第2个回答 2012-03-24
1.f(x)=2sinwxcoswx+2√3cos^2wx-√3=sin2wx+√3cos2wx=2sin(2wx+π/3)由题知函数的最小正周期为T=(π/2)*2=π=2π/2w得w=1
2.由1可知f(x)=2sin(2x+π/3)因为f(a)=2/3=2sin(2a+π/3),sin(2a+π/3)=1/3
1-2sin^2(2a+π/3)=cos(4a+2π/3)=cos(-4a-2π/3)=7/9
cos(-4a+π/3)=cos(-4a-2π/3+π)=-cos(-4a-2π/3)=-7/9
sin(5π/6-4a)=sin(-4a+π/3+π/2)=cos(-4a+π/3)=-7/9