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 80929 by mathocean1 last updated on 08/Feb/20

show that cos(π/7)cos((2π)/7)cos((4π)/7)=−(1/8)

$$\mathrm{show}\:\mathrm{that} \\ $$$${cos}\frac{\pi}{\mathrm{7}}\mathrm{cos}\frac{\mathrm{2}\pi}{\mathrm{7}}\mathrm{cos}\frac{\mathrm{4}\pi}{\mathrm{7}}=−\frac{\mathrm{1}}{\mathrm{8}} \\ $$

Commented by jagoll last updated on 08/Feb/20

let x =(π/7) cos xcos 2xcos 4x ×((2sin x)/(2sin x)) = ((sin 2xcos 2xcos 4x)/(2sin x))= ((sin 4xcos 4x)/(4sin x))= ((sin 8x)/(8sin x)) = ((sin (((8π)/7)))/(8sin ((π/7)))) = ((−sin ((π/7)))/(8sin ((π/7)))) = −(1/8)

$${let}\:{x}\:=\frac{\pi}{\mathrm{7}} \\ $$$$\mathrm{cos}\:{x}\mathrm{cos}\:\mathrm{2}{x}\mathrm{cos}\:\mathrm{4}{x}\:×\frac{\mathrm{2sin}\:{x}}{\mathrm{2sin}\:{x}}\:= \\ $$$$\frac{\mathrm{sin}\:\mathrm{2}{x}\mathrm{cos}\:\mathrm{2}{x}\mathrm{cos}\:\mathrm{4}{x}}{\mathrm{2sin}\:{x}}=\:\frac{\mathrm{sin}\:\mathrm{4}{x}\mathrm{cos}\:\mathrm{4}{x}}{\mathrm{4sin}\:{x}}= \\ $$$$\frac{\mathrm{sin}\:\mathrm{8}{x}}{\mathrm{8sin}\:{x}}\:=\:\frac{\mathrm{sin}\:\left(\frac{\mathrm{8}\pi}{\mathrm{7}}\right)}{\mathrm{8sin}\:\left(\frac{\pi}{\mathrm{7}}\right)}\:= \\ $$$$\frac{−\mathrm{sin}\:\left(\frac{\pi}{\mathrm{7}}\right)}{\mathrm{8sin}\:\left(\frac{\pi}{\mathrm{7}}\right)}\:=\:−\frac{\mathrm{1}}{\mathrm{8}} \\ $$

Commented by mr W last updated on 08/Feb/20

say cos(π/7)cos((2π)/7)cos((4π)/7)=x 2sin (π/7)cos(π/7)cos((2π)/7)cos((4π)/7)=2sin (π/7)×x sin ((2π)/7)cos((2π)/7)cos((4π)/7)=2sin (π/7)×x 2 sin ((2π)/7)cos((2π)/7)cos((4π)/7)=2×2sin (π/7)×x sin ((4π)/7)cos((4π)/7)=4sin (π/7)×x 2 sin ((4π)/7)cos((4π)/7)=2×4sin (π/7)×x sin ((8π)/7)=8 sin (π/7)×x sin (π+(π/7))=8 sin (π/7)×x −sin (π/7)=8 sin (π/7)×x −1=8x −(1/8)=x

