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Question Number 140498 by mnjuly1970 last updated on 08/May/21

........ nice ....... calculus....... simplify: 𝛗(x):= sin((x/2))(1+2Σ_(m=1) ^n cos(mx))

$$\:\:\:\:\:\:........\:{nice}\:\:\:.......\:\:{calculus}....... \\ $$$$\:\:\:\:\:\:{simplify}: \\ $$$$\:\:\:\:\boldsymbol{\phi}\left({x}\right):=\:{sin}\left(\frac{{x}}{\mathrm{2}}\right)\left(\mathrm{1}+\mathrm{2}\underset{{m}=\mathrm{1}} {\overset{{n}} {\sum}}{cos}\left({mx}\right)\right) \\ $$$$\:\:\:\: \\ $$

Answered by Dwaipayan Shikari last updated on 08/May/21

Σ_(m=1) ^n cos(mx)=((cos(((n+1)/2)x)sin(((nx)/2)))/(sin((x/2)))) φ(x)=sin((x/2))+2sin(((nx)/2))cos(((n+1)/2)x)

$$\underset{{m}=\mathrm{1}} {\overset{{n}} {\sum}}{cos}\left({mx}\right)=\frac{{cos}\left(\frac{{n}+\mathrm{1}}{\mathrm{2}}{x}\right){sin}\left(\frac{{nx}}{\mathrm{2}}\right)}{{sin}\left(\frac{{x}}{\mathrm{2}}\right)} \\ $$$$\phi\left({x}\right)={sin}\left(\frac{{x}}{\mathrm{2}}\right)+\mathrm{2}{sin}\left(\frac{{nx}}{\mathrm{2}}\right){cos}\left(\frac{{n}+\mathrm{1}}{\mathrm{2}}{x}\right) \\ $$

Commented by Dwaipayan Shikari last updated on 08/May/21

Σ_(m=1) ^n cos(mx)=cosx+cos2x+..+cosnx =(1/(2sin(x/2)))(2cosxsin(x/2)+..+cosnxsin(x/2)) =(1/(2sin(x/2)))(sin((3x)/2)−sin(x/2)+sin((5x)/2)−sin((3x)/2)+...+sin(n+(1/2))x−sin(n−(1/2))x) =(1/(2sin(x/2)))(sin(n+(1/2))x−sin(x/2))=((2cos(((n+1)/2)x)sin(((nx)/2)))/(2sin(x/2))) =((cos(((n+1)/2)x)sin(((nx)/2)))/(sin((x/2))))

$$\underset{{m}=\mathrm{1}} {\overset{{n}} {\sum}}{cos}\left({mx}\right)={cosx}+{cos}\mathrm{2}{x}+..+{cosnx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}{sin}\frac{{x}}{\mathrm{2}}}\left(\mathrm{2}{cosxsin}\frac{{x}}{\mathrm{2}}+..+{cosnxsin}\frac{{x}}{\mathrm{2}}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}{sin}\frac{{x}}{\mathrm{2}}}\left({sin}\frac{\mathrm{3}{x}}{\mathrm{2}}−{sin}\frac{{x}}{\mathrm{2}}+{sin}\frac{\mathrm{5}{x}}{\mathrm{2}}−{sin}\frac{\mathrm{3}{x}}{\mathrm{2}}+...+{sin}\left({n}+\frac{\mathrm{1}}{\mathrm{2}}\right){x}−{sin}\left({n}−\frac{\mathrm{1}}{\mathrm{2}}\right){x}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}{sin}\frac{{x}}{\mathrm{2}}}\left({sin}\left({n}+\frac{\mathrm{1}}{\mathrm{2}}\right){x}−{sin}\frac{{x}}{\mathrm{2}}\right)=\frac{\mathrm{2}{cos}\left(\frac{{n}+\mathrm{1}}{\mathrm{2}}{x}\right){sin}\left(\frac{{nx}}{\mathrm{2}}\right)}{\mathrm{2}{sin}\frac{{x}}{\mathrm{2}}} \\ $$$$=\frac{{cos}\left(\frac{{n}+\mathrm{1}}{\mathrm{2}}{x}\right){sin}\left(\frac{{nx}}{\mathrm{2}}\right)}{{sin}\left(\frac{{x}}{\mathrm{2}}\right)} \\ $$

Commented by mnjuly1970 last updated on 08/May/21

thanks alot mr payan..

$${thanks}\:{alot}\:{mr}\:{payan}.. \\ $$

Answered by mnjuly1970 last updated on 08/May/21

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