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Question-65723

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Question Number 65723 by bshahid010@gmail.com last updated on 02/Aug/19
Answered by Tanmay chaudhury last updated on 03/Aug/19
7a=2π 4a=2π−3a cos(4a)=cos(2π−3a) (2cos^2 2a−1)=cos3π cosa=c so cos2a=2c^2 −1 cos3a=4c^3 −3c {2(2c^2 −1)^2 −1}=(4c^3 −3c) {2(4c^4 −4c^2 +1)−1}=(4c^3 −3c) 8c^4 −8c^2 +1=4c^3 −3c 8c^4 −4c^3 −8c^2 +3c+1=0 8c^4 −8c^3 +4c^3 −4c^2 −4c^2 +4c−c+1=0 8c^3 (c−1)+4c^2 (c−1)−4c(c−1)−1(c−1)=0 (c−1)(8c^3 +4c^2 −4c−1)=0 8c^3 +4c^2 −4c−1=0 c=cosa=cos(((2π)/7)) [since 7a=2π] cos(((2π)/7)) is a root of 8x^3 +4x^2 −4x−1=0
$$\mathrm{7}{a}=\mathrm{2}\pi \\ $$$$\mathrm{4}{a}=\mathrm{2}\pi−\mathrm{3}{a} \\ $$$${cos}\left(\mathrm{4}{a}\right)={cos}\left(\mathrm{2}\pi−\mathrm{3}{a}\right) \\ $$$$\left(\mathrm{2}{cos}^{\mathrm{2}} \mathrm{2}{a}−\mathrm{1}\right)={cos}\mathrm{3}\pi \\ $$$${cosa}={c}\:\:\:{so}\:\:\:{cos}\mathrm{2}{a}=\mathrm{2}{c}^{\mathrm{2}} −\mathrm{1} \\ $$$${cos}\mathrm{3}{a}=\mathrm{4}{c}^{\mathrm{3}} −\mathrm{3}{c} \\ $$$$\left\{\mathrm{2}\left(\mathrm{2}{c}^{\mathrm{2}} −\mathrm{1}\right)^{\mathrm{2}} −\mathrm{1}\right\}=\left(\mathrm{4}{c}^{\mathrm{3}} −\mathrm{3}{c}\right) \\ $$$$\left\{\mathrm{2}\left(\mathrm{4}{c}^{\mathrm{4}} −\mathrm{4}{c}^{\mathrm{2}} +\mathrm{1}\right)−\mathrm{1}\right\}=\left(\mathrm{4}{c}^{\mathrm{3}} −\mathrm{3}{c}\right) \\ $$$$\mathrm{8}{c}^{\mathrm{4}} −\mathrm{8}{c}^{\mathrm{2}} +\mathrm{1}=\mathrm{4}{c}^{\mathrm{3}} −\mathrm{3}{c} \\ $$$$\mathrm{8}{c}^{\mathrm{4}} −\mathrm{4}{c}^{\mathrm{3}} −\mathrm{8}{c}^{\mathrm{2}} +\mathrm{3}{c}+\mathrm{1}=\mathrm{0} \\ $$$$\mathrm{8}{c}^{\mathrm{4}} −\mathrm{8}{c}^{\mathrm{3}} +\mathrm{4}{c}^{\mathrm{3}} −\mathrm{4}{c}^{\mathrm{2}} −\mathrm{4}{c}^{\mathrm{2}} +\mathrm{4}{c}−{c}+\mathrm{1}=\mathrm{0} \\ $$$$\mathrm{8}{c}^{\mathrm{3}} \left({c}−\mathrm{1}\right)+\mathrm{4}{c}^{\mathrm{2}} \left({c}−\mathrm{1}\right)−\mathrm{4}{c}\left({c}−\mathrm{1}\right)−\mathrm{1}\left({c}−\mathrm{1}\right)=\mathrm{0} \\ $$$$\left({c}−\mathrm{1}\right)\left(\mathrm{8}{c}^{\mathrm{3}} +\mathrm{4}{c}^{\mathrm{2}} −\mathrm{4}{c}−\mathrm{1}\right)=\mathrm{0} \\ $$$$\mathrm{8}{c}^{\mathrm{3}} +\mathrm{4}{c}^{\mathrm{2}} −\mathrm{4}{c}−\mathrm{1}=\mathrm{0} \\ $$$${c}={cosa}={cos}\left(\frac{\mathrm{2}\pi}{\mathrm{7}}\right)\:\:\:\left[{since}\:\mathrm{7}{a}=\mathrm{2}\pi\right] \\ $$$${cos}\left(\frac{\mathrm{2}\pi}{\mathrm{7}}\right)\:{is}\:{a}\:{root}\:{of} \\ $$$$\mathrm{8}{x}^{\mathrm{3}} +\mathrm{4}{x}^{\mathrm{2}} −\mathrm{4}{x}−\mathrm{1}=\mathrm{0} \\ $$
Commented by bshahid010@gmail.com last updated on 03/Aug/19
thank you
$$\mathrm{thank}\:\mathrm{you} \\ $$
Commented by Tanmay chaudhury last updated on 03/Aug/19
most welcome
$${most}\:{welcome} \\ $$
Commented by MJS last updated on 03/Aug/19
but cos (π/7) does solve the equation... the other roots are cos ((3π)/7) and cos ((5π)/7) 8(x−cos (π/7))(x−cos ((3π)/7))(x−cos ((5π)/7))≈ ≈8(x−.900 968 868)(x−.222 520 934)(x+.623 489 801)≈ ≈8x^3 −4x^2 −4x+1
