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


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

$show that : \cos θ + \cos 2 θ + . . . . \cos n θ = \frac{\cos \frac{1}{2} (n + 1) θ \sin \frac{1}{2} n θ}{\sin \frac{1}{2} n θ}$ $Show that : \sin θ + \sin 2 θ + . . . . + \sin n θ = \frac{\sin \frac{1}{2} (n + 1) θ \sin \frac{1}{2} n θ}{\sin \frac{1}{2} n θ}$ $where θ \in R and θ \neq 2 π k, k \in Z$
	Answered by Dwaipayan Shikari last updated on 10/Jul/20

	$c o s θ + c o s 2 θ + . . . . c o s n θ$ $\frac{1}{2 s i n \frac{θ}{2}} (s i n \frac{3 θ}{2} - s i n \frac{θ}{2} + s i n \frac{5 θ}{2} - s i n \frac{3 θ}{2} + . . . . . + s i n \frac{2 n + 1}{2} θ - s i n \frac{2 n - 1}{2} θ)$ $\frac{1}{2 s i n \frac{θ}{2}} (s i n \frac{2 n + 1}{2} θ - s i n \frac{θ}{2}) = \frac{c o s \frac{n + 1}{2} θ s i n \frac{n θ}{2}}{s i n \frac{θ}{2}}$
	Answered by Dwaipayan Shikari last updated on 10/Jul/20

	$s i n θ + s i n 2 θ + . . . . s i n n θ$ $\frac{1}{2 s i n \frac{θ}{2}} (c o s \frac{θ}{2} - c o s \frac{3 θ}{2} + c o s \frac{3 θ}{2} - c o s \frac{5 θ}{2} + . . . . + c o s \frac{2 n - 1}{2} θ - c o s \frac{2 n + 1}{2} θ)$ $\frac{1}{2 s i n \frac{θ}{2}} (c o s \frac{θ}{2} - c o s \frac{2 n + 1}{2} θ)$ $\frac{1}{2 s i n \frac{θ}{2}} . 2 s i n \frac{n + 1}{2} θ s i n \frac{n}{2} θ = \frac{s i n \frac{n + 1}{2} θ s i n \frac{n}{2} θ}{s i n \frac{θ}{2}}$ $S o$ $s i n θ + c o s θ + s i n 2 θ + c o s 2 θ + . . . . . . n$ $= \frac{s i n \frac{n}{2} θ}{s i n \frac{θ}{2}} (c o s \frac{n + 1}{2} θ + s i n \frac{n + 1}{2} θ)$
	Answered by mathmax by abdo last updated on 10/Jul/20
	$A_n =Σ_(k=1) ^n cos(kθ) ⇒ 1+A_n =Σ_(k=0) ^n cos(kθ) =Re(Σ_(k=0) ^n e^(ikθ) ) we have Σ_(k=0) ^n (e^(iθ) )^k =((1−e^(i(n+1)θ) )/(1−e^(iθ) )) =((1−cos(n+1)θ−isin(n+1)θ)/(1−cosθ −isinθ)) =((2sin^2 (((n+1)θ)/2)−2isin(((n+1)θ)/2))cos((((n+1)θ)/2)))/(2sin^2 ((θ/2))−2isin((θ/2))cos((θ/2)))) =((−isin(((n+1)θ)/2)){cos((((n+1)θ)/2)) +isin((((n+1)θ)/2)))/(−isin((θ/2)){cos((θ/2))+isin((θ/2))})) =((sin((((n+1)θ)/2)))/(sin((θ/2))))×(e^(i(((n+1)θ)/2))) /e^((iθ)/2) ) =((sin((((n+1)θ)/2)))/(sin((θ/2))))×e^((inθ)/2) ⇒ 1+A_n =((sin(((n+1)θ)/2))cos(((nθ)/2)))/(sin((θ/2)))) ⇒A_n =((cos(((nθ)/2)))/(sin((θ/2))))sin(n+1)(θ/2) −1 (error in tbe question!)$
	$A_{n} = \sum_{k = 1}^{n} \cos (k θ) \Rightarrow 1 + A_{n} = \sum_{k = 0}^{n} \cos (k θ) = Re (\sum_{k = 0}^{n} e^{ik θ})$ $we have \sum_{k = 0}^{n} {(e^{i θ})}^{k} = \frac{1 - e^{i (n + 1) θ}}{1 - e^{i θ}} = \frac{1 - \cos (n + 1) θ - isin (n + 1) θ}{1 - \cos θ - isin θ}$ $= \frac{{2 \sin}^{2} \frac{(n + 1) θ}{2} - 2 isin (\frac{n + 1) θ}{2}) \cos (\frac{(n + 1) θ}{2})}{{2 \sin}^{2} (\frac{θ}{2}) - 2 isin (\frac{θ}{2}) \cos (\frac{θ}{2})}$ $= \frac{- isin (\frac{n + 1) θ}{2}) {\cos (\frac{(n + 1) θ}{2}) + isin (\frac{(n + 1) θ}{2})}{- isin (\frac{θ}{2}) {\cos (\frac{θ}{2}) + isin (\frac{θ}{2})}}$ $= \frac{\sin (\frac{(n + 1) θ}{2})}{\sin (\frac{θ}{2})} \times \frac{e^{i (\frac{n + 1) θ}{2})}}{e^{\frac{i θ}{2}}} = \frac{\sin (\frac{(n + 1) θ}{2})}{\sin (\frac{θ}{2})} \times e^{\frac{in θ}{2}} \Rightarrow$ $1 + A_{n} = \frac{\sin (\frac{n + 1) θ}{2}) \cos (\frac{n θ}{2})}{\sin (\frac{θ}{2})} \Rightarrow A_{n} = \frac{\cos (\frac{n θ}{2})}{\sin (\frac{θ}{2})} \sin (n + 1) \frac{θ}{2} - 1$ $(error in tbe question!)$
	Commented by Dwaipayan Shikari last updated on 10/Jul/20

	$y e s s i r . i t w i l l b e s i n \frac{θ}{2} i n s t e a d o f s i n \frac{n θ}{2} i t h i n k$
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