Prove that $\frac{\tan x}{x}>\frac{x}{\sin x}, x\in(0,\pi/2)$

Question

Prove that $$\frac{\tan x}{x}>\frac{x}{\sin x},\;\;\; x\in(0,\pi/2).$$

My work

I formulated $$f(x)=\tan x \sin x - x^2$$ in hope that if $f'(x)>0$ i.e. monotonic then I can conclude for $x>0, f(x)>f(0)$ and hence, prove the statement.

However, I got $$f'(x)=\sin x + \sec x \tan x -2x, $$ where I am unable to conclude if $f'(x)>0.$

I also found $$f''(x)=\cos x + 2\sec^3x-\sec x-2,$$

$$f'''(x)=-\sin x (1-6\sec^4x+\sec^2x).$$

But I am not able conclude the sign of any of the higher derivatives either. Am I doing something wrong? Or is there some other way?

Answer 1

I believe the simplest proof is through the Cauchy-Schwarz inequality:

$$\tan(x)\sin(x)=\int_{0}^{x}\frac{d\theta}{\cos^2\theta}\int_{0}^{x}\cos(\theta)\,d\theta\geq\left(\int_{0}^{x}\frac{d\theta}{\sqrt{\cos\theta}}\right)^2\geq\left(\int_{0}^{x}d\theta\right)^2=x^2. $$ In a similar fashion, for any $x\in\left(0,\frac{\pi}{2}\right)$ we have $\frac{\tan x }{x}\geq\left(\frac{x}{\sin x}\right)^2$ by Holder's inequality.

Answer 2

Note that by denoting $f(x) = \tan x \sin x -x^2$, you found that $$f'''(x)=-\sin x (1-6\sec^4x+\sec^2x) = \sin x (1+3\sec^2 x)(2\sec^2 x -1 ) \geq 0 $$ Hence $f''(x)$ is increasing, with $f''(0)=0$, we conclude that $f''(x) \geq 0$.

Hence $f'(x)$ is increasing, with $f'(0)=0$, we conclude that $f'(x) \geq 0$.

Hence $f(x)$ is increasing, with $f(0)=0$, we conclude that $f(x) \geq 0$.

This is what we wish to prove.

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From a more advanced perspective, the inequality follows from the fact that Taylor expansion of $$\tan x \sin x = x^2+\frac{x^4}{6}+\cdots$$ at $x=0$ have all coefficients positive, the radius of convergence of this series is $\pi/2$.

To see why all coefficients are positive, write $$\tan x \sin x = \frac{1}{\cos x} - \cos x$$

The Taylor expansion of $\sec x$ at $x=0$ is $$\sec x = \sum_{n=0}^{\infty} \frac{(-1)^n E_{2n}}{(2n)!} x^{2n}$$ where $E_{2n}$ are Euler number. The fact that $(-1)^n E_{2n}$ is positive follows from the series evaluation: $$\beta(2n+1) = \frac{(-1)^n E_{2n} \pi^{2n+1}}{4^{2n+1} (2n)!}$$ with $\beta(n)$ the Dirichlet beta function.

Also note that we have $|E_{2n}| > 1 $ when $n>1$, hence the power series of $\frac{1}{\cos x}-\cos x$ has all coefficients positive.

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From this, you might want to prove the stronger inequality:

When $0<x<\frac{\pi}{2}$, $$\tan x \sin x > x^2 + \frac{x^4}{6} $$ $$\tan x \sin x > x^2 + \frac{x^4}{6} + \frac{31x^6}{360} $$

Answer 3

We need to prove that $$\frac{\sin^2x}{\cos{x}}>x^2$$ or $f(x)>0$, where $$f(x)=\frac{\sin{x}}{\sqrt{\cos{x}}}-x.$$ Now, let $\cos{x}=t^2$, where $0<t<1$.

