高一函数题目,求助!谢谢

x=0,y=0
f(0+0)+f(0-0)=2*f(0)*f(0)
2*f(0)=2*f(0)*f(0)
f(0)不等于0
f(0)=1

x=π/2,y=π/2
f(π/2+π/2)+f(π/2-π/2)=2*f(π/2)*f(π/2)
f(π)+f(0)=2*0
f(π)=-f(0)=-1

f(2π)=f(π+π)=2*f(π)*f(π)-f(π-π)=2*(-1)*(-1)-1=1
f(3π)=f(2π+π)=2*f(2π)*f(π)-f(2π-π)=2*1*(-1)-(-1)=-1
依此类推

f(x+y)+f(x-y)=2f(x)f(y)
x=y=0
f(0)+f(0)=2f(0)f(0) f(0)=f(0)f(0) f(0)=1
x=y
f(2x)+f(0)=2f(x)f(x)
f(2x)+1=2f(x)f(x)
f(2x)=2f(x)f(x)+1
x=π/2时
f(π)=2f(π/2)f(π/2)+1=1
x=π时
f(2x)=2f(x)f(x)+1
f(2π)=2f(π)f(π)+1=2+1=3

f(x+y)+f(x-y)=2f(x)f(y)
x=y=0
f(0)+f(0)=2f(0)f(0)
f(0)不等于0
f(0)=1

x=y=π/2
f(x+y)+f(x-y)=2f(x)f(y)
f(π)+f(0)=2*0*0=0
f(π)=-1

x=y=π
f(x+y)+f(x-y)=2f(x)f(y)
f(2π)+f(0)=2*f(π)*f(π)
f(2π)+1=2*1
f(2π)=1