Answer
Error $\lt 1.67 \times 10^{-10}$
Work Step by Step
The Taylor series for $\sin x $ can be defined as: $\sin x= x-\dfrac{x^3}{3!}+\dfrac{ x^5}{5!}-....$
and $\sin x= x-\dfrac{x^3}{6}+\dfrac{ x^5}{120}-....$
We need to find $ x $.
$|\dfrac{x^3}{6}| \lt |\dfrac{(10^{-3})^3}{6}| $
or, $|\dfrac{x^3}{6}| \lt |\dfrac{(10^{-3})^3}{6}| = \dfrac{10^{-9}}{6}$
or, Error $\lt |\dfrac{(10^{-3})^3}{3 !}| \approx 1.67 \times 10^{-10}$
or, Error $\lt 1.67 \times 10^{-10}$
