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Question Number 60054 by bhanukumarb2@gmail.com last updated on 17/May/19
Commented by bhanukumarb2@gmail.com last updated on 17/May/19
Answered by tanmay last updated on 17/May/19
![8x^3 −4x^2 −4x+1=0 roots are cos((π/7)),cos(((3π)/7)),cos(((5π)/7)) now cos(((5π)/7))=cos(((7π−2π)/7))=cos(π−((2π)/7))=−cos(((2π)/7)) let c_1 =cos((π/7)) so s_1 =sin((π/7)) c_2 =cos(((2π)/7)) s_2 =sin(((2π)/7)) c_3 =cos(((3π)/7)) s_3 =(((3π)/7)) ★★c_5 =cos(((5π)/7))=−cos(((2π)/7))=−c_2 ★★ our task to find value of (s_1 ^2 /s_2 ^4 )+(s_3 ^2 /s_1 ^4 )+(s_2 ^2 /s_3 ^4 ) 8x^3 −4x^2 −4x+1=0 roots are c_1 ,c_3 ,c_5 [note c_5 =−c_2 ] c_1 +c_3 +c_5 =((−(−4))/8)=(1/2) c_1 −c_2 +c_3 =(1/2) c_1 c_3 +c_1 c_5 +c_3 c_5 =((−4)/8) c_1 c_3 +c_1 (−c_2 )+c_3 (−c_2 )=((−4)/8) c_1 c_3 −c_1 c_2 −c_3 c_2 =((−1)/2) c_1 c_3 c_5 =((−1)/8) c_1 ×c_3 ×−c_2 =((−1)/8) c_1 c_2 c_3 =(1/8) (s_1 ^2 /s_2 ^4 )+(s_3 ^2 /s_1 ^4 )+(s_2 ^2 /s_3 ^4 )=((s_1 ^2 (s_1 ^4 s_3 ^4 )+s_3 ^2 (s_2 ^4 s_3 ^4 )+s_2 ^2 (s_2 ^4 s_1 ^4 ))/((s_1 s_2 s_3 )^4 )) =((s_1 ^6 s_3 ^4 +s_3 ^6 s_2 ^4 +s_2 ^6 s_1 ^(4i) )/((s_1 s_2 s_3 )^4 )) 8c^3 −4c^2 −4c+1=0 c^2 +s^2 =1 8(1−s^2 )(√(1−s^2 )) −4(1−s^2 )−4((√(1−s^2 )))+1=0 4×(√(1−s^2 )) (2−2s^2 −1)=4(1−s^2 )−1 4×(√(1−s^2 )) (1−2s^2 )=(3−4s^2 ) 16(1−s^2 )(1−4s^2 +4s^4 )=9−24s^2 +16s^4 16(1−4s^2 +4s^4 −s^2 +4s^4 −4s^6 )=9−24s^2 +16s^4 −64s^6 +128s^4 −80s^2 +16−16s^4 +24s^2 −9=0 −64s^6 +112s^4 −53s^2 +7=0 if s^2 =t −64t^3 +112t^2 −53t+7=0 s_1 ^2 +s_2 ^2 +s_3 ^2 =((112)/(64)) s_1 ^2 s_2 ^2 +s_1 ^2 s_3 ^2 +s_2 ^2 s_3 ^2 =((53)/(64)) s_1 ^2 s_2 ^2 s_3 ^2 =(7/(64)) wait i am trying...](Q60087.png)
Commented by bhanukumarb2@gmail.com last updated on 17/May/19
Commented by otchereabdullai@gmail.com last updated on 17/May/19
