,}544 (\beta \pi )^{6} -12! \cdot 2{,}036{,}736 (\beta \pi )^{8} + 10! \cdot 622{,}336 (\beta \pi )^{10} \\ \hphantom{H_{1}^{\prime }\biggl( \frac{\pi }{2}\biggr)=}{} -8! \cdot 84{,}864 (\beta \pi )^{12}+ 3{,}525{,}120 (\beta \pi )^{14} -2448 (\beta \pi )^{16} +(\beta \pi )^{18} \biggr), \end{gathered} \end{aligned}
(28)
\begin{aligned}& \textstyle\begin{cases} \bar{H}_{1,\beta _{1}}^{(20)}(0)\approx -1.610 \cdot 10^{8}< 0,\qquad \bar{H}_{1,\beta _{1}}^{\prime }( \frac{\pi }{2})\approx 6.7 \cdot 10^{-7}>0, \\ \bar{H}_{1,\beta _{2}}^{(20)}(0)\approx -2.32\cdot 10^{7}< 0,\qquad \bar{H}_{1,\beta _{2}}^{\prime }( \frac{\pi }{2})\approx 9.85 \cdot 10^{-8}>0, \\ \bar{H}_{1,\beta _{3}}^{(20)}(0)\approx -9.52 \cdot 10^{8}< 0,\qquad \bar{H}_{1,\beta _{3}}^{\prime }( \frac{\pi }{2})\approx 3.97 \cdot 10^{-6} > 0, \\ \bar{H}_{1,\beta _{4}}^{(20)}(0)\approx -2.12 \cdot 10^{7}< 0,\qquad \bar{H}_{1,\beta _{4}}^{\prime }( \frac{\pi }{2})\approx 8.97 \cdot 10^{-8}> 0, \\ \bar{H}_{1,\beta _{5}}^{(20)}(0)\approx -1.02 \cdot 10^{9} < 0,\qquad \bar{H}_{1,\beta _{5}}^{\prime }( \frac{\pi }{2})\approx 4.29 \cdot 10^{-6}>0. \end{cases}\displaystyle \end{aligned}
(29)

Combining Eq. (21) and Eq. (28) with Eq. (29), we obtain

$$\bar{H}_{1,\beta _{i}}(x)< 0, \quad i=1,2,\ldots ,5.$$
(30)

Secondly, it can be verified that

\begin{aligned}& \begin{gathered} \bar{H}_{2,\beta}^{(i)}(0)=0, \quad i=0,1,\ldots ,17, \\ \bar{H}_{2,\beta}^{(18)}(0)= \frac{18!}{\pi ^{18}} \biggl( \frac{\beta ^{18}}{18!}- \frac{\beta ^{16}}{15!\cdot 4 \pi ^{2}} \\ \hphantom{\bar{H}_{2,\beta}^{(18)}(0)=}{}+ \frac{16 \beta ^{14}}{14!\cdot \pi ^{4}} - \frac{16 \beta ^{12}}{10!\cdot 33 \pi ^{6}} + \frac{256 \beta ^{10}}{10!\cdot \pi ^{8}} - \frac{8 \beta ^{8}}{315 \pi ^{10}} + \frac{256 \beta ^{6}}{45 \pi ^{12}} - \frac{2048 \beta ^{4}}{3 \pi ^{14}} \\ \hphantom{\bar{H}_{2,\beta}^{(18)}(0)=}{} + \frac{32{,}768 \beta ^{2}}{\pi ^{16}} + \frac{9!\cdot 2048 \cos ( \frac{\beta \pi }{2})}{ 2835\pi ^{18}}- \frac{9!\cdot 2048}{2835 \pi ^{18}} \biggr), \\ \bar{H}_{2,\beta}\biggl( \frac{\pi }{2}\biggr)=0, \\ \bar{H}_{2,\beta}^{\prime }\biggl( \frac{\pi }{2}\biggr)= - \frac{\beta ^{16} \pi ^{15} }{16!\cdot 16{,}384} + \frac{\beta ^{14} \pi ^{13}}{2048 \cdot 14!}- \frac{273 \beta ^{12} \pi ^{11}}{14\cdot 512} \\ \hphantom{\bar{H}_{2,\beta}^{\prime }\biggl( \frac{\pi }{2}\biggr)=}{}+ \frac{\beta ^{10} \pi ^{9}}{10!\cdot 64} - \frac{5 \beta ^{8} \pi ^{7}}{8!\cdot 64}+ \frac{\beta ^{6} \pi ^{5}}{1920} - \frac{7 \beta ^{4} \pi ^{3}}{96}+ 4 \pi \beta ^{2}+ \beta \sin \biggl( \frac{\beta \pi }{2}\biggr) \\ \hphantom{\bar{H}_{2,\beta}^{\prime }\biggl( \frac{\pi }{2}\biggr)=}{}+ \frac{36 \cos ( \frac{\beta \pi }{2})}{\pi }- \frac{36}{\pi }, \end{gathered} \end{aligned}
(31)
\begin{aligned}& \textstyle\begin{cases} \bar{H}_{2,\beta _{1}}^{(18)}(0)\approx 2.15 \cdot 10^{7}>0,\qquad \bar{H}_{2,\beta _{1}}^{\prime }( \frac{\pi }{2})\approx -0.13 \cdot 10^{-4}< 0, \\ \bar{H}_{2,\beta _{2}}^{(18)}(0)\approx 3.72 \cdot 10^{6} >0,\qquad \bar{H}_{2,\beta _{2}}^{\prime }( \frac{\pi }{2})\approx -2.41 \cdot 10^{-6}< 0, \\ \bar{H}_{2,\beta _{3}}^{(18)}(0)\approx 1.08 \cdot 10^{8}>0,\qquad \bar{H}_{2,\beta _{3}}^{\prime }( \frac{\pi }{2})\approx -0.69 \cdot 10^{-4} < 0, \\ \bar{H}_{2,\beta _{4}}^{(18)}(0)\approx 3.42 \cdot 10^{6}>0,\qquad \bar{H}_{2,\beta _{4}}^{\prime }( \frac{\pi }{2})\approx -2.21 \cdot 10^{-6}< 0, \\ \bar{H}_{2,\beta _{5}}^{(18)}(0)\approx 1.16 \cdot 10^{8}>0,\qquad \bar{H}_{2,\beta _{5}}^{\prime }( \frac{\pi }{2})\approx -0.74 \cdot 10^{-4}< 0. \end{cases}\displaystyle \end{aligned}
(32)

Combining Eq. (21) and Eq. (31) with Eq. (32), we obtain

$$\bar{H}_{2,\beta _{i}}(x)>0, \quad i=1,2,\ldots ,5.$$
(33)

