New Bounds for the Sine Function and Tangent Function

Abstract

Using the power series expansion technique, this paper established two new inequalities for the sine function and tangent function bounded by the functions $x^2{\left(sin(\lambda x)/(\lambda x)\right)}^{\alpha }$ and $x^2{\left(tan(\mu x)/(\mu x)\right)}^{\beta }$. These results are better than the ones in the previous literature.

1. Introduction

Because of the fact the functions $cosx$ and $(sinx)/x$ are less than 1 for $x\in (0,\pi /2)$, in order to determine this relationship $(sinx)/x$ and the weighted geometric mean of $cosx$ and 1, we examine the Taylor expansion of the following function:

$$\begin{array}{ccc}\hfill \frac{sinx}{x}-{\left(cosx\right)}^{\beta }& =& \left(\frac{1}{2}\beta -\frac{1}{6}\right)x^2+\left(-\frac{1}{8}{\beta }^2+\frac{1}{12}\beta +\frac{1}{120}\right)x^4\hfill \\ & & +\left(\frac{1}{48}{\beta }^3-\frac{1}{24}{\beta }^2+\frac{1}{45}\beta -\frac{1}{5040}\right)x^6+O\left(x^8\right).\hfill \end{array}$$

When choosing $\beta$ = $1/3$ in above formula we can obtain the following fact

$$\frac{sinx}{x}-{\left(cosx\right)}^{1/3}=\frac{1}{45}{x}^4+\frac{19}{5670}{x}^6+O(x^8),$$

which will motivate us to prove the following inequality

$$\frac{sinx}{x}>{\left(cosx\right)}^{1/3}$$

(1)

holds for $0<x<\pi /2$. The above inequality was confirmed by Mitrinović and Adamović in [1], so we call it Mitrinović–Adamović inequality. On the other hand, the relationship between $(sinx)/x$ and the weighted arithmetic mean of $cosx$ and 1 has been discussed in Zhu [2] just published, described as the following inequality similarly:

$$\frac{sinx}{x}<\frac{2}{3}+\frac{1}{3}cosx$$

(2)

or

$$\frac{3sinx}{2+cosx}<x.$$

(3)

In 1451, using a geometrical method Nicolaus De Cusa (1401–1464) discovered (3), and in 1664 when considering the estimation of $\pi$ Christian Huygens (1629–1695) confirmed (2). In view of the above historical facts (see [3,4,5,6,7,8,9,10]), we call the inequality (2) Cusa-Huygens inequality.

In 2018, Zhu [11] shown two improvements to (3) as follows: the inequalities

$$\frac{1}{180}{x}^5<x-\frac{3sinx}{2+cosx}$$

(4)

and

$$\frac{1}{2100}{x}^7<x-\frac{3sinx}{2+cosx}\left[1+\frac{{(1-cosx)}^2}{9(3+2cosx)}\right]$$

(5)

hold for all $x\in \left(0,\pi \right]$, where $1/180$ and $1/2100$ are the best constants in previous inequalities, respectively. Two results of previous proposition are corrections of Theorem 3.4.20 from monograph Mitrinović [9]. Malešević et al. [12] made a bilateral supplement to the above two inequalities. Chen and Cheung [13] obtained the bounds for $\left(sinx\right)/x$ in term of ${\left(\left(2+cosx\right)/3\right)}^{\delta }$ as follows

$${\left(\frac{2+cosx}{3}\right)}^{{\theta }_0}<\frac{sinx}{x}<{\left(\frac{2+cosx}{3}\right)}^{{\vartheta }_0}$$

(6)

holds for all $0<x<\pi /2$, where ${\vartheta }_0=1$ and ${\theta }_0=(ln\pi -ln2)/(ln3-ln2)$ are the best possible constants in (6). The double inequality (6) was proved by Bagul [14] and Zhu [15] in different ways. In Zhu [15] some new improvements to inequality (2) can be found:

$$\left(1-\frac{x^3}{{\pi }^3}\right)\frac{2+cosx}{3}<\frac{sinx}{x}<\left(1-\frac{x^4}{180}\right)\frac{cosx+2}{3},$$

