Question-107117

Question Number 107117 by Algoritm last updated on 08/Aug/20

Answered by mr W last updated on 08/Aug/20

\frac{2 \sin 2 ° + 4 \sin 4 ° + \dots + 180 \sin 180 °}{9}

o r

\frac{2 \sin^{2} 2 ° + 4 \sin^{2} 4 ° + \dots + 180 \sin^{2} 180 °}{9} ?

Commented by Algoritm last updated on 08/Aug/20

2 \sin 2 °

Commented by mr W last updated on 08/Aug/20

\frac{2 \sin 2 ° + 4 \sin 4 ° + \dots + 180 \sin 180 °}{9} \approx 572.8996

\frac{2 \sin^{2} 2 ° + 4 \sin^{2} 4 ° + \dots + 180 \sin^{2} 180 °}{9} = 450

Answered by mathmax by abdo last updated on 08/Aug/20

let S = \frac{2 \sin (2^{o}) + 4 \sin (4^{o}) + \dots . + 180 \sin (180^{o})}{9} \Rightarrow

S = \frac{1}{9} \sum_{k = 1}^{90} 2 ksin ({2 k}^{o}) but 2 k^{o} = 2 \frac{k π}{180} = \frac{k π}{90} (radian) \Rightarrow

S = \frac{2}{9} \sum_{k = 0}^{90} k \sin (\frac{k π}{90}) = \frac{2}{9} Im (\sum_{k = 0}^{90} k e^{i (\frac{k π}{90})})

let S_{n} = \sum_{k = 0}^{n} k {(e^{\frac{i π}{90}})}^{k} = f (e^{\frac{i π}{90}}) with f (x) = \sum_{k = 0}^{n} k x^{k} we have

\sum_{k = 0}^{n} x^{k} = \frac{x^{n + 1} - 1}{x - 1} (x \neq 1) \Rightarrow \sum_{k = 1}^{n} {kx}^{k - 1} = \frac{{nx}^{n + 1} - (n + 1) x^{n} + 1}{{(x - 1)}^{2}}

\Rightarrow \sum_{k = 1}^{n} {kx}^{k} = \frac{x}{{(1 - x)}^{2}} {{nx}^{n + 1} - (n + 1) x^{n} + 1} \Rightarrow

\sum_{k = 0}^{n} k {(e^{\frac{i π}{90}})}^{k} = \frac{e^{\frac{i π}{90}}}{{(1 - e^{\frac{i π}{90}})}^{2}} {n {(e^{\frac{i π}{90}})}^{n + 1} - (n + 1) {(e^{\frac{i π}{90}})}^{n} + 1} \Rightarrow

\sum_{k = 0}^{90} k {(e^{\frac{i π}{90}})}^{k} = \frac{e^{\frac{i π}{90}}}{{(1 - \cos (\frac{π}{90}) - isin (\frac{π}{90}))}^{2}} {{90 e}^{i \frac{91}{90} π} + (91) + 1}

= \frac{e^{\frac{i π}{90}} {(1 - \cos (\frac{π}{90}) + isin (\frac{π}{90}))}^{2}}{(1 - \cos {(\frac{π}{90})}^{2} + \sin^{2} (\frac{π}{90})} {90 \cos (\frac{91 π}{90}) + i 90 \sin (\frac{91 π}{90}) + 92}

rest to extract im (of this quantity) \dots . be continued \dots

Commented by Algoritm last updated on 09/Aug/20

thanks