Answer
$$T_{3}(x) = \frac{-1}{2}x^2$$
Work Step by Step
Given $$f(x)=\ln (\cos x), \quad T_{3}, \quad a=0$$
Since
\begin{align*}
f(x)&=\ln (\cos x)\ \ \ \ \ \ \ &f(0)=0\\
f'(x)&=\frac{-\sin x}{\cos x}\ \ \ \ \ \ \ &f'(0)=0\\
f''(x)&= -\sec^2 x\ \ \ \ \ \ \ &f''(0)=-1\\
f'''(x)&= -2\sec^2 x\tan x\ \ \ \ \ \ \ &f'''(0)=0
\end{align*}
Then
\begin{aligned}
T_{3}(x)&= f(a)+f^{\prime}(a)(x-a)+\frac{f^{\prime \prime}(a)}{2 !}(x-a)^{2}+\frac{f^{\prime \prime \prime}(a)}{3 !}(x-a)^{3} \\
&= \frac{-1}{2}x^2
\end{aligned}
