I used the Taylor polynomial for $f(x)=\sin(x)$ centered at $0$, which is $\frac{1}{1!}x-\frac{1}{3!}x^3+\frac{1}{5!}x^5-\frac{1}{7!}x^7+\frac{1}{9!}x^9-\frac{1}{11!}x^{11}+...$.
By replacing $x$ with $1$, I was able to approximate:
$\frac{1}{1!}(1)=1$
$\frac{1}{1!}(1)-\frac{1}{3!}(1)^3=0.\textbf{83333}33333...$
$\frac{1}{1!}(1)-\frac{1}{3!}(1)^3+\frac{1}{5!}(1)^5=0.\textbf{84166}6666...$
$\frac{1}{1!}(1)-\frac{1}{3!}(1)^3+\frac{1}{5!}(1)^5-\frac{1}{7!}(1)^7=0.\textbf{84146}8253...$
$\frac{1}{1!}(1)-\frac{1}{3!}(1)^3+\frac{1}{5!}(1)^5-\frac{1}{7!}(1)^7+\frac{1}{9!}(1)^9=0.\textbf{84147}10097001...$
$\frac{1}{1!}(1)-\frac{1}{3!}(1)^3+\frac{1}{5!}(1)^5-\frac{1}{7!}(1)^7+\frac{1}{9!}(1)^9-\frac{1}{11!}(1)^{11}=0.\textbf{84147}098...$
At this point, I noticed that (5) and (6) have the values for the first five decimal places.
So, I answered a degree of $9$ as the smallest degree that will work.
Continuing the polynomial approximation, I found that degrees $11$ and $13$ have the same first eight decimal places, and answered $11$ as the smallest working degree.
$11$ was correct, but $9$ was not. This made me very confused, since it was clear that $9$ was the first degree that gave the same 5 decimals as the next degree.
Did I make an error on any step?