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)
∵函數(shù)遞增時,有
-π+2kπ≤2x+π/12≤2kπ,k∈Z
即-13π/24+kπ≤x≤-π/24+kπ,k∈Z
函數(shù)遞減時,有
2kπ≤2x+π/12≤π+2kπ,k∈Z
即-π/24+kπ≤x≤11π/24+kπ,k∈Z
∴單調(diào)遞增區(qū)間是[-13π/24+kπ,-π/24+kπ],k∈Z
單調(diào)遞減區(qū)間是[-π/24+kπ,11π/24+kπ],k∈Z