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

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) + \dots + \sin^{2} (n α)

Commented by maxmathsup by imad last updated on 26/Mar/19

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)

\dots

\dots

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 + \dots + 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 + \dots + 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

\dots

\dots

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 \dots

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 \dots

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) + \dots + \sin^{2} (89) + \sin^{2} (90)

to form the question above sir \dots .

Commented by Tawa1 last updated on 19/Mar/19

Correct sir .

\sin^{2} (α) + \sin^{2} (2 α) + \sin^{2} (3 α) + \dots + \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) + \dots + \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) + \dots + \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) + \dots + \sin^{2} (89) + \sin^{2} (90) = \frac{90}{2} - (- \frac{1}{2})

\sin^{2} (1) + \sin^{2} (2) + \sin^{2} (3) + \dots + \sin^{2} (89) + \sin^{2} (90) = 45 + 0.5

\sin^{2} (1) + \sin^{2} (2) + \sin^{2} (3) + \dots + \sin^{2} (89) + \sin^{2} (90) = 45.5

Same as what you got earlier and sir mrW .

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