Question

Chứng minh rằng

sin⁶x + cos⁶x = 1-3sin²x. cos²x

Answer 1

\(sin^6x+cos^6x=\left(sin^2x\right)^3+\left(cos^2x\right)^3=\left(sin^2x+cos^2x\right)^3-3sin^2xcos^2x\left(sin^2x+cos^2x\right)\)

\(=1^3-3sin^2xcos^2x.1=1-3sin^2xcos^2x\)

Answer 2

  sin6x+cos6x=(sin2x)3+(cos2x)3=(sin2x+cos2x)3−3sin2xcos2x(sin2x+cos2x)