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Question Number 161559 by stelor last updated on 19/Dec/21

please show that  (1/2) + cosx + cos2x + cos3x + ... + cosnx = ((sin[(n+1)(x/2)])/(2sin(x/2)))

$${please}\:{show}\:{that} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\:+\:{cosx}\:+\:{cos}\mathrm{2}{x}\:+\:{cos}\mathrm{3}{x}\:+\:...\:+\:{cosnx}\:=\:\frac{{sin}\left[\left({n}+\mathrm{1}\right)\frac{{x}}{\mathrm{2}}\right]}{\mathrm{2}{sin}\frac{{x}}{\mathrm{2}}} \\ $$

Answered by Ar Brandon last updated on 19/Dec/21

P=(1/2)+cosx+cos2x+cos3x+∙∙∙+cosnx  (2sin(x/2))P=sin(x/2)+2sin(x/2)cosx+2sin(x/2)cos2x+∙∙∙+2sin(x/2)cosnx  (2sin(x/2))P=sin(x/2)+(sin((3x)/2)−sin(x/2))+(sin((5x)/2)−sin((3x)/2))+∙∙∙+(sin((2n+1)/2)x−sin((2n−1)/2)x)  (2sin(x/2))P=sin((2n+1)/2)x⇒P=((sin((2n+1)(x/2)))/(2sin(x/2)))

$${P}=\frac{\mathrm{1}}{\mathrm{2}}+\mathrm{cos}{x}+\mathrm{cos2}{x}+\mathrm{cos3}{x}+\centerdot\centerdot\centerdot+\mathrm{cos}{nx} \\ $$$$\left(\mathrm{2sin}\frac{{x}}{\mathrm{2}}\right){P}=\mathrm{sin}\frac{{x}}{\mathrm{2}}+\mathrm{2sin}\frac{{x}}{\mathrm{2}}\mathrm{cos}{x}+\mathrm{2sin}\frac{{x}}{\mathrm{2}}\mathrm{cos2}{x}+\centerdot\centerdot\centerdot+\mathrm{2sin}\frac{{x}}{\mathrm{2}}\mathrm{cos}{nx} \\ $$$$\left(\mathrm{2sin}\frac{{x}}{\mathrm{2}}\right){P}=\mathrm{sin}\frac{{x}}{\mathrm{2}}+\left(\mathrm{sin}\frac{\mathrm{3}{x}}{\mathrm{2}}−\mathrm{sin}\frac{{x}}{\mathrm{2}}\right)+\left(\mathrm{sin}\frac{\mathrm{5}{x}}{\mathrm{2}}−\mathrm{sin}\frac{\mathrm{3}{x}}{\mathrm{2}}\right)+\centerdot\centerdot\centerdot+\left(\mathrm{sin}\frac{\mathrm{2}{n}+\mathrm{1}}{\mathrm{2}}{x}−\mathrm{sin}\frac{\mathrm{2}{n}−\mathrm{1}}{\mathrm{2}}{x}\right) \\ $$$$\left(\mathrm{2sin}\frac{{x}}{\mathrm{2}}\right){P}=\mathrm{sin}\frac{\mathrm{2}{n}+\mathrm{1}}{\mathrm{2}}{x}\Rightarrow{P}=\frac{\mathrm{sin}\left(\left(\mathrm{2}{n}+\mathrm{1}\right)\frac{{x}}{\mathrm{2}}\right)}{\mathrm{2sin}\frac{{x}}{\mathrm{2}}} \\ $$

Commented by stelor last updated on 19/Dec/21

thanks.

$${thanks}. \\ $$

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