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
],则函数y=cos(θ+π 12
)+sin2θ的最小值是( )π 4
A. 0
B. 1
C.
9 8
D.
−
3
2
1 2
答
y=cos(θ+
)+sin2θ=cos(θ+π 4
)-cos(2θ+π 4
)π 2
=-cos2(θ+
)+cos(θ+π 4
)π 4
=-2cos2(θ+
)+cos(θ+π 4
)+1π 4
∵θ∈[-
,π 12
],π 12
∴
≤cos(θ+1 2
)≤π 4
,
3
2
所以设cos(θ+
)=tπ 4
y=-2(t-
)2+1 4
9 8
当t=
时即θ=-
3
2
时有最小值 π 12
−1
3
2
故选D