已知函数f(x)=sin(x+4分之7π)+cos(x-4分之3π),x∈R（1）求函数最小值和最小正周期（2）已知cos（β-α）=5分之4,cos（β+α=-5分之4,（0＜α＜β≤2分之π）求证：【f（β）】方-2=0
问题描述：
已知函数f(x)=sin(x+4分之7π)+cos(x-4分之3π),x∈R（1）求函数最小值和最小正周期
（2）已知cos（β-α）=5分之4,cos（β+α=-5分之4,（0＜α＜β≤2分之π）求证：【f（β）】方-2=0
答
f(x)=sin(x+2π-π/4)+cos(x-3π/4)
=sin(x-π/4)+cos(x-3π/4)
=sin(x-π/4)+cos(x-π/4-π/2) \\公式 cos（x-π/2)=sinx\\
=sin(x-π/4)+sin(x-π/4)
=2sin(β-π/4)
答
f(x)=sin(x+2π-π/4)+cos(x-3π/4)=sin(x-π/4)+cos(x-3π/4)=sin(x-π/4)+cos(x-π/4-π/2) \\公式 cos（x-π/2)=sinx\\=sin(x-π/4)+sin(x-π/4) =2sin(β-π/4)所以最小正周期为2π,函数最小值为 -2（2）f（β）...