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Question Number 165558 by som(math1967) last updated on 03/Feb/22

Prove that sec((2𝛑)/7) +sec((4𝛑)/7)+sec((8𝛑)/7) =βˆ’4

$${Prove}\:{that} \\ $$$$\:\:\boldsymbol{{sec}}\frac{\mathrm{2}\boldsymbol{\pi}}{\mathrm{7}}\:+\boldsymbol{{sec}}\frac{\mathrm{4}\boldsymbol{\pi}}{\mathrm{7}}+\boldsymbol{{sec}}\frac{\mathrm{8}\boldsymbol{\pi}}{\mathrm{7}}\:=βˆ’\mathrm{4} \\ $$

Answered by Rohit143Jo last updated on 04/Feb/22

Ans: Let, x=((2Ο€)/7) ⇔ 7x=2Ο€ LHS, Sec(x)+Sec(2x)+Sec(4x) β‡’ (1/(Cos(x))) + (1/(Cos(2x))) + (1/(Cos(4x))) β‡’ ((Cos(2x).Cos(4x) + Cos(x).Cos(4x) + Cos(x).Cos(2x))/(Cos(x).Cos(2x).Cos(4x))) β‡’ ((2.Cos(2x).Cos(4x)+2.Cos(x).Cos(4x)+2.Cos(x).Cos(2x))/(2.Cos(x).Cos(2x).Cos(4x))) β‡’ ((Sin(x)[Cos(6x)+Cos(2x)+Cos(5x)+Cos(3x)+Cos(3x)+Cos(x)])/(2Sin(x)Cos(x).Cos(2x)Cos(4x))) β‡’ ((2Sin(x)[Cos(2Ο€βˆ’x)+Cos(2x)+Cos(2Ο€βˆ’2x)+Cos(2Ο€βˆ’4x)+Cos(2Ο€βˆ’4x)+Cos(x)])/(2Sin(2x)Cos(2x).Cos(4x))) β‡’ ((4Sin(x)[Cos(x)+Cos(2x)+Cos(2x)+Cos(4x)+Cos(4x)+Cos(x)])/(2Sin(4x)Cos(4x))) β‡’ ((4[2Sin(x)Cos(x)+2Cos(2x)Sin(x)+2Cos(4x)Sin(x)])/(Sin(8x))) β‡’ ((4[Sin(2x)+Sin(3x)βˆ’Sin(x)+Sin(5x)βˆ’Sin(3x)])/(Sin(8x))) β‡’ ((4[βˆ’Sin(x)+Sin(2x)+Sin(2Ο€βˆ’2x)])/(Sin(2Ο€+x))) β‡’ ((4[βˆ’Sin(x)+Sin(2x)βˆ’Sin(2x)])/(Sin(x))) β‡’ ((4[βˆ’Sin(x)])/(Sin(x))) β‡’ βˆ’4, LHS (Proved)

