## University Calculus: Early Transcendentals (3rd Edition)

Error $\lt 1.67 \times 10^{-10}$
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 $\lt 1.67 \times 10^{-10}$