(7)

$$\left[1+\frac{8\left(\pi -3\right)}{{\pi }^3}{x}^2\right]\frac{2+cosx}{3}-\frac{8\left(\pi -3\right)}{{\pi }^3}{x}^2<\frac{sinx}{x}<\left(1+\frac{1}{30}{x}^2\right)\frac{2+cosx}{3}-\frac{1}{30}{x}^2,$$

(8)

and

$$\begin{array}{ccc}& & \left[1+\frac{1}{30}{x}^2+\frac{2\left(240\pi -{\pi }^3-720\right)}{15{\pi }^5}{x}^4\right]\frac{2+cosx}{3}-\left[\frac{1}{30}{x}^2+\frac{2\left(240\pi -{\pi }^3-720\right)}{15{\pi }^5}{x}^4\right]\hfill \\ & <& \frac{sinx}{x}\hfill \\ & <& \left(1+\frac{1}{30}{x}^2+\frac{1}{840}{x}^4\right)\frac{2+cosx}{3}-\left(\frac{1}{30}{x}^2+\frac{1}{840}{x}^4\right)\hfill \end{array}$$

(9)

hold for $0<x<\pi /2$.

Bercu [16] used the truncations of Fourier cosine series to the inequality (2) and obtained an enhanced form of (2):

$$\frac{sinx}{x}-\frac{2+cosx}{3}<-\frac{1}{45}{\left(1-cosx\right)}^2,0<x<\frac{\pi }{2}.$$

(10)

Bagul et al. [17] draw two conclusions about the improvement of inequality (2):

$$-\left(\frac{2}{3}-\frac{2}{\pi }\right)\frac{1}{\left(\pi /2-1\right)}\left(x-sinx\right)<\frac{sinx}{x}-\frac{2+cosx}{3}<-\left(\frac{2}{3}-\frac{2}{\pi }\right)\frac{1}{{\left(\pi /2-1\right)}^2}{\left(x-sinx\right)}^2$$

(11)

and

$$-\left(\frac{2}{3}-\frac{2}{\pi }\right)\left(sinx-xcosx\right)<\frac{sinx}{x}-\frac{2+cosx}{3}<-\left(\frac{2}{3}-\frac{2}{\pi }\right){\left(sinx-xcosx\right)}^2$$

(12)

hold for $0<x<\pi /2$.

Recently, Zhu [2] improved the famous inequality (2) using two different technology paths and draw two results as follows: for $0<x<\pi /2$, the two inequalities

$$\frac{sinx}{x}-\frac{2+cosx}{3}<-\frac{1}{180}{x}^4{\left(\frac{sinx}{x}\right)}^{2/7}$$

(13)

and

$$\frac{sinx}{x}-\frac{2+cosx}{3}<-\left(\frac{2}{3}-\frac{2}{\pi }\right){\left(sinx-xcosx\right)}^{\left({\pi }^2-12\right)/\left(3{\pi }^2-{\pi }^3\right)}$$

(14)

hold with the best constant $-1/180$ and $\left(2/3-2/\pi \right)$ respectively.

Inequalities on two functions $\left(sinx\right)/x$ and $\left(tanx\right)/x$ arouse great enthusiasm of researchers. Interested readers can refer to these literatures [18,19,20,21,22,23,24,25,26,27,28,29,30,31,32,33,34,35,36,37,38,39,40,41,42,43,44,45,46,47,48,49,50,51,52,53,54,55,56,57,58,59,60,61] and monograph [62] which was edited by Rassias and Andrica.

This paper focuses on some new bounds for the functions $sin\left(x\right)/x$ and $tan(x)/x$ and wants to improve the following inequalities:

$$1-\frac{sinx}{x}>0\mathrm{and}\frac{tanx}{x}-1>0.$$

(15)

Recently, Wu and Bercu [63] thought of Fourier series technology to approximate these two functions. They considered the power series expansion of the following function

$$\begin{array}{ccc}& & 1-\frac{sinx}{x}-\left(a+bcosx+ccos2x+dcos3x\right)\hfill \\ & =& -\left(a+b+c+d\right)+x^2\left(\frac{1}{2}b+2c+\frac{9}{2}d+\frac{1}{6}\right)-x^4\left(\frac{1}{24}b+\frac{2}{3}c+\frac{27}{8}d+\frac{1}{120}\right)\hfill \\ & & +x^6\left(\frac{1}{720}b+\frac{4}{45}c+\frac{81}{80}d+\frac{1}{5040}\right)+O\left(x^8\right).\hfill \end{array}$$

