Question Number 196816 by ERLY last updated on 01/Sep/23
Answered by MM42 last updated on 01/Sep/23
$$\left.{a}\right){S}=\frac{\mathrm{1}}{\mathrm{2}}+{cost}+{cos}\mathrm{2}{t}+…+{cosnt} \\ $$$$\mathrm{2}{sin}\frac{{t}}{\mathrm{2}}{S}={sin}\frac{{t}}{\mathrm{2}}+{sin}\frac{\mathrm{3}{t}}{\mathrm{2}}−{sin}\frac{{t}}{\mathrm{2}}+{sin}\frac{\mathrm{5}{t}}{\mathrm{2}}−{sin}\frac{\mathrm{3}{t}}{\mathrm{2}}+…+{sin}\frac{\mathrm{2}{n}+\mathrm{1}}{\mathrm{2}}{t}−{sin}\frac{\mathrm{2}{n}−\mathrm{1}}{\mathrm{2}}{t} \\ $$$${S}=\frac{{sin}\left(\frac{\mathrm{2}{n}+\mathrm{1}}{\mathrm{2}}\right){t}}{\mathrm{2}{sin}\frac{{t}}{\mathrm{2}}}\:\:\checkmark \\ $$$$\left.{b}\right){S}={sint}+{sin}\mathrm{3}{t}+…+{sin}\left(\mathrm{2}{n}+\mathrm{1}\right){t} \\ $$$$\mathrm{2}{sintS}=\mathrm{1}−{cos}\mathrm{2}{t}+{cos}\mathrm{2}{t}−{cos}\mathrm{4}{t}+{cos}\mathrm{4}{t}−{cos}\mathrm{6}{t}+…+{cos}\mathrm{2}\left({n}+\mathrm{1}\right){t}−{cos}\mathrm{2}{nt} \\ $$$$\Rightarrow{S}=\frac{\mathrm{1}−{cos}\mathrm{2}\left({n}+\mathrm{1}\right){t}}{\mathrm{2}{sint}}=\frac{{sin}^{\mathrm{2}} \left({n}+\mathrm{1}\right){t}}{{sint}}\:\:\checkmark \\ $$
Commented by ERLY last updated on 02/Sep/23
$${la}\:{consigne}\:{non}\:{respecte} \\ $$