Somme trigo & binomiale (ter)

Calculer $\cos a+\lp\begin{array}{c}n\\1\enar\rp\cos(a+b)+\lp\begin{array}{c}n\\2\enar\rp\cos(a+2b)+\dots +\lp\begin{array}{c}n\\n\enar\rp\cos(a+nb)$

Correction

$\begin{array}{ll} \cos a+\lp\begin{array}{c}n\\1\enar\rp\cos(a+b)+&\lp\begin{array}{c}n\\2\enar\rp\cos(a+2b)+\dots +\lp\begin{array}{c}n\\n\enar\rp\cos(a+nb) \\[1em] &=\dsp\sum_{k=1}^n \lp\begin{array}{c}n\\k\enar\rp\cos\lp a+kb\rp\\[2em] &=\dsp\Re e\lp\sum_{k=1}^n \lp\begin{array}{c}n\\k\enar\right) e^{i(a+kb)}\right)\\[2em] &=\dsp\Re e\left( e^{ia}\sum_{k=1}^n \left(\begin{array}{c}n\\k\enar\right) e^{ikb}\right)\\[2em] &=\Re e\left( e^{ia}\left(1+e^{ib}\rp^n \rp\\[2em] &=\Re e\left( e^{ia}e^{inb/2}\left( e^{-ib/2}+e^{ib/2}\rp^n\rp\\[2em] &=\Re e\left( e^{i\left( a+nb/2\right)}\left( 2\cos(b/2)\right)^n\right)\\[2em] &=2^n\cos\left( a+nb/2\rp\cos^n(b/2)\\[2em] \enar$

Cacher la correction

Tags:Complexes Sommes

Autres sujets au hasard: