Given an integer n > 1. Consider the sum s of the set S of complex numbers e^2πik/n for integers 0 ≤ k < n.
S = {e^2πik/n | 0 ≤ k < n}.
s = ΣS = Σ_{0 ≤ k < n} e^2πik/n. We want to show that s = 0. We form a new set S' by multiplying each element of S by e^2πi/n.
S' = {e^2πi/nx | x∈S}. ΣS' = e^2πi/nΣS. But S' = S because e^2πin/n = e^2πi = 1 = e^2πi0/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' = e^2πi/ns and
(1 − e^2πi/n)s = 0. (1 − e^2πi/n) is not 0 and thus s = 0.

e^ix = cos(x) + i sin(x).
0 = Σ_{0 ≤ k < n} e^2πik/n
= Σ_{0 ≤ k < n}(cos(2πk/n) + i sin(2πk/n))
= (Σ_{0 ≤ k < n}cos(2πk/n)) + i(Σ_{0 ≤ k < n}sin(2πk/n)).
Thus: Σ_{0 ≤ k < n}cos(2πk/n) = Σ_{0 ≤ k < n}sin(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 e^ix = 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.