Thirdly, it can be verified that

\begin{aligned}& \bar{H}_{3,\beta}^{(i)}(0)=0, \quad i=0,1,\ldots ,16, \\& \bar{H}_{3,\beta}^{(17)}(0)= -\beta ^{17}+ \frac{1088 \beta ^{15}}{\pi ^{2}}- \frac{8! \cdot 68 \beta ^{13}}{3 \pi ^{4}} \\& \hphantom{\bar{H}_{3,\beta}^{(17)}(0)=}{} + \frac{8!\cdot 14{,}144 \beta ^{11}}{\pi ^{6}}- \frac{13! \cdot 1088 \beta ^{9}}{27\pi ^{8}}+ \frac{16!\cdot 1088 \beta ^{7}}{315\pi ^{10}} - \frac{18! \cdot 256 \beta ^{5}}{135 \pi ^{12}} \\& \hphantom{\bar{H}_{3,\beta}^{(17)}(0)=}{} + \frac{18!\cdot 4096 \beta ^{3}}{27 \pi ^{14}} - \frac{17!\cdot 65{,}536 \beta }{\pi ^{16}}+ \frac{17! \cdot 131{,}072 \sin ( \frac{\beta \pi }{2})}{\pi ^{17}}, \end{aligned}
(34)
\begin{aligned}& \bar{H}_{3,\beta}\biggl( \frac{\pi }{2}\biggr)=0, \\& \bar{H}_{3,\beta}^{\prime }\biggl( \frac{\pi }{2}\biggr)= \frac{\beta ^{15} \pi ^{14}}{16!\cdot 512}- \frac{105 \beta ^{13 }\pi ^{12}}{16! \cdot 32} + \frac{117 \beta ^{11} \pi ^{10}}{13!\cdot 128}- \frac{\beta ^{9} \pi ^{8}}{9!\cdot 32} \\& \hphantom{\bar{H}_{3,\beta}^{\prime }\biggl( \frac{\pi }{2}\biggr)=}{} + \frac{5 \beta ^{7} \pi ^{6}}{8!\cdot 4}- \frac{\beta ^{5} \pi ^{4}}{160} + \frac{7 \pi ^{2} \beta ^{3}}{12} -\beta \cos \biggl( \frac{ \beta \pi }{2}\biggr) -16 \beta + \frac{34 \sin ( \frac{ \beta \pi }{2})}{\pi } , \\& \textstyle\begin{cases} \bar{H}_{3,\beta _{1}}^{(17)}(0)\approx -7.9\cdot 10^{6}< 0,\qquad \bar{H}_{3, \beta _{1}}^{\prime }( \frac{\pi }{2})\approx 5.8 \cdot 10^{-5}>0, \\ \bar{H}_{3,\beta _{2}}^{(17)}(0)\approx -1.5 \cdot 10^{6} < 0,\qquad \bar{H}_{3,\beta _{2}}^{\prime }( \frac{\pi }{2})\approx 1.1 \cdot 10^{-5}>0, \\ \bar{H}_{3,\beta _{3}}^{(17)}(0)\approx -3.6 \cdot 10^{7}< 0,\qquad \bar{H}_{3,\beta _{3}}^{\prime }( \frac{\pi }{2})\approx 2.6 \cdot 10^{-4} > 0, \\ \bar{H}_{3,\beta _{4}}^{(17)}(0)\approx -1.3 \cdot 10^{6}< 0,\qquad \bar{H}_{3,\beta _{4}}^{\prime }( \frac{\pi }{2})\approx 1.0 \cdot 10^{-5}> 0, \\ \bar{H}_{3,\beta _{5}}^{(17)}(0)\approx -3.9 \cdot 10^{7}< 0,\qquad \bar{H}_{3,\beta _{5}}^{\prime }( \frac{\pi }{2})\approx 2.8 \cdot 10^{-4}>0. \end{cases}\displaystyle \end{aligned}
(35)

Combining Eq. (21) and Eq. (34) with Eq. (35), we obtain

$$\bar{H}_{3,\beta _{i}}(x)< 0, \quad i=1,2,\ldots ,5.$$
(36)

Finally, it can be verified that

\begin{aligned}& \begin{gathered} \bar{H}_{4,\beta}^{(i)}(0)=0, \quad i=0,1,\ldots ,14, \\ \bar{H}_{4,\beta}^{(15)}(0)= \beta ^{1}5- \frac{ 840 \beta ^{13}}{\pi ^{2}}+ \frac{ 8!\cdot 13 \beta ^{11}}{\pi ^{4}} - \frac{12!\cdot 13 \beta ^{9}}{27 \pi ^{6}}+ \frac{13!\cdot 32 \beta ^{7}}{3 \pi ^{8}} \\ \hphantom{\bar{H}_{4,\beta}^{(15)}(0)=}{}- \frac{14!\cdot 128 \beta ^{5}}{\pi ^{10}}+ \frac{ 16!\cdot 128 \beta ^{3}}{3\pi ^{12}} - \frac{ 16!\cdot 1024 \beta }{\pi ^{14}}+ \frac{16!\cdot 2048 \sin ( \frac{ \beta \pi }{2})}{\pi ^{15}}, \end{gathered} \end{aligned}
(37)
\begin{aligned}& \begin{gathered} \bar{H}_{4,\beta}\biggl( \frac{\pi }{2} \biggr)=0, \\ \bar{H}_{4,\beta}^{\prime }\biggl( \frac{\pi }{2}\biggr)= - \frac{ 105 \beta ^{13} \pi ^{12}}{16!\cdot 64} + \frac{ \beta ^{11} \pi ^{10}}{11!\cdot 256}- \frac{15 \beta ^{9} \pi ^{8}}{10!\cdot 64} + \frac{ \beta ^{7} \pi ^{6}}{8!} \\ \hphantom{\bar{H}_{4,\beta}^{\prime }\biggl( \frac{\pi }{2}\biggr)=}{} - \frac{ \beta ^{5} \pi ^{4}}{192}+ \frac{ \pi ^{2} \beta ^{3}}{2}-\beta \cos \biggl( \frac{ \beta \pi }{2}\biggr) -14 \beta + \frac{ 30 \sin ( \frac{ \beta \pi }{2})}{\pi }, \end{gathered} \end{aligned}
(38)
\begin{aligned}& \textstyle\begin{cases} \bar{H}_{4,\beta _{1}}^{(19)}(0)\approx 1.0 \cdot 10^{6}>0,\qquad \bar{H}_{4, \beta _{1}}^{\prime }( \frac{\pi }{2})\approx -8.7 \cdot 10^{-4}< 0, \\ \bar{H}_{4,\beta _{2}}^{(19)}(0)\approx 2.4\cdot 10^{5} >0,\qquad \bar{H}_{4, \beta _{2}}^{\prime }( \frac{\pi }{2})\approx -1.9\cdot 10^{-4}< 0, \\ \bar{H}_{4,\beta _{3}}^{(19)}(0)\approx 4.3 \cdot 10^{6} >0,\qquad \bar{H}_{4,\beta _{3}}^{\prime }( \frac{\pi }{2})\approx -3.4 \cdot 10^{-3} < 0, \\ \bar{H}_{4,\beta _{4}}^{(19)}(0)\approx 2.3 \cdot 10^{5}>0,\qquad \bar{H}_{4, \beta _{4}}^{\prime }( \frac{\pi }{2})\approx -1.8 \cdot 10^{-4}< 0, \\ \bar{H}_{4,\beta _{5}}^{(19)}(0)\approx 4.5 \cdot 10^{6}>0,\qquad \bar{H}_{4, \beta _{5}}^{\prime }( \frac{\pi }{2})\approx - 3.6 \cdot 10^{-3} < 0. \end{cases}\displaystyle \end{aligned}
(39)

Combining Eq. (21), Eq. (37), and Eq. (38) with Eq. (39), we obtain

$$\bar{H}_{4,\beta _{i}}(x)>0, \quad i=1,2,\ldots ,5.$$
(40)

Combining Eq. (30), Eq. (33), and Eq. (36) with Eq. (40), one obtains Eq. (23), and the proof is completed. □

## 3 Proof of Theorem 1

This is equivalent to proving Eq. (14) in Theorem 1. It can be verified that

\begin{aligned} R_{1}(x,\alpha )&=\bigl(F(x)-G_{1}(x, \alpha )\bigr) \cdot x^{3} \sin \biggl( \frac{\alpha \pi }{2}\biggr) \pi ^{6}\\ & = - x^{3} \sin \biggl( \frac{\alpha \pi }{2}\biggr) \pi ^{6} \cos (x) \\ &\quad {} +\bigl(-\sin (3 x)+3 \sin (x) \bigr) \frac{ \sin ( \frac{\alpha \pi }{2}) \pi ^{6}}{4}-64 x^{6} \sin ( \alpha x). \end{aligned}
(41)

Combining with Eq. (41), Eq. (14) is equivalent to

$$R_{1}(x,\alpha _{2}) > 0 > R_{1}(x,\alpha _{1}), \quad x \in \biggl(0, \frac{\pi }{2}\biggr).$$
(42)

From Lemmas 1 and 2, for $$\forall x \in (0, \frac{\pi }{2})$$ and $$\beta \in (0,2)$$, one has the following lower and upper bounds

$$\begin{gathered} L_{c,1}(x) \triangleq \psi _{g_{1,1},14}(x)< \cos (x)< \psi _{g_{1,1},16}(x) \triangleq U_{c,1}(x), \\ L_{s,1}(x) \triangleq \psi _{g_{1,1},11}(x)< \sin (x)< \psi _{g_{1,1},13}(x) \triangleq U_{s,1}(x), \\ L_{s,3}(x) \triangleq \psi _{g_{2,\beta _{1}},11}(x) < \sin (3 x) < \psi _{g_{2,\beta _{1}},13}(x) \triangleq U_{s,3}(x), \\ L_{s,\beta}(x) \triangleq \psi _{g_{1,\beta},11}(x)< \sin ( \beta x)< \psi _{g_{1,\beta},13}(x) \triangleq U_{s,\beta}(x). \end{gathered}$$
(43)

