Solution:

Multiplying each term by $2\sin\frac{1}{2}$ and using $2\sin k \sin\frac{1}{2} = \cos\left(k - \frac{1}{2}\right) - \cos\left(k + \frac{1}{2}\right),$ the sum telescopes: $$a_n = \frac{\cos\frac{1}{2} - \cos\left(n + \frac{1}{2}\right)}{2\sin\frac{1}{2}}.$$

So $a_n \lt 0$ exactly when $\cos\left(n + \frac{1}{2}\right) \gt \cos\frac{1}{2},$ which happens exactly when $n + \frac{1}{2}$ is within $\frac{1}{2}$ of a multiple of $2\pi:$ $$2\pi m - \tfrac{1}{2} \lt n + \tfrac{1}{2} \lt 2\pi m + \tfrac{1}{2}, \qquad \text{i.e.} \qquad 2\pi m - 1 \lt n \lt 2\pi m.$$ Each interval $(2\pi m - 1,\, 2\pi m)$ has length $1$ and contains exactly one integer, namely $\lfloor 2\pi m \rfloor.$

Hence the $100$th negative term has index $\lfloor 200\pi \rfloor.$ Since $3.14 \lt \pi \lt 3.145,$ we have $628 \lt 200\pi \lt 629,$ so the index is $628.$