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Question Number 112645 by bemath last updated on 09/Sep/20
cos ((π/7))−cos (((2π)/7))+cos (((3π)/7)) =?
$$\mathrm{cos}\:\left(\frac{\pi}{\mathrm{7}}\right)−\mathrm{cos}\:\left(\frac{\mathrm{2}\pi}{\mathrm{7}}\right)+\mathrm{cos}\:\left(\frac{\mathrm{3}\pi}{\mathrm{7}}\right)\:=?\: \\ $$
Answered by john santu last updated on 09/Sep/20
let p = (π/7) and X=cos p−cos 2p+cos 3p multiply by 2sin p 2sin p X = 2sin p(cos p−cos 2p+cos 3p) 2sin p X=sin 2p−2sin pcos 2p+2sin pcos 3p 2sin p X=sin 2p−(sin 3p−sin p)+sin 4p−sin 2p 2sin p X=sin 4p+sin p−sin 3p 2sin p X= 2cos (((7p)/2))sin((p/2))+sin p consider cos (((7p)/2))=cos ((7/2).(π/7))=0 so we get 2sin p X = sin p X = (1/2).
$${let}\:{p}\:=\:\frac{\pi}{\mathrm{7}}\:{and}\:{X}=\mathrm{cos}\:{p}−\mathrm{cos}\:\mathrm{2}{p}+\mathrm{cos}\:\mathrm{3}{p} \\ $$$${multiply}\:{by}\:\mathrm{2sin}\:{p} \\ $$$$\mathrm{2sin}\:{p}\:{X}\:=\:\mathrm{2sin}\:{p}\left(\mathrm{cos}\:{p}−\mathrm{cos}\:\mathrm{2}{p}+\mathrm{cos}\:\mathrm{3}{p}\right) \\ $$$$\mathrm{2sin}\:{p}\:{X}=\mathrm{sin}\:\mathrm{2}{p}−\mathrm{2sin}\:{p}\mathrm{cos}\:\mathrm{2}{p}+\mathrm{2sin}\:{p}\mathrm{cos}\:\mathrm{3}{p} \\ $$$$\mathrm{2sin}\:{p}\:{X}=\mathrm{sin}\:\mathrm{2}{p}−\left(\mathrm{sin}\:\mathrm{3}{p}−\mathrm{sin}\:{p}\right)+\mathrm{sin}\:\mathrm{4}{p}−\mathrm{sin}\:\mathrm{2}{p} \\ $$$$\mathrm{2sin}\:{p}\:{X}=\mathrm{sin}\:\mathrm{4}{p}+\mathrm{sin}\:{p}−\mathrm{sin}\:\mathrm{3}{p} \\ $$$$\mathrm{2sin}\:{p}\:{X}=\:\mathrm{2cos}\:\left(\frac{\mathrm{7}{p}}{\mathrm{2}}\right)\mathrm{sin}\left(\frac{{p}}{\mathrm{2}}\right)+\mathrm{sin}\:{p}\: \\ $$$${consider}\:\mathrm{cos}\:\left(\frac{\mathrm{7}{p}}{\mathrm{2}}\right)=\mathrm{cos}\:\left(\frac{\mathrm{7}}{\mathrm{2}}.\frac{\pi}{\mathrm{7}}\right)=\mathrm{0} \\ $$$${so}\:{we}\:{get}\:\mathrm{2sin}\:{p}\:{X}\:=\:\mathrm{sin}\:{p} \\ $$$${X}\:=\:\frac{\mathrm{1}}{\mathrm{2}}. \\ $$
Answered by 1549442205PVT last updated on 09/Sep/20
P=cos ((π/7))−cos (((2π)/7))+cos (((3π)/7)) =cos(π/7)−(2cos^2 (π/7)−1)+4cos^3 (π/7)−3cos(π/7) =4cos^3 (π/7)−2cos^2 (π/7)−2cos(π/7)+1(1) On the other hands,we have 4cos^3 (π/7)−3cos(π/7)=cos((3π)/7)=−cos(π−((3π)/7)) =−cos((4π)/7)=1−2cos^2 ((2π)/7)=1−2(2cos^2 (π/7)−1)^2 =1−2(4cos^4 (π/7)−4cos^2 (π/7)+1) =−8cos^4 (π/7)+8cos^2 (π/7)−1.Hence,we get 8cos^4 (π/7)+4cos^3 (π/7)−8cos^2 (π/7)−3cos(π/7)+1=0 ⇔(cos(π/7)+1)(8cos^3 (π/7)−4cos^2 (π/7)−4cos(π/7)+1)=0 ⇔8cos^3 (π/7)−4cos^2 (π/7)−4cos(π/7)+1=0 ⇒4cos^3 (π/7)−2cos^2 (π/7)−2cos(π/7)+(1/2)=0(2) From (1)and (2) we get P=(4cos^3 (π/7)−2cos^2 (π/7)−2cos(π/7)+(1/2))+(1/2) =0+(1/2).Thus,P=cos ((𝛑/7))−cos (((2𝛑)/7))+cos (((3𝛑)/7))=(1/( 2))
