Question and Answers Forum

All Questions      Topic List

Trigonometry Questions

Previous in All Question      Next in All Question      

Previous in Trigonometry      Next in Trigonometry      

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

������

Terms of Service

Privacy Policy

Contact: info@tinkutara.com