To obtain a slightly higher precision approximation, they let

$$\left\{\begin{array}{ccc}a+b+c+d& =& 0\\ \frac{1}{2}b+2c+\frac{9}{2}d+\frac{1}{6}& =& 0\\ \frac{1}{24}b+\frac{2}{3}c+\frac{27}{8}d+\frac{1}{120}& =& 0\\ \frac{1}{720}b+\frac{4}{45}c+\frac{81}{80}d+\frac{1}{5040}& =& 0\end{array}\right.$$

to obtain these constants

$$a=\frac{359}{945},b=-\frac{167}{420},c=\frac{2}{105},d=-\frac{1}{756}$$

and find

$$1-\frac{sinx}{x}-\left(\frac{359}{945}-\frac{167}{420}cosx+\frac{2}{105}cos2x-\frac{1}{756}cos3x\right)=\frac{23}{226800}{x}^8+O\left(x^{10}\right),$$

which leads them to prove

$$\begin{array}{ccc}\hfill 1-\frac{sinx}{x}& >& \frac{359}{945}-\frac{167}{420}cosx+\frac{2}{105}cos2x-\frac{1}{756}cos3x\hfill \\ & =& \frac{\left(1-cosx\right)\left(-31cosx+5{cos}^{2}x+341\right)}{945}.\hfill \end{array}$$

This technique can be used to deal with the approximation problem of another function $tan(x)/x-1$, and then they obtain the following results.

Proposition 1

([63]). The following inequalities

$$\begin{array}{ccc}\hfill 1-\frac{sinx}{x}& >& \frac{\left(1-cosx\right)\left(-31cosx+5{cos}^{2}x+341\right)}{945},\hfill \end{array}$$

(16)

$$\begin{array}{ccc}\hfill \frac{tanx}{x}-1& >& \frac{\left(1-cosx\right)\left(604{cos}^{2}x-1817cosx+1843\right)}{945}\hfill \end{array},$$

(17)

$$\begin{array}{cc}\hfill 1-\frac{sinx}{x}& >1-\left(\frac{2+cosx}{3}-\frac{1}{180}{x}^4+\frac{1}{3780}{x}^6\right)\hfill \\ \hfill & >1-\left(1+\frac{\left(1-cosx\right)\left(-5{cos}^{2}x+31cosx-341\right)}{945}\right)\hfill \\ \hfill & >1-\frac{2+cosx}{3}\hfill \end{array}$$

(18)

hold for all $x\in \left(0,\pi /2\right)$.

In this paper, we want to obtain an approximation with appropriate accuracy about these two functions. We examine the power series expansion of function in the following form

$$\begin{array}{ccc}& & 1-\frac{sinx}{x}-a{x}^2{\left(\frac{sinbx}{bx}\right)}^c\hfill \\ & =& -x^2\left(a-\frac{1}{6}\right)+x^4\left(\frac{1}{6}a{b}^{2}c-\frac{1}{120}\right)+x^6\left[a\left(\frac{1}{180}{b}^{4}c-\frac{1}{72}{b}^4{c}^2\right)+\frac{1}{5040}\right]\hfill \\ & & +x^8\left[\frac{1}{45360}a{b}^{6}c\left(35c^2-42c+16\right)-\frac{1}{362880}\right]+O\left(x^{10}\right),\hfill \end{array}$$

and let

$$\left\{\begin{array}{ccc}a-\frac{1}{6}& =& 0\\ \frac{1}{6}a{b}^{2}c-\frac{1}{120}& =& 0\\ a\left(\frac{1}{180}{b}^{4}c-\frac{1}{72}{b}^4{c}^2\right)+\frac{1}{5040}& =& 0\end{array}\right.$$

to determine

$$a=\frac{1}{6},b=\pm \frac{\sqrt{7}}{14},c=\frac{42}{5}.$$

We can obtain that

$$1-\frac{sinx}{x}-\frac{1}{6}{x}^2{\left(\frac{sin\frac{\sqrt{7}}{14}x}{\frac{\sqrt{7}}{14}x}\right)}^{\frac{42}{5}}=\frac{1}{4116000}{x}^8+O\left(x^{10}\right).$$

