Question

cos2x-cos6x=? cosx=1/√3

Answer 1

cos2x-cos6x=-2sin(-2x)sin4x=2sin2xsin4x=2sin2x*2sin2xcos2x=4sin²2xcos2x= =4*8/9*(-1/3)=-32/27 cosx=1/√3 cos2x=2cos²x-1=2*1/3 -1=2/3-1=-1/3 sin²2x=1-cos²2x=1-1/9=8/9

Answer 2

4*8/9 откуда вышло