Answer
$1.58 \times 10^{-2}$
Work Step by Step
The remainder in the n-th order Taylor's polynomial $T_n(x)$ for a function $f(x)$ centered at $a$ can be expressed as:
$|R_n(x)|=|f(x)-T_n(x)| \leq M \dfrac{|x-a|^{n+1}}{(n+1)!}$
We are given that $f(x)=\cos x$ and $T(x)=1-\dfrac{x^2}{2}$ and $n=3$
Thus, we have
$|R_n(x)|\leq \dfrac{|\dfrac{\pi}{4}-0|^{3+1}}{(3+1)!} \approx 1.58 \times 10^{-2}$