Let $$B_{n,i}(x)=C_{i}^{n} ( \frac{\pi }{2}-x )^{n-i} x^{i}$$ satisfying $$B_{n,i}(x)>0$$ for $$x \in (0, \frac{\pi }{2})$$, and

\begin{aligned}& \rho _{0} = \bigl(\pi ^{4}-96\bigr) \approx 1.4 >0, \qquad \rho _{i} = \sin \biggl( \frac{\alpha _{i} \pi }{2}\biggr),\qquad \rho _{2+i}=\cos \biggl( \frac{\alpha _{i} \pi }{2}\biggr), \quad i=1,2, \\& \mu _{1,0} = \frac{32 \pi ^{5} \alpha _{2} }{15} \rho _{4} - \frac{256 \rho _{0} \alpha _{2}}{\pi ^{2}} \approx 1.34 > 0, \\& \mu _{1,1} = \frac{64 \pi ^{5} \alpha _{2} \rho _{4} }{25} - \frac{1536 \rho _{0} \alpha _{2}}{(5 \pi ^{2})} \approx 1.61 > 0, \\& \mu _{1,2} = - \frac{4 \pi ^{5} (23 \pi ^{2}-33{,}264) \alpha _{2} \rho _{4} }{42{,}525}+ \frac{32 \rho _{0} (\pi ^{2} \alpha _{2}^{2}-1584) \alpha _{2}}{135 \pi ^{2}} \approx 1.90 > 0, \\& \begin{aligned} \mu _{1,3} &= - \frac{\alpha _{2} \rho _{4}}{10! \cdot \pi ^{2}} \biggl(29{,}441 \pi ^{9}- \frac{6!\cdot 98{,}564}{5} \pi ^{7}+145{,}152 \pi ^{5}-8! \cdot 384 \pi ^{3}+10! \cdot 128 \pi \\ &\quad {} - \frac{12! \cdot 32}{15}\biggr) + \frac{8 \rho _{0} (\pi ^{2} \alpha _{2}^{2}-528) \alpha _{2}}{9 \pi ^{2}} \\ &\approx 2.21> 0, \end{aligned} \\& \begin{aligned} \mu _{1,4} &= \frac{ \alpha _{2} \rho _{4}}{11!\cdot 140 \pi ^{2}} \biggl(57{,}729 \pi ^{11}-116{,}589{,}880 \pi ^{9}\\ &\quad {}+ \frac{11! \cdot 73{,}931}{105} \pi ^{7}- \frac{11! \cdot 392}{15} \pi ^{5}+ \frac{13! \cdot 160}{9} \pi ^{3} \\ &\quad {} -12! \cdot 6912 \pi +11! \cdot 182{,}272\biggr) - \frac{\rho _{0} (\pi ^{4} \alpha _{2}^{4}-3600 \pi ^{2} \alpha _{2}^{2}+950{,}400) \alpha _{2}}{1575 \pi ^{2}} \\ &\approx 2.52 > 0, \end{aligned} \\& \begin{aligned} \mu _{1,5} &= \frac{ \alpha _{2} \rho _{4}}{12! \cdot 7 \pi ^{2}} \biggl(171{,}867 \pi ^{11}-149{,}683{,}600 \pi ^{9}\\ &\quad {}+ \frac{11!\cdot 58{,}427}{105} \pi ^{7} + \frac{13! \cdot 472}{585} \pi ^{5}+ \frac{12! \cdot 2272}{9} \pi ^{3} \\ &\quad {}-12! \cdot 11{,}776 \pi + \frac{12! \cdot 246{,}016}{9} \biggr) - \frac{4 \rho _{0} (\pi ^{4} \alpha _{2}^{4}-1200 \pi ^{2} \alpha _{2}^{2}+190{,}080) \alpha _{2}}{945 \pi ^{2}} \\ &\approx 2.84> 0, \end{aligned} \\& \begin{aligned} \mu _{1,6} &= \frac{\alpha _{2} \rho _{4}}{10! \cdot 420 \pi ^{2}} \biggl(\pi ^{13}+ \frac{2{,}546{,}325}{11} \pi ^{11}-128{,}711{,}400 \pi ^{9}\\ &\quad {}+ \frac{13! \cdot 9929}{4620} \pi ^{7}+10!\cdot 5768 \pi ^{5}-10! \cdot 9344 \pi ^{3} \\ &\quad {}- \frac{15! \cdot 13{,}184}{3003} \pi +10!\cdot 4{,}118{,}528\biggr) + \frac{\alpha _{2} \rho _{0} }{264{,}600 \pi ^{2}} \biggl(\pi ^{6} \alpha _{2}^{6}-4704 \pi ^{4} \alpha _{2}^{4} \\ &\quad {} +8! \cdot 70 \pi ^{2} \alpha _{2}^{2}- \frac{10! \cdot 1232}{15}\biggr) \\ & \approx 3.15> 0, \end{aligned} \\& \begin{aligned} \mu _{1,7} &= \frac{\alpha _{2} \rho _{4}}{10! \cdot 40 \pi ^{2}} \biggl(\pi ^{13}+ \frac{553{,}530}{11} \pi ^{11}-21{,}018{,}920 \pi ^{9}+ \frac{11! \cdot 47{,}528}{1155} \pi ^{7}+ \frac{10! \cdot 7784}{3} \pi ^{5} \\ &\quad {} -8!\cdot 1{,}512{,}960 \pi ^{3}-10! \cdot 291{,}840 \pi +10! \cdot 914{,}432\biggr) + \frac{\alpha _{2} \rho _{0} }{25{,}200 \pi ^{2}} \biggl(\pi ^{6} \alpha _{2}^{6} \\ &\quad {} -1568 \pi ^{4} \alpha _{2}^{4}+ 8!\cdot 14 \pi ^{2} \alpha _{2}^{2}- \frac{10! \cdot 176}{15} \biggr) \\ &\approx 3.43> 0, \end{aligned} \\& \begin{aligned} \mu _{1,8} &= - \frac{ \alpha _{2} \rho _{4}}{12!\cdot 360 \pi ^{2}} \biggl( \pi ^{15}-7920 \pi ^{13}-6!\cdot 158{,}667 \pi ^{11}\\ &\quad {}+ \frac{11!\cdot 613{,}367}{630} \pi ^{9}- \frac{14!\cdot 82{,}426}{3185} \pi ^{7}- \frac{14!\cdot 31{,}472}{65} \pi ^{5} \\ &\quad {}+13!\cdot 55{,}808 \pi ^{3}+12! \cdot 4{,}608{,}000 \pi -12!\cdot 19{,}111{,}936\biggr) - \frac{\rho _{0} \alpha _{2}}{9!\cdot 45 \pi ^{2}} \\ &\quad {} \cdot \bigl(\pi ^{8} \alpha _{2}^{8} -4320 \pi ^{6} \alpha _{2}^{6}+8! \cdot 84 \pi ^{4} \alpha _{2}^{4}-10!\cdot 224 \pi ^{2} \alpha _{2}^{2}+12! \cdot 96\bigr) \\ &\approx 3.68> 0, \end{aligned} \\& \begin{aligned} \mu _{1,9} &= - \frac{\alpha _{2} \rho _{4}}{12!\cdot 20 \pi ^{2}}\biggl(\pi ^{15}-2640 \pi ^{13}- \frac{9!\cdot 46{,}729}{1680} \pi ^{11}+ \frac{13!\cdot 437}{945} \pi ^{9}\\ &\quad {}- \frac{13!\cdot 1426}{105} \pi ^{7}- \frac{14!\cdot 6352}{65} \pi ^{5} +12!\cdot 160{,}000 \pi ^{3} \\ &\quad {}+12!\cdot 417{,}792 \pi -12!\cdot 2{,}676{,}736\biggr) - \frac{4 \rho _{0} \alpha _{2}}{ 10! \cdot \pi ^{2}} \biggl( \pi ^{8} \alpha _{2}^{8}-1440 \pi ^{6} \alpha _{2}^{6} \\ &\quad {} + \frac{8!\cdot 84}{5} \pi ^{4} \alpha _{2}^{4}-10! \cdot 32 \pi ^{2} \alpha _{2}^{2}+11!\cdot 128 \biggr) \\ &\approx 3.88> 0, \end{aligned} \\& \begin{aligned} \mu _{1,10} &= \frac{ \alpha _{2} \rho _{4}}{14!\cdot 32 \pi ^{2}}\biggl(\pi ^{17}-4368 \pi ^{15}+8!\cdot 143 \pi ^{13}+ \frac{9!\cdot 95{,}719}{15} \pi ^{11}\\ &\quad {}- \frac{14!\cdot 1825}{378} \pi ^{9}- \frac{16!\cdot 839}{175} \pi ^{7} + \frac{16!\cdot 40{,}216}{75} \pi ^{5}-14!\cdot 1{,}193{,}984 \pi ^{3} \\ &\quad {}-14!\cdot 1{,}032{,}192 \pi +14!\cdot 14{,}450{,}688\biggr) + \frac{\rho _{0} \alpha _{2}}{11!\cdot 4 \pi ^{2}} \biggl(\pi ^{10} \alpha _{2}^{10} \\ &\quad {} -2640 \pi ^{8} \alpha _{2}^{8}+ \frac{11!}{21} \pi ^{6} \alpha _{2}^{6}- \frac{11!\cdot 224}{15} \pi ^{4} \alpha _{2}^{4}+12! \cdot 160 \pi ^{2} \alpha _{2}^{2}-12!\cdot 5632 \biggr) \\ &\approx 4.02> 0, \end{aligned} \\& \mu _{2,0} = \frac{64 \alpha _{1} }{63 \pi } \bigl(21 \alpha _{1}^{2}-23 \bigr) \approx -0.62 < 0, \\& \begin{aligned} \mu _{2,1} &= \frac{\alpha _{1}}{882 \pi ^{10}} \biggl(21{,}504 \pi ^{9} \alpha _{1}^{2}-16{,}991 \pi ^{9}- \frac{9!\cdot 81}{70} \pi ^{7}+ \frac{10!\cdot 81}{25} \pi ^{5}-9!\cdot 384 \pi ^{3} \\ &\quad {} +10!\cdot 128 \pi +8!\cdot 8448\biggr) \\ &\approx -0.66< 0, \end{aligned} \\& \begin{aligned} \mu _{2,2} &= - \frac{120 \alpha _{1}}{11!\cdot 7 \pi ^{10}} \biggl(29{,}568 \pi ^{11} \alpha _{1}^{4}+1321 \pi ^{11}- \frac{9!\cdot 224}{135} \pi ^{9} \alpha _{1}^{2}+35{,}011{,}240 \pi ^{9} \\ &\quad {}+ \frac{11!\cdot 405}{7} \pi ^{7} - \frac{12!\cdot 666}{5} \pi ^{5}+12!\cdot 1568 \pi ^{3}-11! \cdot 62{,}464 \pi - \frac{13!\cdot 34{,}304}{117} \biggr) \\ &\approx -0.69 < 0, \end{aligned} \\& \begin{aligned} \mu _{2,3} &= - \frac{ 288 \alpha _{1} }{11!\cdot 7 \pi ^{10}}\biggl( \frac{8!\cdot 11}{10} \pi ^{11} \alpha _{1}^{4}+1981 \pi ^{11}- \frac{11!\cdot 112}{135} \pi ^{9} \alpha _{1}^{2}+ \frac{11!\cdot 2923}{11{,}340} \pi ^{9} \\ &\quad {}+ \frac{11!\cdot 4009}{105} \pi ^{7}- \frac{12!\cdot 3884}{45} \pi ^{5}+ \frac{12!\cdot 8992}{9} \pi ^{3}-12!\cdot 3072 \pi -11! \cdot 35{,}840\biggr) \\ &\approx -0.73 < 0, \end{aligned} \\& \begin{aligned} \mu _{2,4} &= \frac{24 \alpha _{1}}{11!\cdot 7 \pi ^{10}} \biggl(528 \pi ^{1}3 \alpha _{1}^{6}- \frac{11!}{30} \pi ^{11} \alpha _{1}^{4}+11 \pi ^{1}3-59{,}418 \pi ^{11}\\ &\quad {}+ \frac{11!\cdot 112}{9} \pi ^{9} \alpha _{1}^{2}-67{,}745{,}040 \pi ^{9}- \frac{12!\cdot 989}{18} \pi ^{7} \\ &\quad {}+ \frac{12!\cdot 7268}{5} \pi ^{5}-11!\cdot 196{,}352 \pi ^{3}+12!\cdot 42{,}496 \pi +11!\cdot 806{,}912\biggr) \\ &\approx -0.78 < 0, \end{aligned} \\& \begin{aligned} \mu _{2,5} &= \frac{16 \alpha _{1}}{9!\cdot 77\pi ^{10}} \biggl(528 \pi ^{1}3 \alpha _{1}^{6}-8!\cdot 11 \pi ^{11} \alpha _{1}^{4}+11 \pi ^{1}3-19{,}803 \pi ^{11}\\ &\quad {}+8!\cdot 2464 \pi ^{9} \alpha _{1}^{2}-562{,}980 \pi ^{9}- \frac{12!\cdot 1457}{126} \pi ^{7} \\ &\quad {}+ \frac{12!\cdot 4429}{15} \pi ^{5}- \frac{13!\cdot 736}{3} \pi ^{3}+12!\cdot 6016 \pi +11! \cdot 220{,}672\biggr) \\ &\approx -0.83< 0, \end{aligned} \\& \begin{aligned} \mu _{2,6} &= - \frac{\alpha _{1}}{9!\cdot 308 \pi ^{10}} \biggl(88 \pi ^{15} \alpha _{1}^{8}- \frac{9!\cdot 44}{105} \pi ^{13} \alpha _{1}^{6}+\pi ^{15}+ \frac{11!\cdot 8}{5} \pi ^{11} \alpha _{1}^{4}-3168 \pi ^{13} \\ &\quad {} +2{,}851{,}344 \pi ^{11}- \frac{12!\cdot 896}{45} \pi ^{9} \alpha _{1}^{2}- \frac{12!\cdot 4087}{2520} \pi ^{9}+ \frac{14!\cdot 18{,}934}{3185} \pi ^{7} \\ &\quad {}- \frac{16!\cdot 4093}{6825} \pi ^{5} +12!\cdot 266{,}752 \pi ^{3}-12!\cdot 258{,}048 \pi -12!\cdot 2{,}088{,}960\biggr) \\ &\approx -0.89 < 0, \end{aligned} \\& \mu _{2,7} = \frac{512 \rho _{3} \pi ^{7} \alpha _{1}-61{,}440 \pi ^{4} \alpha _{1}+5{,}898{,}240 \alpha _{1}}{\pi ^{10}} \approx -0.96 < 0. \end{aligned}

