Why do choirs work?

A better approach than the one I was taking in the comments is to see the whole process as a question about random variables in the plane. Let us define $$ X_j := \begin{bmatrix}\cos d_j \\ \sin d_j\end{bmatrix}\quad \text{and} \quad S_n := \sum_{j=1}^{n} X_j,$$ where $d_j$ is a family of iid uniform random variables on $[0, 2 \pi]$. Then, $S_n$ is a random vector in $\mathbb{R}^2$ with first and second coordinates given respectively by $$ A_n = \sum_{i=1}^n \cos d_j \quad \text{and} \quad B_n = \sum_{i=1}^n \sin d_j. $$ As mentioned in the comments, we have that $$ C(t) = \sum_{i = 1}^n \sin (t + d_i) = \sin t \Big(\sum_{j=1}^n \cos d_j\Big) + \cos t \Big( \sum_{j=1}^n \sin d_j \Big) = \langle (\cos t, \sin t), (B_n, A_n) \rangle, $$ implying that $$ \sup_{t \in [0,1]} |C(t)| = \sqrt{A_n^2 + B_n^2} = \lVert S_n \rVert $$ Thus, you just want to understand well the behavior of $S_n$, a sum of iid random vectors. Notice that the marginals of $X_j$ have the same distribution, and that $$ \mathbb{E}[\cos d_j] = \mathbb{E}[\sin d_j] = 0 \quad \text{and} \quad \mathbb{E}[\cos^2 d_j] = \mathbb{E}[\sin^2 d_j] = \frac12, $$ by applying the expected value operator to $\sin^2 d_j + \cos^2 d_j = 1$. Also, we have $$ \mathbb{E}[\cos d_j \sin d_j] = \frac12 \mathbb{E}[\sin (2d_j)] = 0. $$ Thus we can apply the multidimensional Central Limit Theorem for the sum $S_n$, and see that $$ \frac{S_n}{\sqrt{n}} \to Z $$ in distribution, where $Z$ is a normal of mean $0$ and covariance matrix $\Sigma = \begin{bmatrix}\frac12 & 0 \\ 0 & \frac12\end{bmatrix}$. I am not very familiar with these multidimensional results but I am quite confident that from here you should be able to derive precise estimates on the distribution of $\lVert S_n \rVert$ and its moments.