Question and Answers Forum

All Questions      Topic List

None Questions

Previous in All Question      Next in All Question      

Previous in None      Next in None      

Question Number 152551 by alisiao last updated on 29/Aug/21

((sin(x)+sin(2x)+....+sin(nx))/(cos(x)+cos(2x)+....+cos(nx))) = ?

$$\frac{{sin}\left({x}\right)+{sin}\left(\mathrm{2}{x}\right)+....+{sin}\left({nx}\right)}{{cos}\left({x}\right)+{cos}\left(\mathrm{2}{x}\right)+....+{cos}\left({nx}\right)}\:=\:? \\ $$

Answered by qaz last updated on 29/Aug/21

(S/C)=((ℑ(e^(ix) +e^(2ix) +...+e^(nix) ))/(ℜ(e^(ix) +e^(2ix) +...+e^(nix) ))) =((ℑ((e^(ix) (1−e^(nix) ))/(1−e^(ix) )))/(ℜ((e^(ix) (1−e^(nix) ))/(1−e^(ix) )))) =((ℑ((e^(ix) (1−e^(nix) )(1−e^(−ix) ))/((1−e^(ix) )(1−e^(−ix) ))))/(ℜ((e^(ix) (1−e^(nix) )(1−e^(−ix) ))/((1−e^(ix) )(1−e^(−ix) ))))) =((ℑ(e^(ix) −1)(e^(nix) −1))/(ℜ(e^(ix) −1)(e^(nix) −1))) =((ℑ(e^((n+1)ix) −e^(ix) −e^(nix) +1))/(ℜ(e^((n+1)ix) −e^(ix) −e^(nix) +1))) =((sin (n+1)x−sin x−sin nx)/(cos (n+1)x−cos x−cos nx+1))

$$\frac{\mathrm{S}}{\mathrm{C}}=\frac{\Im\left(\mathrm{e}^{\mathrm{ix}} +\mathrm{e}^{\mathrm{2ix}} +...+\mathrm{e}^{\mathrm{nix}} \right)}{\Re\left(\mathrm{e}^{\mathrm{ix}} +\mathrm{e}^{\mathrm{2ix}} +...+\mathrm{e}^{\mathrm{nix}} \right)} \\ $$$$=\frac{\Im\frac{\mathrm{e}^{\mathrm{ix}} \left(\mathrm{1}−\mathrm{e}^{\mathrm{nix}} \right)}{\mathrm{1}−\mathrm{e}^{\mathrm{ix}} }}{\Re\frac{\mathrm{e}^{\mathrm{ix}} \left(\mathrm{1}−\mathrm{e}^{\mathrm{nix}} \right)}{\mathrm{1}−\mathrm{e}^{\mathrm{ix}} }} \\ $$$$=\frac{\Im\frac{\mathrm{e}^{\mathrm{ix}} \left(\mathrm{1}−\mathrm{e}^{\mathrm{nix}} \right)\left(\mathrm{1}−\mathrm{e}^{−\mathrm{ix}} \right)}{\left(\mathrm{1}−\mathrm{e}^{\mathrm{ix}} \right)\left(\mathrm{1}−\mathrm{e}^{−\mathrm{ix}} \right)}}{\Re\frac{\mathrm{e}^{\mathrm{ix}} \left(\mathrm{1}−\mathrm{e}^{\mathrm{nix}} \right)\left(\mathrm{1}−\mathrm{e}^{−\mathrm{ix}} \right)}{\left(\mathrm{1}−\mathrm{e}^{\mathrm{ix}} \right)\left(\mathrm{1}−\mathrm{e}^{−\mathrm{ix}} \right)}} \\ $$$$=\frac{\Im\left(\mathrm{e}^{\mathrm{ix}} −\mathrm{1}\right)\left(\mathrm{e}^{\mathrm{nix}} −\mathrm{1}\right)}{\Re\left(\mathrm{e}^{\mathrm{ix}} −\mathrm{1}\right)\left(\mathrm{e}^{\mathrm{nix}} −\mathrm{1}\right)} \\ $$$$=\frac{\Im\left(\mathrm{e}^{\left(\mathrm{n}+\mathrm{1}\right)\mathrm{ix}} −\mathrm{e}^{\mathrm{ix}} −\mathrm{e}^{\mathrm{nix}} +\mathrm{1}\right)}{\Re\left(\mathrm{e}^{\left(\mathrm{n}+\mathrm{1}\right)\mathrm{ix}} −\mathrm{e}^{\mathrm{ix}} −\mathrm{e}^{\mathrm{nix}} +\mathrm{1}\right)} \\ $$$$=\frac{\mathrm{sin}\:\left(\mathrm{n}+\mathrm{1}\right)\mathrm{x}−\mathrm{sin}\:\mathrm{x}−\mathrm{sin}\:\mathrm{nx}}{\mathrm{cos}\:\left(\mathrm{n}+\mathrm{1}\right)\mathrm{x}−\mathrm{cos}\:\mathrm{x}−\mathrm{cos}\:\mathrm{nx}+\mathrm{1}} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com