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Question-65723




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
7a=2π4a=2π3acos(4a)=cos(2π3a)(2cos22a1)=cos3πcosa=csocos2a=2c21cos3a=4c33c{2(2c21)21}=(4c33c){2(4c44c2+1)1}=(4c33c)8c48c2+1=4c33c8c44c38c2+3c+1=08c48c3+4c34c24c2+4cc+1=08c3(c1)+4c2(c1)4c(c1)1(c1)=0(c1)(8c3+4c24c1)=08c3+4c24c1=0c=cosa=cos(2π7)[since7a=2π]cos(2π7)isarootof8x3+4x24x1=0
Commented by bshahid010@gmail.com last updated on 03/Aug/19
thank you
thankyou
Commented by Tanmay chaudhury last updated on 03/Aug/19
most welcome
mostwelcome
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
butcosπ7doessolvetheequationtheotherrootsarecos3π7andcos5π78(xcosπ7)(xcos3π7)(xcos5π7)8(x.900968868)(x.222520934)(x+.623489801)8x34x24x+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
Istilldontunderstandwhyyoudidwhatyoudid.itworksexactlythesanewaywiththegivenvalue7a=π4a=π3acos4a=cos3aa=arccosc8c48c2+1=3c4c38c4+4c38c23c+1=0(c+1)(8c34c24c+1)=0a=π7cosπ7isarootofthegivenequation
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
siryouransweralsocorrectyouhavetaken7a=πandihavetaken7a=2πbotharecorrectequationaredifferentithought7a=2πsoididityouranzwerisaspergivvenconditioninauestiong