f(π/2+x) - f(x) = f[π/4 -(π/2+x)] - f(x)
= f(-π/4 - x) - f(x)
= - f(π/4 + x) - f(x)
= - f[π/4 - (π/4 + x)] - f(x)
= -f(-x) - f(x)
=f(x) - f(x)
= 0f(2π/3)=？f（x）是周期函数吗 如果是 怎么证明是周期函数证明：f(π/2+x) - f(x) = 0 --------请看上面过程 f(π/2+x) = f(x) 周期是π/2的周期函数f(2π/3) = f(2π/3 -π/2 - π/2)-------这里周期函数直接2个周期π= f(π/3)= 1