there is this standard identity:
f(x)=4sinx*sin(60-x)*sin(60+x)=sin3x
putting x=20, we have f(20)=root3/2.....(1)
so taking lcm, we get
(sin^240sin20+sin^280sin40-sin^220sin80)/sin20sin40sin80
from 1 this becomes
8(sin^240sin20+sin^280sin40-sin^220sin80)/root3.....(2)
now we focus on 8(sin^240sin20+sin^280sin40-sin^220sin80)
using identity cos2y=1 – sin^2y and sin(180-z)=sinz and 2sinAcosB=sin(A+B)+sin(A-B), we have
2(3sin20+3sin40 – 3sin80+3sin60)
now, sin80-sin20=2sin30cos50=sin40
so 2(3sin20+3sin40 – 3sin80+3sin60)= 6sin60=3root3
from (2), we have
sin40/sin80 + sin80/sin20 - sin20/sin40= 3root3/root3
= 3
