Question-196816

Question Number 196816 by ERLY last updated on 01/Sep/23

Answered by MM42 last updated on 01/Sep/23

a)S=(1/2)+cost+cos2t+...+cosnt 2sin(t/2)S=sin(t/2)+sin((3t)/2)−sin(t/2)+sin((5t)/2)−sin((3t)/2)+...+sin((2n+1)/2)t−sin((2n−1)/2)t S=((sin(((2n+1)/2))t)/(2sin(t/2))) ✓ b)S=sint+sin3t+...+sin(2n+1)t 2sintS=1−cos2t+cos2t−cos4t+cos4t−cos6t+...+cos2(n+1)t−cos2nt ⇒S=((1−cos2(n+1)t)/(2sint))=((sin^2 (n+1)t)/(sint)) ✓

$$\left.{a}\right){S}=\frac{\mathrm{1}}{\mathrm{2}}+{cost}+{cos}\mathrm{2}{t}+…+{cosnt} \\ $$$$\mathrm{2}{sin}\frac{{t}}{\mathrm{2}}{S}={sin}\frac{{t}}{\mathrm{2}}+{sin}\frac{\mathrm{3}{t}}{\mathrm{2}}−{sin}\frac{{t}}{\mathrm{2}}+{sin}\frac{\mathrm{5}{t}}{\mathrm{2}}−{sin}\frac{\mathrm{3}{t}}{\mathrm{2}}+…+{sin}\frac{\mathrm{2}{n}+\mathrm{1}}{\mathrm{2}}{t}−{sin}\frac{\mathrm{2}{n}−\mathrm{1}}{\mathrm{2}}{t} \\ $$$${S}=\frac{{sin}\left(\frac{\mathrm{2}{n}+\mathrm{1}}{\mathrm{2}}\right){t}}{\mathrm{2}{sin}\frac{{t}}{\mathrm{2}}}\:\:\checkmark \\ $$$$\left.{b}\right){S}={sint}+{sin}\mathrm{3}{t}+…+{sin}\left(\mathrm{2}{n}+\mathrm{1}\right){t} \\ $$$$\mathrm{2}{sintS}=\mathrm{1}−{cos}\mathrm{2}{t}+{cos}\mathrm{2}{t}−{cos}\mathrm{4}{t}+{cos}\mathrm{4}{t}−{cos}\mathrm{6}{t}+…+{cos}\mathrm{2}\left({n}+\mathrm{1}\right){t}−{cos}\mathrm{2}{nt} \\ $$$$\Rightarrow{S}=\frac{\mathrm{1}−{cos}\mathrm{2}\left({n}+\mathrm{1}\right){t}}{\mathrm{2}{sint}}=\frac{{sin}^{\mathrm{2}} \left({n}+\mathrm{1}\right){t}}{{sint}}\:\:\checkmark \\ $$

Commented by ERLY last updated on 02/Sep/23

$${la}\:{consigne}\:{non}\:{respecte} \\ $$

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Question-196816

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