For two exponentials, we have:
$$e^{i\alpha} + e^{i\beta} = 2\exp\left(i\frac{\alpha+\beta}{2}\right)\cos\left(\frac{\alpha-\beta}{2}\right), \tag{1a}$$ $$e^{i\alpha} - e^{i\beta} = 2i\exp\left(i\frac{\alpha+\beta}{2}\right)\sin\left(\frac{\alpha-\beta}{2}\right), \tag{1b}$$
so, in general we have:
$$Ae^{i\alpha} + Be^{i\beta} = e^{i\frac{\alpha+\beta}{2}}\left[(A+B)\cos\left(\frac{\alpha-\beta}{2}\right) + (A-B)i\sin\left(\frac{\alpha-\beta}{2}\right) \right], \tag{1c}$$
Question 1. For three exponentials, can be write this:
$$Ae^{i\alpha} + Be^{i\beta} + Ce^{i\gamma} = e^{i\frac{\alpha+\beta+\gamma}{2}} P_2(\alpha,\beta,\gamma), \tag{2}$$
being $P_2(\alpha,\beta,\gamma)$ a quadratic polynomial in $\sin(\cdot)$ and $\cos(\cdot)$?
Question 2. If the answer to the previous question is "yes", then we can find something similar formula for $n$ terms? Namely:
$$\sum_{k=1}^n A_k e^{i\alpha_k} = e^{i\frac{\sum_k \alpha_k}{2}} P_{n-1}(\alpha_i), \tag{n}$$
being now $P_{n-1}(\alpha_i)$ an homogeneous trigonometric polynomial of degree $n-1$.