2024 AMC 12B 第 11 题

Use $\sin^2\theta = \dfrac{1 - \cos 2\theta}{2}$ to rewrite the sum

The cosine terms $\cos 2^\circ, \cos 4^\circ, \ldots, \cos 180^\circ$ cancel in pairs via $\cos(180^\circ - \theta) = -\cos\theta,$ leaving only $\cos 180^\circ = -1$

Using $\sin^2\theta = \dfrac{1 - \cos 2\theta}{2},$ $$\begin{aligned} \sum_{n=1}^{90} \sin^2(n^\circ) &= \frac{90}{2} \\ &\quad {}- \frac12 \sum_{n=1}^{90}\cos(2n^\circ). \end{aligned}$$ In the cosine sum, the terms for $n$ and $90 - n$ satisfy $\cos(2n^\circ) + \cos(180^\circ - 2n^\circ) = 0,$ and $\cos 90^\circ = 0,$ so everything cancels except $\cos 180^\circ = -1.$

Hence the sum is $45 - \tfrac12(-1) = 45.5,$ and the mean is $\dfrac{45.5}{90} = \dfrac{91}{180}.$

Thus, the correct answer is E.