Question

Cho góc x thỏa mãn \(Cos4x=\dfrac{2}{3}\). Tính giá trị của biểu thức:

A=\(Sin^6xCos^2x+Sin^2xCos^6x\)

Answer 1

\(A=cos^2x.sin^2x\left(sin^4x+cos^4x\right)=\dfrac{1}{4}\left(2sinx.cosx\right)^2\left[\left(sin^2x+cos^2x\right)^2-2sin^2x.cos^2x\right]\)

\(=\dfrac{1}{4}sin^22x\left(1-\dfrac{1}{2}sin^22x\right)=\dfrac{1}{8}\left(1-cos4x\right)\left(1-\dfrac{1}{2}\left(1-cos4x\right)\right)\)

\(=\dfrac{1}{8}\left(1-\dfrac{2}{3}\right)\left(1-\dfrac{1}{2}\left(1-\dfrac{2}{3}\right)\right)=\dfrac{5}{144}\)