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Question Number 198269 by SANOGO last updated on 16/Oct/23
calcul Σ_(k=o) ^n sin(k)
$${calcul} \\ $$$$\underset{{k}={o}} {\overset{{n}} {\sum}}{sin}\left({k}\right) \\ $$
Answered by mr W last updated on 16/Oct/23
B=Σ_(k=0) ^n sin k A=Σ_(k=0) ^n cos k A+iB=Σ_(k=0) ^n (cos k+i sin k)=Σ_(k=0) ^n e^(ik) =((e^(i(n+1)) −1)/(e^i −1))=((cos (n+1)−1+i sin (n+1))/(cos 1−1+i sin 1)) =(([cos (n+1)−1+i sin (n+1)](cos 1−1−i sin 1))/((cos 1−1)^2 +sin^2 1)) =(([cos (n+1)−1](cos 1−1)+sin (n+1) sin 1+i{sin (n+1)(cos 1−1)−[cos (n+1)−1] sin 1})/(2(1−cos 1))) B=((sin (n+1) cos 1−sin (n+1)−cos (n+1) sin 1+sin 1)/(2(1−cos 1))) B=((sin n+sin 1−sin (n+1))/(2(1−cos 1))) ✓
$${B}=\underset{{k}=\mathrm{0}} {\overset{{n}} {\sum}}\mathrm{sin}\:{k} \\ $$$${A}=\underset{{k}=\mathrm{0}} {\overset{{n}} {\sum}}\mathrm{cos}\:{k} \\ $$$${A}+{iB}=\underset{{k}=\mathrm{0}} {\overset{{n}} {\sum}}\left(\mathrm{cos}\:{k}+{i}\:\mathrm{sin}\:{k}\right)=\underset{{k}=\mathrm{0}} {\overset{{n}} {\sum}}{e}^{{ik}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:=\frac{{e}^{{i}\left({n}+\mathrm{1}\right)} −\mathrm{1}}{{e}^{{i}} −\mathrm{1}}=\frac{\mathrm{cos}\:\left({n}+\mathrm{1}\right)−\mathrm{1}+{i}\:\mathrm{sin}\:\left({n}+\mathrm{1}\right)}{\mathrm{cos}\:\mathrm{1}−\mathrm{1}+{i}\:\mathrm{sin}\:\mathrm{1}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\left[\mathrm{cos}\:\left({n}+\mathrm{1}\right)−\mathrm{1}+{i}\:\mathrm{sin}\:\left({n}+\mathrm{1}\right)\right]\left(\mathrm{cos}\:\mathrm{1}−\mathrm{1}−{i}\:\mathrm{sin}\:\mathrm{1}\right)}{\left(\mathrm{cos}\:\mathrm{1}−\mathrm{1}\right)^{\mathrm{2}} +\mathrm{sin}^{\mathrm{2}} \:\mathrm{1}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\left[\mathrm{cos}\:\left({n}+\mathrm{1}\right)−\mathrm{1}\right]\left(\mathrm{cos}\:\mathrm{1}−\mathrm{1}\right)+\mathrm{sin}\:\left({n}+\mathrm{1}\right)\:\mathrm{sin}\:\mathrm{1}+{i}\left\{\mathrm{sin}\:\left({n}+\mathrm{1}\right)\left(\mathrm{cos}\:\mathrm{1}−\mathrm{1}\right)−\left[\mathrm{cos}\:\left({n}+\mathrm{1}\right)−\mathrm{1}\right]\:\mathrm{sin}\:\mathrm{1}\right\}}{\mathrm{2}\left(\mathrm{1}−\mathrm{cos}\:\mathrm{1}\right)} \\ $$$${B}=\frac{\mathrm{sin}\:\left({n}+\mathrm{1}\right)\:\mathrm{cos}\:\mathrm{1}−\mathrm{sin}\:\left({n}+\mathrm{1}\right)−\mathrm{cos}\:\left({n}+\mathrm{1}\right)\:\mathrm{sin}\:\mathrm{1}+\mathrm{sin}\:\mathrm{1}}{\mathrm{2}\left(\mathrm{1}−\mathrm{cos}\:\mathrm{1}\right)} \\ $$$${B}=\frac{\mathrm{sin}\:{n}+\mathrm{sin}\:\mathrm{1}−\mathrm{sin}\:\left({n}+\mathrm{1}\right)}{\mathrm{2}\left(\mathrm{1}−\mathrm{cos}\:\mathrm{1}\right)}\:\checkmark \\ $$

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