sin6A+cos6A+3sin2Acos2A=
0
1
2
3
We have:
sin6A+cos6A+3(sin2A)(cos2A)
=(sin6A)3+(cos2A)3+3(sin2A)(cos2A)×1
=(sin2A)3+(cos2A)3+3(sin2A)(cos2A)(sin2A+cos2A)
=(sin2A+cos2A)
=13=1
Prove that
sin6 A + cos6 A = 1 - 3 sin2 A cos2 A
Prove that sin6A+cos6A=1-3sin2A .cos 2A