Finite-case symmetry leads to infinite-case asymmetry

Question

Formulas for sines or cosines of sums superficially appear to have a certain symmetry, specifically it looks as if sine and cosine play something like symmetrical roles: $$ \begin{align} & \sin(\alpha+\beta+\gamma+\delta+\varepsilon+\zeta) \\[8pt] = & \underbrace{\sin\alpha\ \overbrace{\cos\beta\cos\gamma\cos\delta\cos\varepsilon\cos\zeta}^{\text{5 cosines}}+\cdots\cdots}_{\text{6 terms}} \\[8pt] & \underbrace{{} - \overbrace{\sin\alpha\sin\beta\sin\gamma}^{\text{3 sines}}\ \ \overbrace{\cos\delta\cos\varepsilon\cos\zeta}^{\text{3 cosines}}-\cdots\cdots}_{\text{20 terms}} \\[8pt] & \underbrace{\overbrace{{} + \sin\alpha\sin\beta\sin\gamma\sin\delta\sin\varepsilon}^{\text{5 sines}}\ \ \cos\zeta+ \cdots\cdots}_{\text{6 terms}} \end{align} $$

But that superficial appearance vanishes when there are infinitely many terms: every term has only finitely many sine factors and cofinitely many cosine factors:

$$ \sin\left( \sum_{k=1}^\infty \alpha_k \right) = \sum_{\text{odd }k} (-1)^{(k-1)/2} \sum_{\begin{smallmatrix} A \subseteq\{1,2,3,\ldots\} \\ |A|=k \end{smallmatrix}} \prod_{k\in A} \sin\alpha_k \prod_{k\not\in A}\cos\alpha_k. $$

This is of course an instance of the fact that when one enumerates the finite subsets of a set, then the set of complements is the same as the set enumerated if the set is finite, but quite different if the set is infinite. But until one looks at the infinite case, one doesn't know that sine rather than cosine is the one that will appear only finitely many times.

So I have a somewhat open-ended question: Where else in mathematics is symmetry in finite cases replaced by extreme asymmetry in infinite cases?

Answer 1

I don't know if this is exactly what you are looking for, since you don't really have a defnition for symmetry here. One instance that I can think of is $\sum_{i=1}^{k}a_{i}=\sum_{i=1}^{k}a_{\sigma{(i)}}$ for any finite sequence of real numbers, where $\sigma$ is a permutation. So there is a certain symmetry here; the role played by $a_{1}$ in the sum can be played by any $a_{i}$.

But this isn't true about a (countably infinite) sequence unless that sequence is absolutely convergent.

Answer 2

Alright, does this amount to an "answer" or not?

I ought to have noticed that the "$\pm$" choice in each term is determined by the number of sine factors and not by the number of cosine factors. A term with an odd number of sine factors has a "$+$" sign; a term with a specified number of cosine factors has "$+$" or "$-$" depending on how many terms are added ($\alpha+\beta+\gamma+\cdots\cdots\cdots\cdots$").

(This still leaves unanswered questions$\ldots\ldots$)