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Question Number 86346 by Power last updated on 28/Mar/20

Commented by MJS last updated on 28/Mar/20

−4

$$−\mathrm{4} \\ $$

Commented by jagoll last updated on 28/Mar/20

yes.....i got the same result sir

$$\mathrm{yes}.....\mathrm{i}\:\mathrm{got}\:\mathrm{the}\:\mathrm{same}\:\mathrm{result}\:\mathrm{sir} \\ $$

Answered by TANMAY PANACEA. last updated on 28/Mar/20

7θ=2π 4θ=2π−3θ cos4θ=cos(2π−3θ) 2cos^2 2θ−1=4cos^3 θ−3cosθ 2(2cos^2 θ−1)^2 −1=4cos^3 θ−3cosθ 2(2x^2 −1)^2 −1=4x^3 −3x 2(4x^4 −4x^2 +1)−4x^3 +3x−1=0 8x^4 −4x^3 −8x^2 +3x+1=0 8x^4 −8x^3 +4x^3 −4x^2 −4x^2 +4x−x+1 8x^3 (x−1)+4x^2 (x−1)−4x(x−1)−1(x−1)=0 (x−1)(8x^3 +4x^2 −4x−1)=0 now look 8x^3 +4x^2 −4x−1=0 has three root and one root is cosθ cos(((2π)/7)) another cos(((4π)/7)) and cos(((6π)/7)) cos(((2π)/7))+cos(((4π)/7))+cos(((6π)/7))=((−4)/8)=((−1)/2) Σcos(((2π)/7))cos(((4π)/7))=((−4)/8)=((−1)/2) cos(((2π)/7))cos(((4π)/7))cos(((6π)/7))=(1/8) (1/(cos(((2π)/7))))+(1/(cos(((4π)/7))))+(1/(cos(((8π)/7)))) look... cos(((8π)/7))=cos(π+(π/7))=−cos((π/7)) cos(((6π)/7))=cos(π−(π/7))=−cos((π/7)) so cos(((6π)/7))=cos(((8π)/7)) (1/(cos(((2π)/7))))+(1/(cos(((4π)/7))))+(1/(cos(((8π)/7)))) =(1/(cos(((2π)/7))))+(1/(cos(((4π)/7))))+(1/(cos(((6π)/7)))) =((Σcos(((2π)/7))cos(((4π)/7)))/(cos(((2π)/7))cos(((4π)/7))cos(((6π)/7)))) =(((−1)/2)/(1/8))=−4

