Sum the series: sin^2 (α) + sin^2 (2α) + sin^2 (3α) + ... + sin^2 (nα)


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Question Number 56602 by Tawa1 last updated on 19/Mar/19

$Sum the series :$ $\sin^{2} (α) + \sin^{2} (2 α) + \sin^{2} (3 α) + . . . + \sin^{2} (n α)$
Commented by maxmathsup by imad last updated on 26/Mar/19
$let S_n =Σ_(k=1) ^n sin^2 (kα) ⇒S_n =Σ_(k=1) ^n ((1−cos(2kα))/2) =(n/2) −(1/2)Σ_(k=1) ^n cos(2kα) but Σ_(k=1) ^n cos(2kα) =Σ_(k=0) ^n cos(2kα)−1 =Re(Σ_(k=0) ^n e^(i2kα) )−1 Σ_(k=0) ^n e^(i2kα) =Σ_(k=0) ^n (e^(i2α) )^k =((1−e^(i2(n+1)α) )/(1−e^(i2α) )) (if e^(i2α) ≠1) =((1−cos2(n+1)α−isin2(n+1)α)/(1−cos(2α)−isin(2α))) = ((2sin^2 (n+1)α −2isin((n+1)α) cos((n+1)α))/(2sin^2 α −2i sinα cosα)) =((−i sin(n+1)α{cos(n+1)α +i sin(n+1)α)/(−isinα{ cosα +isinα})) =((sin(n+1)α)/(sinα)) e^(i(n+1)α) e^(−iα) =((sin(n+1)α)/(sinα)) e^(inα) =((sin(n+1)α)/(sinα)){cos(nα)+isin(nα)} ⇒ Σ_(k=1) ^n cos(2kα) =((sin(n+1)α cos(nα))/(sinα)) −1 ⇒ S_n =(n/2) −(1/2)( ((sin(n+1)α cos(nα))/(sinα)) −1) =((n+1)/2) −((sin(n+1)α cos(nα))/(2sinα)) if e^(i2α) ≠1 .$
$l e t S_{n} = \sum_{k = 1}^{n} {s i n}^{2} (k α) \Rightarrow S_{n} = \sum_{k = 1}^{n} \frac{1 - c o s (2 k α)}{2}$ $= \frac{n}{2} - \frac{1}{2} \sum_{k = 1}^{n} c o s (2 k α) b u t$ $\sum_{k = 1}^{n} c o s (2 k α) = \sum_{k = 0}^{n} c o s (2 k α) - 1 = R e (\sum_{k = 0}^{n} e^{i 2 k α}) - 1$ $\sum_{k = 0}^{n} e^{i 2 k α} = \sum_{k = 0}^{n} {(e^{i 2 α})}^{k} = \frac{1 - e^{i 2 (n + 1) α}}{1 - e^{i 2 α}} (i f e^{i 2 α} \neq 1)$ $= \frac{1 - c o s 2 (n + 1) α - i s i n 2 (n + 1) α}{1 - c o s (2 α) - i s i n (2 α)} = \frac{2 {s i n}^{2} (n + 1) α - 2 i s i n ((n + 1) α) c o s ((n + 1) α)}{2 {s i n}^{2} α - 2 i s i n α c o s α}$ $= \frac{- i s i n (n + 1) α {c o s (n + 1) α + i s i n (n + 1) α}{- i s i n α {c o s α + i s i n α}} = \frac{s i n (n + 1) α}{s i n α} e^{i (n + 1) α} e^{- i α}$ $= \frac{s i n (n + 1) α}{s i n α} e^{i n α} = \frac{s i n (n + 1) α}{s i n α} {c o s (n α) + i s i n (n α)} \Rightarrow$ $\sum_{k = 1}^{n} c o s (2 k α) = \frac{s i n (n + 1) α c o s (n α)}{s i n α} - 1 \Rightarrow$ $S_{n} = \frac{n}{2} - \frac{1}{2} (\frac{s i n (n + 1) α c o s (n α)}{s i n α} - 1) = \frac{n + 1}{2} - \frac{s i n (n + 1) α c o s (n α)}{2 s i n α}$ $i f e^{i 2 α} \neq 1 .$
	Answered by tanmay.chaudhury50@gmail.com last updated on 19/Mar/19

