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Question Number 67464 by lalitchand last updated on 27/Aug/19
prove Cos(((2π)/7))+Cos(((4π)/7))+Cos(((8π)/7))=−(1/2)
$$\mathrm{prove}\:\:\:\mathrm{Cos}\left(\frac{\mathrm{2}\pi}{\mathrm{7}}\right)+\mathrm{Cos}\left(\frac{\mathrm{4}\pi}{\mathrm{7}}\right)+\mathrm{Cos}\left(\frac{\mathrm{8}\pi}{\mathrm{7}}\right)=−\frac{\mathrm{1}}{\mathrm{2}} \\ $$
Answered by mind is power last updated on 27/Aug/19
Z^7 −1=0 ⇒(z−1)(1+z+z^2 +z3+z^4 +z^5 +z^6 )=0 z_k =e^(i((2kπ)/7)) 0≤k≤6 1+z+z^2 +z^3 +z^4 +z^5 +z^6 =0⇔1≤k≤6 Σ_(k≤6) e^(i((2kπ)/7)) =−1 ⇒cos(((2π)/7))+cos(((4π)/7))+cos(((6π)/7))+cos(((8π)/7))+cos(((10π)/7))+cos(((12π)/7))=−1 cos(((8π)/7))=cos(((14π−6π)/7))=cos(2π−((6π)/7))=cos(−((6π)/7))=cos(((6π)/7)) same idea give us cos(((2π)/7))=cos(((12π)/7)) cos(((4π)/7))=cos(((10π)/7)) ⇒cos(((2π)/7))+cos(((4π)/7))+cos(((6π)/7))+cos(((8π)/7))+cos(((10π)/7))+cos(((12π)/7))= 2cos(((2π)/7))+2cos(((4π)/7))+2cos(((8π)/7))=−1 cos(((2π)/7))+cos(((4π)/7))+cos(((8π)/7))=−(1/2)
$${Z}^{\mathrm{7}} −\mathrm{1}=\mathrm{0} \\ $$$$\Rightarrow\left({z}−\mathrm{1}\right)\left(\mathrm{1}+{z}+{z}^{\mathrm{2}} +{z}\mathrm{3}+{z}^{\mathrm{4}} +{z}^{\mathrm{5}} +{z}^{\mathrm{6}} \right)=\mathrm{0} \\ $$$${z}_{{k}} ={e}^{{i}\frac{\mathrm{2}{k}\pi}{\mathrm{7}}} \:\:\:\:\mathrm{0}\leqslant{k}\leqslant\mathrm{6} \\ $$$$\mathrm{1}+{z}+{z}^{\mathrm{2}} +{z}^{\mathrm{3}} +{z}^{\mathrm{4}} +{z}^{\mathrm{5}} +{z}^{\mathrm{6}} =\mathrm{0}\Leftrightarrow\mathrm{1}\leqslant{k}\leqslant\mathrm{6} \\ $$$$\sum_{{k}\leqslant\mathrm{6}} {e}^{{i}\frac{\mathrm{2}{k}\pi}{\mathrm{7}}} =−\mathrm{1} \\ $$$$\Rightarrow{cos}\left(\frac{\mathrm{2}\pi}{\mathrm{7}}\right)+{cos}\left(\frac{\mathrm{4}\pi}{\mathrm{7}}\right)+{cos}\left(\frac{\mathrm{6}\pi}{\mathrm{7}}\right)+{cos}\left(\frac{\mathrm{8}\pi}{\mathrm{7}}\right)+{cos}\left(\frac{\mathrm{10}\pi}{\mathrm{7}}\right)+{cos}\left(\frac{\mathrm{12}\pi}{\mathrm{7}}\right)=−\mathrm{1} \\ $$$${cos}\left(\frac{\mathrm{8}\pi}{\mathrm{7}}\right)={cos}\left(\frac{\mathrm{14}\pi−\mathrm{6}\pi}{\mathrm{7}}\right)={cos}\left(\mathrm{2}\pi−\frac{\mathrm{6}\pi}{\mathrm{7}}\right)={cos}\left(−\frac{\mathrm{6}\pi}{\mathrm{7}}\right)={cos}\left(\frac{\mathrm{6}\pi}{\mathrm{7}}\right) \\ $$$${same}\:{idea}\:{give}\:{us} \\ $$$${cos}\left(\frac{\mathrm{2}\pi}{\mathrm{7}}\right)={cos}\left(\frac{\mathrm{12}\pi}{\mathrm{7}}\right) \\ $$$${cos}\left(\frac{\mathrm{4}\pi}{\mathrm{7}}\right)={cos}\left(\frac{\mathrm{10}\pi}{\mathrm{7}}\right) \\ $$$$\Rightarrow{cos}\left(\frac{\mathrm{2}\pi}{\mathrm{7}}\right)+{cos}\left(\frac{\mathrm{4}\pi}{\mathrm{7}}\right)+{cos}\left(\frac{\mathrm{6}\pi}{\mathrm{7}}\right)+{cos}\left(\frac{\mathrm{8}\pi}{\mathrm{7}}\right)+{cos}\left(\frac{\mathrm{10}\pi}{\mathrm{7}}\right)+{cos}\left(\frac{\mathrm{12}\pi}{\mathrm{7}}\right)= \\ $$$$\mathrm{2}{cos}\left(\frac{\mathrm{2}\pi}{\mathrm{7}}\right)+\mathrm{2}{cos}\left(\frac{\mathrm{4}\pi}{\mathrm{7}}\right)+\mathrm{2}{cos}\left(\frac{\mathrm{8}\pi}{\mathrm{7}}\right)=−\mathrm{1} \\ $$$${cos}\left(\frac{\mathrm{2}\pi}{\mathrm{7}}\right)+{cos}\left(\frac{\mathrm{4}\pi}{\mathrm{7}}\right)+{cos}\left(\frac{\mathrm{8}\pi}{\mathrm{7}}\right)=−\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$ \\ $$$$ \\ $$
Commented by Kunal12588 last updated on 27/Aug/19
just AMAZING...
$${just}\:{AMAZING}… \\ $$
Commented by Kunal12588 last updated on 27/Aug/19
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