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I bumped into this question:

Question: Prove that for $x\in \Bigl[0,\dfrac {\pi}{4}\Bigr]$, $$\sin (\tan x) \geq {x}$$

This seems to be an innocent inequality but I am already exhausted trying to prove it. I followed different lines of thought and mention $3$ of them below:

$1.$ I tried using $\sin x<x<\tan x$. Knowing that $\sin x$,$\tan x$ increase in the mentioned interval, I tried applying $\tan$ and $\sin$ on different sides of $\sin x<x<\tan x$ in order to achieve something like that in the question.. But I just could not come to the inequality in the question.

$2.$ I tried using the function $f(x)=\sin(\tan x)-x$. $f(0)=0$. If we prove that the derivative $f'(x)=(\sec ^2 x)(\cos(\tan x))-1$ is positive, then we will be done. But proving even this seems very difficult.

$3.$ Since $\arcsin x$ is one-one and increasing in the mentioned interval, we can take $\arcsin$ on both sides of the inequality $\sin (\tan x) \geq {x}$ to get $\tan x > \arcsin x$. Thus proving $\tan x > \arcsin x$ would be equivalent to solving the problem.

All the three approaches are seeming impossible.. Any suggestions or other methods?

Thanks..

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2  
Did you try letting $u = \tan(x)$? Then you have to show $\sin(u) \geq \arctan(u)$ for $u \in [0, 1]$. –  SpamIAm Jan 18 at 6:35
    
@SpamIAm, thanks for pointing out. user122171's solution uses your substitution.. –  Apurv Jan 18 at 6:48

4 Answers 4

up vote 7 down vote accepted

From $(2)$ it boils down to show that $\cos(\tan(x)) \geq \cos^{2}(x)$. And note that $\cos(x) > 1-\dfrac{1}{2}x^{2}$ which gives that $\cos(\tan(x)) \geq 1-\frac{1}{2}\tan^{2}(x)$. Now it suffices for us to show $$1-\frac{1}{2}\tan^{2}(x) \geq \cos^{2}(x) \qquad \forall \ x \in [0,\frac{\pi}{4}]$$ This can be re-written as $$2\cos^{4}(x) - 3 \cos^{2}(x) + 1 \leq 0$$ or $$(2\cos^2x -1)(\cos^2x-1)\leq0$$ which actually holds for all $x \in \Bigl[0,\dfrac{\pi}{4}\Bigr]$.

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This was neat... Thanks for the solution.. –  Apurv Jan 18 at 6:46
    
@Apurv Thanks, Glad I could be of some help :) –  Chandrasekhar Jan 18 at 6:47

Let $t = \tan x$. Then you need to prove $\sin t \geq \arctan t$ for $t \in [0,1]$.

If you look at $f(t) = \sin t - \arctan t$, we have $f'(t) = \cos t - \frac{1}{1 + t^2}$. But the inequalities $$\frac{1}{1 + t^2} \leq 1 - \frac{1}{2}t^2 \leq \cos t, \quad (0 \leq t \leq 1)$$ are straightforward, showing that $f'(t) \geq 0$. Therefore $f$ is increasing on $[0,1]$. Because $f(0) = 0$, it follows that $f(t) \geq 0$, which is the desired inequality.

To prove the inequality $\cos t \geq 1 - \frac{1}{2}t^2$, let $g(t) = \cos t - 1 + \frac{1}{2}t^2$, and show that $g(0) = 0$, $g'(t) \geq 0$ for $t \geq 0$.

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How did you get $ \dfrac{1}{1+t^2} \le 1 - \dfrac 1 2 t^2 $? Using Binomial expansion, I could get till $ \dfrac{1}{1+t^2} \le 1 - t^2 $, but didn't get that half. –  Parth Thakkar Jan 18 at 6:45
    
Let $u = t^2$. Then $u \in [0,1]$. Multiply both sides of the inequality by $1 + u$. –  user122171 Jan 18 at 6:47
    
Aah! Nice! Thanks. –  Parth Thakkar Jan 18 at 6:49
    
@user122171 , this was a nice solution.. +1 –  Apurv Jan 18 at 6:55
    
@Apurv Thank you. –  user122171 Jan 18 at 7:00

Another method:

Using $\sin(x)=x-\dfrac{x^3}{3!}+\ldots$ we see that it suffices to prove:

$\tan(x)-x-\dfrac{\tan^3(x)} 6 \geq0$. It's derivative is $\dfrac{(\tan^2(x))(2-\sec^2x)}{2}$ which is non negative in the mentioned interval.

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Another approach which is much less satisfactory than what has been given in the previous answers would consist in building the Taylor expansion of $\sin(\tan(x))$ around $x=0$.

This leads to $x + \dfrac {x^3}{ 6} - \dfrac { x^5 }{ 40} - \dfrac {55 x^7} { 1008} - \dfrac {143 x^9 }{ 3456} + \cdots$

Then $\sin(\tan(x)) - x = \dfrac {x^3}{ 6} - \dfrac { x^5 }{ 40} - \dfrac {55 x^7} { 1008} - \dfrac {143 x^9 }{ 3456} + \cdots$ which is an increasing function of x. For $x=\dfrac{\pi}{4}$, the values of the different terms are $0.0807455, 0.0074712, 0.0100585, 0.0047051$.

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Did I miss anything in editing your answer? –  Apurv Jan 18 at 7:41
    
@Apurv. No and I thank you very much for doing it for me. Cheers. –  Claude Leibovici Jan 18 at 7:43
    
Your are welcome, @Claude Leibovici, and your solution was good.. –  Apurv Jan 18 at 7:44

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