Prove that : $ S(n) = \frac{2^{n-1}}{\pi}\int_0^{2\pi}\cos (t)\cos (2t) \dots \cos (nt) dt $

Question

For a positive integer $n$, denote by $S(n)$ the number of choices of the signs "+" or "−" such that $±1 ± 2±\dots±n = 0$. Prove that : $$ S(n) = \frac{2^{n-1}}{\pi}\int_0^{2\pi}\cos (t)\cos (2t) \dots \cos (nt) dt $$ I have no clue on how to approach this problem, but the following hint was given: $$ G_n(x) = \left(x + \frac{1}{x}\right) \left(x^2 + \frac{1}{x^2} \right) \dots \left(x^n + \frac{1}{x^n}\right) $$

Answer 1

The hint pretty much gives away the answer.

$S(n)$ is the term not depending on $x$ in $G_n(x)$. If in the expression $$\left(x + \frac{1}{x}\right) \left(x^2 + \frac{1}{x^2} \right) \dots \left(x^n + \frac{1}{x^n}\right) = S(n) + \sum_{k \neq 0 }c_kx^k$$ we set $x = e^{it}$ and integrate from $0$ to $2\pi$, we obtain $$ \int_0^{2\pi} (2\cos t)(2\cos 2t)\dots(2\cos nt)dt = 2\pi S(n) + 0$$ which gives us the desired formula which is $$ S(n) = \frac{2^{n-1}}{\pi}\int_0^{2\pi}\cos (t)\cos (2t) \dots \cos (nt) dt $$