已知函数f(x)满足下列关系式：（1）对于任意的x,y∈R,恒有2f(x)f(y)=f(π/2-x+y)-f(π/2-x-y) f(π/2)=1

令X=Y=0，代人2f(x)f(y)=f(π/2-x+y)-f(π/2-x-y)
即得f(0)=0
对于任意的x,y∈R，因此定义域对称，
令x=π/2，带入2f(x)f(y)=f(π/2-x+y)-f(π/2-x-y)
化简 f(y)=-f（-y)
所以 为奇函数
令X=0,y=π/2 ,得到f(X)=0
令x=π，y=a+1.5π
带回原式，f(-a)=f(-2π-a)
即T=2π

1. 2f(0)*f(0) = f(π/2)-f(π/2) = 0
f(0)=0

2. 2f(π/2) f(-x) =2f(-x)
2f(π/2)f(-x)
= f(π/2- π/2 -x)-f(π/2-π/2+x)
= f( - x) - f(x)
= 2f(-x)
f( - x) = - f(x)
f(x) 是奇函数

3. 2f(x+2π)*f(π/2)
= f(π -x -2π) - f(-x-2π)
= f(-x -π) - f(-x-2π)
= -f(x+π)+f(x+2π)
= 2f(x+2π)
f(x+ 2π) = -f(x+π)
f(x+π) = -f(x)
f(x+ 2π) = f(x)

1.
将x=y=0代入2f(x)f(y)=f(π/2-x+y)-f(π/2-x-y)
得 f(0)=0
2.
x=π/2代入
得2f(y)=f(y)-f(-y)
f(y)=-f(-y) 又y属于R
故是奇函数
3.
y=π/2代入2f(x)f(y)=f(π/2-x+y)-f(π/2-x-y)

得f(x)=f(π-x)
两边同乘以-1

得f(x-π) = -f(x)
再把x换成x-π
f(x-2π)= -f(x-π)=f(x)

即f(x-2π)=f(x)