They are asking me to prove $$\sin(x) > x - \frac{x^3}{3!},\; \text{for} \, x \, \in \, \mathbb{R}_{+}^{*}.$$ I didn't understand how to approach this kind of problem so here is how I tried: $\sin(x) + x -\frac{x^3}{6} > 0 \\$ then I computed the derivative of that function to determine the critical points. So: $\left(\sin(x) + x -\frac{x^3}{6}\right)' = \cos(x) -1 + \frac{x^2}{2} \\ $ The critical points: $\cos(x) -1 + \frac{x^2}{2} = 0 \\ $ It seems that x = 0 is a critical point. Since $\left(\cos(x) -1 + \frac{x^2}{2}\right)' = -\sin(x) + x \\ $ and $-\sin(0) + 0 = 0 \\$ The function has no local minima and maxima. Since the derivative of the function is positive, the function is strictly increasing so the lowest value is f(0). Since f(0) = 0 and 0 > 0 I proved that $ \sin(x) + x -\frac{x^3}{6} > 0$. I'm not sure if this solution is right. And, in general, how do you tackle this kind of problems?


9 Answers 9


A standard approach is to let $f(x)=\sin x-\left(x-\frac{x^3}{3!}\right)$, and to show that $f(x)\gt 0$ if $x\gt 0$.

Note that $f(0)=0$. We will be finished if we can show that $f(x)$ is increasing in the interval $(0,\infty)$.

Note that $f'(x)=\cos x-1+\frac{x^2}{2!}$. We will be finished if we can show that $f'(x)\gt 0$ in the interval $(0,\infty)$.

Note that $f'(0)=0$. We will be finished if we can show that $f'(x)$ is increasing in $(0,\infty)$.

So we will be finished if we can prove that $f''(x)\gt 0$ in the interval $(0,\infty)$.

We have $f''(x)=-\sin x+x$. Since $f''(0)=0$, we will be finished if we can show that $f'''(x)\ge 0$ on $(0,\infty)$, with equality only at isolated points. This is true.

Or else for the last step we can use the geometrically evident fact that $\frac{\sin x}{x}\lt 1$ if $x\gt 0$.

Remark: It is more attractive to integrate than to differentiate, but we used the above approach because differentiation comes before integration in most calculus courses.

For the integration approach, let $x$ be positive. Since $\sin t\lt t$ on $(0,x)$, we have $\int_0^x (t-\sin t)\,dt\gt 0$. Integrate. We get $\cos x+\frac{x^2}{2}-1\gt 0$ (Mean Value Theorem for integrals), so $\cos t+\frac{t^2}{2}-1\gt 0$ if $t\gt 0$.

Integrate from $0$ to $x$. We get $\sin x+\frac{x^3}{3!}-x\gt 0$, or equivalently $\sin x\gt x-\frac{x^3}{3!}$. Nicer, by a lot.


You just have to prove your inequality when $x\in(0,\pi)$, since otherwise the RHS is below $-1$. Consider that for any $x\in(0,\pi/2)$, $$ \sin^2 x < x^2 \tag{1}$$ by the concavity of the sine function. By setting $x=y/2$, $(1)$ gives: $$ \forall y\in(0,\pi),\qquad \frac{1-\cos y}{2}<\frac{y^2}{4}\tag{2}, $$ so: $$ \cos y > 1-\frac{y^2}{2} \tag{3} $$ for any $y\in(0,\pi)$. By integrating $(3)$ with respect to $y$ over $(0,x)$ we get our inequality.


Take a decreasing sequence of positive real numbers $a_n$ such that $a_n\to 0$.

Now, consider the sequence $b_k=\sum_{n=1}^k (-1)^{n-1}a_n$. The alternating series criterion guarantee us that it converges to some $b$.

Note that $b_1=a_1$, $b_2=b_1-a_2\in(0,b_1)$, $b_3=b_2+a_3\in(b_2,b_1)$, etc. So the limit $b$ is lesser that the terms $b_{2k+1}$ and greater than $b_{2k}$.

Then, if $x<\sqrt 6$, $$\sin x=\sum_{n=1}^\infty(-1)^{n-1} \frac{x^{2n-1}}{(2n-1)!}>\sum_{n=1}^2(-1)^{n-1} \frac{x^{2n-1}}{(2n-1)!}=x-\frac{x^3}{3!}$$

If $x\geq\sqrt 6$, the function $f(x)=x-x^3/6$ is decreasing and $f(\sqrt 6)=0$, so $f(x)<0$ for $x>\sqrt 6$. Since $\sin x>0$ for $0<x<\pi$, we have that $\sin x>f(x)$ for $0<x<\pi$. (Note that $\sqrt 6<\pi$).

Last, for $x\geq \pi$, $\sin x\geq -1$ and $f(x)<f(\pi)<f(3)=3-4.5<-1$.