$$\mathrm{P}=\mathrm{cos}\:\left(\frac{\pi}{\mathrm{7}}\right)−\mathrm{cos}\:\left(\frac{\mathrm{2}\pi}{\mathrm{7}}\right)+\mathrm{cos}\:\left(\frac{\mathrm{3}\pi}{\mathrm{7}}\right) \\ $$$$=\mathrm{cos}\frac{\pi}{\mathrm{7}}−\left(\mathrm{2cos}^{\mathrm{2}} \frac{\pi}{\mathrm{7}}−\mathrm{1}\right)+\mathrm{4cos}^{\mathrm{3}} \frac{\pi}{\mathrm{7}}−\mathrm{3cos}\frac{\pi}{\mathrm{7}} \\ $$$$=\mathrm{4cos}^{\mathrm{3}} \frac{\pi}{\mathrm{7}}−\mathrm{2cos}^{\mathrm{2}} \frac{\pi}{\mathrm{7}}−\mathrm{2cos}\frac{\pi}{\mathrm{7}}+\mathrm{1}\left(\mathrm{1}\right) \\ $$$$\mathrm{On}\:\mathrm{the}\:\mathrm{other}\:\mathrm{hands},\mathrm{we}\:\mathrm{have} \\ $$$$\mathrm{4cos}^{\mathrm{3}} \frac{\pi}{\mathrm{7}}−\mathrm{3cos}\frac{\pi}{\mathrm{7}}=\mathrm{cos}\frac{\mathrm{3}\pi}{\mathrm{7}}=−\mathrm{cos}\left(\pi−\frac{\mathrm{3}\pi}{\mathrm{7}}\right) \\ $$$$=−\mathrm{cos}\frac{\mathrm{4}\pi}{\mathrm{7}}=\mathrm{1}−\mathrm{2cos}^{\mathrm{2}} \frac{\mathrm{2}\pi}{\mathrm{7}}=\mathrm{1}−\mathrm{2}\left(\mathrm{2cos}^{\mathrm{2}} \frac{\pi}{\mathrm{7}}−\mathrm{1}\right)^{\mathrm{2}} \\ $$$$=\mathrm{1}−\mathrm{2}\left(\mathrm{4cos}^{\mathrm{4}} \frac{\pi}{\mathrm{7}}−\mathrm{4cos}^{\mathrm{2}} \frac{\pi}{\mathrm{7}}+\mathrm{1}\right) \\ $$$$=−\mathrm{8cos}^{\mathrm{4}} \frac{\pi}{\mathrm{7}}+\mathrm{8cos}^{\mathrm{2}} \frac{\pi}{\mathrm{7}}−\mathrm{1}.\mathrm{Hence},\mathrm{we}\:\mathrm{get} \\ $$$$\mathrm{8cos}^{\mathrm{4}} \frac{\pi}{\mathrm{7}}+\mathrm{4cos}^{\mathrm{3}} \frac{\pi}{\mathrm{7}}−\mathrm{8cos}^{\mathrm{2}} \frac{\pi}{\mathrm{7}}−\mathrm{3cos}\frac{\pi}{\mathrm{7}}+\mathrm{1}=\mathrm{0} \\ $$$$\Leftrightarrow\left(\mathrm{cos}\frac{\pi}{\mathrm{7}}+\mathrm{1}\right)\left(\mathrm{8cos}^{\mathrm{3}} \frac{\pi}{\mathrm{7}}−\mathrm{4cos}^{\mathrm{2}} \frac{\pi}{\mathrm{7}}−\mathrm{4cos}\frac{\pi}{\mathrm{7}}+\mathrm{1}\right)=\mathrm{0} \\ $$$$\Leftrightarrow\mathrm{8cos}^{\mathrm{3}} \frac{\pi}{\mathrm{7}}−\mathrm{4cos}^{\mathrm{2}} \frac{\pi}{\mathrm{7}}−\mathrm{4cos}\frac{\pi}{\mathrm{7}}+\mathrm{1}=\mathrm{0} \\ $$$$\Rightarrow\mathrm{4cos}^{\mathrm{3}} \frac{\pi}{\mathrm{7}}−\mathrm{2cos}^{\mathrm{2}} \frac{\pi}{\mathrm{7}}−\mathrm{2cos}\frac{\pi}{\mathrm{7}}+\frac{\mathrm{1}}{\mathrm{2}}=\mathrm{0}\left(\mathrm{2}\right) \\ $$$$\mathrm{From}\:\left(\mathrm{1}\right)\mathrm{and}\:\left(\mathrm{2}\right)\:\mathrm{we}\:\mathrm{get} \\ $$$$\mathrm{P}=\left(\mathrm{4cos}^{\mathrm{3}} \frac{\pi}{\mathrm{7}}−\mathrm{2cos}^{\mathrm{2}} \frac{\pi}{\mathrm{7}}−\mathrm{2cos}\frac{\pi}{\mathrm{7}}+\frac{\mathrm{1}}{\mathrm{2}}\right)+\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$=\mathrm{0}+\frac{\mathrm{1}}{\mathrm{2}}.\boldsymbol{\mathrm{Thus}},\boldsymbol{\mathrm{P}}=\boldsymbol{\mathrm{cos}}\:\left(\frac{\boldsymbol{\pi}}{\mathrm{7}}\right)−\boldsymbol{\mathrm{cos}}\:\left(\frac{\mathrm{2}\boldsymbol{\pi}}{\mathrm{7}}\right)+\boldsymbol{\mathrm{cos}}\:\left(\frac{\mathrm{3}\boldsymbol{\pi}}{\mathrm{7}}\right)=\frac{\mathrm{1}}{\:\mathrm{2}} \\ $$

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