Thus, $$f'(x)=\frac{1+\cos^2x}{2\sqrt{\cos^3x}}-1=\frac{(1-t)(1+t+t^2-t^3)}{2t^3}>0,$$ which says $f(x)>f(0)=0$ and we are done!

Answer 4

Note that for $x\in\left(0,\frac{\pi}{2}\right)$

$$\tan x \sin x>\left(x+\frac{x^3}{3}\right)\left(x-\frac{x^3}{6}\right)=x^2+\frac{x^4}{6}-\frac{x^6}{18}>x^2\iff3-x^2>0\iff x < \sqrt 3$$

Answer 5

We can prove a stronger inequality $$\frac{\tan x}{x} > \left(\frac{x}{\sin x}\right)^2$$ for $x \in (0, \pi/2)$. Indeed the function $\tan x \sin^2 x - x^3$ has derivative $\tan^2 x + 2 \sin^2 x - 3 x^2$. Now, from this answer we see that $\frac{\tan x + 2 \sin x }{3} > x$ for $x \in (0, \frac{\pi}{2})$ so $\frac{\tan^2 x + 2 \sin^2 x}{3}> \left(\frac{ \tan x + 2 \sin x }{3}\right)^2 > x^2$ and we are done.

Note that the Taylor series of $1-\frac{x^3}{\tan x \sin^2 x}$ has all coefficients positive $$1-\frac{x^3}{\tan x \sin^2 x}=\frac{x^4}{15} + \frac{4 x^6}{189} + \frac{x^8}{225}+\cdots$$

We will prove a weaker stamement than the above (which was only checked for some coefficients) by proving that the function $$f(x) = \frac{\sin x}{\cos^{\frac{1}{3}} x}$$ has the Taylor series at $0$ with all coefficients positive. For this, we notice that $$f^{(2)}(x) =\frac{4}{9} \frac{\sin^3 x}{\cos^{\frac{7}{3}}x} =\frac{4}{9} \sec^{\frac{4}{3}}x \cdot f^3(x)$$

Now, $\sec x$ at any positive power has all Taylor coefficients positive. This follows from the fact that $\sec x$ has a product expansion with factors $\frac{1}{1- a_k x^2}$ and the series $(1-t)^{-\alpha}$ has all Taylor coefficients positive. Now, the fact that $f$ has all coefficients positive follows by induction, by using the recurrence given by the differential equation.

Answer 6

Since $\frac{x}{2} < \tan \frac{x}{2}$ for $0<x<\frac{\pi}{2}$ we have for such $x$-values: $$ \left( \frac{\sin x}{x}\right)^2 > \left( \frac{2 \sin \frac{x}{2} \cos \frac{x}{2}} {2 \tan \frac{x}{2}}\right)^2 = \left(\cos \frac{x}{2} \right)^4 = \frac14 \left(1+\cos x\right)^2 > \cos x $$ using that the difference between the two last expressions is $\frac14 (1-\cos x)^2>0$. The result follows.

Answer 7

Here is my suggestion using a short Taylor expansion and checking the residual.

Rewriting the inequality, we only need to show

$$\tan{x} \sin{x} \gt x^2 \mbox{ on } x \in (0, \frac{\pi}{2})$$

Now, let's rewrite the left side (you don't need it but I like it) and give it a name $f$; $$\tan{x} \sin{x} = \frac{\sin^2{x}}{\cos{x}} = \frac{1-\cos^2{x}}{\cos{x}} = \frac{1}{\cos{x}} - \cos{x} = f(x)$$

$f$ can be rewritten using the Taylor polynomial around $0$ of degree $2$. With $f(0) = 0$, $f^{\prime}(0)=0$, $f^{\prime\prime}(0)=2$ we get

$$f(x) = x^2 + \frac{f^{(3)}(\theta_x)}{3!}x^3 \mbox{ with } \theta_x \in (0,x)$$

So, let's check the residual on the given interval:

$$f^{(3)}(x) = \frac{\sin{x}}{\cos^4{x}}(5+\sin^2{x}-\cos^4{x}) \gt 0 \mbox{ for } x \in (0, \frac{\pi}{2})$$

Thus,

$$\frac{1}{\cos{x}} - \cos{x} = f(x) \gt x^2 \mbox{ for } x \in (0, \frac{\pi}{2})$$

Done.