$${Ans}: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:{Let},\:{x}=\frac{\mathrm{2}\pi}{\mathrm{7}}\:\:\:\:\Leftrightarrow\:\:\mathrm{7}{x}=\mathrm{2}\pi \\ $$$$\:{LHS},\:\:\:\:\:\:{Sec}\left({x}\right)+{Sec}\left(\mathrm{2}{x}\right)+{Sec}\left(\mathrm{4}{x}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\Rightarrow\:\frac{\mathrm{1}}{{Cos}\left({x}\right)}\:+\:\frac{\mathrm{1}}{{Cos}\left(\mathrm{2}{x}\right)}\:+\:\frac{\mathrm{1}}{{Cos}\left(\mathrm{4}{x}\right)} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\Rightarrow\:\frac{{Cos}\left(\mathrm{2}{x}\right).{Cos}\left(\mathrm{4}{x}\right)\:+\:{Cos}\left({x}\right).{Cos}\left(\mathrm{4}{x}\right)\:+\:{Cos}\left({x}\right).{Cos}\left(\mathrm{2}{x}\right)}{{Cos}\left({x}\right).{Cos}\left(\mathrm{2}{x}\right).{Cos}\left(\mathrm{4}{x}\right)} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\Rightarrow\:\frac{\mathrm{2}.{Cos}\left(\mathrm{2}{x}\right).{Cos}\left(\mathrm{4}{x}\right)+\mathrm{2}.{Cos}\left({x}\right).{Cos}\left(\mathrm{4}{x}\right)+\mathrm{2}.{Cos}\left({x}\right).{Cos}\left(\mathrm{2}{x}\right)}{\mathrm{2}.{Cos}\left({x}\right).{Cos}\left(\mathrm{2}{x}\right).{Cos}\left(\mathrm{4}{x}\right)} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\Rightarrow\:\frac{{Sin}\left({x}\right)\left[{Cos}\left(\mathrm{6}{x}\right)+{Cos}\left(\mathrm{2}{x}\right)+{Cos}\left(\mathrm{5}{x}\right)+{Cos}\left(\mathrm{3}{x}\right)+{Cos}\left(\mathrm{3}{x}\right)+{Cos}\left({x}\right)\right]}{\mathrm{2}{Sin}\left({x}\right){Cos}\left({x}\right).{Cos}\left(\mathrm{2}{x}\right){Cos}\left(\mathrm{4}{x}\right)} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\Rightarrow\:\frac{\mathrm{2}{Sin}\left({x}\right)\left[{Cos}\left(\mathrm{2}\piβˆ’{x}\right)+{Cos}\left(\mathrm{2}{x}\right)+{Cos}\left(\mathrm{2}\piβˆ’\mathrm{2}{x}\right)+{Cos}\left(\mathrm{2}\piβˆ’\mathrm{4}{x}\right)+{Cos}\left(\mathrm{2}\piβˆ’\mathrm{4}{x}\right)+{Cos}\left({x}\right)\right]}{\mathrm{2}{Sin}\left(\mathrm{2}{x}\right){Cos}\left(\mathrm{2}{x}\right).{Cos}\left(\mathrm{4}{x}\right)} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\Rightarrow\:\frac{\mathrm{4}{Sin}\left({x}\right)\left[{Cos}\left({x}\right)+{Cos}\left(\mathrm{2}{x}\right)+{Cos}\left(\mathrm{2}{x}\right)+{Cos}\left(\mathrm{4}{x}\right)+{Cos}\left(\mathrm{4}{x}\right)+{Cos}\left({x}\right)\right]}{\mathrm{2}{Sin}\left(\mathrm{4}{x}\right){Cos}\left(\mathrm{4}{x}\right)} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\Rightarrow\:\frac{\mathrm{4}\left[\mathrm{2}{Sin}\left({x}\right){Cos}\left({x}\right)+\mathrm{2}{Cos}\left(\mathrm{2}{x}\right){Sin}\left({x}\right)+\mathrm{2}{Cos}\left(\mathrm{4}{x}\right){Sin}\left({x}\right)\right]}{{Sin}\left(\mathrm{8}{x}\right)} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\Rightarrow\:\frac{\mathrm{4}\left[{Sin}\left(\mathrm{2}{x}\right)+{Sin}\left(\mathrm{3}{x}\right)βˆ’{Sin}\left({x}\right)+{Sin}\left(\mathrm{5}{x}\right)βˆ’{Sin}\left(\mathrm{3}{x}\right)\right]}{{Sin}\left(\mathrm{8}{x}\right)} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\Rightarrow\:\frac{\mathrm{4}\left[βˆ’{Sin}\left({x}\right)+{Sin}\left(\mathrm{2}{x}\right)+{Sin}\left(\mathrm{2}\piβˆ’\mathrm{2}{x}\right)\right]}{{Sin}\left(\mathrm{2}\pi+{x}\right)} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\Rightarrow\:\frac{\mathrm{4}\left[βˆ’{Sin}\left({x}\right)+{Sin}\left(\mathrm{2}{x}\right)βˆ’{Sin}\left(\mathrm{2}{x}\right)\right]}{{Sin}\left({x}\right)} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\Rightarrow\:\frac{\mathrm{4}\left[βˆ’{Sin}\left({x}\right)\right]}{{Sin}\left({x}\right)} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\Rightarrow\:βˆ’\mathrm{4},\:\:{LHS}\:\:\left({Proved}\right) \\ $$

Commented by som(math1967) last updated on 04/Feb/22

Very nice solution thank you

$$\boldsymbol{{Very}}\:\boldsymbol{{nice}}\:\boldsymbol{{solution}}\:\boldsymbol{{thank}}\:\boldsymbol{{you}} \\ $$

Commented by Rohit143Jo last updated on 04/Feb/22

Welcome......

$${Welcome}...... \\ $$

Commented by peter frank last updated on 06/Feb/22

Thank you

$$\mathrm{Thank}\:\mathrm{you} \\ $$

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