Verified answer

cos6x = cos^2(3x) - sin^2(3x) = (cos3x -sin 3x)(cos3x+sin3x)

=(4cos^3 x -3cosx -3sinx +4sin^3 x)(4cos^3 x -3cosx +3sinx -4sin^3 x)

= [4(cos^3 x +sin^3 x) -3(cosx+sinx)]. [4(cos^3 x -sin^3 x) -3(cosx-sinx)]

=(cosx+sinx)[4(cos^2 x -cosxsinx +sin^2 x)-3](cosx-sinx)[4(cos^2 x +cosxsinx +sin^2 x)-3]

=(cos^2 x -sin^2 x) (1-4cosxsinx)(1+4cosxsinx)

=cos2x (1 -16cos^2 x sin^2 x)

=cos2x(1 -4.2sinxcosx .2sinxcosx)

=cos2x (1-4sin^2 2x)

sin6x= 2sin3xcos3x= 2(3sinx -4sin^3 x)(4cos^3 x -3cosx)

=2sinx(3-4sin^2 x)cosx(4cos^2 x - 3)

=sin2x(4-4sin^2 x -1)(4cos^2 x -4 +1)

=sin2x[4(1-sin^2 x)-1][4(cos^2 x -1) +1]

=sin2x(4cos^2 -1)(1-4sin^2 x)

=sin2x( 4cos^2 x -1 -16sin^2 x cos^2 x +4sin^2 x)

=sin2x[4(cos^2 x +sin^2 x) -4.2sinxcox.2sinxcosx -1]

=sin2x(3 -4sin^2 2x)

chúc bạn học tốt :D

sin6x = sin2.3.x = 2sin3xcos3x = 2(3sinx-4sin^3x)(4cos3x-3cos^3x)=2sinxcosx(3-4sin^2x)(4-3cos^2x)

=sin2x(3-4sin^2x)[4-3(1-sin^2x)]

=sin2x(3-4sin^2x)[1+3sin^2x)]

phân tích típ nha bạn

cos6x làm tương tự

(Sin^2 x)^3+(cos^2x)3 =(sin^2 x+cos^2 x)^3-3sin^2x.cos^3 x. (sin^2 x+cos^2 x)

=1-3sin^2 x.cos^3 x