Question and Answers Forum

All Questions      Topic List

Trigonometry Questions

Previous in All Question      Next in All Question      

Previous in Trigonometry      Next in Trigonometry      

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

$${thank}\:{you}\:{both} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com