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 $|x|\leq \pi/3$.
 A: 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
$$ \frac{x}{1-\frac{4}{\pi^2}x^2}\le 8.$$
A: Let $f(x) = \tan x - x$ for all $x$ such that $|x| < \pi/2.$
Then $f(-x) = -f(x)$ for all $x,$ so in order to prove that
$|f(x)| \leqslant 8x^2$ for all $x$ such that $|x| \leqslant \pi/3,$
it is enough to prove it for all $x$ such that
$0 < x \leqslant \pi/3.$
We have $f'(x) = \sec^2 x - 1 > 0$ for all $x$ such that
$0 < x < \pi/2,$ and $f(0) = 0,$ so $f(x) > 0$ for all $x$ such that
$0 < x < \pi/2.$
Therefore it is enough to prove
$$
f(x) < 8x^2 \quad \left(0 < x \leqslant \frac\pi3\right).
$$
The well-known Maclaurin series (both of which converge for all $x$)
\begin{gather*}
\cos x = \sum_{n=0}^\infty(-1)^n\frac{x^{2n}}{(2n)!}, \\
\sin x = \sum_{n=0}^\infty(-1)^n\frac{x^{2n+1}}{(2n+1)!}
\end{gather*}
are alternating for all $x$ such that $0 < x \leqslant \sqrt2,$
because the negated ratio of successive terms of the cosine series is
$x^2/((2n+1)(2n+2)) \leqslant 1,$ and the calculation for the sine
series is similar.
Because $\pi/3 < \sqrt2,$ we therefore have $\sin x < x$ and
$\cos x > 1 - x^2/2$ for all $x$ such that $0 < x \leqslant \pi/3.$
Therefore for all such $x,$
$$
f(x) = \frac{\sin x}{\cos x} - x < \frac{x}{1 - x^2/2} - x =
\frac{x\cdot x^2}{2 - x^2} \leqslant \frac{(\pi/3)x^2}{2 - (\pi/3)^2} =
\frac{3\pi x^2}{18 - \pi^2} < \frac{5x^2}4,
$$
because $3\pi < \pi^2 < 10.$

Addendum. I think it is worth doing a little more work,
in order to prove the stronger inequality
$$
|\tan x - x| \leqslant \frac{17x^2}{21} \quad
\left(|x| \leqslant \frac\pi3\right).
$$
Proof.
Suppose $0 < x \leqslant \pi/3.$ Then, taking two more terms in the sine series,
$$
f(x) < \frac{x - x^3/6 + x^5/120}{1 - x^2/2} - x =
\frac{x^3/3 + x^5/120}{1 - x^2/2} = Ax^2,
$$
where
$$
A = \frac{x/3 + x^3/120}{1 - x^2/2} =
\frac{x(1 + x^2/40)}{3(1 - x^2/2)}.
$$
Because $\pi^2 < 10$ and $\pi < 22/7,$
$$
A < \frac{22\cdot(1 + 1/36)}{7\cdot9\cdot(1 - 5/9)} =
\frac{22\cdot37}{7\cdot4\cdot36} = \frac{407}{504} <
\frac{408}{504} = \frac{17}{21}.
$$