In the same way, we obtain

$$\frac{tanx}{x}-1-\frac{1}{3}{x}^2{\left(\frac{tan\frac{\sqrt{43}}{7}x}{\frac{\sqrt{43}}{7}x}\right)}^{\frac{294}{215}}=\frac{12416}{18907875}{x}^8+O\left(x^{10}\right).$$

With the above foreshadowing, we can now announce the main work of this paper which established two inequalities of exponential type for the functions $1-\left(sinx\right)/x$ and $\left(tanx\right)/x-1$ bounded by the function $x^2{\left(sin(\lambda x)/(\lambda x)\right)}^{\alpha }$ and $x^2{\left(tan(\mu x)/(\mu x)\right)}^{\beta }$ as follows.

Theorem 1.

Let $0<\left|x\right|<\pi /2$, $\varphi =1$ and

$$\phi ={\left[\frac{{24}^5{\pi }^{27}{\left(\pi -2\right)}^5}{4^{42}{7}^{21}{\left(sin\frac{\sqrt{7}}{28}\pi \right)}^{42}}\right]}^{1/5}>1.$$

Then the double inequality

$$\varphi \left[\frac{1}{6}{x}^2{\left(\frac{sin\frac{1}{2\sqrt{7}}x}{\frac{1}{2\sqrt{7}}x}\right)}^{\frac{42}{5}}\right]<1-\frac{sinx}{x}<\phi \left[\frac{1}{6}{x}^2{\left(\frac{sin\frac{1}{2\sqrt{7}}x}{\frac{1}{2\sqrt{7}}x}\right)}^{\frac{42}{5}}\right]$$

(19)

holds with the best constants ϕ and φ.

Theorem 2.

Let $0<\left|x\right|<\pi /2$. Then

$$\frac{tanx}{x}-1>\frac{1}{3}{x}^2{\left(\frac{tan\frac{\sqrt{43}}{7}x}{\frac{\sqrt{43}}{7}x}\right)}^{\frac{294}{215}}$$

(20)

holds with the best constant $1/3$.

2. Lemmas

The proof of the main conclusions (Theorems 1 and 2) of this paper needs the following lemmas as the basis.

Lemma 1.

Let $n\ge 3$, $n\in \mathbb{N}$,

$$\begin{array}{ccc}\hfill {\sigma }_1& =& \frac{\sqrt{7}+14}{14}\approx 1.1890,\hfill \\ \hfill {\sigma }_2& =& \frac{14-\sqrt{7}}{14}\approx 0.81102,\hfill \\ \hfill {\sigma }_3& =& \frac{\sqrt{7}}{14}\approx 0.18898,\hfill \end{array}$$

(21)

and for $k\ge 4,$

$$\begin{array}{ccc}\hfill a_k& =& -\frac{27}{2}\sqrt{7}\frac{{\sigma }_1^{2k+2}}{\left(2k+2\right)!}+\frac{27}{2}\sqrt{7}\frac{{\sigma }_2^{2k+2}}{\left(2k+2\right)!}-21\frac{{\sigma }_3^{2k}}{\left(2k\right)!}\hfill \\ & & +32\sqrt{7}\frac{{\sigma }_3^{2k+1}}{\left(2k+1\right)!}+\frac{5\sqrt{7}+21}{2}\frac{{\sigma }_2^{2k+1}}{\left(2k+1\right)!}+\frac{21-5\sqrt{7}}{2}\frac{{\sigma }_1^{2k+1}}{\left(2k+1\right)!}.\hfill \end{array}$$

(22)

Then $2a_{2n}-5a_{2n+1}>0$.

Proof. 

