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Question Number 142105 by PRITHWISH SEN 2 last updated on 26/May/21

Sum the series to n terms sin θ−sin 2θ+sin 3θ−........

$$\mathrm{Sum}\:\mathrm{the}\:\mathrm{series}\:\mathrm{to}\:\mathrm{n}\:\mathrm{terms} \\ $$$$\mathrm{sin}\:\theta−\mathrm{sin}\:\mathrm{2}\theta+\mathrm{sin}\:\mathrm{3}\theta−........ \\ $$

Answered by Dwaipayan Shikari last updated on 26/May/21

Σ_(n=1) ^∞ (−1)^n ((sin(nθ))/n)=(1/(2i))log(((1+e^(iθ) )/(1+e^(−iθ) )))=(θ/2) Σ_(n=1) ^∞ (−1)^n sin(nθ)=(1/2) sinθ−sin2θ+sin3θ−...=−(1/2)

$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{{n}} \frac{{sin}\left({n}\theta\right)}{{n}}=\frac{\mathrm{1}}{\mathrm{2}{i}}{log}\left(\frac{\mathrm{1}+{e}^{{i}\theta} }{\mathrm{1}+{e}^{−{i}\theta} }\right)=\frac{\theta}{\mathrm{2}} \\ $$$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{{n}} {sin}\left({n}\theta\right)=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${sin}\theta−{sin}\mathrm{2}\theta+{sin}\mathrm{3}\theta−...=−\frac{\mathrm{1}}{\mathrm{2}} \\ $$

Commented by PRITHWISH SEN 2 last updated on 26/May/21

thank you sir

$$\mathrm{thank}\:\mathrm{you}\:\mathrm{sir} \\ $$

Answered by mathmax by abdo last updated on 26/May/21

S_n =Σ_(k=1) ^n (−1)^(k+1) sin(kθ) ⇒S_n =−Σ_(k=1) ^n (−1)^k sin(kθ) =−Im(Σ_(k=1) ^n (−e^(iθ) )^k ) we have Σ_(k=1) ^n (−e^(iθ) )^k =Σ_(k0) ^n (−e^(iθ) )^k −1 =((1−(−e^(iθ) )^(n+1) )/(1+e^(iθ) ))−1 =((1−(−1)^(n+1) e^(i(n+1)θ) )/(1+e^(iθ) ))−1 =((1−e^(iπ(n+1)+i(n+1)θ) )/(1+e^(iθ) ))−1 =((1−e^(i(n+1)(θ+π)) )/(1+e^(iθ) ))−1 =((1−cos(n+1)(θ+π)−isin(n+1)(θ +π))/(1+cosθ+isinθ))−1 =((2sin^2 ((((n+1)(θ+π))/2))−2isin((((n+1)(θ+π))/2))cos((((n+1)(θ+π))/2)))/(2cos^2 ((θ/2))+2isin((θ/2))cos((θ/2))))−1 =((−isin((((n+1)(θ+π))/2))e^(i((((n+1)(θ +π))/2))) )/(cos((θ/2))e^(i(θ/2)) ))−1 =−i((sin((((n+1)(θ+π))/2)))/(cos((θ/2))))e^(i((((n+1)(θ+π))/2)−(θ/2))) −1 =−i((sin((((n+1)(θ+π))/2)))/(cos((θ/2)))){cos((((n+1)(θ+π)−θ)/2))+isin((((n+1)(θ+π)/(2 )))}−1 ⇒−Im(Σ...)=((sin((((n+1)(θ+π))/2)))/(cos((θ/2))))×cos((((n+1)(θ+π)−θ)/2)) =S_n