$${say}\:{cos}\frac{\pi}{\mathrm{7}}\mathrm{cos}\frac{\mathrm{2}\pi}{\mathrm{7}}\mathrm{cos}\frac{\mathrm{4}\pi}{\mathrm{7}}={x} \\ $$$$\mathrm{2sin}\:\frac{\pi}{\mathrm{7}}{cos}\frac{\pi}{\mathrm{7}}\mathrm{cos}\frac{\mathrm{2}\pi}{\mathrm{7}}\mathrm{cos}\frac{\mathrm{4}\pi}{\mathrm{7}}=\mathrm{2sin}\:\frac{\pi}{\mathrm{7}}×{x} \\ $$$$\mathrm{sin}\:\frac{\mathrm{2}\pi}{\mathrm{7}}\mathrm{cos}\frac{\mathrm{2}\pi}{\mathrm{7}}\mathrm{cos}\frac{\mathrm{4}\pi}{\mathrm{7}}=\mathrm{2sin}\:\frac{\pi}{\mathrm{7}}×{x} \\ $$$$\mathrm{2}\:\mathrm{sin}\:\frac{\mathrm{2}\pi}{\mathrm{7}}\mathrm{cos}\frac{\mathrm{2}\pi}{\mathrm{7}}\mathrm{cos}\frac{\mathrm{4}\pi}{\mathrm{7}}=\mathrm{2}×\mathrm{2sin}\:\frac{\pi}{\mathrm{7}}×{x} \\ $$$$\mathrm{sin}\:\frac{\mathrm{4}\pi}{\mathrm{7}}\mathrm{cos}\frac{\mathrm{4}\pi}{\mathrm{7}}=\mathrm{4sin}\:\frac{\pi}{\mathrm{7}}×{x} \\ $$$$\mathrm{2}\:\mathrm{sin}\:\frac{\mathrm{4}\pi}{\mathrm{7}}\mathrm{cos}\frac{\mathrm{4}\pi}{\mathrm{7}}=\mathrm{2}×\mathrm{4sin}\:\frac{\pi}{\mathrm{7}}×{x} \\ $$$$\mathrm{sin}\:\frac{\mathrm{8}\pi}{\mathrm{7}}=\mathrm{8}\:\mathrm{sin}\:\frac{\pi}{\mathrm{7}}×{x} \\ $$$$\mathrm{sin}\:\left(\pi+\frac{\pi}{\mathrm{7}}\right)=\mathrm{8}\:\mathrm{sin}\:\frac{\pi}{\mathrm{7}}×{x} \\ $$$$−\mathrm{sin}\:\frac{\pi}{\mathrm{7}}=\mathrm{8}\:\mathrm{sin}\:\frac{\pi}{\mathrm{7}}×{x} \\ $$$$−\mathrm{1}=\mathrm{8}{x} \\ $$$$−\frac{\mathrm{1}}{\mathrm{8}}={x} \\ $$

Commented by jagoll last updated on 08/Feb/20

waw great sir

$${waw}\:{great}\:{sir} \\ $$

Commented by peter frank last updated on 08/Feb/20

thank you

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

Commented by mathmax by abdo last updated on 08/Feb/20

let A =cos((π/7)).cos(((2π)/7))cos(((4π)/7)) A ×sin((π/7))=(1/2)sin(((2π)/7))cos(((2π)/7))cos(((4π)/7))=(1/4)sin(((4π)/7))cos(((4π)/7)) =(1/8)sin(((8π)/7)) =(1/8)sin(π+(π/7))=−(1/8)sin((π/7)) by simplification we get A=−(1/8)

$${let}\:{A}\:={cos}\left(\frac{\pi}{\mathrm{7}}\right).{cos}\left(\frac{\mathrm{2}\pi}{\mathrm{7}}\right){cos}\left(\frac{\mathrm{4}\pi}{\mathrm{7}}\right)\: \\ $$$${A}\:×{sin}\left(\frac{\pi}{\mathrm{7}}\right)=\frac{\mathrm{1}}{\mathrm{2}}{sin}\left(\frac{\mathrm{2}\pi}{\mathrm{7}}\right){cos}\left(\frac{\mathrm{2}\pi}{\mathrm{7}}\right){cos}\left(\frac{\mathrm{4}\pi}{\mathrm{7}}\right)=\frac{\mathrm{1}}{\mathrm{4}}{sin}\left(\frac{\mathrm{4}\pi}{\mathrm{7}}\right){cos}\left(\frac{\mathrm{4}\pi}{\mathrm{7}}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{8}}{sin}\left(\frac{\mathrm{8}\pi}{\mathrm{7}}\right)\:=\frac{\mathrm{1}}{\mathrm{8}}{sin}\left(\pi+\frac{\pi}{\mathrm{7}}\right)=−\frac{\mathrm{1}}{\mathrm{8}}{sin}\left(\frac{\pi}{\mathrm{7}}\right)\:{by}\:{simplification} \\ $$$${we}\:{get}\:{A}=−\frac{\mathrm{1}}{\mathrm{8}} \\ $$