	$T_{n} = {s i n}^{2} (n α)$ $T_{n} = \frac{1 - c o s 2 n α}{2}$ $= \frac{1}{2} - \frac{1}{2} (c o s 2 n α)$ $T_{1} = \frac{1}{2} - \frac{1}{2} (c o s 2 α)$ $T_{2} = \frac{1}{2} - \frac{1}{2} (c o s 4 α)$ $T_{3} = \frac{1}{2} - \frac{1}{2} (c o s 6 α)$ $. . .$ $. . .$ $T_{n} = \frac{1}{2} - \frac{1}{2} (c o s 2 n α)$ $S = n \times \frac{1}{2} - \frac{1}{2} (c o s 2 α + c o s 4 α + c o s 6 α + . . . + c o s 2 n α)$ $S = \frac{n}{2} - \frac{p}{2}$ $p = c o s 2 α + c o s 4 α + c o s 6 α + . . . + c o s 2 n α$ $2 c o s 2 α s i n α = s i n 3 α - s i n α$ $2 c o s 4 α s i n α = s i n 5 α - s i n 3 α$ $2 c o s 6 α s i n α = s i n 7 α - s i n 5 α$ $. . .$ $. . .$ $2 c o s 2 n α s i n α = s i n (2 n + 1) α - s i n (2 n - 1) α$ $a d d t h e m$ $2 s i n α (c o s 2 α + c o s 4 α + . . + c o s 2 n α) = s i n (2 n + 1) α - s i n α$ $2 s i n α \times p = s i n (2 n + 1) α - s i n α$ $p = \frac{s i n (2 n + 1) α - s i n α}{2 s i n α}$ $S o S = \frac{n}{2} - \frac{p}{2}$ $= \frac{n}{2} - \frac{s i n (2 n + 1) α - s i n α}{4 s i n α}$ $p l s c h e c k . . .$
	Commented by maxmathsup by imad last updated on 19/Mar/19

	$b e c a r e f u l α h e r e i s i n r a d i a n n o t i n d e g r e . . .$
	Commented by Tawa1 last updated on 19/Mar/19

	$Thanks sir .$
	Commented by Tawa1 last updated on 19/Mar/19

	$Then it will be stated that where α is in degree, can it ?$
	Commented by Tawa1 last updated on 19/Mar/19

	$I used this question \sin^{2} (1) + \sin^{2} (2) + \sin^{2} (3) + . . . + \sin^{2} (89) + \sin^{2} (90)$ $to form the question above sir . . . .$
	Commented by Tawa1 last updated on 19/Mar/19

	$Correct sir .$ $\sin^{2} (α) + \sin^{2} (2 α) + \sin^{2} (3 α) + . . . + \sin^{2} (n α) = \frac{n}{2} - \frac{\sin (2 n + 1) α - \sin (α)}{4 \sin (α)}$ $where α = 1$ $\sin^{2} (1) + \sin^{2} (2) + \sin^{2} (3) + . . . + \sin^{2} (89) + \sin^{2} (90)$ $= \frac{90}{2} - \frac{\sin [2 (90) + 1] - \sin (1)}{4 \sin (1)}$ $\sin^{2} (1) + \sin^{2} (2) + \sin^{2} (3) + . . . + \sin^{2} (89) + \sin^{2} (90)$ $= \frac{90}{2} - \frac{\sin (180) - \sin (1)}{4 \sin (1)}$ $\sin^{2} (1) + \sin^{2} (2) + \sin^{2} (3) + . . . + \sin^{2} (89) + \sin^{2} (90) = \frac{90}{2} - (- \frac{1}{2})$ $\sin^{2} (1) + \sin^{2} (2) + \sin^{2} (3) + . . . + \sin^{2} (89) + \sin^{2} (90) = 45 + 0.5$ $\sin^{2} (1) + \sin^{2} (2) + \sin^{2} (3) + . . . + \sin^{2} (89) + \sin^{2} (90) = 45.5$ $Same as what you got earlier and sir mrW .$
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