<s>[INST] Show that $|\tan x-x|\leq 8x^2$ if $|x|\leq \pi/3$ Show that $|\tan x-x|\leq 8x^2$ if $|x|\leq \pi/3$. I think this is supposed be solved using the maclaurin series.
Let $f(x)=\tan x$ then $f'=1/\cos ^2x$ and $f''=\frac{-2\cos x \sin x}{\cos^4 x}=\frac{-\sin 2x}{\cos^4 x}$. Since $f(0)=0$ and $f'(0)=1$ we have that
$|x +\frac{-\sin 2\theta x}{\cos^4 x\theta}-x|=|\frac{-\sin 2\theta x}{\cos^4 x\theta}|$,where $\theta$ is between $0$ and $1$. Now what is left is to show that $|\frac{-\sin 2\theta x}{\cos^4 x\theta}|\leq 8x^2$ if $|x|\leq \pi/3$.
 [/INST] Let
$$ f(x)=\tan x-x. $$
Then there is $\theta\in(0,1)$ such that
$$ f(x)-f(0)=f'(\theta x)x. $$
Note
$$ f'(\theta x)=\sec^2(\theta x)-1=\tan^2(\theta x)=\frac{\sin^2(\theta x)}{1-\sin^2(\theta x)}. $$
Using
$$ \frac{2}{\pi}\theta\le\sin\theta\le\theta, \theta\in[0,\frac\pi2]$$
one has
$$ \sin^2(\theta x)\le \theta^2x^2\le x^2 $$
and
$$ 1-\sin^2(\theta x)\ge 1-\frac{4}{\pi^2}\theta^2x^2\ge 1-\frac{4}{\pi^2}x^2. $$
So
$$ f(x)\le \frac{x^3}{1-\frac{4}{\pi^2}x^2}=\frac{x}{1-\frac{4}{\pi^2}x^2}\cdot x^2\le 8x^2. $$
It is easy to show
$$ \frac{x}{1-\frac{4}{\pi^2}x^2}\le 8.$$
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