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)=4/5 cos(β+α)=-4/5,0<α<β≤π/2,求证【f(β)】²-2=0
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![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)=4/5 cos(β+α)=-4/5,0<α<β≤π/2,求证【f(β)】²-2=0](/uploads/image/z/981776-56-6.jpg?t=%E5%B7%B2%E7%9F%A5%E5%87%BD%E6%95%B0f%28x%29%3Dsin%28x%2B7%CF%80%2F4%29%2Bcos%28x-3%CF%80%2F4%29%2Cx%E2%88%88R+%281%29%E6%B1%82f%28x%29%E7%9A%84%E6%9C%80%E5%B0%8F%E6%AD%A3%E5%91%A8%E6%9C%9F%E5%92%8C%E6%9C%80%E5%B0%8F%E5%80%BC%282%29%E5%B7%B2%E7%9F%A5cos%28%CE%B2-%CE%B1%29%3D4%2F5+cos%28%CE%B2%2B%CE%B1%29%3D-4%2F5%2C0%EF%BC%9C%CE%B1%EF%BC%9C%CE%B2%E2%89%A4%CF%80%2F2%2C%E6%B1%82%E8%AF%81%E3%80%90f%EF%BC%88%CE%B2%EF%BC%89%E3%80%91%26%23178%3B-2%3D0)
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