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




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−........
Sumtheseriestontermssinθsin2θ+sin3θ..
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)
n=1(1)nsin(nθ)n=12ilog(1+eiθ1+eiθ)=θ2n=1(1)nsin(nθ)=12sinθsin2θ+sin3θ=12
Commented by PRITHWISH SEN 2 last updated on 26/May/21
thank you sir
thankyousir
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
Sn=k=1n(1)k+1sin(kθ)Sn=k=1n(1)ksin(kθ)=Im(k=1n(eiθ)k)wehavek=1n(eiθ)k=k0n(eiθ)k1=1(eiθ)n+11+eiθ1=1(1)n+1ei(n+1)θ1+eiθ1=1eiπ(n+1)+i(n+1)θ1+eiθ1=1ei(n+1)(θ+π)1+eiθ1=1cos(n+1)(θ+π)isin(n+1)(θ+π)1+cosθ+isinθ1=2sin2((n+1)(θ+π)2)2isin((n+1)(θ+π)2)cos((n+1)(θ+π)2)2cos2(θ2)+2isin(θ2)cos(θ2)1=isin((n+1)(θ+π)2)ei((n+1)(θ+π)2)cos(θ2)eiθ21=isin((n+1)(θ+π)2)cos(θ2)ei((n+1)(θ+π)2θ2)1=isin((n+1)(θ+π)2)cos(θ2){cos((n+1)(θ+π)θ2)+isin((n+1)(θ+π2)}1Im(Σ)=sin((n+1)(θ+π)2)cos(θ2)×cos((n+1)(θ+π)θ2)=Sn
Commented by PRITHWISH SEN 2 last updated on 27/May/21
Thank you sir
Thankyousir
Commented by mathmax by abdo last updated on 28/May/21
you arewelcome sir.
youarewelcomesir.

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