Question

已知函数f(x)=2sinx·cos²(φ/2)+cosx·sinφ-sinx,(0＜φ＜π),在x=π处取最小值.

Answer 1

解：1）f(x)
=
2sinxcos
2
(φ/2)
+
cosxsinφ
–
sinx
=
sinx[2cos
2
(φ/2)
-
1]
+
cosxsinφ
=
sinxcosφ
+
cosxsinφ
=
sin(x
+
φ)
，当x
+
φ
=
-π/2
+
2kπ时（k∈Z）取最小值，把x
=
π代入上式可得φ
=
-3π/2
+
2kπ，而且0＜φ＜π，所以当且仅当k
=
1时，
φ
=
π/2

；
2）f(A)
=
sin(A
+
π/2)
=
cosA
=
√3/2
=>
A
=
arccos(√3/2)
=
π/6
，由正弦定理可得
a/sinA
=
b/sinB
=>
sinB
=
√2sin(π/6)/1
=
√2/2
=>
B
=
π/4
或者3π/4
=>
C
=
π
–
A
–
B
=
7π
/12
或者
π
/12

。