设x∈R,函数f(x)=cos^2(wx+p)-1/2(w>0,0
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设x∈R,函数f(x)=cos^2(wx+p)-1/2(w>0,0
![设x∈R,函数f(x)=cos^2(wx+p)-1/2(w>0,0](/uploads/image/z/17469981-45-1.jpg?t=%E8%AE%BEx%E2%88%88R%2C%E5%87%BD%E6%95%B0f%28x%29%3Dcos%5E2%28wx%2Bp%29-1%2F2%28w%3E0%2C0)
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
=(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
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