Question

Chứng minh : 3cos⁴x + sin⁴x = 1 + cos2x + cos²2x

Answer 1

\(3\cdot cos^4x+sin^4x-1\)

\(=cos^4x+sin^4x+2\cdot cos^4x-1\)

\(=\left(cos^2x+sin^2x\right)^2-2\cdot sin^2x\cdot cos^2x+2\cdot cos^4-1\)

\(=-2\cdot sin^2x\cdot cos^2x+2\cdot cos^4x\)

\(=2\cdot cos^2x\left(cos^2x-sin^2x\right)\)

\(cos2x+cos^22x=cos2x\left(cos2x+1\right)\)

\(=\left(cos^2x-sin^2x\right)\left(2\cdot cos^2x-1+1\right)\)

\(=2\cdot cos^2x\cdot\left(cos^2x-sin^2x\right)\)

Do đó: \(3\cdot cos^4x+sin^4x-1=cos2x+cos^22x\)

=>\(3\cdot cos^4x+sin^4x=1+cos2x+cos^22x\)