已知y=cosx-sin^2x+2,(1)若x属于R,求该函数的值域（2）若x属于【0,π/2】,求值域
问题描述：
已知y=cosx-sin^2x+2,(1)若x属于R,求该函数的值域
（2）若x属于【0,π/2】,求值域
答
(1)
y= cosx - (sinx)^2 +2
= (cosx)^2 + cosx +1
= (cosx + 1/2)^2 +3/4
min y = 3/4
max y = (3/2)^2 + 3/4 = 3
值域 [ 3/4, 3]
(2)
x属于【0，π/2】
min y = y(π/2) = 1/4 + 3/4 =1
max y = y(0) =3
值域 [ 1, 3]
答
y=cosx-(1-cos²x)+2
=cos²x+cosx+1
=(cosx+1/2)²+3/4
1、
x∈R
-1