已知函数f(x)=2cos^2x/2-根号3sinx.若α为第二象限的角,且f(a-π/3)=1/3,求cos2a/1+cos2α-sin2α

f(x)=2cos²x/2-√3sinx
=1+cosx-√3sinx
=1+2(1/2cosx-√3/2sinx)
=1+2sin(x+5π/6)
f(a-π/3)=1+2sin(a-π/3+5π/6)
=1+2*sin(a+π/2)
=1+2cosa
=1/3
cosa=-1/3
sina=2√2/3
则 cos2a/(1+cos2α-sin2α)=cos2a/(2cos²a-2sinacosa)
=(cos²a-sin²a)/[2cosa(cosa-sina)]
=(sina+cosa)/(2cosa)
=(2√2-1)/(-2)
=(1-2√2)/2