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Question Number 60682 by maxmathsup by imad last updated on 24/May/19
$${simplify}\:\:\:\:{S}_{{n}} =\sum_{{k}=\mathrm{0}} ^{{n}} \:\:{sin}^{{k}} \left({x}\right){cos}\left({kx}\right)\:\:\: \\ $$
Commented by mathsolverby Abdo last updated on 29/May/19
$${we}\:{have}\:{S}_{{n}} ={Re}\left(\sum_{{k}=\mathrm{0}} ^{{n}} \:\:{sin}^{{k}} {x}\:{e}^{{ikx}} \right) \\ $$$$={Re}\left(\sum_{{k}=\mathrm{0}} ^{{n}} \left({e}^{{ix}} {sinx}\right)^{{k}} \right) \\ $$$$\sum_{{k}=\mathrm{0}} ^{{n}} \left({e}^{{ix}} {sinx}\right)^{{k}} \:=\frac{\mathrm{1}−\left({e}^{{ix}} {sinx}\right)^{{n}+\mathrm{1}} }{\mathrm{1}−{e}^{{ix}} {sinx}} \\ $$$$=\frac{\mathrm{1}−{sin}^{{n}+\mathrm{1}} {x}\:{e}^{{i}\left({n}+\mathrm{1}\right){x}} }{\mathrm{1}−{sinx}\left({cosx}\:+{isinx}\right)} \\ $$$$=\frac{\mathrm{1}−{e}^{{i}\left({n}+\mathrm{1}\right){x}} \:{sin}^{{n}+\mathrm{1}} {x}}{\mathrm{1}−{sinxcosx}−{isin}^{\mathrm{2}} {x}} \\ $$$$=\frac{\left(\mathrm{1}−{sinxcosx}\:+{isin}^{\mathrm{2}} {x}\right)\left(\mathrm{1}−{sin}^{{n}+\mathrm{1}} {x}\left({cos}\left({n}+\mathrm{1}\right){x}+{isin}\left({n}+\mathrm{1}\right){x}\right)\right.}{\left(\mathrm{1}−{sinxcosx}\right)^{\mathrm{2}} \:+{sin}^{\mathrm{4}} {x}} \\ $$$$=...{we}\:{get}\:{Sn}\:{after}\:{extracting}\:{Re}\left({of}\:{this}\right. \\ $$$$\left.{quantity}...\right) \\ $$