！！！三角函数最值求解！！！

y=sin^2θcosθ
=(1-cos^2θ)cosθ
=cosθ-cos^3θ
设cosθ=t
y=t-t^3 t∈[-1,1]
y'=1-3t^2=0
t=±1/√3
y(-1)=0
y(1)=0
y(-1/√3)=-2√3/9
y(1/√3)=2√3/9
-2√3/9<=y<=2√3/9

另解: sin^2θcos^4θ
=sin^2θ*cos^2θ/2*cos^2θ/2*4
(sin^2θ+cos^2θ/2+cos^2θ/2)/3>=(sin^2θ*cos^2θ/2*cos^2θ/2)^(1/3)
(算术平均>=几何平均)
∴sin^2θ*cos^2θ/2*cos^2θ/2<=((sin^2θ+cos^2θ)/3)^3
∴sin^2θcos^4θ<=4(1/3)^3=4/27当且仅当sin^2θ=cos^2θ/2时取等号
∴-2√3/9<=sinθcos^2θ<=2√3/9

y=sin^2θ*cosθ
=(1-cos^2θ)cosθ
=cosθ-cos^3θ
令cosθ=t
y=t-t^3 -1<=t<=1
求导：y'=1-3t^2
令y'=0,t1=1/根号3，t2=-1/根号3
当t<=-1/根号3或t>=1/根号3时，y'<=0；当-1/根号3< t<1/根号3，y'>0
所以当t<=-1/根号3或t>=1/根号3时,函数单调递减；当-1/根号3< t<1/根号3，函数单调递增。
所以t=1/根号3,ymax=2根号3/9

[(sinθ)²cosθ]'
=2sinθ(cosθ)²-(sinθ)³
=sinθ[3(cosθ)²-1]

sinθ=0时：(sinθ)²cosθ=0