Since

$$\begin{array}{ccc}\hfill a_{2n}& =& -\frac{27}{2}\sqrt{7}\frac{{\sigma }_1^{4n+2}}{\left(4n+2\right)!}+\frac{27}{2}\sqrt{7}\frac{{\sigma }_2^{4n+2}}{\left(4n+2\right)!}-21\frac{{\sigma }_3^{4n}}{\left(4n\right)!}\hfill \\ & & +32\sqrt{7}\frac{{\sigma }_3^{4n+1}}{\left(4n+1\right)!}+\frac{5\sqrt{7}+21}{2}\frac{{\sigma }_2^{4n+1}}{\left(4n+1\right)!}+\frac{21-5\sqrt{7}}{2}\frac{{\sigma }_1^{4n+1}}{\left(4n+1\right)!},\hfill \\ \hfill a_{2n+1}& =& -\frac{27}{2}\sqrt{7}\frac{{\sigma }_1^{4n+4}}{\left(4n+4\right)!}+\frac{27}{2}\sqrt{7}\frac{{\sigma }_2^{4n+4}}{\left(4n+4\right)!}-21\frac{{\sigma }_3^{4n+2}}{\left(4n+2\right)!}\hfill \\ & & +32\sqrt{7}\frac{{\sigma }_3^{4n+3}}{\left(4n+3\right)!}+\frac{5\sqrt{7}+21}{2}\frac{{\sigma }_2^{4n+3}}{\left(4n+3\right)!}+\frac{21-5\sqrt{7}}{2}\frac{{\sigma }_1^{4n+3}}{\left(4n+3\right)!},\hfill \end{array}$$

we compute to obtain

$$\left(4n+4\right)!\left(2a_{2n}-5a_{2n+1}\right)=\left[\frac{5}{2}u(n){\sigma }_1^{4n+1}-512v(n){\sigma }_3^{4n}\right]+\frac{5}{2}w(n){\sigma }_2^{4n+1},$$

where

$$\begin{array}{ccc}\hfill u(n)& =& \left(\frac{2688}{5}-128\sqrt{7}\right)n^3-\left(\frac{2304}{5}\sqrt{7}-\frac{5616}{5}\right)n^2-\left(\frac{17559}{35}\sqrt{7}-\frac{3277}{5}\right)n\hfill \\ & & -\left(\frac{19459}{140}\sqrt{7}-\frac{31019}{280}\right),\hfill \\ \hfill v(n)& =& \left(n+1\right)\left(21n^3+\frac{55}{2}{n}^2+\frac{1193}{128}n+\frac{1445}{3584}\right),\hfill \\ \hfill w(n)& =& \left(128\sqrt{7}+\frac{2688}{5}\right)n^3+\left(\frac{2304}{5}\sqrt{7}+\frac{5616}{5}\right)n^2+\left(\frac{17559}{35}\sqrt{7}+\frac{3277}{5}\right)n\hfill \\ & & +\left(\frac{14436973}{109760}\sqrt{7}+\frac{75065}{784}\right)\hfill \end{array}$$

are positive for $n\ge 3$. In order to prove Lemma 1, it suffices to prove that for $n\ge 3$,

$$\frac{5}{2}u(n){\sigma }_1^{4n+1}-512v(n){\sigma }_3^{4n}>0⟺{\left(\frac{{\sigma }_1}{{\sigma }_3}\right)}^{4n}>\frac{1024}{5{\sigma }_1}\frac{v(n)}{u(n)}.$$

To note the fact

$$\frac{{\sigma }_1}{{\sigma }_3}=1+2\sqrt{7}\approx 6.2915,$$

we only need to prove

$$6^{4n}>\frac{1024}{5{\sigma }_1}\frac{v(n)}{u(n)}.$$

(23)

