(1)f(x)=sin²ωx-√3sinωxsin(ωx+π/2)
=1/2-(1/2)cos2ωx-[(√3)/2]sin2ωx
=1/2-sin(2ωx+π/6)
T=2π/2ω=π
ω=1
f(x)=1/2-sin(2x+π/6)
(2)求出函数的单调递增区间为(kπ-π/3,kπ+π/6)
递减区间为(kπ+π/6,kπ+2π/3)
所以当x=2π/3时,有最大值,为 3/2
当x=π/6时,有最小值,为 - 1/2
(1)f(x)=sin²ωx-√3sinωxsin(ωx+π/2)
=1/2-(1/2)cos2ωx-[(√3)/2]sin2ωx
=1/2-sin(2ωx+π/6)
T=2π/2ω=π
ω=1
f(x)=1/2-sin(2x+π/6)
(2)求出函数的单调递增区间为(kπ-π/3,kπ+π/6)
递减区间为(kπ+π/6,kπ+2π/3)
所以当x=2π/3时,有最大值,为 3/2
当x=π/6时,有最小值,为 - 1/2