How to prove: sin^6theta-cos^6theta=(2sin^2theta-1)(cos^2theta+sin^4theta)sin6θcos6θ=(2sin2θ1)(cos2θ+sin4θ) ?

1 Answer
Jun 4, 2018

LHS=sin^6x-cos^6xLHS=sin6xcos6x

=(sin^2x)^3-(cos^2x)^3=(sin2x)3(cos2x)3

=[sin^2x-cos^2x]*[(sin^2x)^2+sin^2x*cos^2x+(cos^2x)^2]=[sin2xcos2x][(sin2x)2+sin2xcos2x+(cos2x)2]

=[sin^2x-(1-sin^2x)][sin^4x+sin^2x*cos^2x+cos^4x]=[sin2x(1sin2x)][sin4x+sin2xcos2x+cos4x]

=[sin^2x-1+sin^2x][sin^4x+cos^2x(sin^2x+cos^2x)]=[sin2x1+sin2x][sin4x+cos2x(sin2x+cos2x)]

=[2sin^2x-1][sin^4x+cos^2x]=RHS=[2sin2x1][sin4x+cos2x]=RHS