# Sum-the-series-to-n-terms-sin-sin-2-sin-3-

Question Number 142105 by PRITHWISH SEN 2 last updated on 26/May/21
$$\mathrm{Sum}\:\mathrm{the}\:\mathrm{series}\:\mathrm{to}\:\mathrm{n}\:\mathrm{terms} \\$$$$\mathrm{sin}\:\theta−\mathrm{sin}\:\mathrm{2}\theta+\mathrm{sin}\:\mathrm{3}\theta−…….. \\$$
Answered by Dwaipayan Shikari last updated on 26/May/21
$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{{n}} \frac{{sin}\left({n}\theta\right)}{{n}}=\frac{\mathrm{1}}{\mathrm{2}{i}}{log}\left(\frac{\mathrm{1}+{e}^{{i}\theta} }{\mathrm{1}+{e}^{−{i}\theta} }\right)=\frac{\theta}{\mathrm{2}} \\$$$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{{n}} {sin}\left({n}\theta\right)=\frac{\mathrm{1}}{\mathrm{2}} \\$$$${sin}\theta−{sin}\mathrm{2}\theta+{sin}\mathrm{3}\theta−…=−\frac{\mathrm{1}}{\mathrm{2}} \\$$
Commented by PRITHWISH SEN 2 last updated on 26/May/21
$$\mathrm{thank}\:\mathrm{you}\:\mathrm{sir} \\$$
Answered by mathmax by abdo last updated on 26/May/21
$$\mathrm{S}_{\mathrm{n}} =\sum_{\mathrm{k}=\mathrm{1}} ^{\mathrm{n}} \left(−\mathrm{1}\right)^{\mathrm{k}+\mathrm{1}} \:\mathrm{sin}\left(\mathrm{k}\theta\right)\:\Rightarrow\mathrm{S}_{\mathrm{n}} =−\sum_{\mathrm{k}=\mathrm{1}} ^{\mathrm{n}} \left(−\mathrm{1}\right)^{\mathrm{k}} \mathrm{sin}\left(\mathrm{k}\theta\right) \\$$$$=−\mathrm{Im}\left(\sum_{\mathrm{k}=\mathrm{1}} ^{\mathrm{n}} \left(−\mathrm{e}^{\mathrm{i}\theta} \right)^{\mathrm{k}} \right)\:\:\mathrm{we}\:\mathrm{have} \\$$$$\sum_{\mathrm{k}=\mathrm{1}} ^{\mathrm{n}} \:\left(−\mathrm{e}^{\mathrm{i}\theta} \right)^{\mathrm{k}} \:=\sum_{\mathrm{k0}} ^{\mathrm{n}} \:\left(−\mathrm{e}^{\mathrm{i}\theta} \right)^{\mathrm{k}} −\mathrm{1}\:=\frac{\mathrm{1}−\left(−\mathrm{e}^{\mathrm{i}\theta} \right)^{\mathrm{n}+\mathrm{1}} }{\mathrm{1}+\mathrm{e}^{\mathrm{i}\theta} }−\mathrm{1} \\$$$$=\frac{\mathrm{1}−\left(−\mathrm{1}\right)^{\mathrm{n}+\mathrm{1}} \mathrm{e}^{\mathrm{i}\left(\mathrm{n}+\mathrm{1}\right)\theta} }{\mathrm{1}+\mathrm{e}^{\mathrm{i}\theta} }−\mathrm{1} \\$$$$=\frac{\mathrm{1}−\mathrm{e}^{\mathrm{i}\pi\left(\mathrm{n}+\mathrm{1}\right)+\mathrm{i}\left(\mathrm{n}+\mathrm{1}\right)\theta} }{\mathrm{1}+\mathrm{e}^{\mathrm{i}\theta} }−\mathrm{1}\:=\frac{\mathrm{1}−\mathrm{e}^{\mathrm{i}\left(\mathrm{n}+\mathrm{1}\right)\left(\theta+\pi\right)} }{\mathrm{1}+\mathrm{e}^{\mathrm{i}\theta} }−\mathrm{1} \\$$$$=\frac{\mathrm{1}−\mathrm{cos}\left(\mathrm{n}+\mathrm{1}\right)\left(\theta+\pi\right)−\mathrm{isin}\left(\mathrm{n}+\mathrm{1}\right)\left(\theta\:+\pi\right)}{\mathrm{1}+\mathrm{cos}\theta+\mathrm{isin}\theta}−\mathrm{1} \\$$$$=\frac{\mathrm{2sin}^{\mathrm{2}} \left(\frac{\left(\mathrm{n}+\mathrm{1}\right)\left(\theta+\pi\right)}{\mathrm{2}}\right)−\mathrm{2isin}\left(\frac{\left(\mathrm{n}+\mathrm{1}\right)\left(\theta+\pi\right)}{\mathrm{2}}\right)\mathrm{cos}\left(\frac{\left(\mathrm{n}+\mathrm{1}\right)\left(\theta+\pi\right)}{\mathrm{2}}\right)}{\mathrm{2cos}^{\mathrm{2}} \left(\frac{\theta}{\mathrm{2}}\right)+\mathrm{2isin}\left(\frac{\theta}{\mathrm{2}}\right)\mathrm{cos}\left(\frac{\theta}{\mathrm{2}}\right)}−\mathrm{1} \\$$$$=\frac{−\mathrm{isin}\left(\frac{\left(\mathrm{n}+\mathrm{1}\right)\left(\theta+\pi\right)}{\mathrm{2}}\right)\mathrm{e}^{\mathrm{i}\left(\frac{\left(\mathrm{n}+\mathrm{1}\right)\left(\theta\:+\pi\right)}{\mathrm{2}}\right)} }{\mathrm{cos}\left(\frac{\theta}{\mathrm{2}}\right)\mathrm{e}^{\mathrm{i}\frac{\theta}{\mathrm{2}}} }−\mathrm{1} \\$$$$=−\mathrm{i}\frac{\mathrm{sin}\left(\frac{\left(\mathrm{n}+\mathrm{1}\right)\left(\theta+\pi\right)}{\mathrm{2}}\right)}{\mathrm{cos}\left(\frac{\theta}{\mathrm{2}}\right)}\mathrm{e}^{\mathrm{i}\left(\frac{\left(\mathrm{n}+\mathrm{1}\right)\left(\theta+\pi\right)}{\mathrm{2}}−\frac{\theta}{\mathrm{2}}\right)} −\mathrm{1} \\$$$$=−\mathrm{i}\frac{\mathrm{sin}\left(\frac{\left(\mathrm{n}+\mathrm{1}\right)\left(\theta+\pi\right)}{\mathrm{2}}\right)}{\mathrm{cos}\left(\frac{\theta}{\mathrm{2}}\right)}\left\{\mathrm{cos}\left(\frac{\left(\mathrm{n}+\mathrm{1}\right)\left(\theta+\pi\right)−\theta}{\mathrm{2}}\right)+\mathrm{isin}\left(\frac{\left(\mathrm{n}+\mathrm{1}\right)\left(\theta+\pi\right.}{\mathrm{2}\:}\right)\right\}−\mathrm{1} \\$$$$\Rightarrow−\mathrm{Im}\left(\Sigma…\right)=\frac{\mathrm{sin}\left(\frac{\left(\mathrm{n}+\mathrm{1}\right)\left(\theta+\pi\right)}{\mathrm{2}}\right)}{\mathrm{cos}\left(\frac{\theta}{\mathrm{2}}\right)}×\mathrm{cos}\left(\frac{\left(\mathrm{n}+\mathrm{1}\right)\left(\theta+\pi\right)−\theta}{\mathrm{2}}\right) \\$$$$=\mathrm{S}_{\mathrm{n}} \\$$
Commented by PRITHWISH SEN 2 last updated on 27/May/21
$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{sir} \\$$
Commented by mathmax by abdo last updated on 28/May/21
$$\mathrm{you}\:\mathrm{arewelcome}\:\mathrm{sir}. \\$$