Prove 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.

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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.

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

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