Show $|-\ln(\cos (x))-\frac{x^2}2| \le \frac23|x^3| \quad\forall x \in[-\frac{\pi}{4},\frac{\pi}{4}] $ 
Let
$$f:(-\frac{\pi}{2},\frac{\pi}{2}) \to \Bbb R$$
  $$f(x)=-\ln(\cos (x))$$
  Show that
  $$\left| f(x)-\frac{x^2}2 \right| \le \frac23\left| x^3 \right| \qquad x
\in[-\frac{\pi}{4},\frac{\pi}{4}]$$

My attempt:
We have $\left| -\ln(\cos (x))-\frac{x^2}2 \right| \le \frac23\left| x^3 \right| \iff  |\ln(\cos
(x))+\frac{x^2}2\left|  \le \frac23 \right|x^3| $
The first thing that comes to my mind is to show that $\cos(\alpha)\ge
\frac{1}{\sqrt2}$ for $\alpha \in[-\frac{\pi}{4},\frac{\pi}{4}]$ (i)
After that, I would have to go on to show that $0 \ge \ln(\beta)$ for
$\beta \in[\frac{1}{\sqrt2},1]$ (ii) which I also don't know how to
prove.
Once that is done we have
$$\left| \frac{x^2}2 \right| \le \frac23\left| x^3 \right|$$ and we are done.
So my question is: how can I go on about (i) and (ii)?
Thanks in advance.
 A: Applying Taylor's theorem with remainder to
$$ \begin{aligned}
 f(x) &= -\ln(\cos(x)) & f(0) &= 0 \\
f'(x) &= \tan(x)  & f'(0) &= 0\\
f''(x) &= 1 + \tan^2(x)  & f''(0) &= 1 \\
f'''(x) &= 2 \tan(x) (1+\tan^2(x)) 
\end{aligned}$$
gives
$$
f(x) = f(0) + f'(0) x + \frac{f''(0)}{2!} x^2 + \frac{f'''(\xi)}{3!} x^3
= \frac{x^2}{2} +  \frac{f'''(\xi)}{6} x^3
$$
for some $\xi$ between $0$ and $x$. The assertion follows because
for $|\xi| \le |x| \le \frac{\pi}{4}$
$$
|\tan(\xi)| \le 1 \Longrightarrow 
|f'''(\xi)| \le 4
$$
and therefore
$$
 \left\lvert f(x) - \frac{x^2}{2} \right\rvert = 
\frac{|f'''(\xi)|}{6} |x|^3 \le \frac 23 |x|^3
$$

Remark: Your approach cannot work because
$$
\left| \frac{x^2}2 \right| \le \frac23\left| x^3 \right|
$$
does not hold for $x$ close to zero.
A: HINTS:
For $x\in [0,\pi/4]$, use $\tan(x)\ge x$ to show that $-\log(\cos(x))-\frac12 x^2 \ge 0$.  
Let $f(x)=-\log(\cos(x))-\frac12 x^2-\frac23 x^3$.  
Use $f(0)=0$ and show that $f'(x)\le 0$.
Exploit symmetry to analyze for $x\in[0,\pi/4]$.
