A. Maths ( F.4)-------20點

Show that sin^6x+cos^6x to be 1-4/3sin^2 2x
sin^6x + cos^6x
= (sin^2x)^3 + (cos^2x)^3
= [(1 - cos2x)/ 2]^3 + [(1 + cos2x)/2]^3
= (1/8)[(1 - cos2x)^3 + (1 + cos2x)^3]
= (1/8)[1 - 3cos2x + 3cos^2 2x - cos^3 2x + 1 + 3cos2x + 3cos^2 2x + cos^3 2x]
= (1/8)[2 + 6cos^2 2x]
= (1/4)[1 + 3cos^2 2x]
= (1/4)[1 + 3(1 - sin^2 2x)]
= (1/4)[4 - 3sin^2 2x]
= 1 - [3/(4sin^2 2x)]
plx check if your answer is correct

∵a^3+b^3=(a+b)(a^2-ab+b^2)
∴sin^6x+cos^6x
=(sin^2x)^3+(cos^2x)^3
=(sin^2x+cos^2x)[(sin^2x)^2-sin^2xcos^2x+(cos^2x)^2]
=1*[(sin^2x)^2+2sin^2xcos^2x+(cos^2x)^2-3sin^2xcos^2x]
=(sin^2x+cos^2x)^2-3sin^2xcos^2x
=1-3*4sin^2xcos^2x/4
=1-3*(2sinxcosx)^2/4
=1-3(sin2x)^2/4