SinXsin2X的最值是什么

y=sinx*sin2x =sinx*2sinx*cosx =2sinx^2*cosx =2*(1-cosx^2)*cosx
u=cosx,
u的范围[-1,1]
y=2u-2u^3
y'=2-6u^2=0 ==>
u^2=1/3==>u=根号3/3,或者-根号3/3
u=根号3/3时,y最大
u=-根号3/3,y最小分别求出即可.

设y=SinXsin2X=2(1-(cosX)^2)cosX
设t=cosX
y=2t-t^3
当X属于[0,90度]时，t=cosX随X单调递减
y'=2-3t^2
当t=sqrt(2/3)或-sqrt(2/3)时取得最大值
为4sqrt(2/3)/3