y=sinx+根号3cosx
=2[sinx*(1/2)+cosx*(√3/2)]
=2[sinx*cos(π/3)+cosx*sin(π/3)]
=2sin(x+π/3)
∵ -2分之π≤x≤2分之π
∴ -π/6≤x+π/3≤5π/6
∴ sin(x+π/3)∈[-1/2,1]
∴ 2sin(x+π/3)∈[-1,2]
即函数y=sinx+根号3cosx的取值范围是[-1,2]=2[sinx*(1/2)+cosx*(√3/2)]这步是怎么得的提取2啊，为了利用两角和的正弦公式=2sin(x+π/3)那这个呢COS哪去了这个就是公式啊sin(A+B)=sinAcosB+cosAsinB