By mathematical induction, we can prove the inequality (23). First, the inequality (23) is obviously true for $n=3$. Assume that (23) holds for $n=m\ge 3$, that is,

$$6^{4m}>\frac{1024}{5{\sigma }_1}\frac{v(m)}{u(m)}$$

holds. In the following, we shall prove that (23) holds for $n=m+1$. Since

$$6^{4\left(m+1\right)}=6^4\times 6^{4m}>1296\left[\frac{1024}{5{\sigma }_1}\frac{v(m)}{u(m)}\right],$$

we can complete the proof of (23) when showing that

$$1296\left[\frac{1024}{5{\sigma }_1}\frac{v(m)}{u(m)}\right]>\frac{1024}{5{\sigma }_1}\frac{v(m+1)}{u(m+1)},$$

or

$$\frac{A}{B}:=\frac{1296v(m)}{u(m)}>\frac{v(m+1)}{u(m+1)}:=\frac{C}{D}$$

In fact,

$$\begin{array}{ccc}\hfill AD-BC& =& 1296v(m)u(m+1)-v(m+1)u(m)\hfill \\ & =& \left(\frac{48280304638987025}{200704}-\frac{7757241843454045}{100352}\sqrt{7}\right)\hfill \\ & & +\left(\frac{12227207591977687}{28672}-\frac{1898338908715519}{14336}\sqrt{7}\right)\left(m-3\right)\hfill \\ & & +\left(\frac{11559549671546923}{35840}-\frac{10809910365147}{112}\sqrt{7}\right){\left(m-3\right)}^2\hfill \\ & & +\left(\frac{17268465300163}{128}-\frac{24831245768359}{640}\sqrt{7}\right){\left(m-3\right)}^3\hfill \\ & & +\left(\frac{674131077463}{20}-\frac{18578001121}{2}\sqrt{7}\right){\left(m-3\right)}^4\hfill \\ & & +\left(5030226642-\frac{6621922062}{5}\sqrt{7}\right){\left(m-3\right)}^5\hfill \\ & & +\left(\frac{2075901072}{5}-104111168\sqrt{7}\right){\left(m-3\right)}^6\hfill \\ & & +\left(14620032-3480960\sqrt{7}\right){\left(m-3\right)}^7\hfill \\ & >& 0\hfill \end{array}$$

holds for all $m\ge 3$. This completes the proof of Lemma 1.  □

It is not difficult to prove the following conclusion in the similar way.

Lemma 2.

Let $n\ge 2$, $n\in \mathbb{N}$,

$$\begin{array}{ccc}\hfill {\rho }_1& =& \frac{2}{7}\sqrt{43}+2\approx 3.8736,\hfill \\ \hfill {\rho }_2& =& \frac{2}{7}\sqrt{43}\approx 1.8736,\hfill \\ \hfill {\rho }_3& =& 2-\frac{2}{7}\sqrt{43}\approx 0.12645,\hfill \end{array}$$

(24)

and for $k\ge 4,$

$$\begin{array}{ccc}\hfill b_k& =& \frac{17\sqrt{43}}{903\times 2}\frac{{\rho }_1^{2k+1}}{\left(2k+1\right)!}-\frac{117\sqrt{43}}{4816}\frac{{\rho }_1^{2k+2}}{\left(2k+2\right)!}\hfill \\ & & +\frac{1}{2}\frac{2^{2k}}{\left(2k\right)!}-\frac{1}{2}\frac{2^{2k+1}}{\left(2k+1\right)!}\hfill \\ & & +\frac{283\sqrt{43}}{3612}\frac{{\rho }_2^{2k+1}}{\left(2k+1\right)!}-\frac{17\sqrt{43}}{903\times 2}\frac{{\rho }_3^{2k+1}}{\left(2k+1\right)!}+\frac{117\sqrt{43}}{4816}\frac{{\rho }_3^{2k+2}}{\left(2k+2\right)!}.\hfill \end{array}$$

(25)

Then $2b_{2n}-5b_{2n+2}>0$.

3. Proofs of Theorems 1 and 2

Proof of Theorem 1.