Answer 8

It is known (see p. 55, Item 1.518 in the monograph [1] below) that \begin{equation*} \ln\sin x=\ln x+\sum_{k=1}^\infty\frac{2^{2k-1}}{k}\bigl[(-1)^kB_{2k}\bigr]\frac{x^{2k}}{(2k)!}, \quad 0<x<\pi, \end{equation*} where $B_k$ denotes the Bernoulli numbers and $(-1)^kB_{2k}<0$ for all $k\ge1$. Then \begin{equation*} \ln x-\ln\sin x=\ln\frac{x}{\sin x}=-\sum_{k=1}^\infty\frac{2^{2k-1}}{k}\bigl[(-1)^kB_{2k}\bigr]\frac{x^{2k}}{(2k)!}>0, \quad 0<x<\pi \end{equation*} and \begin{equation*} (\ln x-\ln\sin x)^{(n)}=\biggl(\ln\frac{x}{\sin x}\biggr)^{(n)} =-\sum_{k=1}^\infty\frac{2^{2k-1}}{k}\bigl[(-1)^kB_{2k}\bigr]\langle2k\rangle_n \frac{x^{2k-n}}{(2k)!}>0, \quad 0<x<\pi. \end{equation*} This means that the function $\ln x-\ln\sin x$ is absolutely monotonic functions on $(0,\pi)$. In other words, the even function \begin{equation*} f(x)=\begin{cases} \dfrac{x}{\sin x}, & x\ne0\\ 1, & x=0 \end{cases} \end{equation*} is logarithmically absolutely monotonic on $(0,\pi)$. Therefore, the function $f(x)$ is absolutely monotonic on $(0,\pi)$. Thus, we acquire $f^{(k)}(0)\ge0$ for all $k\ge0$. Accordingly, the function \begin{equation*} \frac{1}{x}-\frac{1}{\sin x} =\frac1x\biggl(1-\frac{x}{\sin x}\biggr) =-\frac1x\sum_{k=1}^\infty f^{(k)}(0)\frac{x^k}{k!} =-\sum_{k=0}^\infty \frac{f^{(k+1)}(0)}{(k+1)!}x^{k} <0 \end{equation*} and \begin{equation*} \biggl(\frac{1}{x}-\frac{1}{\sin x}\biggr)' =-\frac1{x^2}+\frac{1}{\tan x\sin x} =-\sum_{k=1}^\infty \frac{kf^{(k+1)}(0)}{(k+1)!}x^{k-1} <0 \end{equation*} for $x\in(0,\pi)$. Consequently, we arrive at \begin{equation*} \frac{1}{\tan x\sin x}<\frac1{x^2}, \quad x\in(0,\pi), \end{equation*} which can be rearranged as \begin{equation*} \frac{x}{\sin x}<\frac{\tan x}{x}, \quad x\in\biggl(0,\frac{\pi}2\biggr). \end{equation*} When $x\in\bigl(\frac{\pi}2,\pi\bigr)$, the last inequality is reversed.

References

1. I. S. Gradshteyn and I. M. Ryzhik, Table of Integrals, Series, and Products, Translated from the Russian, Translation edited and with a preface by Daniel Zwillinger and Victor Moll, Eighth edition, Revised from the seventh edition, Elsevier/Academic Press, Amsterdam, 2015; available online at https://doi.org/10.1016/B978-0-12-384933-5.00013-8.
2. Bai-Ni Guo and Feng Qi, A property of logarithmically absolutely monotonic functions and the logarithmically complete monotonicity of a power-exponential function, University Politehnica of Bucharest Scientific Bulletin Series A---Applied Mathematics and Physics 72 (2010), no. 2, 21--30.
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