Question

已知函数f(x)=4cosx·sin(wx+π/4)的最小正周期为π.求w的值

Answer 1

最小公倍数法：
f(x)=4cosx·sin(wx+π/4)
=2{sin[(w+1)x+π/4]+sin[(w-1)x+π/4]}
当w=1时
f(x)=2sin(2x+π/4)+√2
f(x)的最小正周期为π
当w=-1时
f(x)=2sin(-2x+π/4)+√2
f(x)的最小正周期为π
当w≠±1时
∵π为函数f(x)=4cosx·sin(wx+π/4)的周期
∴f(x+π)=f(x)
即
4cos(x+π)sin[w(x+π)+π/4]=4cosx·sin(wx+π/4)
从而
sin[(wx+π/4)+wπ]=-sin(wx+π/4)
∴ w=2k+1 k∈Z且k≠-1、0
f(x)=2{sin[(2k+2)x+π/4]+sin(2kx+π/4)}
sin[(2k+2)x+π/4] 、sin(2kx+π/4)的最小正周期分别为π/|k+1|、π/|k|
易得
f(x)的最小正周期为π/|k+1|和π/|k|的最小公倍数π
综上可得
w=2k+1 k∈Z