trigonom,etry

firstly take π/10=θ

π/2=5θ

4θ=π/2-θ

take sin both sides

sin4θ=sin(π/2-θ)

2sin2θcos2θ=cosθ

4sinθcosθ(1-2sin²Î¸)=cosθ

4sinθ(1-2sin²Î¸)=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

arrey 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 .

sin 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 .