$$\mathrm{S}_{\mathrm{n}} =\sum_{\mathrm{k}=\mathrm{1}} ^{\mathrm{n}} \left(−\mathrm{1}\right)^{\mathrm{k}+\mathrm{1}} \:\mathrm{sin}\left(\mathrm{k}\theta\right)\:\Rightarrow\mathrm{S}_{\mathrm{n}} =−\sum_{\mathrm{k}=\mathrm{1}} ^{\mathrm{n}} \left(−\mathrm{1}\right)^{\mathrm{k}} \mathrm{sin}\left(\mathrm{k}\theta\right) \\ $$$$=−\mathrm{Im}\left(\sum_{\mathrm{k}=\mathrm{1}} ^{\mathrm{n}} \left(−\mathrm{e}^{\mathrm{i}\theta} \right)^{\mathrm{k}} \right)\:\:\mathrm{we}\:\mathrm{have} \\ $$$$\sum_{\mathrm{k}=\mathrm{1}} ^{\mathrm{n}} \:\left(−\mathrm{e}^{\mathrm{i}\theta} \right)^{\mathrm{k}} \:=\sum_{\mathrm{k0}} ^{\mathrm{n}} \:\left(−\mathrm{e}^{\mathrm{i}\theta} \right)^{\mathrm{k}} −\mathrm{1}\:=\frac{\mathrm{1}−\left(−\mathrm{e}^{\mathrm{i}\theta} \right)^{\mathrm{n}+\mathrm{1}} }{\mathrm{1}+\mathrm{e}^{\mathrm{i}\theta} }−\mathrm{1} \\ $$$$=\frac{\mathrm{1}−\left(−\mathrm{1}\right)^{\mathrm{n}+\mathrm{1}} \mathrm{e}^{\mathrm{i}\left(\mathrm{n}+\mathrm{1}\right)\theta} }{\mathrm{1}+\mathrm{e}^{\mathrm{i}\theta} }−\mathrm{1} \\ $$$$=\frac{\mathrm{1}−\mathrm{e}^{\mathrm{i}\pi\left(\mathrm{n}+\mathrm{1}\right)+\mathrm{i}\left(\mathrm{n}+\mathrm{1}\right)\theta} }{\mathrm{1}+\mathrm{e}^{\mathrm{i}\theta} }−\mathrm{1}\:=\frac{\mathrm{1}−\mathrm{e}^{\mathrm{i}\left(\mathrm{n}+\mathrm{1}\right)\left(\theta+\pi\right)} }{\mathrm{1}+\mathrm{e}^{\mathrm{i}\theta} }−\mathrm{1} \\ $$$$=\frac{\mathrm{1}−\mathrm{cos}\left(\mathrm{n}+\mathrm{1}\right)\left(\theta+\pi\right)−\mathrm{isin}\left(\mathrm{n}+\mathrm{1}\right)\left(\theta\:+\pi\right)}{\mathrm{1}+\mathrm{cos}\theta+\mathrm{isin}\theta}−\mathrm{1} \\ $$$$=\frac{\mathrm{2sin}^{\mathrm{2}} \left(\frac{\left(\mathrm{n}+\mathrm{1}\right)\left(\theta+\pi\right)}{\mathrm{2}}\right)−\mathrm{2isin}\left(\frac{\left(\mathrm{n}+\mathrm{1}\right)\left(\theta+\pi\right)}{\mathrm{2}}\right)\mathrm{cos}\left(\frac{\left(\mathrm{n}+\mathrm{1}\right)\left(\theta+\pi\right)}{\mathrm{2}}\right)}{\mathrm{2cos}^{\mathrm{2}} \left(\frac{\theta}{\mathrm{2}}\right)+\mathrm{2isin}\left(\frac{\theta}{\mathrm{2}}\right)\mathrm{cos}\left(\frac{\theta}{\mathrm{2}}\right)}−\mathrm{1} \\ $$$$=\frac{−\mathrm{isin}\left(\frac{\left(\mathrm{n}+\mathrm{1}\right)\left(\theta+\pi\right)}{\mathrm{2}}\right)\mathrm{e}^{\mathrm{i}\left(\frac{\left(\mathrm{n}+\mathrm{1}\right)\left(\theta\:+\pi\right)}{\mathrm{2}}\right)} }{\mathrm{cos}\left(\frac{\theta}{\mathrm{2}}\right)\mathrm{e}^{\mathrm{i}\frac{\theta}{\mathrm{2}}} }−\mathrm{1} \\ $$$$=−\mathrm{i}\frac{\mathrm{sin}\left(\frac{\left(\mathrm{n}+\mathrm{1}\right)\left(\theta+\pi\right)}{\mathrm{2}}\right)}{\mathrm{cos}\left(\frac{\theta}{\mathrm{2}}\right)}\mathrm{e}^{\mathrm{i}\left(\frac{\left(\mathrm{n}+\mathrm{1}\right)\left(\theta+\pi\right)}{\mathrm{2}}−\frac{\theta}{\mathrm{2}}\right)} −\mathrm{1} \\ $$$$=−\mathrm{i}\frac{\mathrm{sin}\left(\frac{\left(\mathrm{n}+\mathrm{1}\right)\left(\theta+\pi\right)}{\mathrm{2}}\right)}{\mathrm{cos}\left(\frac{\theta}{\mathrm{2}}\right)}\left\{\mathrm{cos}\left(\frac{\left(\mathrm{n}+\mathrm{1}\right)\left(\theta+\pi\right)−\theta}{\mathrm{2}}\right)+\mathrm{isin}\left(\frac{\left(\mathrm{n}+\mathrm{1}\right)\left(\theta+\pi\right.}{\mathrm{2}\:}\right)\right\}−\mathrm{1} \\ $$$$\Rightarrow−\mathrm{Im}\left(\Sigma...\right)=\frac{\mathrm{sin}\left(\frac{\left(\mathrm{n}+\mathrm{1}\right)\left(\theta+\pi\right)}{\mathrm{2}}\right)}{\mathrm{cos}\left(\frac{\theta}{\mathrm{2}}\right)}×\mathrm{cos}\left(\frac{\left(\mathrm{n}+\mathrm{1}\right)\left(\theta+\pi\right)−\theta}{\mathrm{2}}\right) \\ $$$$=\mathrm{S}_{\mathrm{n}} \\ $$

Commented by PRITHWISH SEN 2 last updated on 27/May/21

Thank you sir

$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{sir} \\ $$

Commented by mathmax by abdo last updated on 28/May/21

you arewelcome sir.

$$\mathrm{you}\:\mathrm{arewelcome}\:\mathrm{sir}. \\ $$

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