Question Number 102627 by Rio Michael last updated on 10/Jul/20

show that: cosθ + cos2θ + ....cos nθ= ((cos (1/2)(n +1)θ sin(1/2)nθ)/(sin (1/2)nθ))  Show that: sin θ + sin 2θ + ....+ sin nθ = ((sin (1/2)(n + 1)θ sin(1/2)nθ)/(sin (1/2)nθ))  where θ ∈ R and θ ≠2πk , k ∈Z

Answered by Dwaipayan Shikari last updated on 10/Jul/20

cosθ+cos2θ+....cosnθ  (1/(2sin(θ/2)))(sin((3θ)/2)−sin(θ/2)+sin((5θ)/2)−sin((3θ)/2)+.....+sin((2n+1)/2)θ−sin((2n−1)/2)θ)  (1/(2sin(θ/2)))(sin((2n+1)/2)θ−sin(θ/2))=((cos((n+1)/2)θ sin((nθ)/2))/(sin(θ/2)))

Answered by Dwaipayan Shikari last updated on 10/Jul/20

sinθ+sin2θ+....sin nθ  (1/(2sin(θ/2)))(cos(θ/2)−cos((3θ)/2)+cos((3θ)/2)−cos((5θ)/2)+....+cos((2n−1)/2)θ−cos((2n+1)/2)θ)  (1/(2sin(θ/2)))(cos(θ/2)−cos((2n+1)/2)θ)  (1/(2sin(θ/2))).2sin((n+1)/2)θsin(n/2)θ   =((sin((n+1)/2)θ sin(n/2)θ)/(sin(θ/2)))          So   sinθ+cosθ+sin2θ+cos2θ+......  n    =((sin(n/2)θ)/(sin(θ/2)))(cos((n+1)/2)θ+sin((n+1)/2)θ)

Answered by mathmax by abdo last updated on 10/Jul/20

A_n =Σ_(k=1) ^n  cos(kθ) ⇒ 1+A_n =Σ_(k=0) ^n  cos(kθ) =Re(Σ_(k=0) ^n  e^(ikθ) )  we have Σ_(k=0) ^n  (e^(iθ) )^k  =((1−e^(i(n+1)θ) )/(1−e^(iθ) )) =((1−cos(n+1)θ−isin(n+1)θ)/(1−cosθ −isinθ))  =((2sin^2 (((n+1)θ)/2)−2isin(((n+1)θ)/2))cos((((n+1)θ)/2)))/(2sin^2 ((θ/2))−2isin((θ/2))cos((θ/2))))  =((−isin(((n+1)θ)/2)){cos((((n+1)θ)/2)) +isin((((n+1)θ)/2)))/(−isin((θ/2)){cos((θ/2))+isin((θ/2))}))  =((sin((((n+1)θ)/2)))/(sin((θ/2))))×(e^(i(((n+1)θ)/2))) /e^((iθ)/2) ) =((sin((((n+1)θ)/2)))/(sin((θ/2))))×e^((inθ)/2)  ⇒  1+A_n   =((sin(((n+1)θ)/2))cos(((nθ)/2)))/(sin((θ/2)))) ⇒A_n =((cos(((nθ)/2)))/(sin((θ/2))))sin(n+1)(θ/2) −1  (error in tbe question!)

Commented byDwaipayan Shikari last updated on 10/Jul/20

yes sir . it will be sin(θ/2)  instead of sin((nθ)/2)  i think