解：f(x)=sin(πx/4-π/6)-[2cos²(πx/8)-1]
=sin(πx/4-π/6)-cos(πx/4)
=sin(πx/4)cos(π/6)-cos(πx/4)sin(π/6)-cos(πx/4)
=根号3/2sin(πx/4)-（1/2）cos(πx/4)-cos(πx/4)
=根号3/2sin(πx/4)-（3/2）cos(πx/4）
=根号3[1/2sin(πx/4)-（根号3/2）cos(πx/4]
=根号3sin(πx/4-π/3)
=根号3sin(π/4(x-4/3))
T=2π/(π/4)=8,
是F(X)=sin(πx/4-π/6)-2cos(πx/8)^2+1还是F(X)=sin(πx/4-π/6)-2cos^2(πx/8)+1?
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