$$\mathrm{but}\:\mathrm{cos}\:\frac{\pi}{\mathrm{7}}\:{does}\:\mathrm{solve}\:\mathrm{the}\:\mathrm{equation}… \\ $$$$\mathrm{the}\:\mathrm{other}\:\mathrm{roots}\:\mathrm{are}\:\mathrm{cos}\:\frac{\mathrm{3}\pi}{\mathrm{7}}\:\mathrm{and}\:\mathrm{cos}\:\frac{\mathrm{5}\pi}{\mathrm{7}} \\ $$$$\mathrm{8}\left({x}−\mathrm{cos}\:\frac{\pi}{\mathrm{7}}\right)\left({x}−\mathrm{cos}\:\frac{\mathrm{3}\pi}{\mathrm{7}}\right)\left({x}−\mathrm{cos}\:\frac{\mathrm{5}\pi}{\mathrm{7}}\right)\approx \\ $$$$\approx\mathrm{8}\left({x}−.\mathrm{900}\:\mathrm{968}\:\mathrm{868}\right)\left({x}−.\mathrm{222}\:\mathrm{520}\:\mathrm{934}\right)\left({x}+.\mathrm{623}\:\mathrm{489}\:\mathrm{801}\right)\approx \\ $$$$\approx\mathrm{8}{x}^{\mathrm{3}} −\mathrm{4}{x}^{\mathrm{2}} −\mathrm{4}{x}+\mathrm{1} \\ $$
Commented by MJS last updated on 05/Aug/19
I still don′t understand why you did what you did. it works exactly the sane way with the given value 7a=π 4a=π−3a cos 4a =−cos 3a a=arccos c 8c^4 −8c^2 +1=3c−4c^3 8c^4 +4c^3 −8c^2 −3c+1=0 (c+1)(8c^3 −4c^2 −4c+1)=0 a=(π/7) ⇒ cos (π/7) is a root of the given equation
$$\mathrm{I}\:\mathrm{still}\:\mathrm{don}'\mathrm{t}\:\mathrm{understand}\:\mathrm{why}\:\mathrm{you}\:\mathrm{did}\:\mathrm{what}\:\mathrm{you} \\ $$$$\mathrm{did}.\:\mathrm{it}\:\mathrm{works}\:\mathrm{exactly}\:\mathrm{the}\:\mathrm{sane}\:\mathrm{way}\:\mathrm{with}\:\mathrm{the} \\ $$$$\mathrm{given}\:\mathrm{value} \\ $$$$\mathrm{7}{a}=\pi \\ $$$$\mathrm{4}{a}=\pi−\mathrm{3}{a} \\ $$$$\mathrm{cos}\:\mathrm{4}{a}\:=−\mathrm{cos}\:\mathrm{3}{a} \\ $$$${a}=\mathrm{arccos}\:{c} \\ $$$$\mathrm{8}{c}^{\mathrm{4}} −\mathrm{8}{c}^{\mathrm{2}} +\mathrm{1}=\mathrm{3}{c}−\mathrm{4}{c}^{\mathrm{3}} \\ $$$$\mathrm{8}{c}^{\mathrm{4}} +\mathrm{4}{c}^{\mathrm{3}} −\mathrm{8}{c}^{\mathrm{2}} −\mathrm{3}{c}+\mathrm{1}=\mathrm{0} \\ $$$$\left({c}+\mathrm{1}\right)\left(\mathrm{8}{c}^{\mathrm{3}} −\mathrm{4}{c}^{\mathrm{2}} −\mathrm{4}{c}+\mathrm{1}\right)=\mathrm{0} \\ $$$${a}=\frac{\pi}{\mathrm{7}}\:\Rightarrow\:\mathrm{cos}\:\frac{\pi}{\mathrm{7}}\:\mathrm{is}\:\mathrm{a}\:\mathrm{root}\:\mathrm{of}\:\mathrm{the}\:\mathrm{given}\:\mathrm{equation} \\ $$
Commented by Tanmay chaudhury last updated on 05/Aug/19
sir your answer also correct...you have taken 7a=π and i have taken 7a=2π both are correct...equation are different... i thought 7a=2π so i did it... your anzwer is as per givven condition in auestion g
$${sir}\:{your}\:{answer}\:{also}\:{correct}…{you}\:{have}\:{taken} \\ $$$$\mathrm{7}{a}=\pi\:\:\:{and}\:{i}\:{have}\:{taken}\:\mathrm{7}{a}=\mathrm{2}\pi \\ $$$${both}\:{are}\:{correct}…{equation}\:{are}\:{different}… \\ $$$${i}\:{thought}\:\mathrm{7}{a}=\mathrm{2}\pi\:\:\:{so}\:{i}\:{did}\:{it}… \\ $$$${your}\:{anzwer}\:{is}\:{as}\:{per}\:{givven}\:{condition}\:{in}\:{auestion} \\ $$$${g} \\ $$$$ \\ $$

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