Note that $$\mu _{1,i}<0$$ and $$\mu _{2,j}>0$$, $$i=0,1,\ldots ,10$$, $$j=0,1,\ldots ,7$$, combining Lemmas 1 and 2 with Eq. (41), for $$x \in (0, \frac{\pi }{2})$$, one obtains

$$\begin{gathered} R_{1}(x,\alpha _{2}) > - x^{3} \sin \biggl( \frac{\alpha _{2} \pi }{2}\biggr) \pi ^{6} U_{c,1}(x) + \bigl(- U_{s,3}(x)+3 L_{s,1}(x) \bigr) \\ \hphantom{R_{1}(x,\alpha _{2}) >}{} \cdot \frac{ \sin ( \frac{\alpha _{2} \pi }{2}) \pi ^{6}}{4}-64 x^{6} U_{s,\alpha _{2}}( x) \\ \hphantom{R_{1}(x,\alpha _{2}) }= \frac{x^{7} ( \frac{\pi }{2}-x)^{2} }{\rho _{0}} \Biggl( \sum _{i=0}^{10}{\mu _{1,i} B_{10,i}(x) }\Biggr)>0, \\ R_{1}(x,\alpha _{1}) < - x^{3} \sin \biggl( \frac{\alpha _{1} \pi }{2}\biggr) \pi ^{6} L_{c,1}(x) +\bigl(- L_{s,3}(x)+3 U_{s,1}(x) \bigr) \\ \hphantom{R_{1}(x,\alpha _{1}) < }{} \cdot \frac{ \sin ( \frac{\alpha _{1} \pi }{2}) \pi ^{6}}{4}-64 x^{6} L_{s,\alpha _{1}}( x) \\ \hphantom{R_{1}(x,\alpha _{1}) }= \frac{x^{9} ( \frac{\pi }{2}-x) }{\pi ^{10}} \Biggl( \sum _{i=0}^{7}{\mu _{2,i} B_{7,i}(x) }\Biggr)< 0, \end{gathered}$$
(44)

which leads to Eq. (42). Thus, the proof is completed.

