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Question Number 97658 by Vishal Sharma last updated on 09/Jun/20

The value of cos ((2π)/7) +cos ((4π)/7)+cos ((6π)/7)  is

$$\mathrm{The}\:\mathrm{value}\:\mathrm{of}\:\mathrm{cos}\:\frac{\mathrm{2}\pi}{\mathrm{7}}\:+\mathrm{cos}\:\frac{\mathrm{4}\pi}{\mathrm{7}}+\mathrm{cos}\:\frac{\mathrm{6}\pi}{\mathrm{7}}\:\:\mathrm{is} \\ $$

Commented by bemath last updated on 09/Jun/20

x=((2π)/7)   ((cos x+cos 2x+cos 3x)/(2sin x))× 2sin x  ((sin 2x+2sinx cos 2x+2sin xcos 3x )/(2sin x)) =  ((sin 2x+sin 3x−sin x+sin 4x−sin 2x)/(2sin x)) =  ((sin 4x+sin 3x−sin x)/(2sin x)) =  ((2sin (((7x)/2)) cos ((x/2))−sin x)/(2sin x)) = −(1/2)  [ sin (((7x)/2)) = sin π = 0 ]

$${x}=\frac{\mathrm{2}\pi}{\mathrm{7}}\: \\ $$$$\frac{\mathrm{cos}\:{x}+\mathrm{cos}\:\mathrm{2}{x}+\mathrm{cos}\:\mathrm{3}{x}}{\mathrm{2sin}\:{x}}×\:\mathrm{2sin}\:{x} \\ $$$$\frac{\mathrm{sin}\:\mathrm{2}{x}+\mathrm{2sin}{x}\:\mathrm{cos}\:\mathrm{2}{x}+\mathrm{2sin}\:{x}\mathrm{cos}\:\mathrm{3}{x}\:}{\mathrm{2sin}\:{x}}\:= \\ $$$$\frac{\mathrm{sin}\:\mathrm{2}{x}+\mathrm{sin}\:\mathrm{3}{x}−\mathrm{sin}\:{x}+\mathrm{sin}\:\mathrm{4}{x}−\mathrm{sin}\:\mathrm{2}{x}}{\mathrm{2sin}\:{x}}\:= \\ $$$$\frac{\mathrm{sin}\:\mathrm{4}{x}+\mathrm{sin}\:\mathrm{3}{x}−\mathrm{sin}\:{x}}{\mathrm{2sin}\:{x}}\:= \\ $$$$\frac{\mathrm{2sin}\:\left(\frac{\mathrm{7}{x}}{\mathrm{2}}\right)\:\mathrm{cos}\:\left(\frac{{x}}{\mathrm{2}}\right)−\mathrm{sin}\:{x}}{\mathrm{2sin}\:{x}}\:=\:−\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\left[\:\mathrm{sin}\:\left(\frac{\mathrm{7}{x}}{\mathrm{2}}\right)\:=\:\mathrm{sin}\:\pi\:=\:\mathrm{0}\:\right] \\ $$

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