f(x)=(1+cos(2x-π/6))/2 +(1-cos(2x+π/6))/2 -1
=1/2[cos(2x-π/6)-cos(2x+π/6)]
=1/2(√3/2cos2x+1/2sin2x-√3/2cos2x+1/2sin2x)
=1/2sin2x
因此,T=π;
f(-x)=1/2sin2(-x)=-1/2sin2x
所以f(x)为奇函数.
f(x)=(1+cos(2x-π/6))/2 +(1-cos(2x+π/6))/2 -1
=1/2[cos(2x-π/6)-cos(2x+π/6)]
=1/2(√3/2cos2x+1/2sin2x-√3/2cos2x+1/2sin2x)
=1/2sin2x
因此,T=π;
f(-x)=1/2sin2(-x)=-1/2sin2x
所以f(x)为奇函数.