Amaths 三角學問題 sin^6x+cos^6x=A+Bcos4x

L.H.S. = sin6x + cos6x
= (sin2x)3 + (cos2x)3
= (sin2x + cos2x)(sin4x- sin2xcos2x + cos4x)
= (1)[(sin2x)2 - sin2xcos2x+ (cos2x)2]
= (sin2x + cos2x)2 – 3sin2xcos2x
= 1 – 3sin2xcos2x
= 1 – 3/4 (4sin2xcos2x)
= 1 – 3/4 (2sinxcosx)2
= 1 – 3sin22x / 4
= 1 – 3(1 - cos4x) / 8
= 5/8 + 3cos4x / 8
R.H.S. = A + Bcos4x
Comparing coefficients,
A = 5/8
B = 3/8