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Question Number 125833 by bramlexs22 last updated on 14/Dec/20
((4sin (((2π)/7))+sec ((π/(14))))/(cot ((π/7)))) ?
$$\:\:\frac{\mathrm{4sin}\:\left(\frac{\mathrm{2}\pi}{\mathrm{7}}\right)+\mathrm{sec}\:\left(\frac{\pi}{\mathrm{14}}\right)}{\mathrm{cot}\:\left(\frac{\pi}{\mathrm{7}}\right)}\:?\: \\ $$
Answered by Dwaipayan Shikari last updated on 14/Dec/20
((4sin((2π)/7)cos(π/(14))sin(π/7)+sin(π/7))/(cos(π/(14))cos(π/7)))=((2(cos(π/7)−cos((3π)/7))cos(π/(14))+sin(π/7))/(cos(π/(14))cos(π/7))) =((cos(π/(14))−cos((3π)/(14))−cos((5π)/(14))+cos(π/2)+cos((5π)/(14)))/(cos(π/(14))cos(π/7))) sin(π/7)=cos((5π)/(14)) =((cos(π/(14))−cos((3π)/7))/(cos(π/(14))cos(π/7)))=2(((cos(π/(14))cos(π/7))/(cos(π/(14))cos(π/7))))=2
$$\frac{\mathrm{4}{sin}\frac{\mathrm{2}\pi}{\mathrm{7}}{cos}\frac{\pi}{\mathrm{14}}{sin}\frac{\pi}{\mathrm{7}}+{sin}\frac{\pi}{\mathrm{7}}}{{cos}\frac{\pi}{\mathrm{14}}{cos}\frac{\pi}{\mathrm{7}}}=\frac{\mathrm{2}\left({cos}\frac{\pi}{\mathrm{7}}−{cos}\frac{\mathrm{3}\pi}{\mathrm{7}}\right){cos}\frac{\pi}{\mathrm{14}}+{sin}\frac{\pi}{\mathrm{7}}}{{cos}\frac{\pi}{\mathrm{14}}{cos}\frac{\pi}{\mathrm{7}}} \\ $$$$=\frac{{cos}\frac{\pi}{\mathrm{14}}−{cos}\frac{\mathrm{3}\pi}{\mathrm{14}}−{cos}\frac{\mathrm{5}\pi}{\mathrm{14}}+{cos}\frac{\pi}{\mathrm{2}}+{cos}\frac{\mathrm{5}\pi}{\mathrm{14}}}{{cos}\frac{\pi}{\mathrm{14}}{cos}\frac{\pi}{\mathrm{7}}}\:\:\:\:\:\:{sin}\frac{\pi}{\mathrm{7}}={cos}\frac{\mathrm{5}\pi}{\mathrm{14}} \\ $$$$=\frac{{cos}\frac{\pi}{\mathrm{14}}−{cos}\frac{\mathrm{3}\pi}{\mathrm{7}}}{{cos}\frac{\pi}{\mathrm{14}}{cos}\frac{\pi}{\mathrm{7}}}=\mathrm{2}\left(\frac{{cos}\frac{\pi}{\mathrm{14}}{cos}\frac{\pi}{\mathrm{7}}}{{cos}\frac{\pi}{\mathrm{14}}{cos}\frac{\pi}{\mathrm{7}}}\right)=\mathrm{2} \\ $$
Answered by liberty last updated on 14/Dec/20
let (π/(14))=x then ((4sin 4x+sec x)/(cot 2x))= ((4sin 4xcos x+1)/(cos x(((cos 2x)/(sin 2x)))))= ((4sin 4xcos xsin 2x+sin 2x)/(cos x cos 2x)) = ((−2(cos 6x−cos 2x)cos x+sin 2x)/(cos xcos 2x)) = ((2cos 2x cos x−2cos 6xcos x+sin 2x)/(cos xcos 2x)) = ((2cos 2xcos x−cos 7x−cos 5x+sin 2x)/(cos x cos 2x)) [ we know that cos 7x=cos (π/2)=0 ∧cos 5x=cos ((5π)/(14))=sin ((2π)/(14)) ] thus we get cos 7x−cos 5x+sin 2x = 0 so ((2cos 2x cos x+0)/(cos x cos 2x)) = 2.
$${let}\:\frac{\pi}{\mathrm{14}}={x}\:{then}\:\frac{\mathrm{4sin}\:\mathrm{4}{x}+\mathrm{sec}\:{x}}{\mathrm{cot}\:\mathrm{2}{x}}= \\ $$$$\frac{\mathrm{4sin}\:\mathrm{4}{x}\mathrm{cos}\:{x}+\mathrm{1}}{\mathrm{cos}\:{x}\left(\frac{\mathrm{cos}\:\mathrm{2}{x}}{\mathrm{sin}\:\mathrm{2}{x}}\right)}=\:\frac{\mathrm{4sin}\:\mathrm{4}{x}\mathrm{cos}\:{x}\mathrm{sin}\:\mathrm{2}{x}+\mathrm{sin}\:\mathrm{2}{x}}{\mathrm{cos}\:{x}\:\mathrm{cos}\:\mathrm{2}{x}} \\ $$$$\:=\:\frac{−\mathrm{2}\left(\mathrm{cos}\:\mathrm{6}{x}−\mathrm{cos}\:\mathrm{2}{x}\right)\mathrm{cos}\:\:{x}+\mathrm{sin}\:\mathrm{2}{x}}{\mathrm{cos}\:{x}\mathrm{cos}\:\mathrm{2}{x}} \\ $$$$\:=\:\frac{\mathrm{2cos}\:\mathrm{2}{x}\:\mathrm{cos}\:{x}−\mathrm{2cos}\:\mathrm{6}{x}\mathrm{cos}\:{x}+\mathrm{sin}\:\mathrm{2}{x}}{\mathrm{cos}\:{x}\mathrm{cos}\:\mathrm{2}{x}} \\ $$$$\:=\:\frac{\mathrm{2cos}\:\mathrm{2}{x}\mathrm{cos}\:{x}−\mathrm{cos}\:\mathrm{7}{x}−\mathrm{cos}\:\mathrm{5}{x}+\mathrm{sin}\:\mathrm{2}{x}}{\mathrm{cos}\:{x}\:\mathrm{cos}\:\mathrm{2}{x}} \\ $$$$\:\left[\:{we}\:{know}\:{that}\:\mathrm{cos}\:\mathrm{7}{x}=\mathrm{cos}\:\frac{\pi}{\mathrm{2}}=\mathrm{0}\:\wedge\mathrm{cos}\:\mathrm{5}{x}=\mathrm{cos}\:\frac{\mathrm{5}\pi}{\mathrm{14}}=\mathrm{sin}\:\frac{\mathrm{2}\pi}{\mathrm{14}}\:\right] \\ $$$${thus}\:{we}\:{get}\:\mathrm{cos}\:\mathrm{7}{x}−\mathrm{cos}\:\mathrm{5}{x}+\mathrm{sin}\:\mathrm{2}{x}\:=\:\mathrm{0} \\ $$$${so}\:\frac{\mathrm{2cos}\:\mathrm{2}{x}\:\mathrm{cos}\:{x}+\mathrm{0}}{\mathrm{cos}\:{x}\:\mathrm{cos}\:\mathrm{2}{x}}\:=\:\mathrm{2}. \\ $$

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