Question

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

Answer 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

Answer 2

f(x)=sin(π-ωx)cosωx+cos^2ωx
=sinωxcosωx+1/2(1+c
os2
ωx)
=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