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 = 0we 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