For what range of $x$ can $\sin x=x$ approximation be used to a relative accuracy of $\frac{1}{2}10^{-14}$?
The taylor series for $\sin x$ is $$ x-\frac1{3!}x^3+\frac1{5!}x^5-\frac1{7!}x^7+\dots $$ So the absolute value of $\frac1{3!}x^3+\frac1{5!}x^5-\frac1{7!}x^7+\dots$ has to be smaller than $\frac{1}{2}10^{-14}$. I don't know where to go from here. Thanks for the help.
The error in an alternating series is only as large as the first term not in the truncation, i.e. here $$ \left|\frac{x^3}{6}\right| $$ you will be expanding on a symmetric interval around $0$ of the form $(-a,a)$, so you need $$ \left|a^3\right|\leq6\frac{1}{2}10^{-14}=3(10^{-14})\implies |a|\leq\sqrt[3]{3}10^{-14/3} $$
Edit: As noted, perhaps it would be easier to tell how big this is by writing $$ \sqrt[3]{3}10^{-14/3}=\sqrt[3]{30\cdot10^{-15}}=\sqrt[3]{30}\left(10^{-5}\right ) $$