Question

[Lâm Ánh Yên]

Chứng minh các đẳng thức:

\(cos^3xsinx-sin^3xcosx=\dfrac{1}{4}sin4x\)

\(sin^4x+cos^4x=\dfrac{1}{4}\left(3+cos4x\right)\)

Answer 1

[Nguyễn Việt Lâm]

\(cos^3xsinx-sin^3xcosx=sinx.cosx\left(cos^2x-sin^2x\right)=\dfrac{1}{2}sin2x.cos2x=\dfrac{1}{4}sin4x\)

\(sin^4x+cos^4x=\left(sin^2x+cos^2x\right)^2-2sin^2x.cos^2x=1-\dfrac{1}{2}\left(2sinx.cosx\right)^2=1-\dfrac{1}{2}sin^22x\)

\(=1-\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{2}cos4x\right)=\dfrac{3}{4}+\dfrac{1}{4}cos4x=\dfrac{1}{4}\left(3+cos4x\right)\)