设x∈R,函数f(x)=cos(wx+f)(w>0,-π/2
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设x∈R,函数f(x)=cos(wx+f)(w>0,-π/2
(1)解析:∵函数f(x)=cos(wx+f)(w>0,-π/2<f<0)的最小正周期为π
∴w=2π/π=2,f(x)=cos(2x+f)
∵f(π/4)=√3/2
f(π/4)=cos(π/2+f)=-sin(f)=√3/2==>f=-π/3
∴w=2,f=-π/3
(2)解析:
(3)解析:∵f(x)> √2/2
f(x)=cos(2x-π/3)>√2/2
√2/2<cos(2x-π/3)<=1
2kπ-π/4<2x-π/3<2kπ+π/4==>kπ+π/24<x<kπ+7π/24 (k为整数)
∴当kπ+π/24<x<kπ+7π/24时,f(x)> √2/2
∴w=2π/π=2,f(x)=cos(2x+f)
∵f(π/4)=√3/2
f(π/4)=cos(π/2+f)=-sin(f)=√3/2==>f=-π/3
∴w=2,f=-π/3
(2)解析:
(3)解析:∵f(x)> √2/2
f(x)=cos(2x-π/3)>√2/2
√2/2<cos(2x-π/3)<=1
2kπ-π/4<2x-π/3<2kπ+π/4==>kπ+π/24<x<kπ+7π/24 (k为整数)
∴当kπ+π/24<x<kπ+7π/24时,f(x)> √2/2
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