## 4 Proof of Theorem 2

This is equivalent to proving Eq. (15) in Theorem 2. Let

\begin{aligned}& R_{2}(x,\alpha )=\ln \bigl(F(x)\bigr)-\ln \bigl(G_{2}(x,\alpha )\bigr), \\& p_{1} \triangleq p_{1}(\alpha )=- \frac{x (336 \alpha ^{2} x^{2}-92 x^{2})}{168 \alpha ^{2}}, \qquad q_{1} \triangleq q_{1}(\alpha ) = - \frac{x (63 \alpha ^{2}+69 \alpha )}{168 \alpha ^{2}}, \\& p_{2} \triangleq p_{2}(\alpha ) = \frac{x (336 \alpha ^{2} x^{2}-92 x^{2})}{168 \alpha ^{2}}, \qquad q_{2} \triangleq q_{2}(\alpha ) = \frac{x (63 \alpha ^{2}-69 \alpha )}{168 \alpha ^{2}}, \\& p_{3} \triangleq p_{3}(\alpha ) = \frac{23 x}{168 \alpha }- \frac{3 x }{8}, \qquad q_{3}= 0, \\& p_{4} \triangleq p_{4}(\alpha ) = \frac{23 x}{168 \alpha }+ \frac{3 x}{8}, \qquad q_{4}=0, \\& p_{5} \triangleq p_{5}(\alpha ) =- \frac{147 \alpha ^{2}-23}{168 \alpha ^{2}}, \qquad q_{5}=0, \\& p_{6} = 0, \qquad q_{6} \triangleq q_{6}(\alpha ) = \frac{147 \alpha ^{2}-23}{168 \alpha ^{2}}, \\& p_{7}\triangleq p_{7}(\alpha ) = \frac{23 x^{4}}{42 \alpha }+ \frac{x^{4}}{2}- \frac{147 \alpha ^{2}-23}{56 \alpha ^{2}}, \qquad q_{7}=0, \\& p_{8} \triangleq p_{8}(\alpha ) = \frac{23 x^{4}}{42 \alpha }- \frac{x^{4}}{2},\qquad q_{8} \triangleq q_{8}(\alpha ) = \frac{147 \alpha ^{2}-23}{56 \alpha ^{2}}, \\& R_{3}(x,\alpha )=p_{1} \sin (x-\alpha x) + q_{1} \sin (x-\alpha x) + p_{2} \sin (x+\alpha x) + q_{2} \sin (x+ \alpha x) \\& \hphantom{R_{3}(x,\alpha )=}{} +p_{3} \sin (3x+\alpha x) + q_{3} \sin (3x+\alpha x) +p_{4} \sin (3x- \alpha x) + q_{4} \sin (3x- \alpha x) \\& \hphantom{R_{3}(x,\alpha )=}{} +p_{5} \cos (3x+\alpha x) + q_{5} \cos (3x+\alpha x)+p_{6} \cos (3x- \alpha x) + q_{6} \cos (3x- \alpha x) \\& \hphantom{R_{3}(x,\alpha )=}{} + p_{7} \cos (x-\alpha x) + q_{7} \cos (x-\alpha x) + p_{8} \cos (x+ \alpha x) + q_{8} \cos (x+ \alpha x) . \end{aligned}
(45)

It can be verified that

$$\begin{gathered} p_{i}(\alpha _{j}) \geq 0, \qquad q_{i}(\alpha _{j})\leq 0, \quad i=1,2, \ldots ,8, j=3,4, \\ R_{2}'(x,\alpha )= \frac{\partial }{\partial x} R_{2}(x, \alpha ) = \frac{R_{3}(x,\alpha )}{(\sin (x)^{3}-\cos (x) x^{3}) x \sin (\alpha x)}. \end{gathered}$$
(46)

### 4.1 Proof of $$F(x)< G_{2}(x,\alpha _{3})$$

Let $$\beta _{6}=1-\alpha _{3}$$, $$\beta _{7}=1+\alpha _{3}$$, $$\beta _{8}=1-\alpha _{4}$$, and $$\beta _{9}=1+\alpha _{4}$$. Combining Eq. (19) with Eq. (26) in Lemmas 1 and 3, for $$i=2,3,\ldots ,9$$, one has

$$\begin{gathered} \bar{L}_{c,\beta _{i}}(x) \triangleq \varphi _{g_{1,\beta},20}(x) < \cos (\beta _{i} x) < \varphi _{g_{1,\beta _{i}},18}(x) \triangleq \bar{U}_{c,\beta _{i}}(x), \\ \bar{L}_{s,\beta _{i}}(x) \triangleq \varphi _{g_{2,\beta},17}(x) < \sin (\beta _{i} x) < \varphi _{g_{2,\beta _{i}},15}(x) \triangleq \bar{U}_{s,\beta _{i}}(x). \end{gathered}$$
(47)

