问题标题：
已知函数f(x)=sin(x+7π/4)+cos(x-3π/4),x∈R(1)求f(x)的最小正周期和最小值(2)已知cos(β-α)=4/5cos(β+α)=-4/5,0＜α＜β≤π/2,求证【f（β）】²-2=0
问题描述：
已知函数f(x)=sin(x+7π/4)+cos(x-3π/4),x∈R(1)求f(x)的最小正周期和最小值
(2)已知cos(β-α)=4/5cos(β+α)=-4/5,0＜α＜β≤π/2,求证【f（β）】²-2=0
蔡广宇回答：
(1)f(x)=sin(x+7π/4)+cos(x-3π/4)
=sin(x+7π/4)+sin(5π/4-x)
=2sin(3π/2)cos(x+π/4)
=-cos(x+π/4)
最小正周期T=2π/1=2π
(2)已知cos(β-α)=4/5cos(β+α)=-4/5,
两式相加cos(β-α)+cos(β+α)=0
2cosβcosα=0
0＜α＜β≤π/2
cosα≠0
所以cosβ=0
β=π/2
f(β)=f(π/2)=-2cos(π/2+π/4)=2sin(π/4)=√2
所以【f（β）】²-2=(√2)²-2=0
得证
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