How many terms are needed to compute sin 1?

Error bound. $|\sin1-T_n(1)|\le\dfrac1{(n+1)!}$.
- $\frac1{8!}=2.5\cdot10^{-5}$: not enough for $n=7$.
- $\frac1{9!}=2.8\cdot10^{-6}$: not enough for $n=8$ (and $T_8=T_7$ anyway).
- $\frac1{10!}=2.8\cdot10^{-7}<10^{-6}$ ✓ for $n=9$.

Approximation.
$$T_9(1)=1-\frac16+\frac1{120}-\frac1{5040}+\frac1{362\,880}=0.8414710\ldots$$

Check. $\sin1=0.8414709848\ldots$ so the error is about $2.5\cdot10^{-8}$, far better than the guaranteed bound.

Remark. The guaranteed bound is pessimistic by a factor of about $10$ here, because it uses $|\cos c|\le1$ at order $9$, while $T_9=T_{10}$ and the order-$10$ remainder has a factor $\frac1{11!}=2.5\cdot10^{-8}$, the true size of the error.