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
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)所以最小正周期为2π,函数最小值为 -2(2)f(β)...