Answer
$$0.000195$$
Work Step by Step
Since \begin{array}{ll}
{f(x)=\cos x,} & {f(0)=1} \\
{f^{\prime}(x)=-\sin x,} & {f^{\prime}(0)=0} \\
{f^{\prime \prime}(x)=-\cos x,} & {f^{\prime \prime}(0)=-1}
\end{array}
Then \begin{aligned}
T_{2}(x) &=f(a)+\frac{f^{\prime}(a)}{1 !}(x-a)+\frac{f^{\prime \prime}(a)}{2 !}(x-a)^{2} \\
&=1+(0)(x-0)-\frac{1}{2}(x-0)^{2} \\
&=1-\frac{1}{2} x^{2}
\end{aligned}
and
$$\left|f(\pi/12)-T_{2}(\pi/12)\right|=|0.965926-0.965731|=0.000195$$
