Given an integer n > 1. Consider the sum s of the set S of complex numbers e2πik/n for integers 0 ≤ k < n.
S = {e2πik/n | 0 ≤ k < n}.
s = ΣS = Σ0 ≤ k < n e2πik/n. We want to show that s = 0. We form a new set S' by multiplying each element of S by e2πi/n.
S' = {e2πi/nx | x∈S}. ΣS' = e2πi/nΣS. But S' = S because e2πin/n = e2πi = 1 = e2πi0/n; the number that was removed from the set is also the number that was added. The other numbers just moved around by one slot.
Let s' = ΣS'. s = s'. But s' = e2πi/ns and
(1 − e2πi/n)s = 0. (1 − e2πi/n) is not 0 and thus s = 0.

eix = cos(x) + i sin(x).
0 = Σ0 ≤ k < n e2πik/n
= Σ0 ≤ k < n(cos(2πk/n) + i sin(2πk/n))
= (Σ0 ≤ k < ncos(2πk/n)) + i(Σ0 ≤ k < nsin(2πk/n)).
Thus: Σ0 ≤ k < ncos(2πk/n) = Σ0 ≤ k < nsin(2πk/n) = 0.
Q.E.D.

This is unsuitable for a trigonometry class that has not been introduced to complex numbers. If I were teaching trigonometry I would give an unrigorous geometric proof.

When I learned that eix = cos(x) + i sin(x), and why, I discarded by previous vague definitions of sin and cos in favor of their definition by taylor series and never looked back.