Question Number 165558 by som(math1967) last updated on 03/Feb/22
$${Prove}\:{that} \\ $$$$\:\:\boldsymbol{{sec}}\frac{\mathrm{2}\boldsymbol{\pi}}{\mathrm{7}}\:+\boldsymbol{{sec}}\frac{\mathrm{4}\boldsymbol{\pi}}{\mathrm{7}}+\boldsymbol{{sec}}\frac{\mathrm{8}\boldsymbol{\pi}}{\mathrm{7}}\:=β\mathrm{4} \\ $$
Answered by Rohit143Jo last updated on 04/Feb/22
$${Ans}: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:{Let},\:{x}=\frac{\mathrm{2}\pi}{\mathrm{7}}\:\:\:\:\Leftrightarrow\:\:\mathrm{7}{x}=\mathrm{2}\pi \\ $$$$\:{LHS},\:\:\:\:\:\:{Sec}\left({x}\right)+{Sec}\left(\mathrm{2}{x}\right)+{Sec}\left(\mathrm{4}{x}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\Rightarrow\:\frac{\mathrm{1}}{{Cos}\left({x}\right)}\:+\:\frac{\mathrm{1}}{{Cos}\left(\mathrm{2}{x}\right)}\:+\:\frac{\mathrm{1}}{{Cos}\left(\mathrm{4}{x}\right)} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\Rightarrow\:\frac{{Cos}\left(\mathrm{2}{x}\right).{Cos}\left(\mathrm{4}{x}\right)\:+\:{Cos}\left({x}\right).{Cos}\left(\mathrm{4}{x}\right)\:+\:{Cos}\left({x}\right).{Cos}\left(\mathrm{2}{x}\right)}{{Cos}\left({x}\right).{Cos}\left(\mathrm{2}{x}\right).{Cos}\left(\mathrm{4}{x}\right)} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\Rightarrow\:\frac{\mathrm{2}.{Cos}\left(\mathrm{2}{x}\right).{Cos}\left(\mathrm{4}{x}\right)+\mathrm{2}.{Cos}\left({x}\right).{Cos}\left(\mathrm{4}{x}\right)+\mathrm{2}.{Cos}\left({x}\right).{Cos}\left(\mathrm{2}{x}\right)}{\mathrm{2}.{Cos}\left({x}\right).{Cos}\left(\mathrm{2}{x}\right).{Cos}\left(\mathrm{4}{x}\right)} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\Rightarrow\:\frac{{Sin}\left({x}\right)\left[{Cos}\left(\mathrm{6}{x}\right)+{Cos}\left(\mathrm{2}{x}\right)+{Cos}\left(\mathrm{5}{x}\right)+{Cos}\left(\mathrm{3}{x}\right)+{Cos}\left(\mathrm{3}{x}\right)+{Cos}\left({x}\right)\right]}{\mathrm{2}{Sin}\left({x}\right){Cos}\left({x}\right).{Cos}\left(\mathrm{2}{x}\right){Cos}\left(\mathrm{4}{x}\right)} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\Rightarrow\:\frac{\mathrm{2}{Sin}\left({x}\right)\left[{Cos}\left(\mathrm{2}\piβ{x}\right)+{Cos}\left(\mathrm{2}{x}\right)+{Cos}\left(\mathrm{2}\piβ\mathrm{2}{x}\right)+{Cos}\left(\mathrm{2}\piβ\mathrm{4}{x}\right)+{Cos}\left(\mathrm{2}\piβ\mathrm{4}{x}\right)+{Cos}\left({x}\right)\right]}{\mathrm{2}{Sin}\left(\mathrm{2}{x}\right){Cos}\left(\mathrm{2}{x}\right).{Cos}\left(\mathrm{4}{x}\right)} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\Rightarrow\:\frac{\mathrm{4}{Sin}\left({x}\right)\left[{Cos}\left({x}\right)+{Cos}\left(\mathrm{2}{x}\right)+{Cos}\left(\mathrm{2}{x}\right)+{Cos}\left(\mathrm{4}{x}\right)+{Cos}\left(\mathrm{4}{x}\right)+{Cos}\left({x}\right)\right]}{\mathrm{2}{Sin}\left(\mathrm{4}{x}\right){Cos}\left(\mathrm{4}{x}\right)} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\Rightarrow\:\frac{\mathrm{4}\left[\mathrm{2}{Sin}\left({x}\right){Cos}\left({x}\right)+\mathrm{2}{Cos}\left(\mathrm{2}{x}\right){Sin}\left({x}\right)+\mathrm{2}{Cos}\left(\mathrm{4}{x}\right){Sin}\left({x}\right)\right]}{{Sin}\left(\mathrm{8}{x}\right)} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\Rightarrow\:\frac{\mathrm{4}\left[{Sin}\left(\mathrm{2}{x}\right)+{Sin}\left(\mathrm{3}{x}\right)β{Sin}\left({x}\right)+{Sin}\left(\mathrm{5}{x}\right)β{Sin}\left(\mathrm{3}{x}\right)\right]}{{Sin}\left(\mathrm{8}{x}\right)} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\Rightarrow\:\frac{\mathrm{4}\left[β{Sin}\left({x}\right)+{Sin}\left(\mathrm{2}{x}\right)+{Sin}\left(\mathrm{2}\piβ\mathrm{2}{x}\right)\right]}{{Sin}\left(\mathrm{2}\pi+{x}\right)} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\Rightarrow\:\frac{\mathrm{4}\left[β{Sin}\left({x}\right)+{Sin}\left(\mathrm{2}{x}\right)β{Sin}\left(\mathrm{2}{x}\right)\right]}{{Sin}\left({x}\right)} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\Rightarrow\:\frac{\mathrm{4}\left[β{Sin}\left({x}\right)\right]}{{Sin}\left({x}\right)} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\Rightarrow\:β\mathrm{4},\:\:{LHS}\:\:\left({Proved}\right) \\ $$
Commented by som(math1967) last updated on 04/Feb/22
$$\boldsymbol{{Very}}\:\boldsymbol{{nice}}\:\boldsymbol{{solution}}\:\boldsymbol{{thank}}\:\boldsymbol{{you}} \\ $$
Commented by Rohit143Jo last updated on 04/Feb/22
$${Welcome}…… \\ $$
Commented by peter frank last updated on 06/Feb/22
$$\mathrm{Thank}\:\mathrm{you} \\ $$