prove that the equation
x3-2x+1=0
has roots as
√2sin(π/4),2sin(π/10),2sin(3π/10)
q is from adg
1firstly take π/10=θ
π/2=5θ
4θ=π/2-θ
take sin both sides
sin4θ=sin(π/2-θ)
2sin2θcos2θ=cosθ
4sinθcosθ(1-2sin2θ)=cosθ
4sinθ(1-2sin2θ)=1
now for showing 2sin(Ï€/10) to be a root of the give eq
put 2sin(π/10)=2sinθ=x
2(2sinθ)(1-(2sinθ)2/2)=1
2(x)(1-x2/2)=1
x3-2x+1=0
since we get the same eq
HP
similarly take oder 2 roots
1arrey you gave the qustion wrong -- the roots were √2 sin pi / 4 , 2 sin pi /10 and 2 sin 13 pi / 10 , whose sum is 0 .
1sin a/2 sin b/2 sin c/2 = (s -a ) (s - b ) ( s- c ) / abc { by half angle formulas }
so given ( s - a ) ( s - b ) ( s - c ) / abc ≤ 1 / 8
or rearrange to get ( 2s - 2a ) ( 2s - 2b ) ( 2s - 2c ) ≤ abc
or ( a + b - c ) ( a + b - c ) ( b + c -a ) ≤ abc -------------------------------------------- 1
this is what we have to prove --
now a2 ≥ a2 - ( b - c )2 = ( a + b - c ) ( a - b + c )
similarly get two more equations , and then multiply to get
a2 b2 c2 ≥ ( a + b - c )2 ( a + c - b )2 ( b + c - a )2 -------------------------------------- 2
combining 1 and 2 ,
you get the result .
