函数f(x)=sin(π-ωx)cosωx+cos^2ωx最小正周期为π,求ω

ω=1

f(x)=sin(π-ωx)cosωx+cos^2ωx
=sinωxcosωx+cos^2ωx
=1/2sin2ωx+1/2(1+cos2wx)
=2^0.5/2sin(2wx+π/4)+1/2
因为最小正周期为π
所以 2π/(2w)=π w=1

函数可以变形
f(x)=sin(ωx)cosωx+cos^2ωx
=0.5sin(2ωx)+0.5(cos2ωx+1)
=sin(2ωx+π/4）/根号2 +0.5
若f(x)周期为π，f(x)=f(x+π)即2ωπ=正负2π，ω=正负1

f(x)=sin(π-ωx)cosωx+cos^2ωx
=sinωxcosωx+cos^2ωx
=1/2sin2ωx+(2cos2ωx+1)/2
=√2/2sin（2ωx+π/4）+1/2
由题意 2π/2ω = π, 所以 ω=1

f(x)=sin(π-ωx)cosωx+cos^2ωx
=sinωxcosωx+1/2(1+cos2ωx)
=1/2sin(2ωx)+1/2cos2ωx+1/2
=1/2+=√2/2sin(2ωx+π/4)
最小正周期为π,所以最小正周期为2π/2ω=π
因此 ω=1

f(x)=sin(π-ωx)cosωx+cos^2ωx
=sinωxcosωx+(1+cos2ωx)/2
=(sin2ωx)/2+(cos2ωx)/2+1/2
=√2sin(2ωx+π/4）/2+1/2，
2π/(2ω)=π,
ω=1.

FX= （SIN（π）*COS（ωx）-COS（π）*SIN（ωx）)*cosωx+（1+COS（2ωx））/2
=-SIN(2ωx)+1/2cos(2ωx)+1/2
=a*sin(2ωx+B)+1/2

所以2π/（2ω）=π 所以ω=1 （2ω的绝对值 绝对值符号不知道怎么打）