Exercice :
Soient \(x \in \mathbb{R}\) et \(n \in \mathbb{N}\).
Calculer \[\sum_{k=0}^{n} \binom{n}{k} \sin(kx).\] et \[\sum_{k=0}^{n} \binom{n}{k} \cos(kx).\]
Corrigé :
Soient \(x \in \mathbb{R}\) et \(n \in \mathbb{N}\).
On pose\[A_n = \sum_{k=0}^{n} \binom{n}{k} \cos(kx)\quad \text{et} \quad B_n = \sum_{k=0}^{n} \binom{n}{k} \sin(kx).\]
On a :\[\begin{aligned}A_n + iB_n&= \sum_{k=0}^{n} \binom{n}{k}\bigl(\cos(kx) + i\sin(kx)\bigr) \\&= \sum_{k=0}^{n} \binom{n}{k} e^{ikx} \\&= \sum_{k=0}^{n} \binom{n}{k} \left(e^{ix}\right)^k\quad \text{(formule de Moivre)} \\&= \left(1 + e^{ix}\right)^n\quad \text{(formule du binôme de Newton)} \\&= \left(e^{i\frac{x}{2}}\left(e^{i\frac{x}{2}} + e^{-i\frac{x}{2}}\right)\right)^n \\&= \left(2e^{i\frac{x}{2}}\cos\left(\frac{x}{2}\right)\right)^n \\&= 2^n \cos^n\left(\frac{x}{2}\right)e^{i\frac{nx}{2}}.\end{aligned}\]
Ainsi,\[\begin{aligned}A_n&= \operatorname{Re}(A_n+iB_n) \\&= \operatorname{Re}\left(2^n \cos^n\left(\frac{x}{2}\right)e^{i\frac{nx}{2}}\right) \\&= 2^n \cos^n\left(\frac{x}{2}\right)\cos\left(\frac{nx}{2}\right).\end{aligned}\]
De même,\[\begin{aligned}B_n&= \operatorname{Im}(A_n+iB_n) \\&= \operatorname{Im}\left(2^n \cos^n\left(\frac{x}{2}\right)e^{i\frac{nx}{2}}\right) \\&= 2^n \cos^n\left(\frac{x}{2}\right)\sin\left(\frac{nx}{2}\right).\end{aligned}\]