Let

\begin{aligned}& \rho _{5} = \cos \biggl(\frac{\pi \beta _{6}}{2}\biggr),\qquad \rho _{6} = \cos \biggl( \frac{\pi \beta _{7}}{2}\biggr),\qquad \rho _{7} = \cos \biggl(\frac{\pi \beta _{2}}{2}\biggr),\qquad \rho _{8} = \cos \biggl(\frac{\pi \beta _{3}}{2}\biggr), \\& \rho _{9} = \sin \biggl(\frac{\pi \beta _{6}}{2}\biggr),\qquad \rho _{10} = \sin \biggl( \frac{\pi \beta _{7}}{2}\biggr),\qquad \rho _{11} = \sin \biggl( \frac{\pi \beta _{2}}{2}\biggr),\qquad \rho _{12} = \sin \biggl( \frac{\pi \beta _{3}}{2}\biggr), \\& \mu _{3,0}= -\frac{52{,}952 \sqrt{14{,}973} }{5{,}347{,}211{,}336{,}775} \approx -1.21173 \cdot 10^{-6} < 0, \\& \begin{aligned} \mu _{3,1}&= \biggl(\frac{12{,}288 \rho _{11}}{\pi ^{15}}- \frac{12{,}288 \rho _{12}}{\pi ^{15}}\biggr)+ \biggl( \frac{14{,}16{,}929{,}895{,}971{,}358{,}499{,}998{,}076}{19{,}542{,}816{,}039{,}671{,}484{,}181{,}468{,}017{,}546{,}375} \\ &\quad {}- \frac{117{,}523{,}114{,}907{,}613{,}501{,}032}{18{,}349{,}803{,}561{,}143{,}724{,}906{,}661{,}425 \pi ^{2}} \\ &\quad {}+ \frac{1{,}781{,}368{,}211{,}358{,}673{,}664}{4{,}091{,}372{,}031{,}470{,}172{,}777{,}405 \pi ^{4}} \\ &\quad {} -\frac{16{,}534{,}606{,}530{,}917{,}824}{770{,}068{,}140{,}687{,}026{,}685 \pi ^{6}} + \frac{4{,}677{,}807{,}501{,}824}{6{,}561{,}086{,}323{,}365 \pi ^{8}}\\ &\quad {}- \frac{49{,}541{,}496{,}832}{3{,}493{,}036{,}197 \pi ^{10}} + \frac{995{,}086{,}336}{7{,}231{,}959 \pi ^{12}} \\ &\quad {} -\frac{2{,}039{,}808}{4991 \pi ^{14}}+ \frac{94{,}208 \rho _{11}}{651 \pi ^{15}} + \frac{94{,}208 \rho _{12}}{651 \pi ^{15}} \biggr) \sqrt{14{,}973} \\ &\approx 7.938427 \cdot 10^{-7} >0, \end{aligned} \\& \begin{aligned} \mu _{3,2}&= \biggl( \frac{13{,}860{,}984{,}407{,}679{,}042{,}511{,}241{,}872}{80{,}794{,}101{,}399{,}478{,}769{,}763{,}709{,}372{,}232{,}703}\\ &\quad {}+ \frac{389{,}773{,}948{,}131{,}949{,}713{,}869{,}312}{207{,}493{,}932{,}576{,}009{,}812{,}406{,}094{,}575 \pi ^{4}} \\ &\quad {} -\frac{40{,}894{,}464}{31 \pi ^{18}} - \frac{6{,}635{,}520 \rho _{10}}{31 \pi ^{15}} + \frac{6{,}635{,}520 \rho _{9}}{31 \pi ^{15}}+ \frac{49{,}152 \rho _{10}}{\pi ^{17}} -\frac{49{,}152 \rho _{9}}{\pi ^{17}} \\ &\quad {} +\frac{30{,}670{,}848 \rho _{5}}{31 \pi ^{18}}+ \frac{10{,}223{,}616 \rho _{8}}{31 \pi ^{18}}+ \frac{2{,}521{,}825{,}280}{4991 \pi ^{16}}\\ &\quad {}+ \frac{5{,}375{,}227{,}967{,}872{,}815{,}104}{770{,}068{,}140{,}687{,}026{,}685 \pi ^{8}}- \frac{44{,}736{,}004{,}993{,}052{,}672}{177{,}149{,}330{,}730{,}855 \pi ^{10}} \\ &\quad {} + \frac{305{,}099{,}932{,}893{,}184}{52{,}395{,}542{,}955 \pi ^{12}} - \frac{550{,}535{,}757{,}824}{7{,}231{,}959 \pi ^{14}} \\ &\quad {}- \frac{979{,}671{,}812{,}418{,}224{,}756{,}622{,}224}{48{,}493{,}340{,}048{,}812{,}615{,}834{,}908{,}232{,}125 \pi ^{2}}\\ &\quad {} -\frac{8{,}202{,}708{,}703{,}446{,}794{,}527{,}744}{61{,}370{,}580{,}472{,}052{,}591{,}661{,}075 \pi ^{6}}\biggr) \\ &\quad {} +\biggl(-\frac{1{,}665{,}837{,}617{,}047{,}196{,}295{,}904}{1{,}302{,}854{,}402{,}644{,}765{,}612{,}097{,}867{,}836{,}425} \\ &\quad {}- \frac{2{,}810{,}863{,}942{,}243{,}212{,}203{,}024{,}368}{19{,}542{,}816{,}039{,}671{,}484{,}181{,}468{,}017{,}546{,}375 \pi ^{2}} \\ &\quad {} +\frac{32{,}693{,}850{,}502{,}844{,}788{,}384}{2{,}621{,}400{,}508{,}734{,}817{,}843{,}808{,}775 \pi ^{4}}- \frac{3{,}142{,}228{,}027{,}482{,}915{,}584}{4{,}091{,}372{,}031{,}470{,}172{,}777{,}405 \pi ^{6}} \\ &\quad {} + \frac{20{,}431{,}505{,}586{,}135{,}808}{770{,}068{,}140{,}687{,}026{,}685 \pi ^{8}} + \frac{4{,}692{,}888{,}387{,}584}{59{,}049{,}776{,}910{,}285 \pi ^{10}}- \frac{117{,}304{,}573{,}952}{3{,}493{,}036{,}197 \pi ^{12}} \\ &\quad {} +\frac{1{,}440{,}382{,}976}{7{,}231{,}959 \pi ^{14}}+ \frac{8{,}159{,}232}{4991 \pi ^{16}}- \frac{376{,}832 \rho _{9}}{217 \pi ^{17}}- \frac{376{,}832 \rho _{10}}{217 \pi ^{17}} \biggr) \sqrt{14{,}973} \\ &\approx -2.91 \cdot 10^{-7} < 0, \end{aligned} \\& \mu _{3,3}=\biggl(- \frac{55{,}443{,}937{,}630{,}716{,}170{,}044{,}967{,}488}{80{,}794{,}101{,}399{,}478{,}769{,}763{,}709{,}372{,}232{,}703 \pi ^{2}}\\& \hphantom{\mu _{3,3}=}{}+ \frac{3{,}918{,}687{,}249{,}672{,}899{,}026{,}488{,}896}{48{,}493{,}340{,}048{,}812{,}615{,}834{,}908{,}232{,}125 \pi ^{4}} \\& \hphantom{\mu _{3,3}=}{}- \frac{1{,}559{,}095{,}792{,}527{,}798{,}855{,}477{,}248}{207{,}493{,}932{,}576{,}009{,}812{,}406{,}094{,}575 \pi ^{6}}\\& \hphantom{\mu _{3,3}=}{}+ \frac{32{,}810{,}834{,}813{,}787{,}178{,}110{,}976}{61{,}370{,}580{,}472{,}052{,}591{,}661{,}075 \pi ^{8}} - \frac{21{,}500{,}911{,}871{,}491{,}260{,}416}{770{,}068{,}140{,}687{,}026{,}685 \pi ^{10}} \\& \hphantom{\mu _{3,3}=}{}+ \frac{178{,}944{,}019{,}972{,}210{,}688}{177{,}149{,}330{,}730{,}855 \pi ^{12}} - \frac{1{,}220{,}399{,}731{,}572{,}736}{52{,}395{,}542{,}955 \pi ^{14}}\\& \hphantom{\mu _{3,3}=}{}+ \frac{2{,}202{,}143{,}031{,}296}{7{,}231{,}959 \pi ^{16}}- \frac{10{,}087{,}301{,}120}{4991 \pi ^{18}} \\& \hphantom{\mu _{3,3}=}{} -\frac{122{,}683{,}392 \rho _{6}}{31 \pi ^{20}}- \frac{40{,}894{,}464 \rho _{7}}{31 \pi ^{20}}+ \frac{163{,}577{,}856}{31 \pi ^{20}}\biggr) \\& \hphantom{\mu _{3,3}=}{}+\biggl( \frac{29{,}639{,}010{,}266{,}103{,}867{,}934}{1{,}231{,}197{,}410{,}499{,}303{,}503{,}432{,}485{,}105{,}421{,}625} \\& \hphantom{\mu _{3,3}=}{} + \frac{139{,}388{,}160{,}789{,}931{,}088{,}576}{27{,}332{,}609{,}845{,}694{,}383{,}470{,}584{,}639{,}925 \pi ^{2}} \\& \hphantom{\mu _{3,3}=}{}- \frac{85{,}650{,}627{,}306{,}371{,}917{,}376}{148{,}209{,}951{,}840{,}007{,}008{,}861{,}496{,}125 \pi ^{4}} \\& \hphantom{\mu _{3,3}=}{} +\frac{1{,}041{,}424{,}249{,}904{,}576{,}512}{20{,}456{,}860{,}157{,}350{,}863{,}887{,}025 \pi ^{6}} - \frac{2{,}584{,}322{,}860{,}261{,}376}{770{,}068{,}140{,}687{,}026{,}685 \pi ^{8}} \\& \hphantom{\mu _{3,3}=}{} +\frac{27{,}847{,}335{,}952{,}384}{177{,}149{,}330{,}730{,}855 \pi ^{10}}- \frac{84{,}756{,}119{,}552}{17{,}465{,}180{,}985 \pi ^{12}}+ \frac{630{,}456{,}320}{7{,}231{,}959 \pi ^{14}} \\& \hphantom{\mu _{3,3}=}{} - \frac{3{,}407{,}872}{4991 \pi ^{16}} \biggr) \sqrt{14{,}973} \\& \hphantom{\mu _{3,3}}\approx -2.9 \cdot 10^{-10} < 0, \\& \begin{aligned} \mu _{3,4}&= \biggl(\frac{131{,}072 \rho _{5}}{\pi ^{18}}- \frac{131{,}072 \rho _{6}}{\pi ^{18}}\biggr)\\ &\quad {}+ \biggl(- \frac{118{,}556{,}041{,}064{,}415{,}471{,}736}{1{,}231{,}197{,}410{,}499{,}303{,}503{,}432{,}485{,}105{,}421{,}625 \pi ^{2}} \\ &\quad {} + \frac{2{,}001{,}544{,}751{,}520{,}343{,}808}{33{,}987{,}506{,}155{,}950{,}407{,}272{,}118{,}291{,}385 \pi ^{4}}\\ &\quad {} - \frac{105{,}738{,}640{,}249{,}509{,}632}{3{,}866{,}346{,}569{,}739{,}313{,}274{,}647{,}725 \pi ^{6}} \\ &\quad {} +\frac{2{,}242{,}738{,}829{,}640{,}704}{242{,}571{,}464{,}316{,}413{,}405{,}775 \pi ^{8}} - \frac{4{,}806{,}306{,}467{,}840}{2{,}232{,}081{,}567{,}208{,}773 \pi ^{10}} \\ &\quad {} + \frac{1{,}059{,}206{,}610{,}944}{3{,}300{,}919{,}206{,}165 \pi ^{12}}- \frac{12{,}524{,}683{,}264}{455{,}613{,}417 \pi ^{14}} + \frac{344{,}719{,}360}{314{,}433 \pi ^{16}}+ \frac{3{,}014{,}656 \rho _{5}}{651 \pi ^{18}} \\ &\quad {} +\frac{3{,}014{,}656 \rho _{6}}{651 \pi ^{18}}- \frac{6{,}029{,}312}{651 \pi ^{18}} \biggr) \sqrt{14{,}973} \\ &\approx -1.4 \cdot 10^{-14} < 0. \end{aligned} \end{aligned}

