设x∈R,函数f(x)=cos^2(wx+p)-1/2(w>0,0
问题描述：
设x∈R,函数f(x)=cos^2(wx+p)-1/2(w>0,0
答
f(x)=[1+cos(2wx+2p)]/2-1/2
=(1/2)*cos(2wx+2p)
T=2π/|2w|=π,w>0
∴w=1
f(π/8)=(1/2)*cos(π/4+2p)=1/4
∴π/4+2p=±π/3+2kπ,k∈Z
∵p∈(0,π/2)
∴p=π/24
即w=1,p=π/24
f(x)=½cos(2x+π/12)
∵函数递增时,有
-π+2kπ≤2x+π/12≤2kπ,k∈Z
即-13π/24+kπ≤x≤-π/24+kπ,k∈Z
函数递减时,有
2kπ≤2x+π/12≤π+2kπ,k∈Z
即-π/24+kπ≤x≤11π/24+kπ,k∈Z
∴单调递增区间是[-13π/24+kπ,-π/24+kπ],k∈Z
单调递减区间是[-π/24+kπ,11π/24+kπ],k∈Z