$$\mathrm{7}\theta=\mathrm{2}\pi \\ $$$$\mathrm{4}\theta=\mathrm{2}\pi−\mathrm{3}\theta \\ $$$${cos}\mathrm{4}\theta={cos}\left(\mathrm{2}\pi−\mathrm{3}\theta\right) \\ $$$$\mathrm{2}{cos}^{\mathrm{2}} \mathrm{2}\theta−\mathrm{1}=\mathrm{4}{cos}^{\mathrm{3}} \theta−\mathrm{3}{cos}\theta \\ $$$$\mathrm{2}\left(\mathrm{2}{cos}^{\mathrm{2}} \theta−\mathrm{1}\right)^{\mathrm{2}} −\mathrm{1}=\mathrm{4}{cos}^{\mathrm{3}} \theta−\mathrm{3}{cos}\theta \\ $$$$\mathrm{2}\left(\mathrm{2}{x}^{\mathrm{2}} −\mathrm{1}\right)^{\mathrm{2}} −\mathrm{1}=\mathrm{4}{x}^{\mathrm{3}} −\mathrm{3}{x} \\ $$$$\mathrm{2}\left(\mathrm{4}{x}^{\mathrm{4}} −\mathrm{4}{x}^{\mathrm{2}} +\mathrm{1}\right)−\mathrm{4}{x}^{\mathrm{3}} +\mathrm{3}{x}−\mathrm{1}=\mathrm{0} \\ $$$$\mathrm{8}{x}^{\mathrm{4}} −\mathrm{4}{x}^{\mathrm{3}} −\mathrm{8}{x}^{\mathrm{2}} +\mathrm{3}{x}+\mathrm{1}=\mathrm{0} \\ $$$$\mathrm{8}{x}^{\mathrm{4}} −\mathrm{8}{x}^{\mathrm{3}} +\mathrm{4}{x}^{\mathrm{3}} −\mathrm{4}{x}^{\mathrm{2}} −\mathrm{4}{x}^{\mathrm{2}} +\mathrm{4}{x}−{x}+\mathrm{1} \\ $$$$\mathrm{8}{x}^{\mathrm{3}} \left({x}−\mathrm{1}\right)+\mathrm{4}{x}^{\mathrm{2}} \left({x}−\mathrm{1}\right)−\mathrm{4}{x}\left({x}−\mathrm{1}\right)−\mathrm{1}\left({x}−\mathrm{1}\right)=\mathrm{0} \\ $$$$\left({x}−\mathrm{1}\right)\left(\mathrm{8}{x}^{\mathrm{3}} +\mathrm{4}{x}^{\mathrm{2}} −\mathrm{4}{x}−\mathrm{1}\right)=\mathrm{0} \\ $$$${now}\:{look}\:\mathrm{8}{x}^{\mathrm{3}} +\mathrm{4}{x}^{\mathrm{2}} −\mathrm{4}{x}−\mathrm{1}=\mathrm{0} \\ $$$${has}\:{three}\:{root}\:{and}\:{one}\:{root}\:{is}\:{cos}\theta \\ $$$$\boldsymbol{{cos}}\left(\frac{\mathrm{2}\pi}{\mathrm{7}}\right)\:{another}\:{cos}\left(\frac{\mathrm{4}\pi}{\mathrm{7}}\right)\:{and}\:{cos}\left(\frac{\mathrm{6}\pi}{\mathrm{7}}\right) \\ $$$${cos}\left(\frac{\mathrm{2}\pi}{\mathrm{7}}\right)+{cos}\left(\frac{\mathrm{4}\pi}{\mathrm{7}}\right)+{cos}\left(\frac{\mathrm{6}\pi}{\mathrm{7}}\right)=\frac{−\mathrm{4}}{\mathrm{8}}=\frac{−\mathrm{1}}{\mathrm{2}} \\ $$$$\Sigma{cos}\left(\frac{\mathrm{2}\pi}{\mathrm{7}}\right){cos}\left(\frac{\mathrm{4}\pi}{\mathrm{7}}\right)=\frac{−\mathrm{4}}{\mathrm{8}}=\frac{−\mathrm{1}}{\mathrm{2}} \\ $$$${cos}\left(\frac{\mathrm{2}\pi}{\mathrm{7}}\right){cos}\left(\frac{\mathrm{4}\pi}{\mathrm{7}}\right){cos}\left(\frac{\mathrm{6}\pi}{\mathrm{7}}\right)=\frac{\mathrm{1}}{\mathrm{8}} \\ $$$$\frac{\mathrm{1}}{{cos}\left(\frac{\mathrm{2}\pi}{\mathrm{7}}\right)}+\frac{\mathrm{1}}{{cos}\left(\frac{\mathrm{4}\pi}{\mathrm{7}}\right)}+\frac{\mathrm{1}}{{cos}\left(\frac{\mathrm{8}\pi}{\mathrm{7}}\right)} \\ $$$${look}... \\ $$$${cos}\left(\frac{\mathrm{8}\pi}{\mathrm{7}}\right)={cos}\left(\pi+\frac{\pi}{\mathrm{7}}\right)=−{cos}\left(\frac{\pi}{\mathrm{7}}\right) \\ $$$${cos}\left(\frac{\mathrm{6}\pi}{\mathrm{7}}\right)={cos}\left(\pi−\frac{\pi}{\mathrm{7}}\right)=−{cos}\left(\frac{\pi}{\mathrm{7}}\right) \\ $$$${so}\:{cos}\left(\frac{\mathrm{6}\pi}{\mathrm{7}}\right)={cos}\left(\frac{\mathrm{8}\pi}{\mathrm{7}}\right) \\ $$$$\frac{\mathrm{1}}{{cos}\left(\frac{\mathrm{2}\pi}{\mathrm{7}}\right)}+\frac{\mathrm{1}}{{cos}\left(\frac{\mathrm{4}\pi}{\mathrm{7}}\right)}+\frac{\mathrm{1}}{{cos}\left(\frac{\mathrm{8}\pi}{\mathrm{7}}\right)} \\ $$$$=\frac{\mathrm{1}}{{cos}\left(\frac{\mathrm{2}\pi}{\mathrm{7}}\right)}+\frac{\mathrm{1}}{{cos}\left(\frac{\mathrm{4}\pi}{\mathrm{7}}\right)}+\frac{\mathrm{1}}{{cos}\left(\frac{\mathrm{6}\pi}{\mathrm{7}}\right)} \\ $$$$=\frac{\Sigma{cos}\left(\frac{\mathrm{2}\pi}{\mathrm{7}}\right){cos}\left(\frac{\mathrm{4}\pi}{\mathrm{7}}\right)}{{cos}\left(\frac{\mathrm{2}\pi}{\mathrm{7}}\right){cos}\left(\frac{\mathrm{4}\pi}{\mathrm{7}}\right){cos}\left(\frac{\mathrm{6}\pi}{\mathrm{7}}\right)} \\ $$$$=\frac{\frac{−\mathrm{1}}{\mathrm{2}}}{\frac{\mathrm{1}}{\mathrm{8}}}=−\mathrm{4} \\ $$$$ \\ $$

Commented by Power last updated on 28/Mar/20

step by step solution sir

$$\mathrm{step}\:\mathrm{by}\:\mathrm{step}\:\mathrm{solution}\:\mathrm{sir} \\ $$

Commented by jagoll last updated on 28/Mar/20

vieta rule (1/α)+(1/β)+(1/γ) = −(c/d) = −(((−4)/(−1))) = −4 it the answer

$$\mathrm{vieta}\:\mathrm{rule} \\ $$$$\frac{\mathrm{1}}{\alpha}+\frac{\mathrm{1}}{\beta}+\frac{\mathrm{1}}{\gamma}\:=\:−\frac{\mathrm{c}}{\mathrm{d}}\:=\:−\left(\frac{−\mathrm{4}}{−\mathrm{1}}\right)\:=\:−\mathrm{4} \\ $$$$\mathrm{it}\:\mathrm{the}\:\mathrm{answer} \\ $$

Commented by MJS last updated on 28/Mar/20

any steps by yourself Sir Power?

$$\mathrm{any}\:\mathrm{steps}\:\mathrm{by}\:\mathrm{yourself}\:\mathrm{Sir}\:\mathrm{Power}? \\ $$

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