# Proving that $x - \frac{x^3}{3!} < \sin x < x$ for all $x>0$

Prove that $x - \frac{x^3}{3!} < \sin(x) < x$ for all $x>0$

This should be fairly straightforward but the proof seems to be alluding me.

I want to show $x - \frac{x^3}{3!} < \sin(x) < x$ for all $x>0$. I recognize this shouldn't be too difficult but perhaps finals have fried my brain.

• Use the Maclaurin series for $\sin x$, together with what you know about alternating series. May 14, 2013 at 0:38
• You only have to show sin(x) < x when x <= 1, because sin(x) is bounded
– Mark
May 14, 2013 at 0:39
• May 1, 2016 at 16:06
• Of possible historical interest are the references in this answer. Nov 29, 2021 at 7:47

You can use the Taylor expansion for $\sin (x)$ to get a rather quick solution. If you haven't met Taylor yet, then consider the functions $f(x)=x - \sin x$ and $g(x)=\sin (x) -x + \frac{x^3}{3!}$, compute the derivative, and conclude the functions are increasing for suitable $x>0$. Compute $f(0)$ and $g(0)$ to get the result.

• You will have to show $cos(x) - 1 + x^2/2$ is positive which can be done by showing its derivative is positive. Then you will have killed two birds with one stone!
– Mark
May 14, 2013 at 1:00

$\text{Here is a geometric proof for$\sin(\theta) < \theta$. Consider a unit circle as shown in the figure.}$ $\text{We then have that the area of the sector formed by$BC$as }$ $$\color{red}{\dfrac12 \times OB^2 \times \theta = \dfrac12 \times 1^2 \times \theta = \dfrac{\theta}2}$$ $\text{which is greater than the area of triangle$OBC$, which is }$ $$\color{red}{\dfrac12 \times OB \times OC \times \sin(\theta) = \dfrac12 \times 1^2 \times \sin(\theta)}$$ $\text{We hence get that}$ $$\color{blue}{\dfrac{\sin(\theta)}2 < \dfrac{\theta}2 \implies\sin (\theta) < \theta}$$ $\text{The figure was drawn using GeoGebra on Ubuntu. Lets see if some one can come up with a}$ $\text{geometric/pictorial proof for$\color{green}{\theta - \dfrac{\theta^3}{3!} < \sin(\theta)}$.}$

• My 22 April 2008 AP-calculus post at Math Forum discusses this and gives a lot of 19th century references (all freely available on the internet, although you'll have to google for them). Aug 17, 2015 at 20:24

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 $$0 we have, $$\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 $$\color{blue}{\sum_{k=0}^{2n-1}(-1)^k\frac{x^{2k}}{(2k)!}<\cos x<\sum_{k=0}^{2n-1}(-1)^k\frac{x^{2k}}{(2k)!}+\frac{x^{4n}}{(4n)!} }$$ $$\color{blue}{\sum_{k=0}^{2n-1}(-1)^k\frac{x^{2k+1}}{(2k+1)!} <\sin x<\sum_{k=0}^{2n-1}(-1)^k\frac{x^{2k+1}}{(2k+1)!}+\frac{x^{4n+1}}{(4n+1)!}}$$

• All your inequalities can be strict (as asked in question). If a continuous function is positive at some point, its integral is positive. Feb 13, 2018 at 3:44
• @ParamanandSingh sure you are right thanks . Is that a reason why someone gives me a down vote? Feb 13, 2018 at 9:04
• May I ask why there is a down vote here? Feb 14, 2018 at 19:39
• Don't worry about the downvotes! It's the work of some maniac downvoter and is another evidence of the non-mathematical theorem: "every serious user of this website gets a silent undeserved downvote". I have tried to compensate by giving a +1. Feb 15, 2018 at 0:45
• Btw you don't need the restriction of $<1$. You just need all your variables to be positive ie $0<z<x$. Feb 15, 2018 at 0:48

A possible solution without use a Taylor series.Observe that:

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

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

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

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

$\\ \\ \displaystyle \sum_{k=1}^{n}3^{k-1}\sin^3\left(\frac{\phi}{3^k}\right)=\sum_{k=1}^{n}\frac{1}{4}\left(3^{k}\sin\left(\frac{\phi}{3^k}\right)-3^{k-1}\sin\left(\frac{\phi}{3^{k-1}}\right)\right)= \frac{1}{4}\left(\left(3\sin\left(\frac{\phi}{3^k}\right)-\sin\left(\phi\right)\right)+\left(3^{2}\sin\left(\frac{\phi}{3^k}\right)-3\sin\left(\frac{\phi}{3}\right)\right)+...+\left(3^{n}\sin\left(\frac{\phi}{3^n}\right)-3^{n-1}\sin\left(\frac{\phi}{3^{n-1}}\right)\right)\right) =\frac{1}{4}\left(3^{n}\sin\left(\frac{\phi}{3^n}\right)-\sin(\phi)\right)\Rightarrow \sum_{k=1}^{n}3^{k-1}\sin^3\left(\frac{\phi}{3^k}\right)=\frac{1}{4}\left(3^{n}\sin\left(\frac{\phi}{3^n}\right)-\sin(\phi)\right)\\ \\$

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*}

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

You can use the Taylor series of $\sin(x)$ about $x=0$:

$$\sin(x) = \sum_{n=0}^\infty {\frac {(-1)^n x^{2n+1}} {(2n+1)!}}$$

The first few terms are:

$$\sin(x) = x - \frac {x^3} {3!} + \frac {x^5} {5!} - \frac {x^7}{7!} + \dots$$

Thus, set:

$$x - \frac {x^3}{3!} < \sin x\\ x - \frac {x^3}{3!} < x - \frac {x^3} {3!} + \frac {x^5} {5!} - \frac {x^7}{7!} + \dots\\ 0 < \frac{x^5}{5!} - \frac {x^7}{7!} + \frac{x^9}{9!} - \frac{x^{11}}{11!}+ \dots\\$$

For $x>0$, this is true.

Similarly, for $\sin(x) < x$: $$x - \frac {x^3} {3!} + \frac {x^5} {5!} - \frac {x^7}{7!} + \dots < x\\ \frac {x^3} {3!} + \frac {x^5} {5!} - \frac {x^7}{7!} + \dots > 0$$

• To justify the inequalities (due to the alternating signs) you need to mention something about Leibnitz series. May 14, 2013 at 0:44
• It's not obvious to me why your last series is positive
– Mark
May 14, 2013 at 0:46
• Some of the signs need to be flipped in the last series; it should be $$\frac {x^3} {3!} - \frac {x^5} {5!} + \frac {x^7}{7!} - \dots > 0.$$ @Mark The series is positive because it's an alternating series with terms that decrease in absolute value. You can see this by pairing the terms and recalling that $x$ is a small positive number: $$\underbrace{\frac {x^3} {3!} - \frac {x^5} {5!}}_{­>0} + \underbrace{\frac {x^7}{7!}-\frac {x^9}{9!}}_{>0} + \dots > 0$$ Aug 17, 2015 at 19:13
• Note that the Leibniz test applies only when terms decrease. For a general $x>0$ terms will decrease not starting from $x^3/6$ to $x^5/120$ but rather from a specific term depending on value of $x$. Thus this approach can not be utilized here. The right way is to use Lagrange remainder in Taylor's theorem and conclude. Also @Théophile, $x$ is a positive number not necessarily small. Feb 12, 2018 at 19:19
• @ParamanandSingh Oh, indeed. It looks like I didn't put enough thought into my comment. Feb 13, 2018 at 15:21