Let

$$F(x)=ln\left(1-\frac{sinx}{x}\right)-ln\left[\frac{1}{6}{x}^2{\left(\frac{sin\frac{1}{2\sqrt{7}}x}{\frac{1}{2\sqrt{7}}x}\right)}^{\frac{42}{5}}\right],$$

where $0<\left|x\right|<\pi /2$. Since this function $F(x)$ is even, let’s consider the problem on interval $(0,\pi /2)$. We compute to obtain

$$F^{\prime }(x)=\frac{2\sqrt{7}}{35}\frac{f(x)}{x\left(cos{\sigma }_{1}x-cos{\sigma }_{2}x+2xsin{\sigma }_{3}x\right)},$$

where

$$\begin{array}{ccc}\hfill f(x)& =& \frac{27}{2}\sqrt{7}cos{\sigma }_{1}x-\frac{27}{2}\sqrt{7}cos{\sigma }_{2}x-21x^{2}cos{\sigma }_{3}x+32\sqrt{7}xsin{\sigma }_{3}x\hfill \\ & & +\left(\frac{5}{2}\sqrt{7}+\frac{21}{2}\right)x\left(sin{\sigma }_{2}x\right)-\left(\frac{5}{2}\sqrt{7}-\frac{21}{2}\right)x\left(sin{\sigma }_{1}x\right).\hfill \end{array}$$

(26)

and ${\sigma }_i$ $\left(i=1,2,3\right)$ are defined as (21). Substituting

$$sinx=\sum _{k=0}^{\infty }\frac{{\left(-1\right)}^k}{\left(2k+1\right)!}{x}^{2k+1},cosx=\sum _{k=0}^{\infty }\frac{{\left(-1\right)}^k}{\left(2k\right)!}{x}^{2k}$$

(27)

into (26), we can obtain that

$$\begin{array}{ccc}\hfill f(x):=& & \sum _{k=4}^{\infty }{\left(-1\right)}^k{a}_k{x}^{2k+2}=\sum _{n=2}^{\infty }\left(a_{2n}{x}^{4n+2}-a_{2n+1}{x}^{4n+4}\right)\hfill \\ & =& \left(\frac{1}{274400}{x}^{10}-\frac{163}{1521273600}{x}^{12}\right)+\sum _{n=3}^{\infty }\left(a_{2n}{x}^{4n+2}-a_{2n+1}{x}^{4n+4}\right),\hfill \end{array}$$

where $a_k$ is defined by (22). Since

$$\frac{1}{274400}{x}^{10}-\frac{163}{1521273600}{x}^{12}>0$$

for all $x\in \left(0,\pi /2\right)$, we can determine the positive definiteness of the function $f(x)$ on $(0,\pi /2)$ when proving

$$a_{2n}{x}^{4n+2}-a_{2n+1}{x}^{4n+4}>0⟺x^2<\frac{a_{2n}}{a_{2n+1}}$$

(28)

for $n\ge 3$. Since

$$x^2<{\left(\frac{\pi }{2}\right)}^2\approx 2.4674<\frac{5}{2},$$

we can prove (28) when proving that for $n\ge 3$,

$$\frac{5}{2}<\frac{a_{2n}}{a_{2n+1}},$$

or

$$2a_{2n}-5a_{2n+1}>0,$$

which comes from Lemma 1.

So $f(x)>0$ and $F(x)$ is increasing on $(0,\pi /2)$. In view of

$$F(0^{+})=0,F(1^{-})=\frac{1}{5}ln\left(\frac{{24}^5{\pi }^{27}{\left(\pi -2\right)}^5}{4^{42}{7}^{21}{\left(sin\frac{\sqrt{7}}{28}\pi \right)}^{42}}\right)>0,$$

the proof of Theorem 1 is complete.  □

Proof of Theorem 2.

Let

$$G(x)=ln\left(\frac{tanx}{x}-1\right)-ln\left[\frac{1}{3}{x}^2{\left(\frac{tan\frac{\sqrt{43}}{7}x}{\frac{\sqrt{43}}{7}x}\right)}^{\frac{294}{215}}\right],0<x<\frac{\pi }{2}.$$

Then

$$G^{\prime }(x)=\frac{1806\sqrt{43}}{9245}\frac{g(x)}{x\left(tan\frac{\sqrt{43}}{7}x\right)\left(tanx-x\right){cos}^2\frac{\sqrt{43}}{7}x{cos}^{2}x},$$

where

$$\begin{array}{ccc}\hfill g(x)& =& \frac{117\sqrt{43}}{4816}cos{\rho }_{1}x-\frac{117\sqrt{43}}{4816}cos{\rho }_{3}x\hfill \\ & & -\frac{1}{2}xsin2x+\frac{1}{2}{x}^{2}cos2x+\frac{1}{2}{x}^2\hfill \\ & & +\frac{283\sqrt{43}}{3612}xsin{\rho }_{2}x+\frac{17\sqrt{43}}{903\times 2}xsin{\rho }_{1}x-\frac{17\sqrt{43}}{903\times 2}xsin{\rho }_{3}x,\hfill \end{array}$$

