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Đặt $t=\sin x \Rightarrow t\in[0,1]$
Xét hàm: $g(t)=t(1-t^2), t\in[0,1]$
Ta có: $g'(t)=1-3t^2$
$g'(t)=0 \Leftrightarrow t=\frac{1}{\sqrt3}$
Từ đó suy ra:
$\min_{t\in[0,1]} g(t)=\min \{ g(0),g(1),g(\frac{1}{\sqrt3}) \}=g(0)=g(1)=0$
$\max_{t\in[0,1]} g(t)=\max \{ g(0),g(1),g(\frac{1}{\sqrt3}) \}=g(\frac{1}{\sqrt3})=\frac{2}{3\sqrt3}$
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