HMMT February

Solution:
We compute the exact answer as given above. Let $n = 12$, and $\theta_{1}, \theta_{2}, \ldots, \theta_{n}$ be uniformly randomly generated such that $\theta_{1} + \cdots + \theta_{n} = 2\pi$. We are trying to estimate

$$\mathbb{E}\left[\frac{1}{2}\sum_{i = 1}^{n}\sin (\theta_{i})\right] = \frac{1}{2}\sum_{i = 1}^{n}\mathbb{E}[\sin (\theta_{i})].$$

We do this by computing the marginal distribution of $\theta_{i}$, which is proportional to the area of the $n - 2$ dimensional cross section

$$\sum_{j\neq i}\theta_{j} = 2\pi -\theta_{i}.$$

The volume of this cross section is proportional to $(2\pi - \theta_{i})^{n - 2}$. Thus, the marginal probability distribution of $\theta_{i}$ can be written as

$$p(\theta_{i}) = C(2\pi -\theta_{i})^{n - 2}$$

for some constant $C$. We can solve for $C$ because we know $\int_{0}^{2\pi}p(\theta_{i})d\theta_{i} = 1$. The result is that

$$p(\theta_{i}) = \frac{n - 1}{2\pi}\left(1 - \frac{\theta_{i}}{2\pi}\right)^{n - 2}.$$

Thus, the quantity that we seek to estimate is

$$\mathbb{E}\left[\frac{1}{2}\sum_{i = 1}^{n}\sin \theta_{i}\right] = \frac{n}{2}\mathbb{E}[\sin \theta_{i}]$$
$$\qquad = n\int_{0}^{2\pi}p(\theta_{i})\sin \theta_{i}d\theta_{i}$$
$$\qquad = \frac{n(n - 1)}{4\pi}\int_{0}^{2\pi}\left(1 - \frac{\theta}{2\pi}\right)^{n - 2}\sin \theta d\theta$$
$$\qquad = \frac{33}{\pi}\int_{0}^{2\pi}\left(1 - \frac{\theta}{2\pi}\right)^{10}\sin \theta d\theta.$$

We can integrate this using tabular integration by parts.

$$\frac{33}{\pi}\left((1 - \frac{\theta}{2\pi})^{10} - \frac{45}{2\pi^{2}}\left(1 - \frac{\theta}{2\pi}\right)^{8} + \frac{315}{\pi^{4}}\left(1 - \frac{\theta}{2\pi}\right)^{6} - \frac{4725}{2\pi^{6}}\left(1 - \frac{\theta}{2\pi}\right)^{4} \right.$$
$$\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad$$
$$\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \left. + \frac{14175}{2\pi^{8}}\left(1 - \frac{\theta}{2\pi}\right)^{2} + \frac{14175}{4\pi^{10}}\right)\cos \theta \bigg|_{0}^{2\pi}$$

$$\qquad = \frac{33}{\pi}\left(1 - \frac{45}{2\pi^{2}} +\frac{315}{\pi^{4}} -\frac{4725}{2\pi^{6}} +\frac{14175}{2\pi^{8}}\right)$$
$$\qquad = \left[\frac{33}{\pi} -\frac{1485}{2\pi^{3}} +\frac{10395}{\pi^{5}} -\frac{155925}{\pi^{7}} +\frac{467775}{2\pi^{9}}\right]$$