\(\left|sinx\right|;\left|cosx\right|\le1\Rightarrow\left\{{}\begin{matrix}sin^8x\le sin^2x\\cos^8x\le cos^2x\end{matrix}\right.\)
\(\Rightarrow y\le sin^2x+cos^2x-1=0\)
\(y_{max}=0\) khi \(\left(sin^2x;cos^2x\right)=\left(0;1\right)\) và hoán vị
\(y=\left(sin^2x\right)^4+\left(cos^2x\right)^4-1\ge\frac{1}{2}\left[\left(sin^2x\right)^2+\left(cos^2x\right)^2\right]^2-1\)
\(y\ge\frac{1}{2}\left[\frac{1}{2}\left(sin^2x+cos^2x\right)^2\right]^2-1=\frac{1}{8}-1=-\frac{7}{8}\)
\(y_{min}=-\frac{7}{8}\) khi \(sin^2x=cos^2x=\frac{1}{2}\)
