f(x)=√3(cos(x/2))^2+sin(x/2)cos(x/2)
=(√3/2)(cosx+1)+(1/2)sinx
=sin(π/6+x)+√3/2
f(x)=3/5+√3/2
sin(π/6+x)=3/5
π/60
cos(π/6+x)=4/5
sinx=sin(π/6+x-π/6)=sin(π/6+x)cos(π/6)-cos(π/6+x)sin(π/6)
=(3/5)(√3/2) -(4/5)(1/2)
=(3√3-4)/10
cos2x=1-2(sinx)^2=1-2*(3√3-4)^2/100=1-(43-24√3)/50=(7+24√3)/50问的是COS2ｘ啊cos2x=1-2(sinx)^2π/60这步没看懂若负三分之派小于x大于六分之派, π/60