Sin^6+cos^6=1-3sin^2+cos^2

1.	Sin^6+cos^6=1-3sin^2+cos^2
Answer» We have,LHS = sin6A + cos6A{tex}\\Rightarrow{/tex}\xa0LHS = (sin2A)3 + (cos2A)3{tex}\\Rightarrow{/tex}\xa0LHS = (sin2A + cos2A) { (sin2A)2 + (cos2A)2 - sin2A cos2A)} [{tex} \\because{/tex}\xa0a3 + b3 = (a + b) (a2 - ab + b2)]{tex} \\Rightarrow{/tex}\xa0LHS = {(sin2A)2+(cos2A)2 + 2sin2A cos2A - 2 sin2A cos2A - sin2A cos2A}{tex} \\Rightarrow{/tex} LHS = {(sin2A + cos2A)2 - 3sin2A cos2A } = 1 - 3sin2A cos2A = RHS.

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