f(x)=√2sinwxcos(wx+pai/4)+1/2的最小正周期为2pai.求w的值
问题描述:
f(x)=√2sinwxcos(wx+pai/4)+1/2的最小正周期为2pai.求w的值
答
f(x)=√2sinwx(√2/2coswx-√2/2sinwx)+1/2
=sinwxcoswx-sin²wx+1/2
=1/2sin2wx-1/2+1/2cos2wx+1/2
=1/2(sin2wx+cos2wx)
=√2/2sin(2wx+π/4)
T=2π/2w
2π/2w=2π
w=1/2