prove-that-k-0-k-n-cos-2-kx-n-1-2-sin-n-1-x-cos-nx-2-sinx-x-from-R-kpi-k-Z-then-find-the-value-of-integral-0-pi-sin-n-1-x-cos-nx-sinx-dx-

Question Number 26242 by abdo imad last updated on 22/Dec/17

p r o v e t h a t \sum_{k = 0}^{k = n} {c o s}^{2} (k x) = \frac{n + 1}{2} + \frac{s i n ((n + 1) x) c o s (n x)}{2 s i n x}

x f r o m R - {k π . k ε Z} t h e n f i n d t h e v a l u e o f i n t e g r a l

\int_{0}^{π} \frac{s i n ((n + 1) x) c o s (n x)}{s i n x} d x

Commented by abdo imad last updated on 28/Dec/17

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

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

= R e (\frac{1 - e^{i 2 (n + 1) x}}{1 - e^{i 2 x}}) b u t \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} e^{i n x}

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

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

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

\int_{0}^{π} \frac{s i n (n + 1) x c o s (n x)}{2 s i n x} d x = \sum_{k =>}^{n} \int_{0}^{π} {c o s}^{2} (k x) d x - \frac{(n + 1) π}{2}

= \frac{1}{2} \sum_{k = 0}^{k = n} \int_{0}^{π} (1 + c o s (2 k x)) d x - \frac{(n + 1) π}{2}

= \frac{1}{2} \sum_{k = 0}^{k = n} \int_{0}^{π} c o s (2 k x) d x = \frac{π}{2} + \sum_{k = 1}^{k = n} {[\frac{1}{2 k} s i n (2 k x)]}_{k = 0}^{k = π =}

= \frac{π}{2} + 0 = \frac{π}{2} \Rightarrow \int_{0}^{π} \frac{s i n (n + 1) x c o s (n x)}{s i n x} d x = π