若θ∈[-π12,π12],则函数y=cos(θ+π4)+sin2θ的最小值是(  ) A.0 B.1 C.98 D.32−12

问题描述:

若θ∈[-

π
12
π
12
],则函数y=cos(θ+
π
4
)+sin2θ的最小值是(  )
A. 0
B. 1
C.
9
8

D.
3
2
1
2

y=cos(θ+

π
4
)+sin2θ=cos(θ+
π
4
)-cos(2θ+
π
2

=-cos2(θ+
π
4
)+cos(θ+
π
4

=-2cos2(θ+
π
4
)+cos(θ+
π
4
)+1
∵θ∈[-
π
12
π
12
],
1
2
≤cos(θ+
π
4
)≤
3
2

所以设cos(θ+
π
4
)=t
y=-2(t-
1
4
2+
9
8

当t=
3
2
时即θ=-
π
12
时有最小值
3
−1
2

故选D