Question Number 114627 by john santu last updated on 20/Sep/20

sin A+sin 2A+sin 3A+...+sin nA =??

Answered by mathmax by abdo last updated on 20/Sep/20

let S_n (x) =sinx +sin(2x)+...+sin(nx)=Σ_(k=0) ^n  sin(kx)  =Im(Σ_(k=0) ^n  e^(ikx) )  we have Σ_(k=0) ^n  e^(ikx)  =Σ_(k=0) ^n  (e^(ix) )^k  =((1−(e^(ix) )^(n+1) )/(1−e^(ix) ))  =((1−cos(n+1)x−isin(n+1)x)/(1−cosx −isinx))  =((2sin^2 (n+1)(x/2)−2i sin((n+1)(x/2))cos(n+1)(x/2))/(2sin^2 ((x/2))−2isin((x/2))cos((x/2))))  =((−isin(n+1)(x/2){ e^(i(n+1)(x/2)) })/(−isin((x/2))e^((ix)/2) )) =((sin((n+1)(x/2)))/(sin((x/2))))×e^(in(x/2))   =((sin((((n+1)x)/2)))/(sin((x/2))))×{cos(((nx)/2))+isin(((nx)/2))} ⇒  S_n (x) =((sin(((nx)/2))sin((((n+1)x)/2)))/(sin((x/2))))    (x≠2kπ   )

Answered by Dwaipayan Shikari last updated on 20/Sep/20

sinA+sin2A+.....  (1/(2sin(A/2)))(cos(A/2)−cos((3A)/2)+cos((3A)/2)−cos((5A)/2)+cos((5A)/2)−cos((7A)/2)+....cos((2n−1)/2)A−cos((2n+1)/2)A)  (1/(2sin(A/2)))(2sin((n+1)/2)A .sin(n/2)A)=((sin((n+1)/2)A .sin(n/2)A)/(sin(A/2)))