simplify S_n =Σ_(k=0) ^n sin^k (x)cos(kx)


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Question Number 60682 by maxmathsup by imad last updated on 24/May/19

$s i m p l i f y S_{n} = \sum_{k = 0}^{n} {s i n}^{k} (x) c o s (k x)$
Commented by mathsolverby Abdo last updated on 29/May/19

$w e h a v e S_{n} = R e (\sum_{k = 0}^{n} {s i n}^{k} x e^{i k x})$ $= R e (\sum_{k = 0}^{n} {(e^{i x} s i n x)}^{k})$ $\sum_{k = 0}^{n} {(e^{i x} s i n x)}^{k} = \frac{1 - {(e^{i x} s i n x)}^{n + 1}}{1 - e^{i x} s i n x}$ $= \frac{1 - {s i n}^{n + 1} x e^{i (n + 1) x}}{1 - s i n x (c o s x + i s i n x)}$ $= \frac{1 - e^{i (n + 1) x} {s i n}^{n + 1} x}{1 - s i n x c o s x - {i s i n}^{2} x}$ $= \frac{(1 - s i n x c o s x + {i s i n}^{2} x) (1 - {s i n}^{n + 1} x (c o s (n + 1) x + i s i n (n + 1) x)}{{(1 - s i n x c o s x)}^{2} + {s i n}^{4} x}$ $= . . . w e g e t S n a f t e r e x t r a c t i n g R e (o f t h i s$ $q u a n t i t y . . .)$
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