已知函数f（x）＝4COS^4x-2COS2x-1/SIN(TT/4+x)SIN(TT/4-x)求f(-11TT/12)的值
问题描述：
已知函数f（x）＝4COS^4x-2COS2x-1/SIN(TT/4+x)SIN(TT/4-x)求f(-11TT/12)的值
答
f(-11π/12)=4cos^4(11π/12)-2cos(11π/6)+1/sin(7π/6)sin(2π/3)
sin2π/3sin(7π/6)=-sinπ/6sinπ/3=-√3/4
2cos11π/6=2cos(2π-π/6)=2cosπ/6=√3
4cos^4x=(2cos^2x)^2=(1+cos2x)^2
cos2x=cos22π/12=√3/2
(1+√3/2)^2=(1/4)(4+2√3)=1+√3/2
f(-11π/12)=1+√3/2-√3-4/√3
=1-11√3/6