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.

Solution 1:

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.

Solution 2:

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