Pour \(n\) naturel non nul, on pose \(S_n=\sum_{}^{}e^{i(\pm a_1\pm...\pm a_n)}\). \(S_1=e^{ia_1}+e^{-ia_1}=2\cos a_1\) Soit \(n\geq1\). Supposons que \(S_n=2^n\cos a_1...\cos a_n\) alors

\[\begin{align*} S_{n+1}&=\sum_{}^{}e^{i(\pm a_1\pm...\pm a_{n+1})}=e^{ia_{n+1}}\sum_{}^{}e^{i(\pm a_1\pm...\pm a_n)}+e^{-ia_{n+1}}\sum_{}^{}e^{i(\pm a_1\pm...\pm a_n)}\\ &=2\cos(a_{n+1})S_n=2^{n+1}\cos a_1...\cos a_{n+1}. \end{align*}\] On a montré par récurrence que : \(\forall n\geq1,\;S_n=2^n\cos a_1...\cos a_n\). Ensuite, pour \(n\geq1\), \(\sum_{}^{}\cos(\pm a_1\pm...\pm a_n)=\Re(S_n)=2^n\cos a_1...\cos a_n\) (et on obtient aussi \(\sum_{}^{}\sin(\pm a_1\pm...\pm a_n)=\Im(S_n)=0\)).

\(\forall n\in\Nn^*\), \(\sum_{}^{}\cos(\pm a_1\pm...\pm a_n)=2^n\cos a_1...\cos a_n\).

exo7 5069