Prove thatsin6A+cos6A=1-3sin2Acos2A
As we have sin6A+cos6A=1-3sin2Acos2A
Taking LHS sin6A+cos6A=(sin2A)3+(cos2A)3
We know that (a+b)3=a3+b3+3ab(a+b)
(sin2A)3+(cos2A)3=(sin2A+cos2A)3–3sin2Acos2A(sin2A+cos2A)As(sin2A+cos2A=1)So,=13–3sin2Acos2A(1)=1-3sin2Acos2A=R.H.S.
Thus, L.H.S.=R.H.S.
Hence, Proved