Note that $$2 \sqrt{ \mu _{3,0} \mu _{3,2}}-\mu _{3,1} \approx 1.9 \cdot 10^{-6}>0$$, $$\mu _{3,i}<0$$, $$i\neq 1$$ and $$\mu _{3,1}>0$$, combining Eq. (47) with Eq. (46), for $$x \in (0, \frac{\pi }{2})$$, one obtains

\begin{aligned} R_{3}(x,\alpha _{3})&< p_{1} \bar{U}_{s,\beta _{6}}(x) + q_{1} \bar{L}_{s, \beta _{6}}(x) + p_{2} \bar{U}_{s,\beta _{7}}(x) + q_{2} \bar{L}_{s, \beta _{7}}(x) \\ &\quad {} +p_{3} \bar{U}_{s,\beta _{3}}(x) + q_{3} \bar{L}_{s,\beta _{3}}(x) +p_{4} \bar{U}_{s,\beta _{2}}(x) + q_{4} \bar{L}_{s,\beta _{2}}(x) \\ &\quad {} +p_{5} \bar{U}_{c,\beta _{3}}(x) + q_{5} \bar{L}_{c,\beta _{3}}(x) +p_{6} \bar{U}_{c,\beta _{2}}(x) + q_{6} \bar{L}_{c,\beta _{2}}(x) \\ &\quad {} + p_{7} \bar{U}_{c,\beta _{6}}(x) + q_{7} \bar{L}_{c,\beta _{6}}(x) + p_{8} \bar{U}_{c,\beta _{7}}(x) + q_{8} \bar{L}_{c,\beta _{7}}(x) \\ &=x^{14}\bigl(\mu _{3,0}+\mu _{3,1} x^{2} +\mu _{3,2} x^{4} +\mu _{3,3} x^{6} +\mu _{3,4} x^{8}\bigr) < 0. \end{aligned}
(48)

Combining Eq. (45) and Eq. (46) with Eq. (48), $$\forall x \in (0, \frac{\pi }{2})$$, one has

$$R_{2}'(x,\alpha _{3})< 0\quad \text{and}\quad R_{2}(x,\alpha _{3}) < R_{2}(0, \alpha _{3})=0 \quad \text{and}\quad F(x)< G_{2}(x,\alpha _{3}).$$
(49)

### 4.2 Proof of $$F(x)>G_{2}(x,\alpha _{4})$$

First, we prove

$$R_{3}(x,\alpha _{4}) > 0, \quad \forall x \in \biggl(0, \frac{5}{4}\biggr).$$
(50)

Let $$\theta = \frac{5}{4}$$ and $$\theta _{2}=\theta ^{2}= \frac{25}{16}$$, $$\bar{B}_{6,i}(x)=C_{i}^{6}(\theta _{2}-x^{2})^{6-i} x^{2i}$$ such that $$\bar{B}_{6,i}(x)>0$$ for $$\forall x \in (0,\theta )$$. By using Maple software, for $$\forall x \in (0,\theta )$$, it can be verified that

\begin{aligned} R_{3}(x,\alpha _{4})&> p_{1} \bar{L}_{s,\beta _{8}}(x) + q_{1} \bar{U}_{s, \beta _{8}}(x) + p_{2} \bar{L}_{s,\beta _{9}}(x) + q_{2} \bar{U}_{s, \beta _{9}}(x) \\ &\quad {} +p_{3} \bar{L}_{s,\beta _{5}}(x) + q_{3} \bar{U}_{s,\beta _{5}}(x) +p_{4} \bar{L}_{s,\beta _{4}}(x) + q_{4} \bar{U}_{s,\beta _{4}}(x) \\ &\quad {} +p_{5} \bar{L}_{c,\beta _{5}}(x) + q_{5} \bar{U}_{c,\beta _{5}}(x) +p_{6} \bar{L}_{c,\beta _{4}}(x) + q_{6} \bar{U}_{c,\beta _{4}}(x) \\ &\quad {} + p_{7} \bar{L}_{c,\beta _{8}}(x) + q_{7} \bar{U}_{c,\beta _{8}}(x) + p_{8} \bar{L}_{c,\beta _{9}}(x) + q_{8} \bar{U}_{c,\beta _{9}}(x) \\ &=x^{12}\Biggl(\sum_{i=0}^{6}{ \mu _{4,i} \bar{B}_{6,i}(x)}\Biggr) >0, \end{aligned}
(51)

where $$\mu _{4,0} \approx 2.5 \cdot 10^{-6}>0$$, $$\mu _{4,1} \approx 2.1\cdot 10^{-6}>0$$, $$\mu _{4,2} \approx 1.7 \cdot 10^{-6}>0$$, $$\mu _{4,3} \approx 1.2 \cdot 10^{-6} >0$$, $$\mu _{4,4} \approx 8.8 \cdot 10^{-7} >0$$, $$\mu _{4,5} \approx 5.0 \cdot 10^{-7} >0$$, $$\mu _{4,6} \approx 1.3 \cdot 10^{-7} >0$$.

Secondly, let $$\theta _{3}= \frac{\pi ^{2}}{4}$$, $$\hat{B}_{6,i}(x)= \frac{C_{i}^{6}(\theta _{3}-x^{2})^{6-i} (x^{2}-\theta _{2})^{i}}{(\theta _{3}-\theta _{2})^{6}}$$ such that $$\hat{B}_{6,i}(x)>0$$ for $$\forall x \in (\theta , \frac{\pi }{2})$$, and

$$\begin{gathered} \bar{p}_{1} \triangleq \bar{p}_{1}(\alpha )= \frac{x^{2} (23-84 \alpha ^{2})}{14 \alpha ^{2}}, \\ \bar{q}_{1} \triangleq \bar{q}_{1}(\alpha ) = \frac{ x^{4} (21 \alpha ^{2}+2 \alpha -23)}{42 \alpha} + \frac{ (-147 \alpha ^{3}+126 \alpha ^{2}-23)}{(56 \alpha ^{2})}, \\ \bar{p}_{2} \triangleq \bar{p}_{2}(\alpha ) = \frac{ (-147 \alpha ^{3}-126 \alpha ^{2}+23)}{56 \alpha ^{2}}, \\ \bar{q}_{2}\triangleq \bar{q}_{2}(\alpha ) = \frac{ (252 \alpha ^{2}-69) x^{2}}{42 \alpha ^{2}}+ \frac{ (21 \alpha ^{3}-2 \alpha ^{2}-23 \alpha ) x^{4}}{42 \alpha ^{2}} , \\ \bar{p}_{3} \triangleq \bar{p}_{3}(\alpha ) = 0, \qquad \bar{q}_{3} \triangleq \bar{q}_{3}(\alpha ) = \frac{ (49 \alpha ^{3}+126 \alpha ^{2}-23)}{56 \alpha ^{2}}, \\ \bar{p}_{4}\triangleq \bar{p}_{4}(\alpha ) = \frac{ (49 \alpha ^{3}-126 \alpha ^{2}+23)}{56 \alpha ^{2}}, \qquad \bar{q}_{4}=0, \\ \bar{p}_{5} \triangleq \bar{p}_{5}(\alpha ) = \frac{ x (3+\alpha ) (23-63 \alpha )}{168 \alpha}, \qquad \bar{q}_{5}=0, \qquad \\ \bar{p}_{6} \triangleq \bar{p}_{6}(\alpha ) = \frac{ -x (\alpha -3) (23+63 \alpha )}{168 \alpha}, \qquad \bar{q}_{6} =0, \\ \bar{p}_{7}\triangleq \bar{p}_{7}(\alpha ) = \frac{ x^{3} (84 \alpha ^{3}+69 \alpha +23)}{42 \alpha ^{2}}, \quad \bar{q}_{7} \triangleq \bar{q}_{7}( \alpha )= \frac{x (21 \alpha ^{2}+2 \alpha -23)}{56\alpha}, \\ \bar{p}_{8} \triangleq \bar{p}_{8}(\alpha ) = 0, \\ \bar{q}_{8} \triangleq \bar{q}_{8}(\alpha ) = \frac{(63 \alpha ^{3}-6 \alpha ^{2}-69 \alpha ) x}{168 \alpha ^{2}} + \frac{(336 \alpha ^{3}+276 \alpha -92) x^{3}}{168 \alpha ^{2}}. \end{gathered}$$
(52)