Answered by mind is power last updated on 09/Feb/20

z^7 −1=0 z^7 =−1=e^(i(2k)π) z_k =e^(i(((2k)π)/7)) z^7 −1=(z−1)Π_(k=1) ^6 (z−z_k ) =(z−1)Π_(k=1) ^(k=3) (z−z_k )(z−z_k ^− ) =Π_(k=1) ^(k=3) (1−e^(i(((2k)π)/7)) )(1−e^(−((i(2k)π)/7)) ) z=−1⇒ 1=2^6 Π_(k=1) ^3 cos^2 (((kπ)/7)) ⇒(1/2^3 )=cos((π/7))cos(((2π)/7))cos(((3π)/7)) cos(((3π)/7))=−cos(π−((3π)/7))=−cos(((4π)/7))⇒ −cos((π/7))cos(((2π)/7))cos(((4π)/7))=(1/2^3 )=(1/8)⇒ cos((π/7))cos(((2π)/7))cos(((4π)/7))=−(1/8)

$${z}^{\mathrm{7}} −\mathrm{1}=\mathrm{0} \\ $$$${z}^{\mathrm{7}} =−\mathrm{1}={e}^{{i}\left(\mathrm{2}{k}\right)\pi} \\ $$$${z}_{{k}} ={e}^{{i}\frac{\left(\mathrm{2}{k}\right)\pi}{\mathrm{7}}} \\ $$$${z}^{\mathrm{7}} −\mathrm{1}=\left({z}−\mathrm{1}\right)\underset{{k}=\mathrm{1}} {\overset{\mathrm{6}} {\prod}}\left({z}−{z}_{{k}} \right) \\ $$$$=\left({z}−\mathrm{1}\right)\underset{{k}=\mathrm{1}} {\overset{{k}=\mathrm{3}} {\prod}}\left({z}−{z}_{{k}} \right)\left({z}−{z}_{{k}} ^{−} \right) \\ $$$$=\underset{{k}=\mathrm{1}} {\overset{{k}=\mathrm{3}} {\prod}}\left(\mathrm{1}−{e}^{{i}\frac{\left(\mathrm{2}{k}\right)\pi}{\mathrm{7}}} \right)\left(\mathrm{1}−{e}^{−\frac{{i}\left(\mathrm{2}{k}\right)\pi}{\mathrm{7}}} \right) \\ $$$${z}=−\mathrm{1}\Rightarrow \\ $$$$\mathrm{1}=\mathrm{2}^{\mathrm{6}} \underset{{k}=\mathrm{1}} {\overset{\mathrm{3}} {\prod}}{cos}^{\mathrm{2}} \left(\frac{{k}\pi}{\mathrm{7}}\right) \\ $$$$\Rightarrow\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{3}} }={cos}\left(\frac{\pi}{\mathrm{7}}\right){cos}\left(\frac{\mathrm{2}\pi}{\mathrm{7}}\right){cos}\left(\frac{\mathrm{3}\pi}{\mathrm{7}}\right) \\ $$$${cos}\left(\frac{\mathrm{3}\pi}{\mathrm{7}}\right)=−{cos}\left(\pi−\frac{\mathrm{3}\pi}{\mathrm{7}}\right)=−{cos}\left(\frac{\mathrm{4}\pi}{\mathrm{7}}\right)\Rightarrow \\ $$$$−{cos}\left(\frac{\pi}{\mathrm{7}}\right){cos}\left(\frac{\mathrm{2}\pi}{\mathrm{7}}\right){cos}\left(\frac{\mathrm{4}\pi}{\mathrm{7}}\right)=\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{3}} }=\frac{\mathrm{1}}{\mathrm{8}}\Rightarrow \\ $$$${cos}\left(\frac{\pi}{\mathrm{7}}\right){cos}\left(\frac{\mathrm{2}\pi}{\mathrm{7}}\right){cos}\left(\frac{\mathrm{4}\pi}{\mathrm{7}}\right)=−\frac{\mathrm{1}}{\mathrm{8}} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com