当-π/2≤x≤π/2时,求函数f(x)=sinx+√3cosx的值域.请问我这么做有问题吗?f(x)=sinx+√3cosx=2sin(x+π/3)-π/6≤x+π/3≤5π/6所以-1/2≤sin(x+π/3)≤1/2进而函数值域为[-1,1]答案是[-1,2],可是sin5π/6明明等于1/2的…
问题描述：
当-π/2≤x≤π/2时,求函数f(x)=sinx+√3cosx的值域.
请问我这么做有问题吗?
f(x)=sinx+√3cosx=2sin(x+π/3)
-π/6≤x+π/3≤5π/6
所以-1/2≤sin(x+π/3)≤1/2
进而函数值域为[-1,1]
答案是[-1,2],可是sin5π/6明明等于1/2的…
答
解从-π/6≤x+π/3≤5π/6开始错了由-π/6≤x+π/3≤5π/6,知当x+π/3=-π/6是,sin（x+π/3）有最小值=-1/2当x+π/3=π/2是,sin（x+π/3）有最大值=1即由-π/6≤x+π/3≤5π/6得到-1/2≤sin(x+π/3)≤1即-1≤2sin(x+...