By using Maple software, for $$\forall x \in ( \frac{5}{4}, \frac{\pi }{2})$$, it can be verified that

\begin{aligned} R_{4}(x,\alpha )&=R_{3}'(x, \alpha ) =\bar{p}_{1} \sin (x-\alpha x) + \bar{q}_{1} \sin (x-\alpha x) \\ &\quad {} + \bar{p}_{2} \sin (x+\alpha x) + \bar{q}_{2} \sin (x+\alpha x) + \bar{p}_{3} \sin (3x+\alpha x) \\ &\quad {} + \bar{q}_{3} \sin (3x+\alpha x) +\bar{p}_{4} \sin (3x-\alpha x) + \bar{q}_{4} \sin (3x-\alpha x) \\ &\quad {} +\bar{p}_{5} \cos (3x+\alpha x) + \bar{q}_{5} \cos (3x+\alpha x)+ \bar{p}_{6} \cos (3x-\alpha x) \\ &\quad {} + \bar{q}_{6} \cos (3x-\alpha x) + \bar{p}_{7} \cos (x-\alpha x) + \bar{q}_{7} \cos (x-\alpha x) \\ &\quad {} + \bar{p}_{8} \cos (x+\alpha x) + \bar{q}_{8} \cos (x+\alpha x) \\ &< p_{1} \bar{U}_{s,\beta _{8}}(x) + q_{1} \bar{L}_{s,\beta _{8}}(x) + p_{2} \bar{U}_{s,\beta _{9}}(x) + q_{2} \bar{L}_{s,\beta _{9}}(x) \\ &\quad {} +p_{3} \bar{U}_{s,\beta _{5}}(x) + q_{3} \bar{L}_{s,\beta _{5}}(x) +p_{4} \bar{U}_{s,\beta _{4}}(x) + q_{4} \bar{L}_{s,\beta _{4}}(x) \\ &\quad {} +p_{5} \bar{U}_{c,\beta _{5}}(x) + q_{5} \bar{L}_{c,\beta _{5}}(x) +p_{6} \bar{U}_{c,\beta _{4}}(x) + q_{6} \bar{L}_{c,\beta _{4}}(x) \\ &\quad {} + p_{7} \bar{U}_{c,\beta _{8}}(x) + q_{7} \bar{L}_{c,\beta _{8}}(x) + p_{8} \bar{U}_{c,\beta _{9}}(x) + q_{8} \bar{L}_{c,\beta _{9}}(x) \\ &=x^{11}\Biggl(\sum_{i=0}^{6}{ \mu _{5,i} \bar{B}_{6,i}(x)}\Biggr) < 0, \end{aligned}
(53)

where $$\mu _{5,0} \approx -0.12 \cdot 10^{-4}<0$$, $$\mu _{5,1} \approx -0.15 \cdot 10^{-4}<0$$, $$\mu _{5,2} \approx -0.17\cdot 10^{-4}<0$$, $$\mu _{5,3} \approx -0.19\cdot 10^{-4}<0$$, $$\mu _{5,4} \approx -0.20\cdot 10^{-4}<0$$, $$\mu _{5,5} \approx -0.19 \cdot 10^{-4}<0$$, $$\mu _{5,6} \approx -0.17 \cdot 10^{-4}<0$$.

Note that $$R_{3}( \frac{5}{4},\alpha _{4})\approx 2.4 \cdot 10^{-6}>0$$ and $$R_{3}( \frac{\pi }{2},\alpha _{4})\approx -2.2 \cdot 10^{-4}<0$$, combining Eq. (46) and Eq. (52) with Eq. (53), there exists $$x_{0} \in ( \frac{5}{4}, \frac{\pi }{2})$$ such that

$$\textstyle\begin{cases} R_{3}(x,\alpha _{4}) > 0 \quad \text{and}\quad R_{2}'(x,\alpha _{4})>0,\quad \forall x \in (0,x_{0}), \\ R_{3}(x,\alpha _{4}) < 0 \quad \text{and}\quad R_{2}'(x,\alpha _{4})< 0,\quad \forall x \in (x_{0}, \frac{\pi }{2}). \end{cases}$$
(54)

Combining with Eq. (54), one obtains

$${\textstyle\begin{cases} R_{2}(x_{0},\alpha _{4}) > R_{2}(x,\alpha _{4}) > R_{2}(0,\alpha _{4})=0, \quad \forall x \in (0,x_{0}), \\ R_{2}(x_{0},\alpha _{4}) > R_{2}(x,\alpha _{4}) > R_{2}( \frac{\pi }{2},\alpha _{4})=0, \quad \forall x \in (x_{0}, \frac{\pi }{2}), \end{cases}\displaystyle }$$
(55)

$$R_{2}(x,\alpha _{4}) > 0, \quad \forall x \in \biggl(0, \frac{\pi }{2}\biggr).$$
(56)

From Eq. (56), we obtain

$$F(x) > G_{2}(x,\alpha _{4}), \quad \forall x \in \biggl(0, \frac{\pi }{2}\biggr).$$
(57)

Combining Eq. (57) with Eq. (49), one obtains Eq. (15), and the proof is completed.

## 5 Discussions and conclusions

Figure 1 shows the error plots of different bounds, i.e., $$F(x)-Z_{i}(x)$$ from Eq. (8) (in solid black) and $$F(x)-G_{1}(x,\alpha _{i})$$ from Eq. (14) (in dashed red), $$i=1,2$$. the maximum errors of lower bounds $$F(x)-Z_{1}(x)$$ and $$F(x)-G_{1}(x,\alpha _{2})$$ are ≈0.0024817 and ≈0.0008217, while $$F(x) \leq G_{1}(x,\alpha _{1}) \leq Z_{2}(x)$$, which means the approximation effect from Eq. (14) is much better than that from Eq. (8). Similarly, Fig. 2 shows the error plots of different bounds, i.e., $$F(x)-Z_{i+2}(x)$$ from Eq. (9) (in solid black) and $$F(x)-G_{2}(x,\alpha _{i})$$ from Eq. (15) (in dashed red), $$i=1,2$$. Again, it shows that the approximation effect from Eq. (15) is much better than that from Eq. (9).

This paper provides new bounds of Mitrinović–Adamović inequalities, which achieve much better approximation accuracy. It also proposes a new method for proving the inequalities by combining the classical mathematical method with Maple software, the corresponding idea is coincident with those in [6, 23], which tends to automatically prove the inequalities on mixed trigonometric polynomial functions. In principle,for a positive integer number n and $$x \in [0,\frac{\pi}{2n}]$$, $$\cos (n x)$$ and $$\sin (n x)$$ can be bounded by using two polynomials that satisfy arbitrary precision, which can be proved in a similar way to Lemma 1. Thus, any mixed trigonometric polynomial function can be bounded by piecewise polynomials. Note that there are many methods for proving inequalities in polynomial form, it is promising to extend the idea in this paper to automatically prove inequalities on mixed trigonometric polynomial functions, which will be one of our future studies.

Not applicable.

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## Acknowledgements

The authors would like to thank the editor and the anonymous referees for their valuable suggestions and comments that helped us to improve this paper greatly.

## Funding

This research work was partially supported by the National Natural Science Foundation of China (61972120), the Zhejiang Basic Public Welfare Research Project (Grant No. LGG20F020001) and the Key Lab of Film and TV Media Technology of Zhejiang Province.

## Author information

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### Contributions

QG deduced the conclusions in Theorems 1 and 2; both QG and CX verified and proved these two theorems. All authors contributed equally to the manuscript and read and approved the final manuscript.

### Corresponding author

Correspondence to Xiao-Diao Chen.

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