Exercice :
Soient \( n \in \mathbb{N}\) et \( \theta \in \mathbb{R}.\)
Calculer : \( \displaystyle \sum_{k=0}^{n} \cos (k \theta)\text{ et } \sum_{k=0}^{n} \sin (k \theta)\)
Corrigé :
On a :\[\begin{aligned}\sum_{k=0}^{n} \cos(k\theta)&= \operatorname{Re}\left(\sum_{k=0}^{n} e^{ik\theta}\right) \\&= \operatorname{Re}\left(\sum_{k=0}^{n} \left(e^{i\theta}\right)^k\right).\end{aligned}\]Et\[\sum_{k=0}^{n} \sin(k\theta)= \operatorname{Im}\left(\sum_{k=0}^{n} e^{ik\theta}\right).\]
Si \(\theta \equiv 0 \ [2\pi]\), alors\[\sum_{k=0}^{n} \cos(k\theta) = n+1\quad \text{et} \quad\sum_{k=0}^{n} \sin(k\theta) = 0.\]
Si \(\theta \not\equiv 0 \ [2\pi]\), alors\[\begin{aligned}\sum_{k=0}^{n} \cos(k\theta)&= \operatorname{Re}\left(\frac{\left(e^{i\theta}\right)^{n+1}-1}{e^{i\theta}-1}\right) \\&= \operatorname{Re}\left(\frac{e^{i(n+1)\theta}-1}{e^{i\theta}-1}\right) \\&= \operatorname{Re}\left(e^{i\frac{n\theta}{2}}\frac{e^{i\frac{(n+1)\theta}{2}}-e^{-i\frac{(n+1)\theta}{2}}}{e^{i\frac{\theta}{2}}-e^{-i\frac{\theta}{2}}}\right) \\&= \operatorname{Re}\left(e^{i\frac{n\theta}{2}}\frac{2i\sin\left(\frac{(n+1)\theta}{2}\right)}{2i\sin\left(\frac{\theta}{2}\right)}\right) \\&= \frac{\cos\left(\frac{n\theta}{2}\right)\sin\left(\frac{(n+1)\theta}{2}\right)}{\sin\left(\frac{\theta}{2}\right)}.\end{aligned}\]Et\[\begin{aligned}\sum_{k=0}^{n} \sin(k\theta)&= \operatorname{Im}\left(\sum_{k=0}^{n} \left(e^{i\theta}\right)^k\right) \\&= \operatorname{Im}\left(e^{i\frac{n\theta}{2}}\frac{\sin\left(\frac{(n+1)\theta}{2}\right)}{\sin\left(\frac{\theta}{2}\right)}\right) \\&= \frac{\sin\left(\frac{n\theta}{2}\right)\sin\left(\frac{(n+1)\theta}{2}\right)}{\sin\left(\frac{\theta}{2}\right)}.\end{aligned}\]