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Question Number 97135 by  M±th+et+s last updated on 06/Jun/20

prove that: sin(16x) cot(x)=1+2cos(2x)+2cos(4x)+2cos(6x)+...+2cos(16x)

$${prove}\:{that}: \\ $$$${sin}\left(\mathrm{16}{x}\right)\:{cot}\left({x}\right)=\mathrm{1}+\mathrm{2}{cos}\left(\mathrm{2}{x}\right)+\mathrm{2}{cos}\left(\mathrm{4}{x}\right)+\mathrm{2}{cos}\left(\mathrm{6}{x}\right)+...+\mathrm{2}{cos}\left(\mathrm{16}{x}\right) \\ $$

Commented by prakash jain last updated on 07/Jun/20

cos x=(1/2)(e^(ix) +e^(−ix) ) RHS=1+Σ_(k=1) ^8 (e^(2kix) +e^(−2kix) ) =e^(−16ix) +e^(−14kx) +...+e^(16ix) =((e^(−16ix) (e^(34ix) −1))/(e^(2ix) −1))=((e^(18ix) −e^(−16ix) )/(e^(2ix) −1)) sin 16xcot x=((e^(16ix) −e^(−16ix) )/2)×((e^(ix) +e^(−ix) )/(e^(ix) −e^(−ix) )) =((e^(16x) −e^(−16x) )/2)×((e^(2ix) +1)/(e^(2ix) −1)) =((e^(18ix) −e^(−14ix) +e^(16ix) −e^(−16ix) )/(2(e^(2ix) −1))) Equality does not seem be valid. I will recheck my calculation again.

$$\mathrm{cos}\:{x}=\frac{\mathrm{1}}{\mathrm{2}}\left({e}^{{ix}} +{e}^{−{ix}} \right) \\ $$$$\mathrm{RHS}=\mathrm{1}+\underset{{k}=\mathrm{1}} {\overset{\mathrm{8}} {\sum}}\left({e}^{\mathrm{2}{kix}} +{e}^{−\mathrm{2}{kix}} \right) \\ $$$$={e}^{−\mathrm{16}{ix}} +{e}^{−\mathrm{14}{kx}} +...+{e}^{\mathrm{16}{ix}} \: \\ $$$$=\frac{{e}^{−\mathrm{16}{ix}} \left({e}^{\mathrm{34}{ix}} −\mathrm{1}\right)}{{e}^{\mathrm{2}{ix}} −\mathrm{1}}=\frac{{e}^{\mathrm{18}{ix}} −{e}^{−\mathrm{16}{ix}} }{{e}^{\mathrm{2}{ix}} −\mathrm{1}} \\ $$$$ \\ $$$$\mathrm{sin}\:\mathrm{16}{x}\mathrm{cot}\:{x}=\frac{{e}^{\mathrm{16}{ix}} −{e}^{−\mathrm{16}{ix}} }{\mathrm{2}}×\frac{{e}^{{ix}} +{e}^{−{ix}} }{{e}^{{ix}} −{e}^{−{ix}} } \\ $$$$=\frac{{e}^{\mathrm{16}{x}} −{e}^{−\mathrm{16}{x}} }{\mathrm{2}}×\frac{{e}^{\mathrm{2}{ix}} +\mathrm{1}}{{e}^{\mathrm{2}{ix}} −\mathrm{1}} \\ $$$$=\frac{{e}^{\mathrm{18}{ix}} −{e}^{−\mathrm{14}{ix}} +{e}^{\mathrm{16}{ix}} −{e}^{−\mathrm{16}{ix}} }{\mathrm{2}\left({e}^{\mathrm{2}{ix}} −\mathrm{1}\right)} \\ $$$$\mathrm{Equality}\:\mathrm{does}\:\mathrm{not}\:\mathrm{seem}\:\mathrm{be}\:\mathrm{valid}. \\ $$$$\mathrm{I}\:\mathrm{will}\:\mathrm{recheck}\:\mathrm{my}\:\mathrm{calculation}\:\mathrm{again}. \\ $$

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