Question

已知函数f(x)=cos(2x+π/3)+sin^2x.

已知函数f(x)=cos(2x+π/3)+sin^2x.
（1）求函数f(x)的单调递减区间及最小正周期
（2）设锐角△ABC的三个内角a,b,c的对边分别为a,b,c,若c=√6,cosB=1/3,f(C/2)=-1/4,求b

Answer 1

解：f(x)=cos(2x+π/3)+sin²x
=1/2cos2x-√3/2sin2x+(1-cos2x)/2
=-√3/2sin2x+1/2
(1)T=π
单减区间：2kπ-π/2≤2x≤2kπ+π/2 (k∈Z)
=>kπ-π/4≤x≤kπ+π/4 (k∈Z)
∴单减区间为：x∈[kπ-π/4,kπ+π/4] (k∈Z)
(2)f(C/2)=-√3/2sinC+1/2=-1/4
=>√3/2sinC=3/4
=>sinC=√3/2
∵cosB=1/3
∴sinB=(2√2) /3
又a/sinA=b/sinB=c/sinC
∴b=csinB/sinC=√6*[(2√2) /3] / (√3/2) =8/3