(29)

where ${\rho }_i$ $\left(i=1,2,3\right)$ are defined as (24). Substituting (27) into (29), we can obtain that

$$g(x)=:\sum _{k=4}^{\infty }{\left(-1\right)}^k{b}_k{x}^{2k+2},$$

where $b_k$ is defined by (25). We can rewrite $g(x)$ as

$$g(x)=\sum _{n=2}^{\infty }\left(b_{2n}{x}^{4n+2}-b_{2n+1}{x}^{4n+4}\right).$$

Then, we determine the positive definiteness of the function $g(x)$ on $(0,\pi /2)$ when proving

$$b_{2n}{x}^{4n+2}-b_{2n+1}{x}^{4n+4}>0⟺x^2<\frac{b_{2n}}{b_{2n+1}}$$

(30)

for $n\ge 3$. Since

$$x^2<{\left(\frac{\pi }{2}\right)}^2=\frac{{\pi }^2}{4}\approx 2.4674<\frac{5}{2},$$

we can prove (30) when proving that for $n\ge 3,$

$$\frac{5}{2}<\frac{b_{2n}}{b_{2n+1}},$$

which is the result of Lemma 2. So $g(x)>0$ and $G(x)$ is strictly increasing on $(0,\pi /2)$. Therefore $G(x)>G(0^{+})=0$. Considering the reason

$$\underset{x\to 0^{+}}{lim}\frac{\frac{tanx}{x}-1}{x^2{\left(\frac{tan\frac{\sqrt{43}}{7}x}{\frac{\sqrt{43}}{7}x}\right)}^{\frac{294}{215}}}=\frac{1}{3},$$

the proof of Theorem 2 is completed.  □

4. Comparison of New and Old Results

When letting $\varphi =1$ in (19) we can obtain

$$1-\frac{sinx}{x}>\frac{1}{6}{x}^2{\left(\frac{sin\frac{1}{2\sqrt{7}}x}{\frac{1}{2\sqrt{7}}x}\right)}^{\frac{42}{5}}.$$

(31)

By using the similar proof method of Theorem 1 it is not difficult to prove the following results:

$$\begin{array}{ccc}\hfill \frac{1}{6}{x}^2{\left(\frac{sin\frac{1}{2\sqrt{7}}x}{\frac{1}{2\sqrt{7}}x}\right)}^{\frac{42}{5}}& >& 1-\left({\displaystyle \frac{2+cosx}{3}}-\frac{1}{180}{x}^4+\frac{1}{3780}{x}^6\right),\hfill \end{array}$$

(32)

$$\begin{array}{ccc}\hfill \frac{1}{3}{x}^2{\left(\frac{tan\frac{\sqrt{43}}{7}x}{\frac{\sqrt{43}}{7}x}\right)}^{\frac{294}{215}}& >& \frac{\left(1-cosx\right)\left(604{cos}^{2}x-1817cosx+1843\right)}{945}\hfill \end{array}$$

(33)

hold for all $x\in \left(0,\pi /2\right)$. So new results (19) and (20) are better that the old ones (16) and (17), respectively. In addition, there are deeper conclusions:

$$\begin{array}{ccc}\hfill 1-\frac{sinx}{x}& >& \frac{1}{6}{x}^2{\left(\frac{sin\frac{1}{2\sqrt{7}}x}{\frac{1}{2\sqrt{7}}x}\right)}^{\frac{42}{5}}\hfill \\ & >& 1-\left(\frac{2+cosx}{3}-\frac{1}{180}{x}^4+\frac{1}{3780}{x}^6\right)\hfill \\ & >& \frac{\left(1-cosx\right)\left(5{cos}^{2}x-31cosx+341\right)}{945}>1-\frac{2+cosx}{3}.\hfill \end{array}$$

(34)