Answer: 241

Small Hint:

Multiply numerator and denominator by $2\sin\frac{1}{2}^\circ$ and telescope using product-to-sum identities

Big Hint:

Sum-to-product turns the result into $\cot 22.5^\circ,$ which the half-angle formula evaluates exactly

Solution:

Multiply numerator and denominator by $2\sin\frac{1}{2}^\circ.$ Since $2\cos n^\circ \sin\frac{1}{2}^\circ$ $= \sin\left(n + \frac{1}{2}\right)^\circ$ $- \sin\left(n - \frac{1}{2}\right)^\circ$ and $2\sin n^\circ \sin\frac{1}{2}^\circ$ $= \cos\left(n - \frac{1}{2}\right)^\circ$ $- \cos\left(n + \frac{1}{2}\right)^\circ,$ both sums telescope: $$\begin{aligned} x &= \frac{\sin 44.5^\circ - \sin 0.5^\circ}{\cos 0.5^\circ - \cos 44.5^\circ} \\ &= \frac{2\cos 22.5^\circ \sin 22^\circ}{2\sin 22.5^\circ \sin 22^\circ} \\ &= \cot 22.5^\circ, \end{aligned}$$ using the sum-to-product identities in the last step.

By the half-angle formula, $\cot 22.5^\circ = \frac{1 + \cos 45^\circ}{\sin 45^\circ} = \sqrt{2} + 1.$ Since $1.41^2 \lt 2 \lt 1.42^2,$ we have $241 \lt 100\sqrt{2} + 100 \lt 242.$ Therefore the greatest integer not exceeding $100x$ is $241.$