To find the sum `sin^(2)""(2pi)/7 + sin^(2)""(4pi)/7 + sin^(2)""(8pi)/7` , we follow the following method . Put `7 theta = 2npi`, where n is any integer . Then `sin 4theta = sin (2npi - 3 theta)=-sin3 theta .......(i)` This means that `sin theta` that takes the values `0,pm sin(2pi//7),pmsin(4pi//7), and pm sin(8pi//7)` From Eq . (i) we now get `2 sin 2theta cos 2 theta = 4sin^(3) theta" or " 4 sin theta sin theta cos theta (1-2 sin^(2) theta)=(4sin^2theta - 3) sin theta ` Rejecting the value `sin theta = 0 ` we get `4 cos theta(1-2sin^2 theta)=4 sin^2 theta-3` or `16 cos^2 theta (1 - 2 sin^(2) theta)^(2) = (4 sin^(2) theta -3)^(2) " or " 16 ( 1- sin^(2) theta) (1-4sin^(2) theta+4sin^(4)theta)= 16 sin^(4) theta- 24 sin^(2) theta + 9` or `64 sin^(6) theta - 112 sin^(4) theta - 56 sin^(2) theta - 7 =0` , and this is cubic in `sin^2 theta` with the roots `sin^(2)((2pi)/7) , sin^(2)((4pi)/7) and sin^(2)((8pi)/7)` The sum of these roots in `sin^(2)""(2pi)/7 + sin^(2)""(4pi)/7 + sin^(2)""(8pi)/7 = 112/64 = 7/4` . The value of `tan^2""(pi)/7tan^2""(2pi)/7tan^2""(3pi)/7`