1-simplify-S-n-x-k-1-n-sin-2-kx-2-simplify-A-n-k-1-n-sin-2-kpi-n-

Question Number 39332 by abdo mathsup 649 cc last updated on 05/Jul/18

1) s i m p l i f y S_{n} (x) = \sum_{k = 1}^{n} {s i n}^{2} (k x)

2) s i m p l i f y A_{n} = \sum_{k = 1}^{n} {s i n}^{2} (\frac{k π}{n})

Commented by math khazana by abdo last updated on 09/Jul/18

1) w e h a v e S_{n} (x) = \sum_{k = 0}^{n} \frac{1 - c o s (2 k x)}{2}

= \frac{n + 1}{2} - \frac{1}{2} \sum_{k = 0}^{n} c o s (2 k x) b u t

\sum_{k = 0}^{n} c o s (2 k x) = R e (\sum_{k = 0}^{n} e^{i 2 k x})

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

= \frac{1 - c o s 2 (n + 1) x - i s i n 2 (n + 1) x}{1 - c o s (2 x) - i s i n (2 x)}

= \frac{2 {s i n}^{2} (n + 1) x - 2 i s i n (n + 1) x c o s (n + 1) x}{2 {s i n}^{2} x - 2 i s i n x c o s x}

= \frac{- i s i n (n + 1) x {c o s (n + 1) x + i s i n (n + 1) x}}{- i s i n x {c o s x + i s i n x}}

= \frac{s i n (n + 1) x}{s i n x} e^{i (n + 1) x} e^{- i x}

= \frac{s i n (n + 1) x}{s i n x} {c o s (n x) + i s i n x} \Rightarrow

\sum_{k = 0}^{n} c o s (2 k x) = \frac{s i n (n + 1) x}{s i n x} c o s (n x) \Rightarrow

S_{n} (x) = \frac{n + 1}{2} - \frac{s i n (n + 1) x c o s (n x)}{2 s i n x} .

2) A_{n} = S_{n} (\frac{π}{n}) = \frac{n + 1}{2} - \frac{s i n (n + 1) \frac{π}{n} c o s (n \frac{π}{n})}{2 s i n (\frac{π}{n})}

= \frac{n + 1}{2} + \frac{s i n (π + \frac{π}{n})}{2 s i n (\frac{π}{n})} = \frac{n + 1}{2} + \frac{- s i n (\frac{π}{n})}{2 s i n (\frac{π}{n})}

= \frac{n + 1}{2} - \frac{1}{2}

= \frac{n}{2} \Rightarrow A_{n} = \frac{n}{2} .