题出错了sin(π-φ)-否则没法算
f(x)=sin2xcosφ-2cos^2xsin(π-φ)-cos(π/2+φ)（-π/2＜φ＜π/2）
=sin2xcosφ-sinφ(2cos^2x-1)
=sin2xcosφ-cos2xsinφ
=sin(2x-φ)
x=π/6时取得最大值（-π/2＜φ＜π/2）
π/3-φ=π/2+2kπ,k是整数
φ=-π/6
g(x)=sin(x+π/6)
cos(a)=sin(a+π/6)=1/3
α∈（-π/2,0）
α+π/6∈（-π/3,π/6)
cos(α+π/6)=√(1-1/9)=2√2/3
cosα=cos((α+π/6)-α)
=cos(α+π/6)cosπ/6+sin(α+π/6)sinπ/6
=(2√6+1)/6