Chapter 9 - Section 9.8 - Taylor and Maclaurin Series - Exercises - Page 537: 41

$L(x)=0$ and $Q(x)=\dfrac{-x^2}{2!}$

We have $f(x) =\ln (\cos x)$ and $f'(x)=-\tan x ; \\f''(x)=-\sec^2 x $ $\implies f(0)=0; f'(0)=0; f'(0)=-1$ Therefore, $L(x)=f(0)+xf'(0)=0$ and $Q(x) =f(0) x +xf'(0) +\dfrac{x^2}{2!}f''(0)=\dfrac{-x^2}{2!}$ So, $L(x)=0$ and $Q(x)=\dfrac{-x^2}{2!}$