Lời giải:
Vì $\sin x, \cos x\leq 1$ nên:
$\sin ^7x\leq \sin ^2x$
$\cos ^8x\leq \cos ^2x$
$\Rightarrow y=\sin ^7x+\cos ^8x\leq \sin ^2x+\cos ^2x=1$
Vậy $y_{\max}=1$
Vì $\sin x, \cos x\geq -1$ nên:
$\sin ^7x\geq -\sin ^2x$
$\cos ^8x\geq -\cos ^2x$
$\Rightarrow y=\sin ^7x+\cos ^8x\geq -(\sin ^2x+\cos ^2x)=-1$
Vậy $y_{\min}=-1$