  • $\begingroup$ There is a caveat which this answer does not address. The sequence $$a_n=\frac{x^{2n-1}}{(2n-1)!}$$ needs not be decreasing. For example, for $x=100$ one has $$a_n=\left(100, \frac{10^6}{6}, \frac{10^{10}}{120},\ldots\right).$$The sequence becomes decreasing eventually, but it is not obvious that you can apply your reasoning. $\endgroup$ May 20, 2014 at 18:31
  • $\begingroup$ @Giuseppe Thanks. I hope this fixes the caveat. $\endgroup$
    – ajotatxe
    May 21, 2014 at 7:27
  • $\begingroup$ Sorry I had forgot to come back here. I see your edit now. It is perfectly fine to me (Minus an extremely small detail: you should turn the line "Then, for $x<\sqrt{6}$..." into "Then, for $0<x<\sqrt{6}$...". For $x=0$ strict inequality does not hold. Good solution, BTW. $\endgroup$ May 25, 2014 at 15:40

Observe that:

$\\ \\ \displaystyle \sin(3\gamma)=\sin(2\gamma)\cos(\gamma)+\sin(\gamma)\cos(2\gamma)=2\sin(\gamma)\cos^2(\gamma)+\sin(\gamma)(1-2\sin^2(\gamma))=2\sin(\gamma)(1-\sin^2\gamma)+\sin(\gamma)(1-2\sin^2(\gamma))=3\sin(\gamma)-4\sin^3(\gamma)\Rightarrow \sin^3(\gamma)=\frac{1}{4}\left(3\sin(\gamma)-\sin(3\gamma)\right)$ \begin{equation} \sin^3(\gamma)=\frac{1}{4}\left(3\sin(\gamma)-\sin(3\gamma)\right) \end{equation} Do it $\displaystyle \gamma=\frac{\phi}{3^k}$:

\begin{equation} \sin^3\left(\frac{\phi}{3^k}\right)=\frac{1}{4}\left(3\sin\left(\frac{\phi}{3^k}\right)-\sin\left(\frac{\phi}{3^{k-1}}\right)\right) \end{equation} Multiplying by $\displaystyle 3^{k-1}$:

\begin{equation} 3^{k-1}\sin^3\left(\frac{\phi}{3^k}\right)=\frac{1}{4}\left(3^{k}\sin\left(\frac{\phi}{3^k}\right)-3^{k-1}\sin\left(\frac{\phi}{3^{k-1}}\right)\right) \end{equation}

Applying summation on both sides of equality, we will have:

$\\ \\ \displaystyle \sum_{k=1}^{n}3^{k-1}\sin^3\left(\frac{\phi}{3^k}\right)=\sum_{k=1}^{n}\frac{1}{4}\left(3^{k}\sin\left(\frac{\phi}{3^k}\right)-3^{k-1}\sin\left(\frac{\phi}{3^{k-1}}\right)\right) =\frac{1}{4}\left(3^{n}\sin\left(\frac{\phi}{3^n}\right)-\sin(\phi)\right)\Rightarrow \sum_{k=1}^{n}3^{k-1}\sin^3\left(\frac{\phi}{3^k}\right)=\frac{1}{4}\left(3^{n}\sin\left(\frac{\phi}{3^n}\right)-\sin(\phi)\right)\\ \\$

Take the limit: \begin{equation*} \lim_{n\rightarrow \infty}\sum_{k=1}^{n}3^{k-1}\sin^{3}\left(\frac{\phi}{3^{k}}\right)=\frac{1}{4}\left(\phi-\sin(\phi)\right) \end{equation*}

On the other hand, using the inequality $ \displaystyle \sin x \leq x $ and using the infinite geometric progression formula, it follows that: $\\ \displaystyle \sin\left(\frac{\phi}{3^{k}}\right) \leq \frac{\phi}{3^{k}}\Rightarrow \sin^{3}\left(\frac{\phi}{3^{k}}\right) \leq \frac{\phi^3}{3^{3k}} \Rightarrow 3^{k-1}\sin^{3}\left(\frac{\phi}{3^{k}}\right) \leq \frac{\phi^3}{3^{2k+1}}\Rightarrow \sum_{k=1}^{\infty}3^{k-1}\sin^{3}\left(\frac{\phi}{3^{k}}\right) \leq \sum_{k=1}^{\infty} \frac{\phi^3}{3^{2k+1}}=\frac{\phi^3}{3}\sum_{k=1}^{\infty} \frac{1}{3^{2k}}=\frac{\phi^3}{3\times 8}\Rightarrow \sum_{k=1}^{\infty}3^{k-1}\sin^{3}\left(\frac{\phi}{3^{k}}\right) \leq \frac{\phi^3}{3\times 8} \Rightarrow \frac{1}{4}\left(\phi-\sin(\phi)\right) \leq \frac{\phi^3}{3\times 8} \Rightarrow \phi-\frac{\phi^3}{6}\leq \sin(\phi) $

  • $\begingroup$ A nice approach $\endgroup$
    – Yuriy S
    Dec 5, 2016 at 7:09
  • $\begingroup$ thank s for your comment @Yuri $\endgroup$ Feb 28 at 21:03

(This started as a comment, but it is too big so I made it into an answer).

That method is quite general, indeed. It will work if you do computations with enough care.

