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

Question

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.

Answer 1

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

Answer 2

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.

Answer 3

$\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 $\color{green}{\theta - \dfrac{\theta^3}{3!} < \sin(\theta)}$.}$

Answer 4

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

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

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

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) $

Answer 5

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.

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