Question

Tìm GTLN,GTNN

y=(cos⁴x-sin⁴x).sinx.cosx

Answer 1

\(y=\left(cos^2x+sin^2x\right)\left(cos^2x-sin^2x\right).sinx.cosx\)

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

\(=\dfrac{1}{4}sin4x\)

Do \(-1\le sin4x\le1\Rightarrow-\dfrac{1}{4}\le y\le\dfrac{1}{4}\)

\(y_{min}=-\dfrac{1}{4}\) khi \(sin4x=1\)

\(y_{max}=\dfrac{1}{4}\) khi \(sin4x=1\)