Somebody is suggesting the use of a Taylor polynomial. I think that with an approach like that, the only thing that you can prove easily is weaker than the one you are seeking. For example, to quickly prove that there exists a $\delta >0$ such that $$\tag{1}\sin x \ge \frac{1}{2}\left(x-\frac{x^3}{3!}\right), \qquad \forall x\in [0, \delta),$$ you can note that $$\frac{\sin x}{x-\frac{x^3}{3!}} = 1 +O(x^4), $$ so there exists a constant $C>0$ such that in a neighborhood of $x=0$ one has $$\frac{\sin x}{x-\frac{x^3}{3!}}\ge 1 - Cx^4$$ and the right hand side is bigger than $\frac{1}{2}$ for $x\le \delta=\left( \frac{1}{2C}\right)^{\frac{1}{4}}$.

Note that this value of $\delta$ is completely devoid of significance, it is just a small number that can in principle be very small. Also, in formula (1) you have a multiplicative factor of $\frac{1}{2}$ in the right hand side. All of that make this result way weaker than the one you are trying to prove; however, this proof is very quick and sometimes you do not really need the full-powered inequality.


Check that $$\int_0^x (x-t)^2 \sin^2\frac{t}{2} d\,t= \sin x - (x - \frac{x^3}{6})$$

WA link


Your solution is right. There is no problem or fault in your solution. If you know something about trigonometric functions convergent series Sin(x) can be written as x-x^3/6+x^5/5!.... which is >x-x^3/6. For a beginner the method of yours is first to strike and it will always lead you to your result but if we know trigonometric functions convergent series we can conclude also

  • $\begingroup$ @DSinghvi 0 down vote I would try this? sin(x)=x−x36+x5120−x75040+...− easy to prove now. shareeditflag answered 27 mins ago bobbym 964211 1 upvote Could the down voter please elaborate? – bobbym 21 mins ago 1 I'm not the downvoter but using this approach you would need to show that x5/5!−x7/7!+x9/9!⋯>0 for x>0 is which seems non trivial. – Dan 15 mins ago math.stackexchange.com/questions/803127/… $\endgroup$
    – BCLC
    May 20, 2014 at 17:31
  • $\begingroup$ I have already edited my post and the dots are for continuation. Well these functions are trancedental functions not polynomial functions. they can be expressed in the infinite series following some pattern which will be a convergent series $\endgroup$
    – DSinghvi
    May 20, 2014 at 17:32
  • $\begingroup$ It can be shown by theory of alternating series. $\endgroup$
    – DSinghvi
    May 20, 2014 at 17:43
  • 1
    $\begingroup$ Sorry I did not see the dots before. But I agree with BCLC: this answer tricks the reader into thinking that the result is obvious, which is not true. The only thing that is actually obvious is a local result in a neighborhood of $x=0$, as I said in my answer. For that, I am not removing my downvote for now. I will remove it if you edit again and show precisely how to prove the sought inequality. $\endgroup$ May 20, 2014 at 18:22
  • $\begingroup$ P.S.: See my comment to ajotatxe's answer. I am still convinced that the method of Taylor series does not lead to the solution that easily. $\endgroup$ May 20, 2014 at 18:34

A trick is to get rid of the transcendental function.

As the sine is monotonic in $(0,\pi/2)$,

$$x>\arcsin\left(x-\frac{x^3}3\right),$$ and, as the two members coincide for $x=0$, by differentiation


Then it suffices that for $x<1$,


$$-\frac{x^6}9-\frac{x^4}3+x^2>0,$$ which is true.


Recursive integration We know that $$0\le\cos a\le 1\implies \sin t = \int_0^t\cos s ds < t$$ for $0\lt t\lt z\lt x$. Integrating over $\color{blue}{(0,z)}$ we get

$$1-\cos z=\int_0^z\sin tdt < \int_0^ztdt= \frac{z^2}{2}$$ that is for all $0<z<x$ we have, $$\color{blue}{1-\frac{z^2}{2}< \cos z\le 1}$$ integrating again over $\color{blue}{(0,x)}$ we get

$$\color{red}{x-\frac{x^3}{6} = \int_0^x 1-\frac{z^2}{2} dz< \int_0^x\cos z dz=\sin x}$$ that is $$\color{blue}{x-\frac{x^3}{6} <\sin x< x}$$

continuing with this process you get, $$\color{blue}{1-\frac{x^2}{2}< \cos x< 1-\frac{x^2}{2}+\frac{x^4}{24} }$$ $$\color{blue}{x-\frac{x^3}{6} <\sin x< x-\frac{x^3}{6} +\frac{x^5}{5!}}$$

More generally for $n\geq1$, by induction we get $$\color{blue}{\sum_{k=0}^{2n-1}(-1)^k\frac{x^{2k}}{(2k)!}<\cos x<\sum_{k=0}^{2n-1}(-1)^k\frac{x^{2k}}{(2k)!}+\frac{x^{4n}}{(4n)!} }$$ $$\color{blue}{\sum_{k=0}^{2n-1}(-1)^k\frac{x^{2k+1}}{(2k+1)!} <\sin x<\sum_{k=0}^{2n-1}(-1)^k\frac{x^{2k+1}}{(2k+1)!}+\frac{x^{4n+1}}{(4n+1)!}}$$


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