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?


  • 3
    $\begingroup$ Did you try letting $u = \tan(x)$? Then you have to show $\sin(u) \geq \arctan(u)$ for $u \in [0, 1]$. $\endgroup$ Jan 18, 2014 at 6:35
  • $\begingroup$ @SpamIAm, thanks for pointing out. user122171's solution uses your substitution.. $\endgroup$
    – Apurv
    Jan 18, 2014 at 6:48

4 Answers 4


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 $t \geq 0$.

  • $\begingroup$ 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. $\endgroup$ Jan 18, 2014 at 6:45
  • $\begingroup$ Let $u = t^2$. Then $u \in [0,1]$. Multiply both sides of the inequality by $1 + u$. $\endgroup$
    – user122171
    Jan 18, 2014 at 6:47

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 $x \in \Bigl[0,\dfrac{\pi}{4}\Bigr]$.


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.


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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