Выражение (2 sin3xcosx - sin2x) / (cos2x - cos6x)

Обозначим S = sin(pi/7)sin(2pi/7)sin(3pi/7) и 
C = cos(pi/7)cos(2pi/7)cos(3pi/7)

Тогда S*C = sin(pi/7)sin(2pi/7)sin(3pi/7)cos(pi/7)cos(2pi/7)cos(3pi/7) = 
(sin(pi/7)cos(pi/7))*(sin(2pi/7)cos(2pi/7))*(sin(3pi/7)cos(3pi/7)) = 
(1/2*sin(2pi/7))*(1/2*sin(4pi/7))(1/2*sin(6pi/7)) = 
1/8*sin(2pi/7)*sin(4pi/7)*sin(6pi/7) = 1/8*sin(2pi/7)*sin(3pi/7)*sin(pi/7) = 1/8*S

T.e. S*C = 1/8*S, S не ноль, следовательно C = 1/8
Мы доказали, что 
cos(pi/7)cos(2pi/7)cos(3pi/7) = 1/8

Теперь решим пример:

cos(2pi/7)+cos(4pi/7)+cos(6pi/7) = (cos(2pi/7)+cos(4pi/7)) + cos(2*(3pi/7)) = 
2cos(3pi/7)cos(pi/7) + 2cos(3pi/7)cos(3pi/7) - 1 = 
2cos(3pi/7)*(cos(pi/7) + cos(3pi/7)) - 1 = 2cos(3pi/7)*2cos(2pi/7)cos(pi/7) - 1 = 
4cos(pi/7)cos(2pi/7)cos(3pi/7) - 1 = 4*1/8 - 1 = -1/2