Question Number 86346 by Power last updated on 28/Mar/20 | ||
Commented by MJS last updated on 28/Mar/20 | ||
$$−\mathrm{4} \\ $$ | ||
Commented by jagoll last updated on 28/Mar/20 | ||
$$\mathrm{yes}.....\mathrm{i}\:\mathrm{got}\:\mathrm{the}\:\mathrm{same}\:\mathrm{result}\:\mathrm{sir} \\ $$ | ||
Answered by TANMAY PANACEA. last updated on 28/Mar/20 | ||
$$\mathrm{7}\theta=\mathrm{2}\pi \\ $$$$\mathrm{4}\theta=\mathrm{2}\pi−\mathrm{3}\theta \\ $$$${cos}\mathrm{4}\theta={cos}\left(\mathrm{2}\pi−\mathrm{3}\theta\right) \\ $$$$\mathrm{2}{cos}^{\mathrm{2}} \mathrm{2}\theta−\mathrm{1}=\mathrm{4}{cos}^{\mathrm{3}} \theta−\mathrm{3}{cos}\theta \\ $$$$\mathrm{2}\left(\mathrm{2}{cos}^{\mathrm{2}} \theta−\mathrm{1}\right)^{\mathrm{2}} −\mathrm{1}=\mathrm{4}{cos}^{\mathrm{3}} \theta−\mathrm{3}{cos}\theta \\ $$$$\mathrm{2}\left(\mathrm{2}{x}^{\mathrm{2}} −\mathrm{1}\right)^{\mathrm{2}} −\mathrm{1}=\mathrm{4}{x}^{\mathrm{3}} −\mathrm{3}{x} \\ $$$$\mathrm{2}\left(\mathrm{4}{x}^{\mathrm{4}} −\mathrm{4}{x}^{\mathrm{2}} +\mathrm{1}\right)−\mathrm{4}{x}^{\mathrm{3}} +\mathrm{3}{x}−\mathrm{1}=\mathrm{0} \\ $$$$\mathrm{8}{x}^{\mathrm{4}} −\mathrm{4}{x}^{\mathrm{3}} −\mathrm{8}{x}^{\mathrm{2}} +\mathrm{3}{x}+\mathrm{1}=\mathrm{0} \\ $$$$\mathrm{8}{x}^{\mathrm{4}} −\mathrm{8}{x}^{\mathrm{3}} +\mathrm{4}{x}^{\mathrm{3}} −\mathrm{4}{x}^{\mathrm{2}} −\mathrm{4}{x}^{\mathrm{2}} +\mathrm{4}{x}−{x}+\mathrm{1} \\ $$$$\mathrm{8}{x}^{\mathrm{3}} \left({x}−\mathrm{1}\right)+\mathrm{4}{x}^{\mathrm{2}} \left({x}−\mathrm{1}\right)−\mathrm{4}{x}\left({x}−\mathrm{1}\right)−\mathrm{1}\left({x}−\mathrm{1}\right)=\mathrm{0} \\ $$$$\left({x}−\mathrm{1}\right)\left(\mathrm{8}{x}^{\mathrm{3}} +\mathrm{4}{x}^{\mathrm{2}} −\mathrm{4}{x}−\mathrm{1}\right)=\mathrm{0} \\ $$$${now}\:{look}\:\mathrm{8}{x}^{\mathrm{3}} +\mathrm{4}{x}^{\mathrm{2}} −\mathrm{4}{x}−\mathrm{1}=\mathrm{0} \\ $$$${has}\:{three}\:{root}\:{and}\:{one}\:{root}\:{is}\:{cos}\theta \\ $$$$\boldsymbol{{cos}}\left(\frac{\mathrm{2}\pi}{\mathrm{7}}\right)\:{another}\:{cos}\left(\frac{\mathrm{4}\pi}{\mathrm{7}}\right)\:{and}\:{cos}\left(\frac{\mathrm{6}\pi}{\mathrm{7}}\right) \\ $$$${cos}\left(\frac{\mathrm{2}\pi}{\mathrm{7}}\right)+{cos}\left(\frac{\mathrm{4}\pi}{\mathrm{7}}\right)+{cos}\left(\frac{\mathrm{6}\pi}{\mathrm{7}}\right)=\frac{−\mathrm{4}}{\mathrm{8}}=\frac{−\mathrm{1}}{\mathrm{2}} \\ $$$$\Sigma{cos}\left(\frac{\mathrm{2}\pi}{\mathrm{7}}\right){cos}\left(\frac{\mathrm{4}\pi}{\mathrm{7}}\right)=\frac{−\mathrm{4}}{\mathrm{8}}=\frac{−\mathrm{1}}{\mathrm{2}} \\ $$$${cos}\left(\frac{\mathrm{2}\pi}{\mathrm{7}}\right){cos}\left(\frac{\mathrm{4}\pi}{\mathrm{7}}\right){cos}\left(\frac{\mathrm{6}\pi}{\mathrm{7}}\right)=\frac{\mathrm{1}}{\mathrm{8}} \\ $$$$\frac{\mathrm{1}}{{cos}\left(\frac{\mathrm{2}\pi}{\mathrm{7}}\right)}+\frac{\mathrm{1}}{{cos}\left(\frac{\mathrm{4}\pi}{\mathrm{7}}\right)}+\frac{\mathrm{1}}{{cos}\left(\frac{\mathrm{8}\pi}{\mathrm{7}}\right)} \\ $$$${look}... \\ $$$${cos}\left(\frac{\mathrm{8}\pi}{\mathrm{7}}\right)={cos}\left(\pi+\frac{\pi}{\mathrm{7}}\right)=−{cos}\left(\frac{\pi}{\mathrm{7}}\right) \\ $$$${cos}\left(\frac{\mathrm{6}\pi}{\mathrm{7}}\right)={cos}\left(\pi−\frac{\pi}{\mathrm{7}}\right)=−{cos}\left(\frac{\pi}{\mathrm{7}}\right) \\ $$$${so}\:{cos}\left(\frac{\mathrm{6}\pi}{\mathrm{7}}\right)={cos}\left(\frac{\mathrm{8}\pi}{\mathrm{7}}\right) \\ $$$$\frac{\mathrm{1}}{{cos}\left(\frac{\mathrm{2}\pi}{\mathrm{7}}\right)}+\frac{\mathrm{1}}{{cos}\left(\frac{\mathrm{4}\pi}{\mathrm{7}}\right)}+\frac{\mathrm{1}}{{cos}\left(\frac{\mathrm{8}\pi}{\mathrm{7}}\right)} \\ $$$$=\frac{\mathrm{1}}{{cos}\left(\frac{\mathrm{2}\pi}{\mathrm{7}}\right)}+\frac{\mathrm{1}}{{cos}\left(\frac{\mathrm{4}\pi}{\mathrm{7}}\right)}+\frac{\mathrm{1}}{{cos}\left(\frac{\mathrm{6}\pi}{\mathrm{7}}\right)} \\ $$$$=\frac{\Sigma{cos}\left(\frac{\mathrm{2}\pi}{\mathrm{7}}\right){cos}\left(\frac{\mathrm{4}\pi}{\mathrm{7}}\right)}{{cos}\left(\frac{\mathrm{2}\pi}{\mathrm{7}}\right){cos}\left(\frac{\mathrm{4}\pi}{\mathrm{7}}\right){cos}\left(\frac{\mathrm{6}\pi}{\mathrm{7}}\right)} \\ $$$$=\frac{\frac{−\mathrm{1}}{\mathrm{2}}}{\frac{\mathrm{1}}{\mathrm{8}}}=−\mathrm{4} \\ $$$$ \\ $$ | ||
Commented by Power last updated on 28/Mar/20 | ||
$$\mathrm{step}\:\mathrm{by}\:\mathrm{step}\:\mathrm{solution}\:\mathrm{sir} \\ $$ | ||
Commented by jagoll last updated on 28/Mar/20 | ||
$$\mathrm{vieta}\:\mathrm{rule} \\ $$$$\frac{\mathrm{1}}{\alpha}+\frac{\mathrm{1}}{\beta}+\frac{\mathrm{1}}{\gamma}\:=\:−\frac{\mathrm{c}}{\mathrm{d}}\:=\:−\left(\frac{−\mathrm{4}}{−\mathrm{1}}\right)\:=\:−\mathrm{4} \\ $$$$\mathrm{it}\:\mathrm{the}\:\mathrm{answer} \\ $$ | ||
Commented by MJS last updated on 28/Mar/20 | ||
$$\mathrm{any}\:\mathrm{steps}\:\mathrm{by}\:\mathrm{yourself}\:\mathrm{Sir}\:\mathrm{Power}? \\ $$ | ||