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Question Number 86716 by john santu last updated on 30/Mar/20
cos ((π/(17)))×cos (((2π)/(17)))×cos (((4π)/(17)))×cos (((8π)/(17))) =
$$\mathrm{cos}\:\left(\frac{\pi}{\mathrm{17}}\right)×\mathrm{cos}\:\left(\frac{\mathrm{2}\pi}{\mathrm{17}}\right)×\mathrm{cos}\:\left(\frac{\mathrm{4}\pi}{\mathrm{17}}\right)×\mathrm{cos}\:\left(\frac{\mathrm{8}\pi}{\mathrm{17}}\right)\:= \\ $$
Commented by jagoll last updated on 30/Mar/20
((cos x cos 2x cos 4x cos 8x)/(2sin x)) × 2sin x ((sin 2x cos 2x cos 4x cos 8x)/(2sin x)) = ((sin 4x cos 4x cos 8x)/(4sin x)) = ((sin 8x cos 8x)/(8sin x)) = ((sin 16x)/(16sin x)) = ((sin (π/(17)))/(16 sin (π/(17)))) = (1/(16)) [ sin (π/(17)) = sin (π−(π/(17))) = sin (((16π)/(17))) ]
$$\frac{\mathrm{cos}\:\mathrm{x}\:\mathrm{cos}\:\mathrm{2x}\:\mathrm{cos}\:\mathrm{4x}\:\mathrm{cos}\:\mathrm{8x}}{\mathrm{2sin}\:\mathrm{x}}\:×\:\mathrm{2sin}\:\mathrm{x}\:\: \\ $$$$\frac{\mathrm{sin}\:\mathrm{2x}\:\mathrm{cos}\:\mathrm{2x}\:\mathrm{cos}\:\mathrm{4x}\:\mathrm{cos}\:\mathrm{8x}}{\mathrm{2sin}\:\mathrm{x}}\:=\: \\ $$$$\frac{\mathrm{sin}\:\mathrm{4x}\:\mathrm{cos}\:\mathrm{4x}\:\mathrm{cos}\:\mathrm{8x}}{\mathrm{4sin}\:\mathrm{x}}\:=\: \\ $$$$\frac{\mathrm{sin}\:\mathrm{8x}\:\mathrm{cos}\:\mathrm{8x}}{\mathrm{8sin}\:\mathrm{x}}\:=\:\frac{\mathrm{sin}\:\mathrm{16x}}{\mathrm{16sin}\:\mathrm{x}}\:=\:\frac{\mathrm{sin}\:\frac{\pi}{\mathrm{17}}}{\mathrm{16}\:\mathrm{sin}\:\frac{\pi}{\mathrm{17}}}\: \\ $$$$=\:\frac{\mathrm{1}}{\mathrm{16}} \\ $$$$\left[\:\mathrm{sin}\:\frac{\pi}{\mathrm{17}}\:=\:\mathrm{sin}\:\left(\pi−\frac{\pi}{\mathrm{17}}\right)\:=\:\mathrm{sin}\:\left(\frac{\mathrm{16}\pi}{\mathrm{17}}\right)\:\right] \\ $$
Commented by mathmax by abdo last updated on 30/Mar/20
let A =cos((π/(17)))cos(((2π)/(17)))cos(((4π)/(17)))cos(((8π)/(17))) we have sin((π/(17)))A = (1/2)sin(((2π)/(17)))cos(((2π)/(17)))cos(((4π)/(17)))cos(((8π)/(17))) =(1/4)sin(((4π)/(17)))cos(((4π)/(17)))cos(((8π)/(17))) =(1/8)sin(((8π)/(17)))cos(((8π)/(17))) =(1/(16))sin(((16π)/(17))) =(1/(16))sin(π−(π/(17))) =(1/(16))sin((π/(17))) ⇒ A =(1/(16))
$${let}\:{A}\:={cos}\left(\frac{\pi}{\mathrm{17}}\right){cos}\left(\frac{\mathrm{2}\pi}{\mathrm{17}}\right){cos}\left(\frac{\mathrm{4}\pi}{\mathrm{17}}\right){cos}\left(\frac{\mathrm{8}\pi}{\mathrm{17}}\right)\:{we}\:{have} \\ $$$${sin}\left(\frac{\pi}{\mathrm{17}}\right){A}\:=\:\frac{\mathrm{1}}{\mathrm{2}}{sin}\left(\frac{\mathrm{2}\pi}{\mathrm{17}}\right){cos}\left(\frac{\mathrm{2}\pi}{\mathrm{17}}\right){cos}\left(\frac{\mathrm{4}\pi}{\mathrm{17}}\right){cos}\left(\frac{\mathrm{8}\pi}{\mathrm{17}}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}{sin}\left(\frac{\mathrm{4}\pi}{\mathrm{17}}\right){cos}\left(\frac{\mathrm{4}\pi}{\mathrm{17}}\right){cos}\left(\frac{\mathrm{8}\pi}{\mathrm{17}}\right)\:=\frac{\mathrm{1}}{\mathrm{8}}{sin}\left(\frac{\mathrm{8}\pi}{\mathrm{17}}\right){cos}\left(\frac{\mathrm{8}\pi}{\mathrm{17}}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{16}}{sin}\left(\frac{\mathrm{16}\pi}{\mathrm{17}}\right)\:=\frac{\mathrm{1}}{\mathrm{16}}{sin}\left(\pi−\frac{\pi}{\mathrm{17}}\right)\:=\frac{\mathrm{1}}{\mathrm{16}}{sin}\left(\frac{\pi}{\mathrm{17}}\right)\:\:\Rightarrow \\ $$$${A}\:=\frac{\mathrm{1}}{\mathrm{16}} \\ $$
Answered by redmiiuser last updated on 30/Mar/20
(1/(16))
$$\frac{\mathrm{1}}{\mathrm{16}} \\ $$
Answered by TANMAY PANACEA. last updated on 30/Mar/20
p=cosacos2acos4acos8a [a=(π/(17))] 2psina=(2sinacosa)cos2acos4acos8a =sin2acos2acos4acos8a 2^2 psina=(2sin2acos2a)cos4acos8a 2^2 psina=sin4acos4acos8a 2^3 psina=(2sin4acos4a)cos8a 2^3 psina=sin8acos8a 2^4 psina=(2sin8acos8a) 2^4 psina=sin16a 16×p×sina=sin(17a−a) p=(1/(16sina))×sin(π−a) p=((sina)/(16sina))=(1/(16))
$${p}={cosacos}\mathrm{2}{acos}\mathrm{4}{acos}\mathrm{8}{a}\:\:\left[{a}=\frac{\pi}{\mathrm{17}}\right] \\ $$$$\mathrm{2}{psina}=\left(\mathrm{2}{sinacosa}\right){cos}\mathrm{2}{acos}\mathrm{4}{acos}\mathrm{8}{a} \\ $$$$={sin}\mathrm{2}{acos}\mathrm{2}{acos}\mathrm{4}{acos}\mathrm{8}{a} \\ $$$$\mathrm{2}^{\mathrm{2}} {psina}=\left(\mathrm{2}{sin}\mathrm{2}{acos}\mathrm{2}{a}\right){cos}\mathrm{4}{acos}\mathrm{8}{a} \\ $$$$\mathrm{2}^{\mathrm{2}} {psina}={sin}\mathrm{4}{acos}\mathrm{4}{acos}\mathrm{8}{a} \\ $$$$\mathrm{2}^{\mathrm{3}} {psina}=\left(\mathrm{2}{sin}\mathrm{4}{acos}\mathrm{4}{a}\right){cos}\mathrm{8}{a} \\ $$$$\mathrm{2}^{\mathrm{3}} {psina}={sin}\mathrm{8}{acos}\mathrm{8}{a} \\ $$$$\mathrm{2}^{\mathrm{4}} {psina}=\left(\mathrm{2}{sin}\mathrm{8}{acos}\mathrm{8}{a}\right) \\ $$$$\mathrm{2}^{\mathrm{4}} {psina}={sin}\mathrm{16}{a} \\ $$$$\mathrm{16}×{p}×{sina}={sin}\left(\mathrm{17}{a}−{a}\right) \\ $$$${p}=\frac{\mathrm{1}}{\mathrm{16}{sina}}×{sin}\left(\pi−{a}\right) \\ $$$${p}=\frac{{sina}}{\mathrm{16}{sina}}=\frac{\mathrm{1}}{\mathrm{16}} \\ $$
Commented by peter frank last updated on 31/Mar/20
thank you both
$${thank}\:{you}\:{both} \\ $$

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