三角函数求周期和奇偶性

解：用Pi表示圆周率.

f(x)=sin^2(x+Pi/4)-sin^2(x-Pi/4)
=[sin(x+Pi/4)+sin(x-Pi/4)]*[sin(x+Pi/4)-sin(x-Pi/4)]
(利用和差化积公式)
=[2*sinx*cos(Pi/4)][2*cosx*sin(Pi/4)]
=2*sinx*cosx
=sin(2x).
所以周期就是Pi,是奇函数.

f(x)=sin^2(x+π/4)-sin^2(x-π/4)
=cos^2(π/2-x-π/4)-sin^2(x-π/4)
=cos^2(x-π/4)-sin^2(x-π/4)
=cos(2x-π/2)

f(-x)=cos(-2x-π/2)
=cos(2x+π/2)
=-cos(2x-π/2)

奇函数
周期π